User minhyong kim - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T04:59:49Z http://mathoverflow.net/feeds/user/1826 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets Why is a topology made up of 'open' sets? Minhyong Kim 2010-03-23T22:25:01Z 2013-01-21T20:07:25Z <p>I'm ashamed to admit it, but I don't think I've ever been able to genuinely motivate the definition of a topological space in an undergraduate course. Clearly, the definition distills the essence of many examples, but it's never been obvious to me how it came about, compared, for example, to the rather intuitive definition of a metric space. In some ways, the sparseness of the definition is startling as it tries to capture, apparently successfully, the barest notion of 'space' imaginable.</p> <p>I can try to make this question more precise if necessary, but I'd prefer to leave it slightly vague, and hope that someone who has discussed this successfully in a first course, perhaps using a better understanding of history, might be able to help me out.</p> <p>Added 24 March: </p> <p>I'm grateful to everyone for their thoughtful answers so far. I'll have to think over them a bit before I can get a sense of the 'right' answer for myself. In the meanwhile, I thought I'd emphasize again the obvious fact that the standard concise definition has been <em>tremendously successful.</em> For example, when you classify two-manifolds with it, you get equivalence classes that agree exactly with intuition. Then in as divergent a direction as the study of equations over finite fields, there is the etale topology*, which explains very clearly surprising and intricate patterns in the behaviour of solution sets. </p> <p>*If someone objects that the etale topology goes beyond the usual definition, I would argue that the logical essence is the same. It is notable that the standard definition admits such a generalization so naturally, whereas some of the others do not. (At least not in any obvious way.) </p> <p>For those who haven't encountered one before, a <em>Grothendieck topology</em> just replaces subsets of a set $X$ by maps $$Y\rightarrow X.$$ The collection of maps that defines the topology on $X$ is required to satisfy some obvious axioms generalizing the usual ones.</p> <p>Added 25 March: </p> <p>I hope people aren't too annoyed if I admit I don't quite see a satisfactory answer yet. But thank you for all your efforts. Even though Sigfpe's answer is undoubtedly interesting, invoking the notion of measurment, even a fuzzy one, just doesn't seem to be the best approach. As Qiaochu has pointed out, a topological space is genuinely supposed to be more general than a metric space. If we leave aside the pedagogical issue for a moment and speak as working mathematicians, a general concept is most naturally justified in terms of its consequences. As pointed out earlier, topologies that have no trace of a metric interpretation have been consequential indeed.</p> <p>When topologies were naturally generalized by Grothendieck, a good deal of emphasis was put on the notion of an <em>open covering</em>, and not just the open sets themselves. I wonder if this was true for Hausdorff as well. (Thanks for the historical information, Donu!) We can see the reason as we visualize a two-manifold. Any sufficiently fine open covering captures a combinatorial skeleton of the space by way of the intersections. Note that this is not true for a closed covering. In fact, I'm not sure what a sensible condition might be on a closed covering of a reasonable space that would allow us to compute homology with it. (Other than just saying they have to be the simplices of a triangulation. Which also reminds me to point out that homology can be computed for ordinary objects without any notion of topology.)</p> <p>To summarize, a topology relates to <em>analysis</em> with its emphasis on functions and their continuity, and to <em>metric geometry</em>, with its measurements and distances. However, it also interpolates between these and something like <em>combinatorial geometry</em>, where continuous functions and measurements play very minor roles indeed.</p> <p>For myself, I'm still confused.</p> <p>Another afterthought: I see what I was trying to say above is that open sets in topology provide an abstract framework for describing local properties of functions. However, an open <em>cover</em> is also able to encode global properties of spaces. It seems the finite intersection property is important for this, but I'm not able to say for sure. And then, when I try to return to the pedagogical question with all this, I'm totally at a loss. There are very few basic concepts that trouble me as much in the classroom.</p> http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/106658#106658 Answer by Minhyong Kim for Philosophy behind Mochizuki's work on the ABC conjecture Minhyong Kim 2012-09-08T08:31:45Z 2012-09-09T15:59:15Z <p>I would have preferred not to comment seriously on Mochizuki's work before much more thought had gone into the very basics, but judging from the internet activity, there appears to be much interest in this subject, especially from young people. It would obviously be very nice if they were to engage with this circle of ideas, regardless of the eventual status of the main result of interest. That is to say, the current sense of urgency to understand something seems generally a good thing. So I thought I'd give the flimsiest bit of introduction imaginable at this stage. On the other hand, as with many of my answers, there's the danger I'm just regurgitating common knowlege in a long-winded fashion, in which case, I apologize.</p> <p>For anyone who wants to really get going, I recommend as starting point some familiarity with two papers, 'The Hodge-Arakelov theory of elliptic curves (HAT)' and 'The Galois-theoretic Kodaira-Spencer morphism of an elliptic curve (GTKS).' [It has been noted here and there that the 'Survey of Hodge Arakelov Theory I,II' papers might be reasonable alternatives.][I've just examined them again, and they really might be the better way to begin.] These papers depart rather little from familiar language, are essential prerequisites for the current series on IUTT, and will take you a long way towards a grasp at least of the motivation behind Mochizuki's imposing collected works. This was the impression I had from conversations six years ago, and then Mochizuki himself just pointed me to page 10 of IUTT I, where exactly this is explained. The goal of the present answer is to decipher just a little bit those few paragraphs.</p> <p>The beginning of the investigation is indeed the function field case (over $\mathbb{C}$, for simplicity), where one is given a family $$f:E \rightarrow B$$ of elliptic curves over a compact base, best assumed to be semi-stable and non-isotrivial. There is an exact sequence $$0\rightarrow \omega_E \rightarrow H^1_{DR}(E) \rightarrow H^1(O_E)\rightarrow0,$$ which is moved by the logarithmic Gauss-Manin connection of the family. (I hope I will be forgiven for using standard and non-optimal notation without explanation in this note.) That is to say, if $S\subset B$ is the finite set of images of the bad fibers, there is a log connection $$H^1_{DR}(E) \rightarrow H^1_{DR}(E) \otimes \Omega_B(S),$$ which <em>does not preserve</em> $\omega_E$. This fact is crucial, since it leads to an $O_B$-linear Kodaira-Spencer map $$KS:\omega \rightarrow H^1(O_E)\otimes \Omega_B(S),$$ and thence to a non-trivial map $$\omega_E^2\rightarrow \Omega_B(S).$$ From this, one easily deduces Szpiro's inequality: $$\deg (\omega_E) \leq (1/2)( 2g_B-2+|S|).$$ At the most simple-minded level, one could say that Mochizuki's programme has been concerned with replicating this argument over a number field $F$. Since it has to do with differentiation on $B$, which eventually turns into $O_F$, some philosophical connection to $\mathbb{F}_1$-theory begins to appear. I will carry on using the same notation as above, except now $B=Spec(O_F)$.</p> <p>A large part of HAT is exactly concerned with the set-up necessary to implement this idea, where, roughly speaking, the Galois action has to play the role of the GM connection. Obviously, $G_F$ doesn't act on $H^1_{DR}(E)$. But it does act on $H^1_{et}(\bar{E})$ with various coefficients. The comparison between these two structures is the subject of $p$-adic Hodge theory, which sadly works only over local fields rather than a global one. But Mochizuki noted long ago that something like $p$-adic Hodge theory should be a key ingredient because over $\mathbb{C}$, the comparison isomorphism $$H^1_{DR}(E)\simeq H^1(E(\mathbb{C}), \mathbb{Z})\otimes_{\mathbb{Z}} O_B$$ allows us to completely recover the GM connection by the condition that the topological cohomology generates the flat sections.</p> <p>In order to get a global arithmetic analogue, Mochizuki has to formulate a <em>discrete non-linear</em> version of the comparison isomorphism. What is non-linear? This is the replacement of $H^1_{DR}$ by the universal extension $$E^{\dagger}\rightarrow E,$$ (the moduli space of line bundles with flat connection on $E$) whose tangent space is $H^1_{DR}$ (considerations of this nature already come up in usual p-adic Hodge theory). What is discrete is the \'etale cohomology, which will just be $E[\ell]$ with global Galois action, where $\ell$ can eventually be large, on the order of the height of $E$ (that is $\deg (\omega_E)$). The comparison isomorphism in this context takes the following form: $$\Xi: A_{DR}=\Gamma(E^{\dagger}, L)^{&lt;\ell}\simeq L|E[\ell]\simeq (L|e_{E})\otimes O_{E[\ell]}.$$ (I apologize for using the notation $A_{DR}$ for the space that Mochizuki denotes by a calligraphic $H$. I can't seem to write calligraphic characters here.) Here, $L$ is a suitably chosen line bundle of degree $\ell$ on $E$, which can then be pulled back to $E^{\dagger}$. The inequality refers to the polynomial degree in the fiber direction of $E^{\dagger} \rightarrow E$. The isomorphism is effected via evaluation of sections at $$E^{\dagger}[\ell]\simeq E[\ell].$$ Finally, $$L|E[\ell]\simeq (L|e_{E})\otimes O_{E[\ell]}$$ comes from Mumford's theory of theta functions. The interpretation of the statement is that it gives an isomorphism between the space of functions of some bounded fiber degree on non-linear De Rham cohomology and the space of functions on discrete \'etale cohomology. This kind of statement is entirely due to Mochizuki. One sometimes speaks of $p$-adic Hodge theory with finite coefficients, but that refers to a theory that is not only local, but deals with linear De Rham cohomology with finite coefficients.</p> <p>Now for some corrections: As stated, the isomorphism is not true, and must be modified at the places of bad reduction, the places dividing $\ell$, and the infinite places. This correction takes up a substantial portion of the HAT paper. That is, the isomorphism is generically true over $B$, but to make it true everywhere, the integral structures must be modified in subtle and highly interesting ways, while one must consider also a comparison of metrics, since these will obviously figure in an arithmetic analogue of Szpiro's conjecture. The correction at the finite bad places can be interpreted via coordinates near infinity on the moduli stack of elliptic curves as the subtle phenomenon that Mochizuki refers to as 'Gaussian poles' (in the coordinate $q$). Since this is a superficial introduction, suffice it to say for now that these Gaussian poles end up being a major obstruction in this portion of Mochizuki's theory.</p> <p>In spite of this, it is worthwhile giving at least a small flavor of Mochizuki's Galois-theoretic KS map. The point is that $A_{DR}$ has a Hodge filtration defined by</p> <p>$F^rA_{DR}= \Gamma(E^{\dagger}, L)^{ &lt; r}$</p> <p>(the direction is unconventional), and <em>this is moved around by the Galois action induced by the comparison isomorphism.</em> So one gets thereby a map $$G_F\rightarrow Fil (A_{DR})$$ into some space of filtrations on $A_{DR}$. This is, in essence, the Galois-theoretic KS map. That, is if we consider the equivalence over $\mathbb{C}$ of $\pi_1$-actions and connections, the usual KS map measures the extent to which the GM connection moves around the Hodge filtration. Here, we are measuring the same kind of motion for the $G_F$-action.</p> <p>This is already very nice, but now comes a very important variant, essential for understanding the motivation behind the IUTT papers. In the paper GTKS, Mochizuki modified this map, producing instead a 'Lagrangian' version. That is, he assumed the existence of a Lagrangian Galois-stable subspace $G^{\mu}\subset E[l]$ giving rise to another isomorphism $$\Xi^{Lag}:A_{DR}^{H}\simeq L\otimes O_{G^{\mu}},$$ where $H$ is a Lagrangian complement to $G^{\mu}$, which I believe does not itself need to be Galois stable. $H$ is acting on the space of sections, again via Mumford's theory. This can be used to get another KS morphism to filtrations on $A_{DR}^{H}$. But the key point is that </p> <p><em>$\Xi^{Lag}$, in contrast to $\Xi$, is free of the Gaussian poles</em> </p> <p>via an argument I can't quite remember (If I ever knew).</p> <p>At this point, it might be reasonable to see if $\Xi^{Lag}$ contributes towards a version of Szpiro's inequality (after much work and interpretation), except for one small problem. A subspace like $G^{\mu}$ has no reason to exist in general. This is why GTKS is mostly about the universal elliptic curve over a formal completion near $\infty$ on the moduli stack of elliptic curves, where such a space does exists. What Mochizuki explains on IUTT page 10 is exactly that the scheme-theoretic motivation for IUG was to enable the move to a single elliptic curve over $B=Spec(O_F)$, via the intermediate case of an elliptic curve 'in general position'.</p> <p>To repeat:</p> <p><em>A good 'nonsingular' theory of the KS map over number fields requires a global Galois invariant Lagrangian subspace $G^{\mu}\subset E[l]$.</em></p> <p>One naive thought might just be to change base to the field generated by the $\ell$-torsion, except one would then lose the Galois action one was hoping to use. (Remember that Szpiro's inequality is supposed to come from <em>moving</em> the Hodge filtration inside De Rham cohomology.) On the other hand, such a subspace does often exist <em>locally</em>, for example, at a place of bad reduction. So one might ask if there is a way to globally extend such local subspaces.</p> <p>It seems to me that this is one of the key things going on in the IUTT papers I-IV. As he say in loc. cit. he works with various categories of collections of local objects that <em>simulate</em> global objects. It is crucial in this process that many of the usual scheme-theoretic objects, local or global, are encoded as suitable categories with a rich and precise combinatorial structure. The details here get very complicated, the encoding of a scheme into an associated Galois category of finite \'etale covers being merely the trivial case. For example, when one would like to encode the Archimedean data coming from an arithmetic scheme (which again, will clearly be necessary for Szpiro's conjecture), the attempt to come up with a category of about the same order of complexity as a Galois category gives rise to the notion of a <em>Frobenioid</em>. Since these play quite a central role in Mochizuki's theory, I will quote briefly from his first Frobenioid paper:</p> <p>'Frobenioids provide a single framework [cf. the notion of a "Galois category"; the role of monoids in log geometry] that allows one to capture the essential aspects of both the Galois and the divisor theory of number fields, on the one hand, and function fields, on the other, in such a way that one may continue to work with, for instance, global degrees of arithmetic line bundles on a number field, but which also exhibits the new phenomenon [not present in the classical theory of number fields] of a "Frobenius endomorphism" of the Frobenioid associated to a number field.'</p> <p>I believe the Frobenioid associated to a number field is something close to the finite \'etale covers of $Spec(O_F)$ (equipped with some log structure) together with metrized line bundles on them, although it's probably more complicated. The Frobenious endomorphism for a prime $p$ is then something like the functor that just raises line bundles to the $p$-th power. This is a functor that would come from a map of schemes if we were working in characteristic $p$, but obviously not in characteristic zero. But this is part of the reason to start encoding in categories: </p> <p><em>We get more morphisms and equivalences.</em></p> <p>Some of you will notice at this point the analogy to developments in algebraic geometry where varieties are encoded in categories, such as the derived category of coherent sheaves. There as well, one has reconstruction theorems of the Orlov type, as well as the phenomenon of non-geometric morphisms of the categories (say actions of braid groups). Non-geometric morphisms appear to be very important in Mochizuki's theory, such as the Frobenius above, which allows us to simulate characteristic $p$ geometry in characteristic zero. Another important illustrative example is a non-geometric isomorphism between Galois groups of local fields (which can't exist for global fields because of the Neukirch-Uchida theorem). In fact, I think Mochizuki was rather fond of Ihara's comment that the positive proof of the anabelian conjecture was somewhat of a disappointment, since it destroys the possibility that encoding curves into their fundamental groups will give rise to a richer category. Anyways, I believe the importance of non-geometric maps of categories encoding rather conventional objects is that </p> <p><em>they allow us to glue together several standard categories in nonstandard ways.</em></p> <p>Obviously, to play this game well, some things need to be encoded in rigid ways, while others should have more flexible encodings.</p> <p>For a very simple example that gives just a bare glimpse of the general theory, you might consider a category of pairs $$(G,F),$$ where $G$ is a profinite topological group of a certain type and $F$ is a filtration on $G$. It's possible to write down explicit conditions that ensure that $G$ is the Galois group of a local field and $F$ is its ramification filtration in the upper numbering (actually, now I think about it, I'm not sure about 'explicit conditions' for the filtration part, but anyways). Furthermore, it is a theorem of Mochizuki and Abrashkin that the functor that takes a local field to the corresponding pair is fully faithful. So now, you can consider triples $$(G,F_1, F_2),$$ where $G$ is a group and the $F_i$ are <em>two</em> filtrations of the right type. If $F_1=F_2$, then this 'is' just a local field. But now you can have objects with $F_1\neq F_2$, that correspond to strange amalgams of two local fields.</p> <p>As another example, one might take a usual global object, such as $$(E, O_F, E[l], V)$$ (where $V$ denotes a collection of valuations of $F(E[l])$ that restrict bijectively to the valuations $V_0$ of $F$), and associate to it a collection of local categories indexed by $V_0$ (something like Frobenioids corresponding to the $E_v$ for $v\in V_0$). One can then try to glue them together in non-standard ways along sub-categories, after performing a number of non-standard transformations. My rough impression at the moment is that the 'Hodge theatres' arise in this fashion. [This is undoubtedly a gross oversimplification, which I will correct in later amendments.] You might further imagine that some construction of this sort will eventually retain the data necessary to get the height of $E$, but also have data corresponding to the $G^{\mu}$, necessary for the Lagrangian KS map. In any case, I hope you can appreciate that a good deal of 'dismantling' and 'reconstructing,' what Mochizuki calls <em>surgery</em>, will be necessary.</p> <p>I can't emphasize enough times that much of what I write is based on faulty memory and guesswork. At best, it is superficial, while at worst, it is (not even) wrong. [<s>In particular, I am no longer sure that the GTKS map is used in an entirely direct fashion.</s>] I have not yet done anything with the current papers than give them a cursory glance. If I figure out more in the coming weeks, I will make corrections. But in the meanwhile, I do hope what I wrote here is mostly more helpful than misleading.</p> <p>Allow me to make one remark about set theory, about which I know next to nothing. Even with more straightforward papers in arithmetic geometry, the question sometimes arises about Grothendieck's universe axiom, mostly because universes appear to be used in SGA4. Usually, number-theorists (like me) neither understand, nor care about such foundational matters, and questions about them are normally met with a shrug. The conventional wisdom of course is that any of the usual theorems and proofs involving Grothendieck cohomology theories or topoi do not actually rely on the existence of universes, except general laziness allows us to insert some reference that eventually follows a trail back to SGA4. However, this doesn't seem to be the case with Mochizuki's paper. That is, universes and interactions between them seem to be important actors rather than conveniences. How this is really brought about, and whether more than the universe axiom is necessary for the arguments, I really don't understand enough yet to say. In any case, for a number-theorist or an algebraic geometer, I would guess it's still prudent to acquire a reasonable feel for the 'usual' background and motivation (that is, HAT, GTKS, and anabelian things) before worrying too much about deeper issues of set theory.</p> http://mathoverflow.net/questions/48864/is-grothendieck-a-computer Is Grothendieck a computer? Minhyong Kim 2010-12-10T02:00:58Z 2012-07-06T00:20:28Z <p>I can't resist asking this companion question to the <a href="http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking" rel="nofollow"> one of Gowers</a>. There, Tim Dokchitser suggested the idea of Grothendieck topologies as a fundamentally new insight. But Gowers' original motivation is to probe the boundary between a human's way of thinking and that of a computer. I argued, therefore, that Grothendieck topologies might be more natural to computers, in some sense, than to humans. It seems Grothendieck always encouraged people to think of an object in terms of the category that surrounds it, rather than its internal structure. That is, even the most lovable mathematical structure might be represented simply as a symbol $A$, and its special properties encoded in arrows $A\rightarrow B$ and $C\rightarrow A$, that is, a grand combinatorial network. I'm tempted to say that the idea of a Grothendieck topology is something of an obvious corollary of this framework. It's not something I've devoted much thought to, but it seems this is exactly the kind of reasoning more agreeable to a computer than to a woolly, touchy-feelly thinker like me.</p> <p>So the actual question is, what other mathematical insights do you know that might come more naturally to a computer than to a human? I won't try here to define computers and humans, for lack of competence. I don't think having a deep knowledge of computers is really a prerequisite for the question or for an answer. But it would be nice if your examples were connected to substantial mathematics. </p> <p>I see that this question is subjective (but not argumentative in intent), so if you wish to close it on those grounds, that's fine.</p> <p>Added, 11 December: Being a faulty human, I had an inexplicable attachment to the past tense. But, being weak-willed on top of it all, I am bowing to peer pressure and changing the title.</p> http://mathoverflow.net/questions/24971/practical-applications-of-algebraic-number-theory Practical applications of algebraic number theory? Minhyong Kim 2010-05-17T06:26:33Z 2012-05-13T15:35:44Z <p>I'm interested in learning about any applications, the more worldly the better*. Pointing to a nice reference on the number field sieve, for example, would be fine. However, let me mention one direction I would be especially grateful to learn about.</p> <p>In my introductory course, I like to spend some time on the perspective that algebraic number theory is the study of sophisticated multiplications on $\mathbb{Q}^n$ (an algebraic number field $F$ of degree $n$) and on $\mathbb{Z}^n$ (the ring of algebraic integers in $F$). This is in part because I still find it amazing that a little bit of abstract algebra (of irreducible polynomials) enables us to construct such things systematically**. But I also believe at least half-seriously that this is the view through which the general public will gradually learn about algebraic numbers, until the time they're taught in primary schools several thousand years hence. After all, we have ourselves witnessed the remarkable ascent of multiplication on $\mathbb{F}_2^n$, a set whose initial practical use was entirely devoid of algebraic content, as a powerful tool for information processing.</p> <p>After such grandiose reflection, I can't help but wonder: are multiplicative structures on $n$-tuples of integers provided by algebraic number theory already of some practical use? A superficial google search uncovered nothing. But surely, there must be something? I would love to be able to mention some examples to my students.</p> <p>As I write, one class of examples occurs to me. Algebraic integers can be used to construct arithmetic groups, which I understand can be applied in a number of ways. Perhaps someone can comment knowledgeably on that. But something direct that could at least vaguely be explained in an undergraduate course would be even better.</p> <p>Added: </p> <p>Such was the depth of my ignorance that I didn't even know about number field codes until Victor Protsak pointed to them in his answer. Thanks to him, I stumbled upon a <a href="https://openaccess.leidenuniv.nl/bitstream/1887/3823/1/346_082.pdf" rel="nofollow">short survey</a> by Lenstra. To get the gist of it, one need only read this quote: </p> <p>'The new codes are the analogues, for number fields, of the codes constructed by Goppa and Tsfasman from curves over finite fields.'</p> <p>The time-worn analogy continues to prosper.</p> <p>Added again: </p> <p>In order to avoid misleading people with the word 'prosper,' I should say that Lenstra has many negative things to say about these codes. For example,</p> <p>'If the generalized Riemann hypothesis is true our codes are, asymptotically speaking, not as good as those of Goppa and Tsfasman Also, the latter codes are linear and non-mixed.'</p> <p>My original question still stands.</p> <hr> <p>*I do not wish, however, to give the impression of a firm belief in the division between pure and applied mathematics.</p> <p>** To appreciate this, one need only spend a little time on a direct approach using the multi-linear algebra of structure constants.</p> http://mathoverflow.net/questions/31538/non-abelian-class-field-theory-and-fundamental-groups Non-abelian class field theory and fundamental groups Minhyong Kim 2010-07-12T11:49:23Z 2011-11-30T05:58:00Z <p>Over the years, I've been somewhat in the habit of asking questions in this vein to experts in the Langlands programme. </p> <p>As is well known, given an algebraic number field $K$, they propose to replace the reciprocity map $$A_K^*/K^*\rightarrow Gal(K^{ab}/K)$$ of abelian class field theory by a correspondence between the $n$-dimensional representations $\rho$ of $Gal(\bar{K}/K)$ and certain automorphic representations $\pi_{\rho}$ of $GL_n(A_K)$. (We'll skip the Weil group business for this discussion.) Substantial arithmetic information is carried on either side by the $L$-functions, which are supposed to be equal.</p> <p>This involves deep and beautiful mathematics whenever something can be proved, and there are many applications, such as the Sato-Tate conjecture or this recent paper of Chenevier and Clozel:</p> <p><a href="http://www.math.polytechnique.fr/~chenevier/articles/galoisQautodual2.pdf" rel="nofollow">http://www.math.polytechnique.fr/~chenevier/articles/galoisQautodual2.pdf</a></p> <p>(I mention this one because it is in some ways very close to the point of this question.)</p> <p>However, there are elementary consequences of abelian class field theory that seem not to have obvious non-abelian analogues. The one I wish to mention today has to do with the fundamental group. Given a number field $K$ (assume it's totally imaginary to avoid some silly issues), how can we tell if it has non-trivial abelian unramified extensions? Class field theory says we can look at the class group, which is quite computable in principle, and even in practice for small discriminants. But now, suppose we go on to ask the non-abelian question: which number fields have $$\pi_1(Spec(O_K))=0?$$ That is to say, when does $K$ have no unramified extension at all, abelian or not? As far as I know, there is no easy answer to this question. Niranjan Ramachandran has pointed out that there are at least ten examples, $K=\mathbb{Q}$ (oops, that's real) and $K$ an imaginary quadratic field of class number one. I know of no others. Of course I would be happy to collect some more, if someone else has them lying around. </p> <p>But the question I really wanted to ask is: Suppose we are in a Langlands paradise where everything reasonably conjectured by the programme is a theorem. Does this give a way to algorithmically (as we run over fields $K$) resolve this question as in the abelian case? Otherwise, is there a sensible refinement of the usual formulation that would subsume such applications?</p> <p>Added: </p> <p>I'm embarrassed to admit I hadn't followed the question mentioned by David Hansen (even after commenting on it). Thanks to David for pointing it out. Of course my main question still stands. I've changed the title following Andy Putman's suggestion. The original title evolved from a (humorously) provocative version that I normally use only among friends who already know I'm a Langlands fan: 'What is the Langlands programme good for?'</p> <p>Regarding jnewton's very natural thought: in addition to other difficulties, one would also need to bound $n$.</p> <hr> <p>Added, 13 July:</p> <p>Here is one more remark concerning jnewton's suggestion. Of course in the realm of classical holomorphic cusp forms, there are infinitely many of level one. More generally, it is shown in the paper </p> <p><a href="http://www.math.uchicago.edu/~swshin/Plancherel.pdf" rel="nofollow">http://www.math.uchicago.edu/~swshin/Plancherel.pdf</a></p> <p>that whenever $G$ is a split reductive group over $\mathbb{Q}$, there are infinitely many cuspidal automorphic representations that are unramified everywhere and belong to the discrete series at $\infty$. (I presume there are other results of this sort. This one I just happen to know from a talk last Fall.) According to Clozel's conjecture as you might find in</p> <p><a href="http://seven.ihes.fr/IHES/Scientifique/asie/textes/Clozel-juil06.pdf" rel="nofollow">http://seven.ihes.fr/IHES/Scientifique/asie/textes/Clozel-juil06.pdf</a></p> <p>(conjecture (2.1)), algebraic ones among them should correspond to motivic Galois representations (after we choose a representation of the dual group)*. I don't have the expertise to recognize algebraicity in such constructions, in addition to the danger that I'm misunderstanding something more elementary. But it seems to me quite a task to show directly that there are none corresponding to Artin representations. (The only case I could do myself is the classical one.)</p> <p>Now, I would like very much to be corrected on all this. But such families do seem to indicate that a 'purely automorphic' approach to the the $\pi_1$ question is somewhat unlikely, at least within the current framework of the Langlands correspondence. </p> <p>I suppose I'm sabotaging my own question.</p> <hr> <p>*Note that in these situations, the Galois representations don't have to be unramified, since there is the choice of a coefficient field $\mathbb{Q}_p$. In general, they should only be <em>crystalline</em> at $p$.</p> <hr> <p>Added 14 July:</p> <p>Matthew: Since I didn't really expect a complete answer to my question, if you could write your extremely informative series of comments as an answer, I will accept it. (Barring the highly unlikely possibility that someone will write something better between the time you submit your answer and the time I look at it.)</p> http://mathoverflow.net/questions/47315/extension-of-g-bundles/47482#47482 Answer by Minhyong Kim for extension of $G$-bundles Minhyong Kim 2010-11-27T04:42:44Z 2011-11-24T18:05:25Z <p>Nick Shepherd-Barron once asked me this question, and I think I can remember what was eventually concluded. </p> <p>The short answer to the original question is negative for the additive group $\mathbb{G_a}$. Map $SL_2(\mathbb{C})$ to $X=\mathbb{C}^2-(0,0)$ by letting a matrix act on the vector $(1,0)^T$. This realizes the group as a torsor over $X$ for the group of unipotent matrices $\begin{pmatrix} 1 &amp; a \\ 0&amp; 1 \end{pmatrix}.$ We can trivialize it over the vectors $(v_1,v_2)^T$ with $v_1\neq 0$ with the section $\begin{pmatrix} v_1 &amp; 0 \\ v_2&amp; v_1^{-1} \end{pmatrix},$ while it trivializes on the set $v_2\neq 0$ via the section $\begin{pmatrix} v_1 &amp; -v_2^{-1} \\ v_2&amp; 0 \end{pmatrix}.$ The transition function on the overlap is easliy computed to be $\begin{pmatrix} 1 &amp; (v_1v_2)^{-1}\\ 0&amp; 1 \end{pmatrix}.$ This represents the standard non-trivial generator of $H^1(O_X)$, and hence, the bundle is non-trivial. On the other hand, if you could extend it to $\mathbb{C}^2$, it would trivialize, since there are no non-trivial $\mathbb{G}_a$-bundles on an affine variety.</p> <p>There is indeed a correspondence between principal bundles and tensor functors from representations to vector bundles. But if I recall correctly, the functor is required to be exact. In the case at hand, the extension of vector bundles is the direct image with respect to the inclusion $X\hookrightarrow \mathbb{C}^2$, which fails this.</p> <hr> <p>Added: OK, I see this was just an elaborate way to say: take any principal $\mathbb{G}_a$-bundle corresponding to a non-zero element of $H^1(O_X)$. Note, anyways, that the derived group is trivial in this example. Certainly the statement is false for general connected groups, contrary to some of the comments.</p> <hr> <p>Added, 25, November, 2011:</p> <p>This question came back to me today while I was thinking about something unrelated. It occurred to me then to point out that for the example above, if we work in the analytic category, we have $$H^1(X, \mathbb{G}_a)\simeq H^1(X, \mathbb{G}_m),$$ via the exponential sequence. On the other hand, $$H^1(\mathbb{C}^2, \mathbb{G}_a)=H^1(\mathbb{C}^2, \mathbb{G}_m)=0.$$</p> <p>So the desired extension property is false on analytic spaces even for reductive structure groups.</p> http://mathoverflow.net/questions/81259/on-the-periods-in-the-periodic-table-or-why-is-a-noble-gas-stable On the periods in the periodic table (or Why is a noble gas stable?) Minhyong Kim 2011-11-18T16:23:35Z 2011-11-21T15:02:11Z <p>Added 22, November:</p> <p>I've succeeded in making the question entirely unintelligible with all my additions. So I thought I would summarize it in the simplest form I could manage and add it to the title. The question is thereby somewhat narrower in scope than my original query, but I would be happy to have this focused version answered.</p> <p>The main point is that the loose answer </p> <p>(A) because the outer shell is filled</p> <p>frequently heard is definitely wrong. If we take the usual definition of a shell, all the noble gases but Helium have just the s and p subshells (corresponding to the representations $V_0\otimes S$ and $V_1\otimes S$) filled in the outermost shell, and this is not a full shell once the atom is bigger than Argon. The <a href="http://en.wikipedia.org/wiki/Noble_gas" rel="nofollow">wikipedia article on noble gases</a> offers an amusing formulation whereby, for a noble gas, 'the outer shell of valence electrons is <em>considered</em> to be "full"' (my emphasis).</p> <p>All this led to my initial confusion: I thought that the term 'shell' must mean something else for multi-electron systems in a manner adapted to the answer (A). Such an alternative definition does not seem to exist, and it is not at all obvious how to come up with one in a non-tautological way. So I believe the essence of what I am asking is whether or not there is </p> <p><em>a natural mathematical explanation for the stability of noble gases</em></p> <p>that is more or less independent of experiments confirmed by difficult computations using approximation schemes.</p> <p>Anyone interested in the background and details is invited to read the incoherent paragraphs below, or to look up a proper reference like Atkin's book on physical chemistry.</p> <hr> <p>Added, 21 November:</p> <p>Having read some more here and there, I should make some terminological corrections, in case I mislead anyone with my ignorance. I will do so here, and leave the text below as written, in case the temporary confusion is helpful to other interested non-experts. </p> <p>Firstly, as far as I can tell, the word 'orbital' does seem to refer to a wavefunction, not a representation. So the eigenfunctions in the second occurrence of the representation $V_1\otimes S$ will be called the $3p$-orbitals. That is, the individual wavefunction is an orbital, while the representation itself might be referred to as the so-and-so orbitals, in the plural. The definitive term for any given occurrence of an irreducible representation in $L^2\otimes S$ seems to be <em>subshell</em>.</p> <p>Secondly, I finally read the <a href="http://en.wikipedia.org/wiki/Electron_shell" rel="nofollow">wikipedia article on shells</a>. It doesn't help much with my questions, but it does describe the convention regarding the term 'shell'. As far as I can tell, the shells are simply the direct sums of the following form:</p> <p>K-shell: $1s$ </p> <p>L-shell: $2s\oplus 2p$ </p> <p>M-shell: $3s \oplus 3p \oplus 3d$ </p> <p>N-shell: $4s\oplus 4p \oplus 4d\oplus 4f$ </p> <p>O-shell: $5s\oplus 5p \oplus 5d\oplus 5f\oplus 5g$ </p> <p>and so on. (In case you're wondering, the representations after $V_3\otimes S=f$ are labeled consecutively in the alphabet.) The key point is that, in this form, a shell does not consist of a grouping of subshells of similar energies when more than a few electrons are present. Rather, the conventional description of the phenomena says that more than one shell can be incomplete in an atom. For example, in the fourth row of the periodic table, starting with potassium (K) and up to copper (Cu), they say both the $M$-shell and the $N$-shell are incomplete. If this is confusing to you, I suggest you don't worry about it. I just wanted to point out that my question what is a shell?' has a clear-cut answer in this usage. If we don't want to go against this convention, we need to stick to the version</p> <p>What determines a period?'</p> <p>I'm very sorry for all the confusion.</p> <hr> <p>Added, 19 November:</p> <p>After receiving the nice answers from Jeff and Antoine, I realized that I should sharpen the question somewhat more. I like Neil's formulation, but I think I am asking something much more naive. Allow me to start by providing more background for mathematicians whose knowledge is as hazy as mine. As mentioned by Jeff and Antoine, the Hamiltonian for a system of $N$ electrons moving around a nucleus of charge $Z$ looks like</p> <p>$$H=-\frac{\hbar^2}{2m}\sum_{i=1}^N \Delta_i-\sum_{i=1}^N \frac{Ze^2}{r_i}+ \sum \frac{e^2}{r_{ij}}.$$</p> <p>In principle, one would like to understand the structure of discrete spectrum solutions to the eigenvalue equation $$H\psi=E\psi.$$ One characterizes solutions using labels called 'quantum numbers'. In general, there is an <em>objective label</em> often denoted $l$, the <em>angular momentum quantum number</em>. This comes from the fact that there is an $SO(3)\times SU(2)$ symmetry, breaking up solutions into the irreducible representations $V_l\otimes S$, mentioned earlier. All the vectors in a given irreducible representation must have the same energy level $E$, because a basis for the space can be obtained from a highest weight vector by applying elements of $LieSO(3)$. Any given $V_l\otimes S$ is called a <em>subshell</em> or an <em>orbital</em> (I'm a bit unclear about this because the term 'orbital' is also used for the individual eigenfunctions) and is described using the (historical) labels</p> <p>$$V_0\otimes S \leftrightarrow s$$ $$V_1\otimes S \leftrightarrow p$$ $$V_2\otimes S \leftrightarrow d$$ $$V_3\otimes S \leftrightarrow f$$</p> <p>The <em>principal quantum number</em> is determined as follows. We order all the orbitals by energy levels, and the $n$-th time $V_0 \otimes S$ occurs, we call that subspace $ns$. However, the $n$-th time that $V_1\otimes S$ occurs, we call it $(n+1)p$. In general, the $n$-th time $V_l\otimes S$ occurs, the label is </p> <p>$(n+l+1)$(whatever letter $l$ corresponds to). </p> <p>For example, the orbital $3d$ is the first occurrence of the representation $V_2\otimes S$. When there is just one electron in a $1/r$ potential, these $n$ really label the order of the energy levels, and $d$ really occurs the first time in the third energy level. This is the reason for the shift in labeling, even though this correspondence with energy levels breaks down for multi-electron systems. In fact, in neutral atoms with $N=Z$, the energy levels are ordered like $$1s&lt;2s&lt;2p&lt;3s&lt;3p&lt;4s&lt;3d &lt;4p&lt;5s&lt;4d&lt;5p&lt;6s&lt; \ldots$$ (It's not supposed to be obvious how to fill in the $\ldots$. However, I've been told that knowing $4s&lt;3d$ is essential to passing A-level chemistry in England.) I hope it is clear from this that the orbitals (or the subshells) are completely well-defined and have a labeling scheme that is a bit odd, but makes sense when the obvious translation is combined with history. So the real question is</p> <p><em>What determines a shell?</em></p> <p>Some energy-non-decreasing consecutive sequence of subshells is a shell. The shells then determine the rows of the periodic table (if we ignore the added complication that $Z$ is also increasing as we move right and down). </p> <p>Now, it is tempting to conclude that the standard convention is a bit arbitrary. After all, the chemical similarity of the columns could possibly be explained simply by the fact that</p> <p><em>they have the same representation of $SO(3)\times SU(2)$ at the uppermost energy level, and the same number of electrons in this representation</em></p> <p>without any reference to an outermost shell at all. This pattern can be easily seen by the arrangement into blocks shown <a href="http://upload.wikimedia.org/wikipedia/commons/e/e5/Periodic_Table_structure.svg" rel="nofollow"> here</a>. And then, unlike the $N=1$ case, the energy levels in one shell are not even the same, just rather close to each other.</p> <p>Unfortunately, the grouping into rows represents real phenomena. This comes out very clearly, for example, in the <a href="http://upload.wikimedia.org/wikipedia/commons/3/32/Ionization_energies.svg" rel="nofollow">graph of ionization energies</a>, with the noticeable peaks at the end of the rows immediately followed by precipitous drops. So I believe a bit of thought reduces a good deal of my original question to two parts:</p> <ol> <li><p>Is there some reason for a big gap in ionization energy when one moves to an $s$-orbital?</p></li> <li><p>Can one show that two orbitals of the same type are necessarily separated by a huge energy gap, so that it is highly unlikely for them to occur in the same shell?</p></li> </ol> <p>1 and 2 together make it natural that the dimensions we see in each shell will be of the form</p> <p>2, 2+6, 2+6+10, 2+6+10+14, etc,</p> <p>even in the general case. Of course, this doesn't say anything about how many times each combination is likely to occur.</p> <p>In any case, I hope there is a mathematical answer to these two questions that doesn't involve a full-blown programme in hard analysis.</p> <p>At the more speculative end, one might analyze a family of operators like $$H_{\epsilon}=-\frac{\hbar^2}{2m}\sum_{i=1}^N \Delta_i-\sum_{i=1}^N\frac{Ze^2}{r_i}+\epsilon \sum \frac{e^2}{r_{ij}}.$$ Is there a way to see the numbers we see occurring via the spectral flow of this family as we go from $\epsilon =0$ to $\epsilon=1$? But maybe this is just as difficult as giving a full account of the structures using analysis.</p> <p>By the way, I certainly wouldn't like to complain, but it is a bit puzzling to me why some people regard this question as inappropriate for the site. Since I don't keep too well in touch with cultural trends in the mathematical world, maybe I am unaware of how much things have changed since I was a Ph.D. student. In those days, a programme like</p> <p><em>Prove the stability of matter, based only on the Schroedinger equation</em></p> <p>was regarded as an example of an important mathematical problem motivated by atomic structure, tackled by people like Fefferman and Lieb. The questions I ask here are hopefully much easier, but still research-level mathematics to my mind.</p> <hr> <p>Original question:</p> <p><a href="http://mathoverflow.net/questions/80146/is-the-mendeleev-table-explained-in-quantum-mechanics" rel="nofollow">This question</a> prompts me to ask something more specific about the periodic table. As far as I know, the main significance of the periodic table is that </p> <p><em>The elements in the same column have similar chemical properties.</em></p> <p>For example, the noble gases at the far right are all pretty stable on their own. The explanation for this is that they have (in the neutral state) a full outermost shell. Now my question is</p> <p><em>Is there an explanation, at some reasonable level of mathematical rigor, of when a new shell starts?</em></p> <p>That is, where do the lengths of the periods</p> <p>2, 8, 8, 18, 18, 32, 32</p> <p>come from? I confess I've been puzzled by this ever since my university physics course.</p> <p>Allow me to pinpoint my confusion a bit more. I understand that there are some numbers that are important in atomic structure, and these are 2, 6, 10, 14, and so on. This is because of the occurrence of the representations </p> <p>$$V_l\otimes S$$</p> <p>inside the Hilbert space for a single particle moving in a central potential. These are the <em>orbitals</em> one hears about in physical chemistry courses. Here, the $V_l$ are the representations of $SO(3)$ of odd-dimension $2l+l$, while $S$ is the standard two-dim representation of $SU(2)$. Since all the states in a single orbital have the same energy, it is natural that the breaks will occur after some collection of orbitals are all filled.</p> <p>Thus, for a hydrogen atom, the successive shells have dimensions</p> <p>$2n^2$</p> <p>for $n=1, 2, 3, \ldots$ because the representations $V_l\otimes S$ for $l=1, 2,\ldots, n-1$ occur each with multiplicity one inside the $n$-th shell.</p> <p>So if the periods in the table were of lengths</p> <p>2, 8, 18, 32,</p> <p>I would have vaguely assumed that the pattern of shells even in general looks like that of the hydrogen atom. But of course, among the known elements, each period length that occurs in the hydrogen atom is repeated twice, except for the first one. So a more precise question is</p> <p><em>Is there a reasonable mathematical explanation of this multiplicity two' of the periods?</em></p> <p>The little I recall of discussions in standard textbooks were quite unclear. There are various rules by the name of Hund's rule, the <a href="http://en.wikipedia.org/wiki/Aufbau_principle" rel="nofollow"> Aufbau principle</a>, and so on, but I couldn't gather from any of them</p> <p>*Where the breaks should occur. *</p> <p>What I do see is that periods end when the orbitals 1p, 2p, 3p, etc. get filled following the Aufbau principle. (Here, p is the chemist's label for $V_1\otimes S$.) So perhaps another version of the question is</p> <p><em>Is there some reason that the p-orbitals mark the end of the periods?</em></p> <p>To be honest, I've never understood the Aufbau principle either, because I don't know the rationale behind the principal quantum number for the larger atoms. That is, the number 4 in the orbital 4p refers to the 4-th energy level in the case of the hydrogen atom. But for larger nuclei, the $n$ in orbital '$np$' does not refer to the energy level. (This discrepancy is in fact implied by the Aufbau principle.) So what is the significance of the $n$ in general that enables them to play some role in a physical principle?</p> <p>I realize this question is becoming incoherent already. Nevertheless, I would very much appreciate clarification on any sensible version of it at a level of mathematical rigor of your choice. (I am not asking for any axiomatics.) Pointers to an accessible reference would be equally welcome.</p> <hr> <p>As with the earlier question, a word of explanation is in order on the decision to post this on Math Overflow. I will draw upon an analogy I read long ago in an article of George Mackey's that went something like this: Say my mother tongue is Korean. If I would really like to use English fluently, it is probably best eventually to learn from native speakers of English. On the other hand, if I would like a <em>good translation</em> into Korean of English literature, it is better to consult an educated Korean who knows a lot about English. Of course answers from real physicists will be very gratefully received, especially if they bear in mind that the query comes from someone who struggles against a serious language handicap.</p> http://mathoverflow.net/questions/77635/what-exactly-is-the-relation-between-string-theory-and-conformal-field-theory What exactly is the relation between string theory and conformal field theory? Minhyong Kim 2011-10-10T00:12:37Z 2011-10-13T01:49:08Z <p>Maybe it would be helpful for me to summarize the little bit I think know. A 2D CFT assigns a Hilbert space ${\cal H}$ to a circle and an operator $$A(X): {\cal H}^{\otimes n}\rightarrow {\cal H}^{\otimes m}$$ to a Riemann surface $X$ with $n$ incoming boundaries and $m$ outgoing boundaries. This data is subject to natural conditions arising from the sewing of surfaces.</p> <p>Here is how I understand the relation to string theory. The Hilbert space ${\cal H}$ might be the space of functions on the configuration space of a string sitting in a manifold $M$. So ${\cal H}=L^2(Maps(S^1,M))$ with some suitable restrictions on the maps. It is natural then that the functions on the configuration space of $n$ circles is ${\cal H}^{\otimes n}$. Now we consider $n$ strings evolving into $m$ strings. There are many ways to do this, one for each Riemann surface $X$ as above. When $X$ is fixed, $A(X)$ is the evolution operator, usually described in terms of some path integral over maps from $X$ into $M$ involving a conformally invariant functional. </p> <p>All this makes a modicum of sense. So ${\cal H}$ is the Hilbert space arising from quantization of the cotangent bundle of $Map(S^1,M)$, while $A$ describes time evolution. So in this sense, such a conformal field theory appears to be the quantization of the classical string. I guess what is missing in my description up to here is the prescription for turning functions on $T^*Map(S^1,M)$ into operators on ${\cal H}$. To my deficient understanding, maybe this situation corresponds to having quantized only the Hamiltonian.</p> <p>Now, what I was really wondering was this: When I was in graduate school, I remember frequently hearing the phrase:</p> <p><em>CFT theory is the space of classical solutions to string theory.</em></p> <p>Does this make some sense? And if so, what does it mean? This phrase has been hindering my understanding of conformal field theory ever since, making me feel like my grasp of physics is all wrong. According to the paragraphs above, my naive formulation would have been:</p> <p><em>Quantization of a string theory gives rise to a CFT.</em></p> <p>What is wrong with this naive point of view? If you could provide some enlightenment on this, you'll have resolved a long-standing cognitive itch in the back of my mind.</p> <p>Thanks in advance.</p> <hr> <p>Added:</p> <p>As Jose suggests, I could simply be remembering incorrectly, or misunderstood before what I heard. That, in fact, is what I had hoped to be the case. But read, for example, the first page of Moore and Seiberg's famous paper "Classical and quantum conformal field theory":</p> <p><a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.cmp/1104178762" rel="nofollow">http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.cmp/1104178762</a></p> <hr> <p>Added again:</p> <p>To quote Moore and Seiberg more precisely, the second sentence of the paper reads 'Conformal field theories are classical solutions of the string equations of motion.' Now, I might attempt to understand this as follows. When the Riemann surface is $$S^1\times [0,t]$$(with the conformal structure induced by the standard metric) one interprets $$A(S^1\times [0,t])$$ as $$e^{itH}.$$ Thus, when applied to a vector $\psi_0\in {\cal H}$, the theory would generate a solution to Schroedinger's equation $$\frac{d}{dt}\psi =iH \psi$$ with initial condition $\psi_0$ as $t$ varies. So one might think of the various $A(X)$ as $X$ varies as being 'generalized solutions' to Schroedinger's equation for a quantized string. I suppose I could get used to such an idea (if correct). But then, the question remains: why do they (and others) say <em>classical</em> solutions? Is there some kind of second quantization in mind with this usage?</p> <hr> <p>Added, 11 October:</p> <p>Even at the risk of boring the experts, I will have one more go. Jeff Harvey seems to indicate the following. We can think of $Map(X, M)$ as the fields in a non-linear sigma model on $X$, provisionally thought of as 1+1 dimensional spacetime. However, it seems that one can also associate to the situation a space of fields <em>on $M$</em> (the string fields?). If we denote by ${\cal F}$ this space of fields, it seems that there is a functional $S$ on ${\cal F}$ with the property that the extrema of $S$ (the 'string equations of motion') can be interpreted as the $A(X)$. From this perspective, my main question might then be 'what is ${\cal F}$?' Since I think of fields on $M$ as being sections of some bundle on $M$, I can't see how to get such a thing out of maps from $X$ <em>to</em> $M$. </p> <p>Thank you very much for your patience with these ignorant questions.</p> <hr> <p>Final addition, 11 October:</p> <p>Thanks to the kind guidance of Jose, Aaron, and especially Jeff, I think I have some kind of an understanding of the situation. I will attempt to summarize it now, superficial as my knowledge obviously is. I don't wish to waste more of the experts' time on this question. However, I am hoping that truly egregious errors will offend their sensibilities enough to elicit at least a cry of outrage, enabling me to improve my poor understanding. I apologize in advance for putting down even more statements that are either trivial or wrong.</p> <p>As far as I can tell, the sense of Moore and Seiberg's sentence is as in my second addition: it is referring to second quantization. Recall that in this process, the single particle wave functions become the classical fields, and Schroedinger's equation is the classical equation of motion. Now the truly elementary point that I was missing (as I feared), is that </p> <p><em>quantization of a 'single particle' string theory cannot give you a conformal field theory.</em></p> <p>At most, a single string will propagate though space, giving us exactly the operators $A(S^1\times [0,t])$. If we want operators $$A(X):{\cal H}^{\otimes n}\rightarrow {\cal H}^{\otimes m}$$ corresponding to a Riemann surface with many boundaries, then we are already requiring a theory where particle numbers can change, that is, a quantum field theory coming from second quantization. WIth such a theory in place, of course, the $A(S^1\times [0,t])$ are exactly the solutions to the classical equations of motion, while the general $A(X)$ can be viewed either as 'generalized classical solutions' (I hope this expression is reasonable) or contributions to a perturbation series, as in the field theory of a point particle. So this, I think. already answers my original question. To repeat, because of the changing 'particle number'</p> <p><em>the operators of conformal field theory cannot be the quantization of a 'single particle' theory. They must be construed as classical evolution operators of some kind of quantum (string) field theory.</em></p> <p>The part I'm still far, far from understanding even superficially is this: The classical fields in the case of strings would be something like functions on $Map(S^1, M)$. I haven't the vaguest idea of how to get from this to fields on spacetime. The difficulty surrounding this issue seems to be discussed in the beginning pages of Zwiebach's paper referred to by Jeff, which is quite heavy reading for a pure mathematician like me. Some mention is made of infinitely many fields arising from the situation (alluded to also by Jeff), which perhaps is some way of turning the data of a function on loop space to fields on space(-time).</p> http://mathoverflow.net/questions/23264/are-c-and-barq-p-isomorphic Are $C$ and $\bar{Q_p}$ isomorphic? Minhyong Kim 2010-05-02T16:57:10Z 2011-08-13T04:47:29Z <p>There is a famous passage on the third page of Deligne's second paper on the Weil conjectures where he expresses his dislike of the axiom of choice, as manifested in the isomorphism between $C$ and $\bar{Q_p}$. The proof of said isomorphism runs as follows. Both $C$ and $\overline{Q_p}$ have transcendence bases, $S$ and $T$. Then $C\simeq \overline{Q(S)}$ and $\overline{Q_p}\simeq \overline{Q(T)}$. But $C$ and $\overline{Q_p}$ have the same cardinality, and hence, so do $S$ and $T$. Therefore, $Q(S)\simeq Q(T)$ and, from there, $C\simeq \overline{Q_p}$. </p> <p>For myself, this proof is quite convincing. Recently, Torsten Ekedahl expressed his opinion to the contrary, and this led to the following exchange:</p> <p><a href="http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice/22944#22944" rel="nofollow">http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice/22944#22944</a></p> <p>So I wondered about other expert opinions on this matter. Do you find the isomorphism unbelievable and, if so, why? </p> http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture/69714#69714 Answer by Minhyong Kim for Why should I believe the Mordell Conjecture? Minhyong Kim 2011-07-07T12:36:50Z 2011-07-07T23:31:20Z <p>Since summer is here, I will bore people a bit with my own ahistorical view of Mordell. The main point is that Serge Lang's strategy, mentioned by several people, is essentially correct. Nevertheless, we might add that</p> <p><em>you should believe the Mordell conjecture because most local principal bundles should not extend to global principal bundles.</em></p> <p>I've spoken about this in one form or another many times, but maybe this particular sentence hasn't received enough emphasis. So I will explain it. </p> <p>Jean-Benoit Bost has pointed out that this story is not entirely devoid of historical context, since Andre Weil's famous 1938 paper on vector bundles can be construed as giving motivation roughly of this nature.</p> <hr> <p>Recall that Lang's idea is to consider the diagram</p> <p>$$\begin{array}{ccc} X(F)&amp;\longrightarrow &amp; X(\mathbb{C}) \end{array}$$ $$\begin{array}{lcr} \downarrow \ \ \ \ &amp; &amp; \ \ \ \ \downarrow\end{array}$$ $$\begin{array}{ccc} J(F) &amp;\longrightarrow &amp; J(\mathbb{C})\ \end{array}$$</p> <p>for a smooth projective curve $X$ of genus $\geq 2$ with Jacobian $J$ and an embedding $F\hookrightarrow \mathbb{C}$ of the number field $F$ into the complex numbers. Lang suggested that</p> <p>$$X(\mathbb{C})\cap J(F)\subset J(\mathbb{C})$$</p> <p>should be finite. This is indeed plausible and turns out to be true after Faltings proof. Of course we are considering its status as motivation rather than corollary, following the original question posted.</p> <p>In fact, the plausibility is strengthened when we replace the diagram above by a refinement $$\begin{array}{ccc} X(F)\ \ \ \ \ \ \ \ \ \ &amp;\longrightarrow &amp;\ \ \ \ \ \ \ X(F_v) \end{array}$$ $$\begin{array}{ccc} \downarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ &amp; &amp;\ \ \ \ \ \ \ \ \ \ \ \downarrow j \end{array}$$ $$\begin{array}{ccc} H^1_f(G_S, \pi_1(\bar{X}, b)) &amp;\stackrel{loc}{\longrightarrow} &amp; H^1_f(G_v, \pi_1(\bar{X}, b)) \end{array}$$ and try to prove that $$Im(j)\cap Im(loc) \subset H^1_f(G_v, \pi_1(\bar{X}, b))$$ is finite. Here, $$\pi_1(\bar{X}, b)$$ is some $\mathbb{Q}_p$-algebraic fundamental group of $\bar{X}$ with base-point $b\in X(F)$. The completion $F_v$ should be taken to have degree one over $\mathbb{Q}_p$. The $H^1$'s are moduli spaces of (locally constant) principal bundles for $\pi_1(\bar{X}, b)$, a global one$$H^1_f(G_S, \pi_1(\bar{X}, b))$$ consisting of principal bundles over some $Spec(O_F[1/S])$ and a local one $$H^1_f(G_v, \pi_1(\bar{X}, b)) \simeq \mathbb{A}^N_{\mathbb{Q}_p},$$ (almost naturally) isomorphic to affine space, consisting of principal bundles on $Spec(F_v)$ (satisfying some technical condition).</p> <p>Thus, we are replacing $\mathbb{C}$ by a non-Archimedean completion and the Jacobian (a moduli space of line bundles) by a moduli space of principal bundles. The vertical maps assign to a point $x$ the principal bundle of paths $$\pi_1(\bar{X};b,x).$$ This framework turns out to refine considerably the intuition that $J(F)$ and $X(\mathbb{C})$ have very different natures inside $J(\mathbb{C})$.</p> <hr> <p>Now, <em>why should the Mordell conjecture be true?</em></p> <p>There are two steps.</p> <p>A. The easy one: The map $j$, being a non-Archimedean period map, is highly transcendental, and maps $X(F_v)$ to a Zariski-dense compact analytic curve in $H^1_f(G_v, \pi_1(\bar{X}, b))$, which therefore meets any proper subvariety in finitely many points. One proves this by showing that certain transcendental functions on $X(F_v)$ (the coordinates of the map) are algebraically independent. Meanwhile, the localization map $$loc: H^1_f(G_S, \pi_1(\bar{X}, b))\longrightarrow H^1_f(G_v, \pi_1(\bar{X}, b)),$$ is algebraic, and hence, has constructible image in the Zariski topology. These are the different natures alluded to in the previous paragraph: $$\begin{array}{ccccc} H^1_f(G_S, \pi_1(\bar{X},b)) &amp;\stackrel{\scriptstyle \mbox{algebraic}}{\longrightarrow} &amp; H^1_f(G_v, \pi_1(\bar{X},b)) &amp; \stackrel{\scriptstyle \mbox{dense analytic}}{\longleftarrow}&amp; X(F_v)\end{array}$$</p> <p>Therefore, it suffices to show:</p> <p>B. The hard step: $Im(loc)$ is not Zariski dense, that is, the localization map is not dominant.</p> <p>Now, why should this be true? Well, the moduli space $$H^1_f(G_v, \pi_1(\bar{X}, b))$$ consists of <em>local</em> principal bundles, while $$H^1_f(G_S, \pi_1(\bar{X}, b))$$ is a moduli space of <em>global</em> principal bundles. So it makes sense that most local bundles should not extend to global ones when the group $\pi_1(\bar{X}, b)$ is sufficiently large and non-abelian. (Perhaps you will disagree...)</p> <p>These thoughts were actually inspired by Yang-Mills theory: We have something like local solutions to the Yang-Mills equation, that is, on a small annulus (or a handle-body) embedded in a Riemann surface (or a three-manifold). It seems they should not all extend to global solutions. Natural enough, but quite hard to prove in general. Galois cohomology, on the whole, appears to be harder than Yang-Mills theory. One motivation for posting this answer is the hope that some bright young person will have an idea.</p> <hr> <p>Added: As mentioned by Matthew Emerton, the strategy outlined above is an extension of Chabuaty's method, which quite likely inspired Lang's conjecture as well (according to one reading of the notes to <em>Fundamentals of Diophantine Geometry</em>). There, the analogue of $H^1_f(G_v, \pi_1(\bar{X},b))$ is $$T_eJ(F_v),$$ the Lie algebra of the Jacobian, while the role of the global moduli space is played by $$J(F)\otimes_{\mathbb{Z}}F_v.$$ My own feeling is that the 'local vs. global perspective' that emerges out of the principal bundle interpretation is somehow critical to understanding the Mordell conjecture, and constitutes a natural generalization not just of Chabauty's method, but of the arithmetic theory of curves of genus zero and one. In this sense, the essential motivation for Mordell's conjecture should not just be probabilistic, but something rather precise coming out of <em>class field theory.</em> It's fairly clear that this couldn't have been Mordell's reason for believing in it, but it is plausible, as mentioned above, that it was Weil's reason, in spite of his eventual non-committal assessment.</p> http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/68444#68444 Answer by Minhyong Kim for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications Minhyong Kim 2011-06-21T22:20:30Z 2011-06-22T15:55:44Z <p>I believe the following is an elementary example: Let $X$ be an affine smooth curve of geometric genus at least one. Let $L$ be a non-trivial algebraic line bundle on $X$ (easy to produce such things). Then $L$ is analytically trivial because $X$ is a Stein space ($H^1(X, O_X)=0$) with trivial integral $H^2$. Hence, there is an analytic isomorphism $L\simeq X\times A^1$. We see that there is no algebraic isomorphism by noting:</p> <hr> <p>Suppose there is an isomorphism $$f: L\simeq X\times A^1$$ of algebraic varieties. Then $L$ and $X\times A^1$ are isomorphic as algebraic line bundles.</p> <hr> <p>Proof: For any fiber $L_y$ of $L$, if we consider the composite $$p\circ f: L_y\rightarrow X\times A^1 \stackrel{p}{\rightarrow} X$$ of $f$ with the projection, it must be constant, since $X$ is not rational. Hence, we have $$f(L_y)\subset z\times A^1$$ for some point $z$. Since the map $L_y\rightarrow A^1$ thus obtained is injective, it must be of the form $ax+b$ for non-zero $a$, that is, $f$ induces an isomorphism $$L_y\simeq z\times A^1.$$ Also by injectivity, we see that $y\neq y'$ implies $f(L_y)=z\times A^1$ and $f(L_{y'})=z'\times A^1$ for $z\neq z'$. Now let $s:X\rightarrow L$ be the zero section. Then $$\phi=p\circ f\circ s: X\rightarrow X$$ is an injective map, and hence, an automorphism. Thus, $$(\phi^{-1}\times 1)\circ f:L\rightarrow X\times A^1 \stackrel{\phi^{-1}\times 1}{\rightarrow} X\times A^1$$ is a map preserving the fibers of the projections to $X$. Now let $$q:X\times A^1 \rightarrow A^1$$ be the other projection, and $$h=q\circ (\phi^{-1}\times 1)\circ f\circ s.$$ So $h(y)$ is the image in $A^1$ under $(\phi^{-1}\times 1)\circ f$ of the origin of $L_y$. We use this function to get a fiber-preserving isomorphism $g: X\times A^1\simeq X\times A^1$ that sends $(y,\lambda )$ to $(y, \lambda-h(y))$. So finally, $$g\circ (\phi^{-1}\times 1)\circ f: L\simeq X\times A^1$$ is a fiber-preserving isomorphism of varieties that furthermore preserves the origins of each fiber. It must then be fiber-wise an isomorphism of vector spaces. Thus, it is an isomorphism of line bundles.</p> <hr> <p>Added: The argument above can be easily modified to show that if $X$ is an irrational smooth curve and $L$ and $M$ are line bundles on $X$, then any isomorphism of algebraic varieties $$f:L\simeq M$$ is of the form $$f=T_s\circ \tilde{\phi} \circ g$$ where $$\tilde{\phi}:\phi^*M\rightarrow M$$ is the base-change map for an automorphism $\phi$ of $X$, $$g:L\simeq \phi^*M$$ is an isomorphism of line bundles, and $$T_s:M\rightarrow M$$ is translation by a section $s:X\rightarrow M$ of $M$.</p> <p>Since the algebraic automorphism group of an affine irrational curve is finite, we see, by varying $L$, that for $X$ as above, there is in fact a </p> <p><em>continuum</em> of distinct algebraic structures </p> <p>on the analytic space $X\times A^1$.</p> http://mathoverflow.net/questions/62401/logic-in-mathematics-and-philosophy/63020#63020 Answer by Minhyong Kim for Logic in mathematics and philosophy Minhyong Kim 2011-04-26T09:19:42Z 2011-04-26T15:32:58Z <p>There was some comment in the meta thread to the effect that answers from non-experts are useless. Nevertheless, I thought to indulge myself with a few remarks (that actually address only a small part of the question). The objective is to get experts to correct me, if they can be troubled to do so. With such a contribution, perhaps this answer can be useful to others as well. </p> <p>My answer to the original question is that there are no substantial relations at this point. This is because</p> <p><em>mathematical logic is concerned mainly with mathematics;</em></p> <p>while</p> <p><em>philosophical logic is concerned mainly with philosophy.</em></p> <p>This characterization is obviously an oversimplification, but I wonder if it does not capture the distinction in all essentials.</p> <p>Here is a short Wikipedia description of the research of A. C. Grayling, who, I gather, is a rather distinguished person in philosophical logic:</p> <p>His principal interests in technical philosophy lie at the intersection of theory of knowledge, metaphysics, and philosophical logic, through which he attempts to define the relationship between mind and world, thereby challenging philosophical scepticism. Grayling uses philosophical logic to counter the arguments of the sceptic in order to try to shed light on the traditional ideas of the realism debate and developing associated views on truth and meaning.'</p> <p>This sounds like philosophy to me.</p> <p>On the other hand, if we examine the work of leading people in model theory (the only branch of mathematical logic with which I have some passing acquaintance) like Udi Hrushovski, Angus Macintyre, Anand Pillay, and Boris Zilber, it is hard not to think it looks like generalized algebraic geometry.' Indeed, applications to algebraic geometry and number theory form a mainstay of their work.</p> <p>As to the reasons, I gathered some insight from some amusing passages in the preface of Van Dalen's textbook on logic (which I do not have on hand right now). He writes of the 'sacred' tradition of mathematical logic closely related to Hilbert's programme and the incompleteness phenomena, where foundations were handled with great care and awe. He then goes on to describe his own encounter with recursion theory lectures by Hartley Rogers, where logic was treated like any other branch of mathematics, say linear algebra or complex analysis. This he refers to as the 'profane' tradition, obviously more distant from philosophical origins. I wonder if it isn't the case that mathematical logicians simply became bored with the sacred tradition (in keeping with the twentieth century trend to find many sacred things boring). In any case, it seems relatively clear that the profane tradition is more dominant among current day practitioners. One way to see this, according to an old conversation with Hrushovski, is that papers in mathematical logic contain as many mistakes as those in algebraic geometry.</p> <p>Because of my ignorance, I may be missing the possibility that Proof Theory and Set Theory are still somewhat close to philosophy. But the simple-minded answer still seems to be a reasonable one.</p> <hr> <p>Just after posting the above, I noticed an obvious flaw in my own argument. I could have written, for example,</p> <p><em>mathematical gauge theory is concerned mainly with mathematics;</em></p> <p>while</p> <p><em>physical gauge theory is concerned mainly with physics.</em></p> <p>But it would be ridiculous to claim that there are no substantial relations between the two. So if my conclusion is correct, it would require a more elaborate discussion. Oh well, perhaps later.</p> <hr> <p>Here is a response to my own objection. The difference between the two cases mentioned above has little to do with logic and gauge theory in particular. That is, mathematical logic and philosophical logic have little in common simply because mathematics and philosophy have little in common. Therefore, <em>difference of purpose</em> is enough to produce a divergence in methods and ideas. Physical problems, on the other hand, are resolved in the language of mathematics. Hence, <em>commonality of origin</em> becomes enough to maintain a tight thread between, say, the two gauge theories.</p> <p>This superficial analysis is all I have time for now, but maybe it's plausible.</p> http://mathoverflow.net/questions/57657/p-adic-integrals-and-cauchys-theorem/57781#57781 Answer by Minhyong Kim for $p$-adic integrals and Cauchy's theorem Minhyong Kim 2011-03-08T06:09:51Z 2011-03-08T13:12:27Z <p>There is an important difference, relevant to the original question, between the two kinds of $p$-adic integrals mentioned by Kevin in his comments. Because I see frequent confusion on this issue, I thought I'd comment.</p> <p>The 'usual' $p$-adic integrals as you might see in, say, Tate's thesis on L-functions or the adelic theory of automorphic forms, are <em>volume</em> integrals, with respect to a measure, typically on some group. This kind of volume integral can also be easily defined on arbitary varieties, and you will see plenty in Weil's book on Tamagawa numbers, or in papers on motivic integration. Coleman integration, on the other hand, is a $p$-adic analogue of <em>line integrals</em>, and comes up most naturally in discussing the holonomy of vector bundles with connection on a variety over a $p$-adic field (often interpreted as isocrytals). These, therefore, should be the right quantities to relate to a Cauchy formula. However, unfortunately (and fortunately), it doesn't work. The reason is that Coleman integration is a line integral along a <em>canonical path</em> between two points on a variety over the $p$-adics. So there is a canonical holonomy in the theory, at least if you just want to compute it for a bundle with unipotent connection, that is, one that has a strictly upper-triangular connection form. This is where a mysterious 'crystalline' structure on the space of paths is used, whereby there is a unique path invariant under the action of the Frobenius. The notion of a path, by the way, uses the Tannakian formalism in this context. For a very quick overview of this approach, you can look at section 2 of this paper: <a href="http://www.ucl.ac.uk/~ucahmki/siegelinv.pdf" rel="nofollow">http://www.ucl.ac.uk/~ucahmki/siegelinv.pdf</a></p> <p>Breuil's paper linked from Chandan's answer should provide a more systematic overview.</p> <p>Anyways, because of the canonical paths in Coleman's theory, there can be no holonomy around a loop, and hence, no Cauchy formula. I was told quite a few years ago by Berkovich that he has a theory of line integrals on Berkovich spaces that are path dependent in interesting ways, but I've never looked into it.</p> <p>Added: I realize I didn't mention above the connection between holonomy and usual integration of a one-form $A$. You get this by considering the connection $$d+\begin{bmatrix}0&amp; A; \ 0&amp; 0\end{bmatrix}$$</p> <p>on the trivial bundle of rank two. One view of Coleman integration is that the holonomy $H_a^b$ from $a$ to $b$ is defined first. And then, the naive integral is defined by the fomula $$H_a^b=\begin{bmatrix}1&amp; \int_a^bA ;\ 0&amp; 1\end{bmatrix}$$</p> http://mathoverflow.net/questions/49960/are-class-numbers-encoded-in-the-absolute-galois-group-of-mathbb-q/49967#49967 Answer by Minhyong Kim for Are class numbers encoded in the absolute Galois group of ${\mathbb Q}$? Minhyong Kim 2010-12-20T15:05:28Z 2010-12-20T17:51:39Z <p>Dear Tim,</p> <p>As you're probably aware, this is part of the 'anabelian' etcetera. </p> <p>It suffices to recover all intertia subgroups $I_v\subset H$, because their union will then be a normal subgroup $N$ such that $H/N$ is the Galois group of the maximal extension of $K$ unramified everywhere. We can get the ideal class group then by (topological) abelianization. The fact that we can get all the decomposition groups $D_v\subset G$ is Neukirch's theorem (together with Artin-Schreier at infinity). This says the maximal subgroups isomorphic to a local Galois group are exactly the decomposition groups. If you want to make this purely group-theoretic for the finite places, you invoke the theorem of Jannsen-Wingberg that lays out a presentation for all local Galois groups and consider maximal elements in the lattice of subgroups isomorphic to such an explicit presentation. Once you have the $D_v$, there is a standard group-theoretic recipe for $I_v$, which escapes me for the moment. But I'll get back to you with it, if you don't figure it out in the meanwhile.</p> <p>Added:</p> <p>OK, so here is the easy part. Now let $F$ be a finite extension of $\mathbb{Q}_p$ and $D=Gal(\bar{F}/F)$. We know that $D^{ab}$ fits into an exact sequence $$0\rightarrow U_F\rightarrow D^{ab}\rightarrow \hat{\mathbb{Z}}\rightarrow 0,$$ so we recover $p$ as the unique prime such that the topological $\mathbb{Z}_p$-rank of $D^{ab}$ is $r_D\geq 2$. The order $q_D$ of the residue field is 1 greater than the order of the prime-to-$p$ torsion subgroup of $D^{ab}$. Also, we know $r_D=1+[F:\mathbb{Q}_p]$. Now we apply the same reasoning to the subgroups of finite index in $D$ to figure out those corresponding to unramified extensions. That is, consider the subgroups $E$ of finite index such that $q^{r_D-1}_E=q_D^{r_E-1}$. Then the inertia subgroup of $D$ is the intersection of all of these.</p> http://mathoverflow.net/questions/48622/has-the-mathematical-content-of-grothendiecks-recoltes-et-semailles-been-used/48946#48946 Answer by Minhyong Kim for Has the mathematical content of Grothendieck's "Récoltes et Semailles" been used? Minhyong Kim 2010-12-10T17:04:23Z 2010-12-14T08:51:03Z <p>Begging your pardon for indulging in some personal history (perhaps personal propaganda), I will explain how I ended up applying R\'ecollte et Semaille. I do apologize in advance for interpreting the question in such a self-centered fashion!</p> <p>I didn't come anywhere near to reading the whole thing, but I did spend many hours dipping into various portions while I was a graduate student. Serge Lang had put his copy into the mathematics library at Yale, a very cozy place then for hiding among the shelves and getting lost in thoughts or words. Even the bits I read of course were hard going. However, one thing was clear even to my superficial understanding: Grothendieck, at that point, was dissatisfied with motives. Even though I wasn't knowledgeable enough to have an opinion about the social commentary in the book, I did wonder quite a bit if some of the discontent could have a purely mathematical source. </p> <p>A clue came shortly afterwards, when I heard from Faltings Grothendieck's ideas on anabelian geometry. I still recall my initial reaction to the section conjecture: Surely there are more splittings than points!' to which Faltings replied with a characteristically brief question:' Why?' Now I don't remember if it's in R&amp;S as well, but I did read somewhere or hear from someone that Grothendieck had been somewhat pleased that the proof of the Mordell conjecture came from outside of the French school. Again, I have no opinion about the social aspect of such a sentiment (assuming the story true), but it is interesting to speculate on the mathematical context. </p> <p>There were in Orsay and Paris some tremendously powerful people in arithmetic geometry. Szpiro, meanwhile, had a very lively interest in the Mordell conjecture, as you can see from his writings and seminars in the late 70's and early 80's. But somehow, the whole thing didn't come together. One suspects that the habits of the Grothendieck school, whereby the six operations had to be established first in every situation where a problem seemed worth solving, could be enormously helpful in some situations, and limiting in some others. In fact, my impression is that Grothendieck's discussion of the operations in R&amp;S has an ironical tinge. [This could well be a misunderstanding due to faulty French or faulty memory.] Years later, I had an informative conversation with Jim McClure at Purdue on the demise of sheaf theory in topology. [The situation has changed since then.] But already in the 80's, I did come to realize that the motivic machinery didn't fit in very well with homotopy theory. </p> <p>To summarize, I'm suggesting that the mathematical content of Grothendieck's strong objection to motives was inextricably linked with his ideas on homotopy theory as appeared in 'Pursuing Stacks' and the anabelian letter to Faltings, and catalyzed by his realization that the motivic philosophy had been of limited use (maybe even a bit of an obstruction) in the proof of the Mordell conjecture. More precisely, motives were inadequate for the study of points (the most basic maps between schemes!) in any non-abelian setting, but Faltings' pragmatic approach using all kinds of Archimedean techniques may not have been quite Grothendieck's style either. Hence, arithmetic homotopy theory.</p> <p>Correct or not, this overall impression was what I came away with from the reading of R&amp;S and my conversations with Faltings, and it became quite natural to start thinking about a workable approach to Diophantine geometry that used homotopy groups. Since I'm rather afraid of extremes, it was pleasant to find out eventually that one had to go back and find some <a href="http://www.ucl.ac.uk/~ucahmki/siegelinv.pdf" rel="nofollow"> middle ground</a> between the anabelian and motivic philosophies to get definite results.</p> <p>This is perhaps mostly a story about inspiration and inference, but I can't help feeling like I did apply R&amp;S in some small way. (For a bit of an update, see my paper with Coates <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;page=current&amp;handle=euclid.kjm" rel="nofollow">here</a>.)</p> <hr> <p>Added, 14 December: I've thought about this question on and off since posting, and now I'm quite curious about the bit of R&amp;S I was referring to, but I no longer have access to the book. So I wonder if someone knowledgeable could be troubled to give a brief summary of what it is Grothendieck really says there about the six operations. I do remember there was a lot, and this is a question of mathematical interest.</p> http://mathoverflow.net/questions/46815/which-curves-have-infinitely-many-rational-points/49058#49058 Answer by Minhyong Kim for Which curves have infinitely many rational points Minhyong Kim 2010-12-11T17:57:42Z 2010-12-11T22:57:04Z <p>Dear Tim: </p> <p>How right you are to worry about computing the Jacobian of a genus one curve! Years ago, in the midst of many discussions with Bill McCallum, Alex Perlis, Nick Shepherd-Barron, and John Tate, I convinced myself this could be done, and it was eventually written up in Alex's Arizona thesis:</p> <p><a href="http://math.arizona.edu/~aprl/publications/dissertation/" rel="nofollow">http://math.arizona.edu/~aprl/publications/dissertation/</a></p> <p>Unfortunately, my own argument had many waves of the hand, and I didn't actually read the final product. But hopefully, it's all worked out. Alex really was a pretty careful guy, with substantial computational saavy.</p> <p>Possibly, you might also enjoy this unpublished note of mine:</p> <p><a href="http://www.ucl.ac.uk/~ucahmki/jockusch.pdf" rel="nofollow">http://www.ucl.ac.uk/~ucahmki/jockusch.pdf</a></p> <p>where these issues are discussed somewhat informally.</p> <p>Added: I only read the comments above just now. I'm not entirely sure that the algorithm described in Perlis' thesis is any more efficient than what Pete suggests. Let me know if it turns out to be so.</p> http://mathoverflow.net/questions/47950/comparison-of-de-rham-cohomology-and-etale-cohomology/47967#47967 Answer by Minhyong Kim for comparison of de Rham cohomology and etale cohomology Minhyong Kim 2010-12-01T23:50:16Z 2010-12-02T14:03:12Z <p>Often, yes. What always has the same dimension as $H^n_{et}(X,Q_l)$ is the rational crystalline cohomology $H^n_{cr}(X)\otimes K$ with coefficients in the fraction field $K$ of the Witt vectors $W$ of $k$. $H^n_{cr}(X)$ itself will have coefficients in $W$, and of course, have rank equal to the dimension of $H^n_{cr}(X)\otimes K$. But it might have torsion in general. On the other hand, there is an exact sequence $$0\rightarrow H^n_{cr}(X)\otimes_W k\rightarrow H^n(X,\Omega_X^{\cdot})\rightarrow H^{n+1}_{cr}(X)[p]\rightarrow 0$$ as in the universal coefficient theorem. This is because crystalline cohomology can be taken with any of the torsion coefficients $W/p^n$, and when you take it with coefficients in $W/p=k$, you get exactly De Rham cohomology. (One of the most important things to learn at the beginning about crystalline cohomology with $W/p^n$ coefficients is that it can be computed using the divided power De Rham complex associated to a smooth embedding over $W/p^n$, which reduces to the De Rham complex of $X$ itself when the coefficients are $W/p$.)</p> <p>So you will get the same dimensions you want if enough of crystalline cohomology is torsion-free. All this is explained in introductory books, such as the one by Berthelot and Ogus, except the comparison with \'etale cohomology. That is perhaps explained in a paper by Katz and Messing from the 70's.</p> http://mathoverflow.net/questions/42406/why-certain-diophantine-equations-are-interesting-and-others-are-not/42633#42633 Answer by Minhyong Kim for Why certain diophantine equations are interesting (and others are not) ? Minhyong Kim 2010-10-18T13:28:13Z 2010-10-25T16:12:24Z <p>Added 22, October:</p> <p>While reflecting on this question and the subsequent discussion, I fell again into a bit of a sermonizing mood. I hope I will be forgiven for inserting two words of caution.</p> <ol> <li><p>There is a difference, probably significant, between lack of interest and <em>aggressive</em> lack of interest. Most people will know what I mean by this, I think. The latter is probably best avoided. Now, I wouldn't be too strong about doing away with it altogether since some people do seem to derive substantial creative energy by being <em>against</em> something, both in mathematics and in the world at large. But for most of us, the less contemplative version of disinterest is, I believe, more destructive than constructive.</p></li> <li><p>Since I've already expounded on the mainstream view, it might be alright to vary on it a bit. It's good to educate oneself about fashion and probably sensible to follow it a fair amount. But then there are the true cliches about independent thinking. I'm old enough to have witnessed first-hand the fervor surrounding the wonderful proof of FLT. At this point, I'm sure many of us can recite by heart all the reasons it is more important than Goldbach, for example, and the significance of the relation to modularity, etc. This may indeed be a reasonable point of view. However, to be honest, I rarely got the impression then that the young expert in the common room was doing more than just that: reciting the viewpoint. I suppose I'm just repeating the platitude that fashion might be quite sensible, but slavish adherence to it is not. So if you feel strongly about some specific Diophantine equation that doesn't quite fit the standing paradigms, my own advice is to to think about it frequently enough to see if some real ideas develop. I hope I'm not misrepresenting him, but Swinnerton-Dyer once claimed that very few people were interested in L-functions around the time he was first experimenting with points on elliptic curves. Even now, he will speak with considerable passion about a single equation, or at least, about a single special family. (Nonetheless, I hope these paragraphs don't strike you as regurgitation of some superficial faith in 'diversity'. I have some of that as well, but some mathematics is clearly better than others.)</p></li> </ol> <p>Regarding Goldbach, I have the curious impression that it's about to gain substantially in respectability, especially with the remarkable ascent of additive number theory related to the work of Gowers, Green-Tao, et. al. I try to think about it myself every now and then, partly because of the influence of Shinichi Mochizuki, who insists on viewing the connection between additive and multiplicative structures in arithmetic through the prism of non-abelian fundamental groups.</p> <hr> <p>Oops! I'd better clarify right away that my remarks on Fermat and Goldbach are meant in no way as criticism of Jordan's nice answer.</p> <hr> <p>Original answer:</p> <p>A proper answer might require at least an essay, but here is an abridged attempt.</p> <p>Two classes of equations have already been discussed in the other answers:</p> <p>(1) Some equations are 'just interesting' for their special or exotic properties. I quite like the classical mathematics generated by the equation $$x^3+y^5=z^2$$ mentioned in Mike Bennett's comment. Smooth cubic surfaces like $$x^3+y^3+z^3+w^3=0$$ are also nice with their twenty-seven lines that eventually do influence their arithmetic. Fermat equations might be appreciated for their similarly high degree of symmetry that induces the complex multiplication on their Jacobians. By the way, regardless of their age, I find Calabi-yau varieties quite fascinating myself, since the interplay between Hodge-theoretic and Diophantine properties is a subtle phenomenon deserving of study.</p> <p>There is obviously no objective criterion being offered here, but still an equation may appear as especially interesting in the same way certain spaces are interesting or some animals are interesting. They excite a certain desire to know about them in considerable detail. Investigation with affection is usually richly rewarded in these cases.</p> <p>(2) Equations that come up while studying some other problem. Pell's equations were mentioned above, and one could consider other norm equations while studying number fields. Keith mentioned also the relation between the far-reaching ABC conjecture and Mordell's equation. One other nice class of elliptic curves of this nature are $$y^2=x^3-n^2x,$$ which were famously connected to the congruence number problem on the area of right angle triangles with rational sides and areas. A spectacular example in the ABC vein is Mazur's study of points on modular curves that gave rise to uniform bounds on the torsion subgroup of <em>all</em> elliptic curves over $\mathbb{Q}$. And then, integral points on Siegel moduli spaces were bounded by Faltings in his proof of the Mordell conjecture. In short, one can even apply one kind of equation to the study of another (family). In any case, for this class, one presumes the equations are as interesting as the motivating problem.</p> <p>However, the perspective I actually wished to mention sidesteps the question somewhat. The view is perhaps the most mainstream and reactionary possible in this context, but closest to mathematical practice as I see it. It says most equations have or lack interest not in and of themselves. Rather, the main issue is the questions we ask about them. I will remind you of three examples:</p> <p>(i) Consider the various conjectures on $L$-functions. Say the conjectures of Birch and Swinnerton-Dyer. To oversimplify the case a bit, suppose you could prove the conjecture in full up to the last detail for the single elliptic curve $$498208y^2=x^3+309208472x^2+1204948278x+3920984$$ or with some other choice of coefficients as random as you want. There will be little disagreement that this would be highly interesting.</p> <p>In case you think that elliptic curves are already deserving of special attention, choose a random collection of homogeneous equations $$f_1=0, f_2=0, \ldots, f_n=0$$ in $m$ variables. Most of the time, they will define a smooth projective variety $X$. Bloch and Beilinson have conjectured that the order of vanishing of $L(H^{2i-1}(X),s)$ at $s=i$ is equal to the rank of the Chow group of algebraic cycles of codimension $i$ homologically equivalent to zero. Being able to prove that statement for any given $X$ chosen randomly would be highly interesting. Of course because one expects this to be so difficult, people concentrate rather on special $X$'s. [In case you are wondering about their relevance, algebraic cycles on $X$ should rightly be thought of as 'generalized solutions'.]</p> <p>(ii) Continuing with the same notation, suppose $X$ happens to be a Fano variety, which will often happen if the degrees of the polynomials add up to something rather small compared to the number of variables. In that case, Manin has conjectured that the rational solutions in some fixed number field $F$ (depending on the equations) will be Zariski dense. That is, this is a set of equations with very simple geometry, because of which it should be consistently easy to find many, many solutions, as soon as some obvious obstruction is overcome. Once again, you are free to attempt this after choosing the $f_i$'s in as arbitrary and as unaesthetic a manner as possible. As with the Beilinson-Bloch conjecture, the more random your choice is, the more impressive your result will be, in some sense.</p> <p>(iii) Close to my own heart is the effective Mordell conjecture, which asks for an algorithm to find all rational solutions to a generic equation $$f(x,y)=0$$ with degree at least 4. As a consequence of the fact that such an algorithm is unknown, it becomes of considerable interest to be able to list full solution sets in any given case. Sometimes, it's easy to show that there are none, such as $$x^4+y^4=-1,$$ just to be absurdly simple. However, once such silly reasons for triviality are excluded, for example, if you happen to notice already one solution, it is notoriously difficult to list the whole solution set. Here is an example due to Bjorn Poonen: $$y^2 = x^6 - 2x^4 + 2x^3 + 5x^2 + 2x + 1.$$ You will easily see the solutions $(0,\pm 1)$. However, it requires quite a bit of technology to show that $$(0,\pm 1), (-1,\pm 1), (1,\pm 3)$$ is the full solution set. You can see that this particular equation does seem pretty random. On other hand, because of an enduring focus on the difficult and structurally demanding question of an algorithm, any example of this sort generates quite a bit of interest.</p> <p>Many people have suggested that the variety of <em>techniques</em> that come up in attacking a problem are as important as the problem itself. There is something to this, in as much as we would like the problem to tell us as much as possible about the mathematical landscape in general, which is, after all, the ultimate object of our investigation. On the other hand, once certain overarching questions have already been established as powerful probes for this process, being able resolve them for any specific object is interesting regardless of how pretty or ugly someone may find the object on its own. Obviously, this is <em>the</em> raison d'etre for good conjectures.</p> <hr> <p>Added 26, October:</p> <p>Eventually, I stumbled on to the 'box equation' I referred to in the comments. It is $$a_1^2+a_2^2=b_3^2;$$ $$a_1^2+a_3^2=b_2^2;$$ $$a_2^2+a_3^2=b_1^2;$$ $$a_1^2+a_2^2+a_3^2=c^2;$$ defining a surface in $\mathbb{P}^6$. A rational solution with $a_1a_2a_3\neq 0$ corresponds to a 'rational box' having all edges, face diagonals, and long diagonal rational. Apparently, the existence of such a thing is still unknown. There is a nice discussion in this <a href="http://arxiv.org/PS_cache/arxiv/pdf/1009/1009.0388v1.pdf" rel="nofollow">paper of Stoll and Testa</a>. Of course, you have to decide for yourself if it's interesting. The flavor of it is somewhat reminiscent of the congruent number problem, and I think that was why it caught my attention. That is, given my own bias, I had to consider for a minute or two if there were a sneaky connection to a 'conceptually sophisticated' problem. Stoll and Testa relate it, in fact, to the Bombieri-Lang conjecture.</p> http://mathoverflow.net/questions/38527/the-fundamental-lemma-and-the-conjecture-of-birch-and-swinnerton-dyer The fundamental lemma and the conjecture of Birch and Swinnerton-Dyer Minhyong Kim 2010-09-13T02:51:12Z 2010-10-12T05:57:08Z <p>Here is a rather pathetic question. In a <a href="http://gowers.wordpress.com/2010/08/21/icm2010-ngo-laudatio/#comment-9720" rel="nofollow">comment</a> on Tim Gower's weblog, I tentatively stated that the fundamental lemma was necessary for the <a href="http://www.mathunion.org/ICM/ICM2006.2/Main/icm2006.2.0473.0500.ocr.pdf" rel="nofollow">work of Skinner and Urban</a> relating ranks of Selmer groups of elliptic curves to the vanishing of their $p$-adic $L$-functions. Now, I believe it is correct that some endoscopic version of transfer from a unitary group to a general linear group is necessary for the construction of their $\Lambda$-adic representations. However, having a really poor understanding of the actual techniques, I don't know which version is crucial. That is to say, it's entirely likely that some earlier special case is sufficient for Skinner-Urban. Could I trouble some expert to give a brief outline of the situation? </p> <p>The pathetic part of this is that the journalist I mentioned in the comment will call in about 4 hours, so it would be nice to know before that. Of course I shouldn't have agreed to speak about something I know so little about, but it was hard to refuse under the circumstances. Oh, in case you're worried that I'm going to discuss Skinner-Urban with the fellow, don't. I just want to bone up on the background.</p> <hr> <p>Added: </p> <p>For people who like the idea of linguistic diversity in mathematics, I am including a link to a <a href="http://math.postech.ac.kr/~minhyong/kmsfields.pdf" rel="nofollow">report </a> written (with Sugwoo Shin) for the Korean Mathematical Society that expands on the comment to the journalist.</p> http://mathoverflow.net/questions/546/galois-groups-vs-fundamental-groups/24068#24068 Answer by Minhyong Kim for Galois Groups vs. Fundamental Groups Minhyong Kim 2010-05-10T09:00:27Z 2010-10-04T16:25:48Z <p>I saw this question a while ago and felt something in the way of a (probably misguided) missionary zeal to make at least a few elementary remarks. But upon reflection, it became clear that even that would end up rather long, so it was difficult to find the time until now.</p> <p>The point to be made is a correction: fundamental groups in arithmetic geometry are not the same as Galois groups, per se. Of course there is a long tradition of parallels between Galois theory and the theory of covering spaces, as when Takagi writes of being misled by Hilbert in the formulation of class field theory essentially on account of the inspiration from Riemann surface theory. And then, Weil was fully aware that homology and class groups are somehow the same, while speculating that a sort of non-abelian number theory informed by the full theory of the 'Poincare group' would become an ingredient of many serious arithmetic investigations.</p> <p>A key innovation of Grothendieck, however, was the formalism for refocusing attention on the <em>base-point</em>. In this framework, which I will review briefly below, when one says $$\pi_1(Spec(F), b)\simeq Gal(\bar{F}/F),$$ the base-point in the notation is the choice of separable closure $$b:Spec(\bar{F})\rightarrow Spec(F).$$ That is, </p> <p><em>Galois groups are fundamental groups with generic base-points.</em></p> <p>The meaning of this is clearer in the Galois-theoretic interpretation of the fundamental group of a smooth variety $X$. There as well, the choice of a separable closure $k(X)\hookrightarrow K$ of the function field $k(X)$ of $X$ can be viewed as a base-point $$b:Spec(K)\rightarrow X$$ of $X$, and then $$\pi_1(X,b)\simeq Gal(k(X)^{ur}/k(X)),$$ the Galois group of the maximal sub-extension $k(X)^{ur}$ of $K$ unramified over $X$. However, it would be quite limiting to take this last object as the <em>definition</em> of the fundamental group.</p> <p>We might recall that even in the case of a path-connected pointed topological space $(M,b)$ with universal covering space $$M'\rightarrow M,$$ the isomorphism $$Aut(M'/M)\simeq \pi_1(M,b)$$ is <em>not</em> canonical. It comes rather from the choice of a base-point lift $b'\in M'_b$. Both $\pi_1(M,b)$ and $Aut(M'/M)$ act on the fiber $M'_b$, determining bijections $$\pi_1(M,b)\simeq M'_b\simeq Aut(M'/M)$$ via evaluation at $b'$. It is amusing to check that the isomorphism of groups obtained thereby is independent of $b'$ if and only if the fundamental group is abelian. The situation here is an instance of the choice involved in the isomorphism $$\pi_1(M,b_1)\simeq \pi_1(M,b_2)$$ for different base-points $b_1$ and $b_2$. The practical consequence is that when fundamental groups are equipped with natural extra structures coming from geometry, say Hodge structures or Galois actions, different base-points give rise to enriched groups that are are often genuinely non-isomorphic.</p> <p>A more abstract third group is rather important in the general discussion of base-points. This is $$Aut(F_b),$$ the automorphism group of the functor $$F_b:Cov(M)\rightarrow Sets$$ that takes a covering $$N\rightarrow M$$ to its fiber $N_b$. So elements of $Aut(F_b)$ are compatible collections $$(f_N)_N$$ indexed by coverings $N$ with each $f_N$ an automorphism of the set $N_b$. Obviously, newcomers might wonder where to get such compatible collections, but lifting loops to paths defines a natural map $$\pi_1(M,b)\rightarrow Aut(F_b)$$ that turns out to be an isomorphism. To see this, one uses again the fiber $M'_b$ of the universal covering space, on which both groups act compatibly. The key point is that while $M'$ is not actually universal in the category-theoretical sense, $(M',b')$ <em>is</em> universal among <em>pointed</em> covers. This is enough to show that an element of $Aut(F_b)$ is completely determined by its action on $b'\in M'_b$, leading to another bijection $$Aut(F_b)\simeq M'_b.$$ Note that the map $\pi_1(M,b)\rightarrow Aut(F_b)$ is entirely canonical, even though we have used the fiber $M'_b$ again to prove bijectivity, whereas the identification with $Aut(M'/M)$ requires the use of $(M'_b,b')$ just for the definition.</p> <p>Among these several isomorphic groups, it is $Aut(F_b)$ that ends up most relevant for the definition of the etale fundamental group.</p> <p>So for any base-point $b:Spec(K)\rightarrow X$ of a connected scheme $X$ (where $K$ is a separably closed field, a 'point' in the etale theory), Grothendieck defines the 'homotopy classes of etale loops' as $$\pi^{et}_1(X,b):=Aut(F_b),$$ where $$F_b:Cov(X) \rightarrow \mbox{Finite Sets}$$ is the functor that sends a finite etale covering $$Y\rightarrow X$$ to the fiber $Y_b$. Compared to a construction like $Gal(k(X)^{ur}/k(X))$, there are three significant advantages to this definition.</p> <p>(1) One can easily consider small base-points, such as might come from a rational point on a variety over $\mathbb{Q}$.</p> <p>(2) It becomes natural to study the <em>variation</em> of $\pi^{et}_1(X,b)$ with $b$.</p> <p>(3) There is an obvious extension to <em>path spaces</em> $$\pi^{et}_1(X;b,c):=Isom(F_b,F_c),$$ making up a two-variable variation.</p> <p>This last, in particular, has no analogue at all in the Galois group approach to fundamental groups. When $X$ is a variety over $\mathbb{Q}$, it becomes possible, for example, to study $\pi^{et}_1(X,b)$ and $\pi^{et}_1(X;b,c)$ as sheaves on $Spec(\mathbb{Q})$, which encode rich information about rational points. This is a long story, which would be rather tiresome to expound upon here (cf. <a href="http://www.ucl.ac.uk/~ucahmki/cambridgews.pdf" rel="nofollow">lecture at the INI</a> ). However, even a brief contemplation of it might help you to appreciate the arithmetic perspective that general $\pi^{et}_1$'s are substantially more powerful than Galois groups. Having read thus far, it shouldn't surprise you that I don't quite agree with the idea explained, for example, in <a href="http://mathoverflow.net/questions/2791/understanding-gal-bar-q-q/2812#2812" rel="nofollow"> this post</a> that a Galois group is only a 'group up to conjugacy'. To repeat yet again, the usual Galois groups are just fundamental groups with specific large base-points. The dependence on these base-points as well as a generalization to small base-points is of critical interest.</p> <p>Even though the base-point is very prominent in Grothendieck's definition, a curious fact is that it took quite a long time for even the experts to fully metabolize its significance. One saw people focusing mostly on base-point independent constructions such as traces or characteristic polynomials associated to representations. My impression is that the initiative for allowing the base-points a truly active role came from Hodge-theorists like Hain, which then was taken up by arithmeticians like Ihara and Deligne. Nowadays, it's possible to give entire lectures just about base-points, as Deligne has actually done on several occasions.</p> <p>Here is a puzzle that I gave to my students a while ago: It has been pointed out that $Gal(\bar{F}/F)$ already refers to a base-point in the Grothendieck definition. That is, the choice of $F\hookrightarrow \bar{F}$ gives at once a universal covering space <em>and</em> a base-point. Now, when we turn to the manifold situation $M'\rightarrow M$, a careful reader may have noticed a hint above that there is a base-point implicit in $Aut(M'/M)$ as well. That is, we would like to write $$Aut(M'/M)\simeq \pi_1(M,B)$$ <em>canonically</em> for some base-point $B$. What is $B$?</p> <p>Added: </p> <p>-In addition to the contribution of Hodge-theorists, I should say that Grothendieck himself urges attention to many base-points in his writings from the 80's, like 'Esquisse d'un programme.'</p> <p>-I also wanted to remark that I don't really disagree with the point of view in JSE's answer either. </p> <p>Added again: </p> <p><a href="http://mathoverflow.net/questions/24241/functoriality-of-fundamental-group-via-deck-transformations" rel="nofollow">This question</a> reminds me to add another very basic reason to avoid the Galois group as a definition of $\pi_1$. It's rather tricky to work out the functoriality that way, again because the base-point is de-emphasized. In the $Aut(F_b)$ approach, functoriality is essentially trivial.</p> <p>Added, 27 May:</p> <p>I realized I should fix one possible source of confusion. If you work it out, you find that the bijection $$\pi_1(M,b)\simeq M'_b\simeq Aut(M'/M)$$ described above is actually an <em>anti</em>-isomorphism. That is, the order of composition is reversed. Consequently, in the puzzle at the end, the canonical bijection $$Aut(M'/M)\simeq \pi_1(M,B)$$ is an anti-isomorphism as well. However, another simple but amusing exercise is to note that the various bijections with Galois groups, like $$\pi_1(Spec(F), b)\simeq Gal(\bar{F}/F),$$ are actually isomorphisms.</p> <p>Added, 5 October:</p> <p>I was asked by a student to give away the answer to the puzzle. The crux of the matter is that any continuous map $$B:S\rightarrow M$$ from a simply connected set $S$ can be used as a base-point for the fundamental group. One way to do this to use $B$ to get a fiber functor $F_B$ that associates to a covering $$N\rightarrow M$$the set of splittings of the covering $$N_B:=S\times_M N\rightarrow S$$ of $S$. If we choose a point $b'\in S$, any splitting is determined by its value at $b'$, giving a bijection of functors $F_B=F_{b'}=F_b$ where $b=B(b')\in M$. Now, when $$B:M'\rightarrow M$$ is the universal covering space, I will really leave it as a (tautological) exercise to exhibit a canonical anti-isomorphism $$Aut(F_B)\simeq Aut(M'/M).$$ The 'point' is that $$F_B(M')$$ has a canonical base-point that can be used for this bijection.</p> http://mathoverflow.net/questions/39828/how-do-you-decide-whether-a-question-in-abstract-algebra-is-worth-studying/39938#39938 Answer by Minhyong Kim for How do you decide whether a question in abstract algebra is worth studying? Minhyong Kim 2010-09-25T08:31:25Z 2010-09-26T16:23:35Z <p>Dear Alex,</p> <p>It seems to me that the general question in the background of your query on algebra really is the better one to focus on, in that we can forget about irrelevant details. That is, as you've mentioned, one could be asking the question about motivation and decision in any kind of mathematics, or maybe even life in general. In that form, I can't see much useful to write other than the usual cliches: there are safer investments and riskier ones; most people stick to the former generically with occasional dabbling in the latter, and so on. This, I think, is true regardless of your status. Of course, going back to the corny financial analogy that Peter has kindly referred to, just <em>how</em> risky an investment is depends on how much money you have in the bank. We each just make decisions in as informed a manner as we can.</p> <p>Having said this, I rather like the following example: <a href="http://en.wikipedia.org/wiki/Kac%E2%80%93Moody_algebra" rel="nofollow">Kac-Moody algebras</a> could be considered 'idle' generalizations of finite-dimensional simple Lie algebras. One considers the construction of simple Lie algebras by generators and relations starting from a Cartan matrix. When a positive definiteness condition is dropped from the matrix, one arrives at general Kac-Moody algebras. I'm far from knowledgeable on these things, but I have the impression that the initial definition by Kac and Moody in 1968 really was somewhat just for the sake of it. Perhaps indeed, the main (implicit) justification was that the usual Lie algebras were such successful creatures. Other contributors here can describe with far more fluency than I just how dramatically the situation changed afterwards, accelerating especially in the 80's, as a consequence of the interaction with conformal field theory and string theory. But many of the real experts here seem to be rather young and perhaps regard vertex operator algebras and the like as being just so much bread and butter. However, when I started graduate school in the 1980's, this story of Kac-Moody algebras was still something of a marvel. There must be at least a few other cases involving a rise of comparable magnitude. </p> <p>Meanwhile, I do hope some expert will comment on this. I fear somewhat that my knowledge of this story is a bit of the fairy-tale version.</p> <p>Added: In case someone knowledgeable reads this, it would also be nice to get a comment about further generalizations of Kac-Moody algebras. My vague memory is that some naive generalizations have not done so well so far, although I'm not sure what they are. Even if one believes it to be the purview of masters, it's still interesting to ask if there is a pattern to the kind of generalization that ends up being fruitful. Interesting, but probably hopeless.</p> <p>Maybe I will add one more personal comment, in case it sheds some darkness on the question. I switched between several supervisors while working towards my Ph.D. The longest I stayed was with Igor Frenkel, a well-known expert on many structures of the Kac-Moody type. I received several personal tutorials on vertex operator algebras, where Frenkel expressed his strong belief that these were really fundamental structures, 'certainly more so than, say, Jordan algebras.' I stubbornly refused to share his faith, foolishly, as it turns out (so far).</p> <p>Added again:</p> <p>In view of Andrew L.'s question I thought I'd add a few more clarifying remarks.</p> <p>I explained in the comment below what I meant with the story about vertex operator algebras. Meanwhile, I can't genuinely regret the decision not to work on them because I quite like the mathematics I do now, at least in my own small way. So I think what I had in mind was just the platitude that most decisions in mathematics, like those of life in general, are mixed: you might gain some things and lose others.</p> <p>To return briefly to the original question, maybe I do have some practical remarks to add. It's obvious stuff, but no one seems to have written it so far on this page. Of course, I'm not in a position to give anyone advice, and your question didn't really ask for it, so you should read this with the usual reservations. (I feel, however, that what I write <em>is</em> an answer to the original question, in some way.)</p> <p>If you have a strong feeling about a structure or an idea, of course keep thinking about it. But it may take a long time for your ideas to mature, so keep other things going as well, enough to build up a decent publication list. The part of work that belongs to quotidian maintenance is part of the trade, and probably a helpful routine for most people. If you go about it sensibly, it's really not that hard either. As for the truly original idea, I suspect it will be of interest to many people at some point, if you keep at it long enough. Maybe the real difference between starting mathematicians and established ones is the length of time they can afford to invest in a strange idea before feeling like they're running out of money. But by keeping a suitably interesting business going on the side, even a young person can afford to dream. Again, I suppose all this is obvious to you and many other people. But it still is easy to forget in the helter-skelter of life.</p> <p>By the way, I object a bit to how several people have described this question of community interest as a two-state affair. Obviously, there are many different degrees of interest, even in the work of very famous people. </p> http://mathoverflow.net/questions/38639/thinking-and-explaining/38694#38694 Answer by Minhyong Kim for Thinking and Explaining Minhyong Kim 2010-09-14T15:22:51Z 2010-09-16T02:37:41Z <p>Final addition: </p> <p>Since I've produced many rambles, I thought I'd close my (anti-)contribution with a distilled version of the example I've attempted below. It's still something very standard, but, I hope, in the spirit of the original question. I'll describe it as if it were a personal thing.</p> <p>Almost always, I think of an integer as a function of the primes. So for 20, say,</p> <p>20(2)= 0</p> <p>20(3)=2</p> <p>20(5)=0</p> <p>20(7)=6</p> <p>. .</p> <p>20(19)=1</p> <p>20(23)=20</p> <p>20(29)=20</p> <p>20(31)=20</p> <p>20(37)=20</p> <p>. . .</p> <p>It's quite a compelling image, I think, an integer as a function that varies in this way for a while before eventually leveling off. But, for a number of reasons, I rarely mention it to students or even to colleagues. Maybe I should.</p> <hr> <p>Original answer:</p> <p>It's unclear if this is an appropriate kind of answer, in that I'm not putting forward anything very specific. But I'll take the paragraph in highlight at face value.</p> <p>I find it quite hard to express publicly my <em>vision</em> of mathematics, and I think this is a pretty common plight. Part of the reason is the difficulty of putting into words a sense of things that ultimately stems from a view of the landscape, as may be suggested by the metaphor. But another important reason is the disapprobation of peers. To appeal to hackneyed stereotypes, each of us has in him/her a bit of Erdos, a bit of Thurston, and perhaps a bit of Grothendieck, of course in varying proportions depending on education and temperament. I think I saw somewhere on this site the sentiment that 'a bad Erdos still might be an OK mathematician, but a bad Grothendieck is really terrible,' or something to that effect. This opinion is surrounded by a pretty broad consensus, I think. If I may be allowed some cliches now from the world of finance, it's almost as though definite mathematical results are money in the bank. After you've built up some savings, you can afford to spend a bit by philosophizing. But then, you can't let the balance get too low because people will start looking at you in funny, suspicious ways. I know that on the infrequent occasions* that I get carried away and convey at any length my vision of how a certain area of mathematics should work, what should be true and why, compelling analogies, and so on, I feel rather embarrassed for a little while. It feels like I am indeed running out of money and will need to back up the highfalutin words with some theorems (or at least lemmas) relatively soon. (And then, so many basically sound ideas are initially mistaken for trivial reasons.)</p> <p>Now, I wish to make it clear that unlike Grothendieck (see the beginning paragraphs of this <a href="http://www.math.jussieu.fr/~leila/grothendieckcircle/GtoF.pdf" rel="nofollow">letter to Faltings</a>) I find this quite sensible a state of affairs. For myself, it seems to be pretty healthy that my tendency to philosophize is held in check by the demand of the community that I have something to show for it. I grant that this may well be because my own visions are so meagre in comparison to Grothendieck's. In any case, the general phenomenon itself is interesting to observe, in myself and in others. </p> <p>Incidentally, I find the peer pressure in question remarkably democratic. Obviously, a well-established mathematician typically has more money than average in the bank, so to speak. But it's not a few times I've observed eminent people during periods of slowdown, being gradually ignored or just tolerated in their musings by many young people, even students.</p> <p>Meanwhile, if you're an energetic youngster with some compelling vision of an area of mathematics, it may not be so bad to let loose. If you have a really good business idea, it may even make sense to take out a large loan. And provided you have the right sort of personality, the pressure to back up your philosophical bravado with results may spur you on to great things. This isn't to say you won't have to put up with perfectly reasonable looks of incredulity, even from me, possibly for years.</p> <hr> <p>*Maybe it seems frequent to my friends.</p> <hr> <p>Added:</p> <p>Since I commented above on something quite general, here is an attempt at a specific contribution. It's not at all personal in that I'm referring to a well-known point of view in Diophantine geometry, whereby solutions to equations are <em>sections of fiber bundles</em>. Some kind of a picture of the fiber bundle in question was popularized by Mumford in his Red Book. I've discovered a reproduction on <a href="http://www.neverendingbooks.org/index.php/mumfords-treasure-map.html" rel="nofollow">this page</a>. The picture there is of $Spec(\mathbb{Z}[x])$, but interesting equations even in two variables will conjure up a more complicated image of an arithmetic surface fibered over the 'arithmetic curve' $Spec(\mathbb{Z})$. A solution to the equation will then be a section of the bundle cutting across the fibers, also in a complicated manner. Much interesting work in number theory is concerned with how the sections meet the singular fibers. </p> <p>Over the years, I've had many different thoughts about this perspective. For me personally, it was truly decisive, in that I hadn't been very interested in number theory until I realized, almost with a shock, that the study of solutions to equations had been 'reduced' to the study of maps between spaces of a quite rigid sort. In recent years, I think I've also reconciled myself with the more classical view, whereby numbers are some kinds of algebraic gadgets. That is, thinking about matters purely algebraically does seem to provide certain flexible modes that can be obscured by the insistence on geometry. I've also discovered that there is indeed a good deal of variation in how compelling the inner picture of a fiber bundle can be, even among seasoned experts in arithmetic geometry. Nevertheless, it's clear that the geometric approach is important, and informs a good deal of important mathematics. For example, there is an elementary but key step in Faltings' proof of the Mordell conjecture referred to as the 'Kodaira-Parshin trick,' whereby you (essentially) get a compact curve $X$ of genus at least two to <em>parametrize</em> a smooth family of curves $$Y\rightarrow X.$$ Then, whenever you have a rational point $$P:Spec(\mathbb{Q})\rightarrow X$$ of $X$, you can look at the fiber $Y_P$ of $Y$ above $P$, which is itself a curve. The argument is that if you have too many points $P$, you get too many good curves over $\mathbb{Q}$. What is good about them? Well, they all spread out to arithmetic surfaces over the spectrum of $\mathbb{Z}$ that are singular only over a fixed set of places. This part can be made obvious by spreading out both $Y$, $X$, and the map between them over the integers as well, right at the outset. If you don't have that picture in mind, the goodness of the $Y_P$ is not at all easy to explain.</p> <p>Anyways, what I wanted to say is that the picture of solutions as sections to fiber bundles is really difficult to explain to people without a certain facility in scheme theory. Because it seems so important, and because it is a crucial ingredient in my own thinking, I make an attempt every now and then in an exposition at the colloquium level, and fail miserably. I notice almost none of my colleagues even try to explain it in a general talk.</p> <p>Now, I've mentioned already that this is far from a personal image of a mathematical object. But it still seems to be a good example of a very basic picture that you refrain from putting into words most of the time. If it really had been only a personal vision, it may even have been all but maddening, the schism between the clarity of the mental image and what you're able to say about it. Note that the process of putting the whole thing into words in a convincing manner in fact took thousands of pages of foundational work.</p> <hr> <p>Added again:</p> <p>Professor Thurston: To be honest, I'm not sure about the significance of competing mental images in this context. If I may, I would like to suggest another possibility. It isn't too well thought out, but I don't believe it to be entirely random either. </p> <p>Many people from outside the area seem to have difficulty understanding the picture I mentioned <em>because they are intuitively suspicious of its usefulness</em>. Consider a simpler picture of the real algebraic curve that comes up when one studies cubic equations like $$E: y^2=x^3-2.$$ There, people are easily convinced that geometry is helpful, especially when I draw the tangent line at the point $P=(3,5)$ to produce another rational point. What is the key difference from the other picture of an arithmetic surface and sections? My feeling is it has mainly to do with the suggestion that the point itself has a complicated geometry encapsulated by the arrow $$P:Spec(\Bbb{Z})\rightarrow E.$$ That is, spaces like $Spec(\Bbb{Q})$ and $Spec(\Bbb{Z})$ are problematic and, after all, are quite radical.</p> <p>In $Spec(\Bbb{Q})$, one encounters the absurdity that the space $Spec(\Bbb{Q})$ itself is just a point. So one has to go into the whole issue that the point is equipped with a ring of functions, which happens to be $\Bbb{Q}$, and so on. At this point, people's eyes frequently glaze over, but not, I think, because this concept is too difficult or because it competes with some other view. Rather, the typical mathematician will be unable to <em>see the point</em> of looking at these commonplace things in this way. The temptation arises to resort to persuasion by authority then (such and such great theorem uses this language and viewpoint, etc.), but it's obviously better if the audience can really appreciate the ideas through some first-hand experience, even of a simple sort. I do have an array of examples that might help in this regard, provided someone is kind enough to be still interested. But how helpful they really are, I'm quite unsure.</p> <p>At the University of Arizona, we once had a study seminar on random matrices and number theory, to which I was called upon to contribute a brief summary of the analogous theory over finite fields. Unfortunately, this does involve some mention of sheaves, arithmetic fundamental groups, and some other strange things. Afterwards, my colleague Hermann Flaschka, an excellent mathematician with whom I felt I could speak easily about almost anything, commented that he couldn't tell if the whole language just consisted of word associations or if some actual geometry was going on. Now, I'm sure this was due in part to my poor powers of exposition. But further conversation gave me the strong impression that the question that really went through his mind was: 'How could it possibly be useful to think about these objects in this way?'</p> <p>To restate my point, I think a good deal of conceptual inhibition comes from a kind of intuitive utilitarian concern. Matters are further complicated by the important fact that this kind of conceptual conservatism is perfectly sensible much of the time.</p> <p>By the way, my choice of example was somewhat motivated by the fact that it is quite likely to be difficult for people outside of arithmetic geometry, including many readers of this forum. This gives it a different flavor from the situations where we all understand each other more or less well, and focus therefore on pedagogical issues referring to classroom practice.</p> <hr> <p>Yet again:</p> <p>Forgive me for being a bore with these repeated additions. </p> <p>The description of your approach to lectures seems to confirm the point I made, or at least had somewhat in mind: When someone can't understand what we try to explain, it's maybe in his or her best interest (real or perceived) not to. It's hard not to feel that this happens in the classroom as well oftentimes. This then brings up the obvious point that what we try to say is best informed by some understanding of who we're speaking to as well as some humility*. As a corollary, what we <em>avoid</em> saying might equally well be thus informed. </p> <p>My own approach, by the way, is almost opposite to yours. Of course I can't absorb technical details just sitting there, but I try my best to concentrate for the whole hour or so, almost regardless of the topic. (Here in Korea, it's not uncommon for standard seminar lectures to be <em>two</em> hours.) If I may be forgiven a simplistic generalization, your approach strikes me as common among deeply creative people, while perennial students like me tend to follow colloquia more closely. I intend neither flattery nor modesty with this remark, but only observation. Also, I am trying to create a complex picture (there's that word again) of the problem of communication.</p> <p>As to $Spec(\Bbb{Z})$, perhaps there will be occasion to bore you with that some other time. Why don't you post a question (assuming you are interested)? Then you are likely to get a great many perspectives more competent than mine. It might be an interesting experiment relevant to your original question.</p> <hr> <p>*I realize it's hardly my place to tell anyone else to be humble.</p> http://mathoverflow.net/questions/36362/characteristic-power-series-for-maps-of-e-infty-ring-spectra Characteristic power series for maps of E_{\infty} ring spectra Minhyong Kim 2010-08-22T06:16:37Z 2010-08-22T16:26:06Z <p>Let me admit right at the outset that I have a very superficial outsider's knowledge of homotopy theory. Nevertheless, I was trying to gain some understanding of Hopkins' ICM lecture <a href="http://arxiv.org/abs/math/0212397" rel="nofollow">'algebraic topology and modular forms.'</a> </p> <p>In section 6, he mentions two constructions. To a map </p> <p>$$\phi: MSpin\rightarrow KO$$</p> <p>of $E_{\infty}$ ring spectra, he associates a characteristic power series $$K_{\phi}(x)\in \mathbb{Q}[[x]].$$ Similarly, to an $E_{\infty}$-map </p> <p>$$\psi: MO\langle 8\rangle \rightarrow tmf,$$</p> <p>he associates a power series $$K_{\psi}(x)\in MF_{\mathbb{Q}}[[x]],$$ where $tmf$ is the topological modular form spectrum and $MF_{\mathbb{Q}}=MF\otimes _{\mathbb{Z}}\mathbb{Q}$ is the ring of modular forms with rational coefficients.</p> <p>I wonder if someone could give a brief outline of how these associations are carried out. I presume it is something elementary having to do with the homotopy groups of $MSpin$ and $MO\langle 8\rangle$, but I don't quite have the resources right now to track these down. </p> <p>As usual with questions of this sort, I'm sure my level of ignorance is incongruous with the words I am employing already, but thank you in advance for any tolerant answers or references.</p> <p>Added:</p> <p>Maybe I should summarize the point of this question for fellow number-theorists who are too busy to look into the paper. In the notation above, one associates to $\phi$ a characteristic sequence </p> <p>$$b(\phi)=(b_2, b_4, b_6,\ldots)$$</p> <p>via the formula</p> <p>$$\log(K_{\phi}(x))=-2\sum_{n>0} b_n\frac{x^n}{n!}.$$</p> <p>Incredibly, this procedure sets up a bijection:</p> <p>homotopy classes of $E_{\infty}$ maps from $MSpin$ to $KO$ $\leftrightarrow$ the set of sequences of rational numbers $(b_i)$ as above that satisfy</p> <p>(1) $b_n\equiv B_{n}/n \ \ \mod \mathbb{Z}$, where the $B_n$ are the Bernouilli numbers;</p> <p>(2) for each odd prime $p$ and $p$-adic unit $c$,</p> <p>$$m\equiv n \ \mod p^k(p-1) \Rightarrow (1-c^n)(1-p^{n-1})b_n \equiv (1-c^m)(1-p^{m-1})b_m \ \mod p^{k+1};$$</p> <p>(3) for each $2$-adic unit $c$,</p> <p>$$m\equiv n \ \mod 2^k \Rightarrow (1-c^n)(1-2^{n-1})b_n \equiv (1-c^m)(1-2^{m-1})b_m \ \mod 2^{k+2}.$$</p> <p>In the case of the homotopy classes of maps from $MO\langle 8\rangle$ to $tmf$, one gets similar congruences involving Eisenstein series instead of their constant terms. Incidentally, perhaps these congruences imply the ones above?</p> http://mathoverflow.net/questions/5611/the-class-number-formula-the-bsd-conjecture-and-the-kronecker-limit-formula/5655#5655 Answer by Minhyong Kim for The class number formula, the BSD conjecture, and the Kronecker limit formula Minhyong Kim 2009-11-15T22:50:44Z 2010-07-31T05:47:01Z <p>I apologize for in advance for making just a few superificial remarks. These are:</p> <ol> <li><p>The question is not uninteresting. Just because Sha doesn't appear in the definition of L easily, there's no reason one shouldn't ask about manifestations more fundamental than the usual one.</p></li> <li><p>An approach might be to think about the p-adic L-function rather than the complex one. I'm far from an expert on this subject, but the algebraic L-function is supposed to be a characteristic element of a dual Selmer group over some large extension of the ground field. The Selmer group (over the ground field) of course does break up into cosets indexed by Sha. Perhaps one could examine carefully the papers of Rubin, where various versions of the Iwasawa main conjectures are proved for CM elliptic curves.]</p></li> </ol> <p>Added, 8 July:</p> <p>This old question came back to me today and I realized that I had forgotten to make one rather obvious remark. However, I still won't answer the original question.</p> <p>You see, instead of the $L$-function of an elliptic curve $E$, we can consider the zeta function $\zeta({\bf E},s)$ of a regular minimal model ${\bf E}$ of $E$, which, in any case, is the better analogue of the Dedekind zeta function. One definition of this zeta function is given the product $$\zeta({\bf E},s)=\prod_{x\in {\bf E}_0} (1-N(x)^{-s})^{-1},$$ where ${\bf E}_0$ denotes the set of closed points of ${\bf E}$ and $N(x)$ counts the number of elements in the residue field at $x$. It is not hard to check the expression $$\zeta({\bf E},s)=L(E,s)/\zeta(s)\zeta(s-1)$$ in terms of the usual $L$-function and the Riemann zeta function.</p> <p>The product expansion, which converges on a half-plane, can also be written as a Dirichlet series $$\zeta({\bf E},s)=\sum_{D}N(D)^{-s},$$ where $D$ now runs over the <em>effective zero cycles</em> on ${\bf E}$. This way, you see the decomposition $$\zeta({\bf E},s)=\sum_{c\in CH_0({\bf E})}\zeta_c({\bf E},s),$$ in a manner entirely analogous to the Dedekind zeta. Here, $CH_0({\bf E})$ denotes the rational equivalence classes of zero cycles, and we now have the partial zetas $$\zeta_c({\bf E},s)=\sum_{D\in c}N(D)^{-s}.$$ It is a fact that $CH_0({\bf E})$ is finite. I forget alas to whom this is due, although the extension to arbitrary schemes of finite type over $\mathbb{Z}$ can be found in the papers of Kato and Saito.</p> <p>It's not entirely unreasonable to ask at this point if the group $CH_0({\bf E})$ is related to $Sha (E)$. At least, this formulation seems to give the original question some additional structure.</p> <p>Added, 31, July, 2010:</p> <p>This question came back yet again when I realized two errors, which I'll correct explicitly since such things can be really confusing to students. The expression for the zeta function in terms of $L$-functions above should be inverted: $$\zeta({\bf E},s)=\zeta(s)\zeta(s-1)/L(E,s).$$ The second error is slightly more subtle and likely to cause even more confusion if left uncorrected. For this precise equality, ${\bf E}$ needs to be the Weierstrass minimal model, rather than the regular minimal model. I hope I've got it right now.</p> http://mathoverflow.net/questions/33808/how-connected-are-you How connected are you? Minhyong Kim 2010-07-29T16:51:05Z 2010-07-29T23:06:19Z <p>I apologize if this question seems frivolous, but the motivation for it is quite serious. When I encounter the endless topic of the 'relevance' of mathematics, I am rather fond of referring to a <em>network</em> of knowledge. Before explaining this term, I should say that I'm not one who considers it very difficult to make a plausible direct case for the importance of mathematics to the general public, provided the context and tone of such a discussion is chosen with some care. Nevertheless, a little bit of thought makes it clear that the realistic value of any research assessed over time depends very heavily on the integrity of a network of related activity within which it sits. That is to say, a pure mathematician might occasionally speak to an applied mathematician who will speak to a physicist who will speak to an engineer who will speak to a biologist who will speak to a computer scientist who will speak to a mathematician, and so on. If we find sufficient coherence in the overall network of interaction, the individual components will frequently take on deeper significance than may have been noticed in isolated observation. Clearly, this isn't to imply that any obscure activity is as important as any other, but it seems to me an awareness of the network is critical in any discussion of relevance.</p> <p>After making this point recently to some graduate students, it occurred to me to investigate more than casually a rather amusing measure of connectedness, the collaboration distance calculator. So here is a small but somewhat surprising list of (finite!) distances from myself I've found using mathscinet:</p> <p><a href="http://en.wikipedia.org/wiki/Einstein" rel="nofollow"> Albert Einstein</a> 5</p> <p><a href="http://en.wikipedia.org/wiki/James_Clerk_Maxwell" rel="nofollow">James Clerk Maxwell</a> 6</p> <p><a href="http://en.wikipedia.org/wiki/Feynman" rel="nofollow">Richard Feynman</a> 5</p> <p><a href="http://en.wikipedia.org/wiki/Stephen_Hawking" rel="nofollow">Stephen Hawking</a> 4</p> <p><a href="http://en.wikipedia.org/wiki/J_B_S_Haldane" rel="nofollow">J.B.S. Haldane</a> 7</p> <p><a href="http://en.wikipedia.org/wiki/Noam_chomsky" rel="nofollow">Noam Chomsky</a> 4</p> <p><a href="http://en.wikipedia.org/wiki/Willard_Van_Orman_Quine" rel="nofollow">Willard Van Orman Quine</a> 6</p> <p><a href="http://en.wikipedia.org/wiki/Nelson_Goodman" rel="nofollow">Nelson Goodman</a> 5</p> <p><a href="http://en.wikipedia.org/wiki/Amartya_Sen" rel="nofollow">Amartya Sen</a> 7</p> <p><a href="http://en.wikipedia.org/wiki/Ilya_Prigogine" rel="nofollow">Ilya Prigogine</a> 4</p> <p><a href="http://en.wikipedia.org/wiki/Jean_Piaget" rel="nofollow">Jean Piaget</a> 6</p> <p>Needless to say, this array of celebrity intellectuals says nothing about me in particular. Among mathematicians of comparable seniority, I have relatively few collaborators, and examining the specific bridges that make up the paths will quickly reveal that they have nothing much to do with the signficance of my own research. It's obvious then that the diversity of this list reflects nothing less than the centrality of mathematics, and perhaps the coherence of human scholarly endeavour as a whole. </p> <p>Now to the question: Could you investigate a bit yourself now, and let me know of interesting research connections you find to people working outside mathematics, using mathscinet or otherwise? For the purposes of this question, what I would like to know about are concrete sequences of research links as might appear in the collaboration distance calculator, rather than an anecdotal discussion of some application of mathematics. Perhaps you could also add a word about your own area of research, so I can get some sense of surprise (or lack of it). By the way, I work on arithmetical algebraic geometry.</p> <p>As mentioned at the beginning, I do think my motivation is serious: In grandiose terms, a general awareness of the connections is pretty important not just for reassuring students, but for cultivating ourselves a reasonably sophisticated sense of where we stand in the scheme of things.</p> <hr> <p>Added: </p> <p>As mentioned by Sonia Balagopalan in the comments below, it would be also interesting to hear of unlikely connections Mathematician A--Non-mathematician B. However, I did think people would be pleasantly surprised to experiment a bit and find out specifically about their personal connections. That is, I thought it would be fun to concretely illustrate my own generic answer to </p> <p>'How connected are you?': </p> <p>More than you expect! </p> <p>I agree with most of the other comments, especially what Professor Milne writes about inaccuracy. But even with the few weak links that show up, the paths still seem to be interesting, and mostly illustrate my point. (Of course I can think of examples where they would be nearly uninteresting, such as a link created by two articles appearing in an encyclopedia.)</p> <p>Felipe: Actually, I have the impression that my more unusual connections go through you!</p> http://mathoverflow.net/questions/33190/galoisian-sets-and-the-langlands-programme Galoisian sets and the Langlands programme Minhyong Kim 2010-07-24T12:09:43Z 2010-07-26T12:23:56Z <p>Note: I've revised the question just a little bit in the hope of making it easier.</p> <hr> <p>Given an algebraic number field $F$, which we may as well take to be Galois over $\mathbb{Q}$, we denote by $S_F$ the set of rational primes that split in $F$. Sets of the form $S_F$ are called Galoisian. At some point, there was a discussion</p> <p><a href="http://mathoverflow.net/questions/11688/why-do-congruence-conditions-not-suffice-to-determine-which-primes-split-in-non-a" rel="nofollow">http://mathoverflow.net/questions/11688/why-do-congruence-conditions-not-suffice-to-determine-which-primes-split-in-non-a</a></p> <p>of the fact that the <em>abelian Galoisian sets</em>, that is, $S_F$ corresponding to $F$ abelian over $\mathbb{Q}$, are exactly the sets of primes defined by congruence conditions.</p> <p>A while later, Matthew Emerton gave this nice answer</p> <p><a href="http://mathoverflow.net/questions/11747/galoisian-sets-of-prime-numbers" rel="nofollow">http://mathoverflow.net/questions/11747/galoisian-sets-of-prime-numbers</a></p> <p>to a question of Chandan Singh Dalawat about non-abelian Galoisian sets. </p> <p>I made a comment there I thought I would upgrade to a question. As Matthew points out, Neukirch's remark that the Langlands program provides a <em>characterization</em> of all Galoisian sets is probably meant as a metaphor for some other process. However, I couldn't help but hope that the characterization could be taken literally, at least for some special families. For example, we will refer to a number field $F$ as being of $GL_2$ type if it is the fixed field of $Ker(\rho)$, where $$\rho: Gal(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow GL_2(\mathbb{C})$$ is an irreducible two-dimensional Artin representation of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$*. Now call a set of primes *a $GL_2$ Galoisian set* if it is of the form $S_F$ for some extension $F$ of $GL_2$-type.</p> <p>The question then is: can one use the Langlands program (or anything else) to give a sensible characterization of $GL_2$ Galoisian sets?</p> <p>One could obviously change this question in any way that would make it more tractable. One could try to characterize, for example:</p> <p>-Solvable $GL_2$ Galoisian sets, where the $GL_2$-field $F$ is further required to be solvable;</p> <p>-Odd $GL_2$ Galoisian sets: $S_F$ where $F$ is the fixed field of a representation $\rho_f$ arising from a holomorphic modular form $f$ of weight one;</p> <p>-Odd $GL_2$ Galoisan sets of conductor $N$, where we further require the form $f$ to have level $N$;</p> <p>and so on. The last case probably admits a tautological answer of sorts, in that we can in principle list the finitely many forms (sorted by Dirichlet characters $\epsilon$), and then make some statement about the $p$'s where $$X^2-a_pX+\epsilon(p)=(X-1)^2.$$ Is it entirely unreasonable to hope for something more compact?</p> <hr> <p>*The idea that we should simply organize fields in this manner corresponding to representations is perhaps a valuable perspective coming out of the Langlands program.</p> <hr> <p>Added, 25 July:</p> <p>Having thought about it a bit more, it occurs to me that this is yet another situation where Langlands urges us to go beyond a classical framework in seeking answers to non-abelian questions. For example, when we associate to an odd Artin represention </p> <p>$$\rho:Gal(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow Aut(V)$$</p> <p>of dimension two an Artin $L$-function $$L(\rho,s)=\sum a_n/n^s,$$ we can perfectly sensibly assert that the $a_n$'s follows a pattern. When asked what that pattern is, the answer, satisfying to some and mysterious to others, is that $$\sum a_nq^n$$ is a modular form. This is the kind of thing that comes out of Langlands. </p> <p>Now, if we want to 'characterize,' say, odd $GL_2$ Galoisian sets, we can say the following: Enumerate the normalized holomorphic Hecke (new) eigenforms $f$ of weight one sorted by level $N$ and character $\epsilon$. For each such form, run over the prime numbers $p$ not dividing $N$, and take the number $a_p$ defined by the equation $$T_pf=a_pf.$$ for the $p$-th Hecke operator $T_p$. Now look at the set $S_f$ of primes $p$ such that $$(p,N)=1, \epsilon (p)=1, a_p=2.$$ These $S_f$'s are exactly the odd $GL_2$ Galoisian sets.</p> <p>Perhaps it's unreasonable to want more from the Langlands' programme. Whether or not this is the final word on all such questions, well, that's a different matter.</p> http://mathoverflow.net/questions/33190/galoisian-sets-and-the-langlands-programme/33388#33388 Answer by Minhyong Kim for Galoisian sets and the Langlands programme Minhyong Kim 2010-07-26T12:23:56Z 2010-07-26T12:23:56Z <p>Here is an extremely naive answer to a case of my own question, which is surely obvious to experts. For me, even coming up with this silly version required quite a bit of conversation with Sugwoo Shin (who is of course blameless of any errors).</p> <p>We give a description of the odd $GL_2$-Galoisian sets of level $N$ in terms of 'higher congruence conditions.' The point is to consider the $\mathbb{Q}$-Hecke algebra $H(N)$ determined by the Hecke operators acting on modular forms of weight 1, level $N$. The maximal ideals in $H(N)$ are in correspondence with Galois conjugacy classes of normalized new weight one eigenforms of level $N$. There is also a map $$p\mapsto T_p$$ from primes not dividing $N$ to $H(N)$.</p> <p>Any given maximal ideal $m$ determines a Dirichlet character $\epsilon_m$, and one considers the set of primes $S(m)$ defined by the 'congruence conditions' $$(p,N)=1, \ \epsilon_m(p)=1, \ T_p\equiv 2 \ \ \mod \ m$$ These $S_m$ are exactly the odd $GL_2$ Galoisian sets of level $N$. With a bit more care, one should be able to give a similar description that doesn't refer to the level beforehand.</p> <p>Of course this is no different from what I wrote before, but focussing on the Hecke algebra seems to allow a formulation that's rather analogous to the classical one. That is, one can forget about modular forms for a moment and examine sets of primes determined by congruences in the Hecke algebra.</p> http://mathoverflow.net/questions/29334/k-2-of-rings-of-algebraic-integers/29346#29346 Answer by Minhyong Kim for K_2 of rings of algebraic integers Minhyong Kim 2010-06-24T07:00:23Z 2010-07-08T17:20:25Z <p>It's a theorem of Garland that $K_2(R)$ is finite. Perhaps the best way to get a handle on it is to use Quillen's localization sequence $$0\rightarrow K_2(R)\rightarrow K_2(F)\stackrel{T}{\rightarrow} \oplus_v k(v)^*\rightarrow 0,$$ where $F$ is the fraction field and the $k(v)$ are the residue fields. The map $T$ is the sum of the tame symbols, which is surjective by a theorem of Matsumoto. The injectivity on the left follows from the vanishing of $K_2$ for finite fields. </p> <p>This isn't much of an answer, but considering $K_2(R)$ as a subgroup of $K_2(F)$ seems a reasonable way to start some concrete considerations. For a detailed discussion of an algorithm that proceeds essentially along these lines ('Tate's method), see the paper </p> <p>Belabas, Karim; Gangl, Herbert Generators and relations for $K_2( O_F)$. $K$-Theory 31 (2004), no. 3, 195--231.</p> <p>Added, 8 July:</p> <p>I'm sure most people know this, but I forgot to mention (for newcomers) the fact that $$K_2(F) = F^\times\otimes_{\mathbf Z} F^\times/\langle a\otimes(1-a)\mid a\not=0,1\rangle,$$ which I suppose motivates the original question, and makes it worthwhile to view $K_2(R)$ as a subgroup.</p> http://mathoverflow.net/questions/14714/what-do-heat-kernels-have-to-do-with-the-riemann-roch-theorem-and-the-gauss-bonne/26679#26679 Answer by Minhyong Kim for What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem? Minhyong Kim 2010-06-01T07:57:47Z 2010-06-02T08:12:43Z <p>Added 2 June:</p> <p>Since the summary below is already a bit long, I thought I'd add a few lines at the beginning as a guide. The proofs all proceed as follows:</p> <ol> <li><p>Identify the quantity of interest (like the Euler characteristic) as the index of an operator going from an 'even' bundle to an 'odd' bundle.</p></li> <li><p>Use Hodge theory to write the index in terms of the dimensions of harmonic sections, i.e., kernels of Laplacians.</p></li> <li><p>Use the heat evolution operator for the Laplacians and 'supersymmetry' to rewrite this as a 'supertrace.'</p></li> <li><p>Write the heat evolution operator in terms of the heat kernel to express the supertrace as the integral of a local density.</p></li> <li><p>Use the eigenfunction expansion of the heat kernel to identify the constant (in time) part of the local density.</p></li> </ol> <p>Most of this is general nonsense, and the difficult step is 5. By and large, the advances made after the seventies all had to do with finding interpretations of this last step that employed intuition arising from physics.</p> <hr> <p>I suffered over this proof quite a bit in my pre-arithmetic youth and wrote up a number of summaries. A condensed and extremely superficial version is given here, mostly for my own review. If by chance someone finds it at all useful, of course I will be delighted. I apologize that I don't say anything about physical intuition (because I have none), and for repeating parts of the previous nice answers. It's been years since I've thought about these matters, so I will forgo all attempts at even a semblance of analytic rigor. In fact, the main pedagogical reason for posting is that a basic outline of the proof is possible to understand with almost <em>no</em> analysis.</p> <p>The usual setting has a compact Riemannian manifold $M$, two hermitian bundles $E^+$ and $E^-$, and a linear operator $$P:H^+\rightarrow H^-,$$ where $H^{\pm}:=L^2(E^{\pm})$. With suitable assumptions (ellipticity), $ker(P)$ and $coker(P)$ have finite dimension, and the number of interest is the index: $$Ind(P)=dim(ker(P))-dim(coker(P)).$$ This can also be expressed as $$dim(ker(P))-dim(ker(P^{*})),$$ where $$P^{*}:H^-\rightarrow H^+$$ is the Hilbert space adjoint. A straightforward generalization of the Hodge theorem allows us also to write this in terms of Laplacians $\Delta^+=P^* P$ and $\Delta^-=PP^*$ as $$dim(ker(\Delta^+))-dim(ker(\Delta^-)).$$ Things get a bit more tricky when we try to identify the index with the expression ('supertrace,' so-called) $$Tr(e^{-t\Delta^+})-Tr(e^{-t\Delta^-}).$$ The operator $$e^{-t\Delta^{\pm}}:H^{\pm}\rightarrow H^{\pm}$$ sends a section $f$ to the solution of the heat equation $$\frac{\partial}{\partial t} F(t,x)+\Delta^{\pm}F(t,x)=0$$ ($x$ denoting a point of $M$) at time $t$ with intial condition $F(0,x)=f(x).$ One important part of this is that there are discrete Hilbert direct sum decompositions $$H^+=\oplus_{\lambda} H^+(\lambda)$$ and $$H^-=\oplus_{\mu} H^-(\mu)$$ in terms of finite-dimensional eigenspaces for the Laplacians with non-negative eigenvalues. And then, the identities $$\Delta^-P=PP^{*}P=P\Delta^+$$ and $$\Delta^+P^{*}=P^{*}PP^{*}=P^{*}\Delta^-$$ show that the (supersymmetry) operators $P$ and $P^{*}$ can be used to define isomorphisms between all non-zero eigenspaces of the two Laplacians with a correspondence for eigenvalues as well. Thus, once you believe that the exponential operators are trace class, it's easy to see that the only contributions to the trace are from the kernels of the plus and minus Laplacians. This is the 'easy cancellation' that occurs in this proof. But on the zero eigenspaces, the heat evolution operators are clearly the identity, allowing us to identify the supertrace with the index. To summarize up to here, we have $$Ind(P)=Tr(e^{-t\Delta^+})-Tr(e^{-t\Delta^-}).$$ This identity also makes it obvious that the supertrace is in fact independent of $t>0$.</p> <p>The proofs under discussion all have to do with identifying this supertrace in terms of local expressions that relate naturally to characteristic classes. The beginning of this process involves first writing the operator $e^{-t\Delta^+}$ in terms of an integral kernel $$K^+_t(x,y)=\sum_i e^{-t\lambda_i } \phi^+_i(x)\otimes \phi^+_i(y)$$ where the $\phi^+_i$ make up an orthonormal basis of eigenvectors for the Laplacian. That is, $$e^{-t\Delta^+}f=\int_M K^+_t(x,y)f(y)dvol(y)=\sum_i e^{-t\lambda_i } \int_M \phi^+_i(x) \langle \phi^+_i(y),f(y)\rangle dvol(y).$$ Formally, this identity is obvious, and the real work consists of the global analysis necessary to justify the formal computation. Obviously, there is a parallel discussion for $\Delta^-$. Now, by an infinite-dimensional version of the formula that expresses the trace of a matrix as a sum of diagonals, we get that $$Tr(e^{-t\Delta^+})=\int_M Tr(K^+_t(x,x))dvol(x)=\int_M \sum_ie^{-t\lambda_i}||\phi^+_i(x)||^2 dvol(x),$$ an integral of local (point-wise) traces, and similarly for $Tr(e^{-t\Delta^-})$. One needs therefore, techniques to evaluate the density</p> <p>$$\sum_ie^{-t\lambda_i}||\phi^+_i(x)||^2 dvol(x)-\sum_ie^{-t\mu_i}||\phi^-_i(x)||^2 dvol(x).$$</p> <p>More analysis gives an asymptotic expansion for the plus and minus densities of the form $$a^{ \pm }_{-d/2}(x) t^{-d/2}+a^{ \pm }_{d/2+1}(x) t^{-d/2+1}+\cdots$$ where $d$ is the dimension of $M$.</p> <p>Up to here the discussion was completely general, but then the proof begins to involve special cases, or at least, broad division into classes of cases. But note that even for the special cases mentioned in the original question, one would essentially carry out the procedure outlined above for a specific operator $P$.</p> <p>The breakthrough in this line of thinking came from Patodi's incredibly complicated computations for the operator $d+d^*$ going from even to odd differential forms, where one saw that the $$a^{+}_i(x)$$ and $$a^{-}_i(x)$$ canceled each other out locally, that is, for each point $x$, for all the terms with negative $i$. I think it was fashionable to refer to this cancellation as 'miraculous,' which it is, compared to the easy cancellation above. At this point, Patodi could take a limit $$\lim_{t\rightarrow 0}[\sum_ie^{-t\lambda_i}||\phi^+_i(x)||^2 dvol(x)-\sum_ie^{-t\mu_i}||\phi^-_i(x)||^2 dvol(x)],$$ that he identified with the Euler form. This important calculation set a pattern that recurred in all other versions of the heat kernel approach to index theorems. One proves the existence of an analogous limit as $t\rightarrow 0$ and identifies it. The identification as a precise differential form representative for a characteristic class is referred to sometimes as a <em>local index theorem</em>, a statement more refined than the topological formula for the global index. There is even a beautiful version of a local <em>families index theorem</em> that relates eventually to deep work in arithmetic intersection theory and Vojta's proof of the Mordell conjecture.</p> <p>As I understand it, Gilkey's contribution was an invariant theory argument that tremendously simplified the calculation and allowed a differential form representative for the $\hat{A}$ genus to emerge naturally in the case of the Dirac operator. And then, I believe there is a $K$-theory argument that deduces the index theorem for a general elliptic operator from the one for the twisted Dirac operator.</p> <p>Experts can correct me if I'm wrong, but from a purely mathematical point of view, essentially all the work on the heat kernel proof was done at this point. Subsequent <em>interpretations</em> of the proof (more precisely, the supertrace), in terms of supersymmetry, path integrals, loop spaces, etc., were tremendously influential to many areas of mathematics and physics, but the mathematical core of the index theorem itself appears to have remained largely unchanged for almost forty years. In particular, the terminology I've used myself above, the super- things, didn't occur at all in the original papers of Patodi, Atiyah-Bott-Patodi, or Gilkey.</p> <p>Added: </p> <p>Here is just a little bit of geometric-physical intuition regarding the heat kernel in the Gauss-Bonnet case, which I'm sure is completely banal for most people. The density $$\sum_ie^{-t\lambda_i}||\phi^+_i(x)||^2 dvol(x)-\sum_ie^{-t\mu_i}||\phi^-_i(x)||^2 dvol(x)$$ expresses the heat kernel in terms of orthonormal bases for the even and odd forms. When $t\rightarrow \infty$ all terms involving the positive eigenvalues decay to zero, leaving only contributions from orthonormal <em>harmonic</em> forms. This is one way to to see that the integral of this density, which is independent of $t$, must be the Euler characteristic. On the other hand, as $t\rightarrow 0$, the operator $$K^+_t(x,y)dvol(y)=[\sum_i e^{-t\lambda_i } \phi^+_i(x)\otimes \phi^+_i(y)]dvol(y)$$ literally approaches the identity operator on all even forms (except for the fact that it diverges). That is, the heat kernel interpolates between the identity and the projection to the harmonic forms, in some genuine sense expressing the diffusion of heat from a point distribution to a harmonic steady-state. A similar discussion holds for the odd forms as well. I can't justify this next point even vaguely at the moment, but one should therefore think of $$[K^+_t(x,y)-K^-_t(x,y)]dvol(y)$$ as regularizing the current on $M\times M$ given by the diagonal $M\subset M\times M$. Thus, the integral of $$[TrK^{+}_t(x,x)-TrK^-_t(x,x)]dvol(x)$$ ends up computing a deformed self-intersection number of the diagonal in $M\times M$. From this perspective, it shouldn't be too surprising that the Euler class, representing exactly this self-intersection, shows up.</p> <p>Added:</p> <p>I forgot to mention that the Riemann-Roch case is where $$P=\bar{\partial}+\bar{\partial}^*$$ going from the even to the odd part of the Dolbeault resolution associated to a holomorphic vector bundle. The limit of the local density is a differential form representing the top degree portion of the Chern character of the bundle multiplied by the Todd class of the tangent bundle. Perhaps it's worth stressing that these special cases all go through the general argument outlined above.</p> http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism Equality vs. isomorphism vs. specific isomorphism Minhyong Kim 2010-05-12T12:02:18Z 2010-05-14T10:00:41Z <p><a href="http://mathoverflow.net/questions/24318/are-there-situations-when-regarding-isomorphic-objects-as-identical-leads-to-mist" rel="nofollow">This question</a> prompted a reformulation: </p> <p><em>What is a really good example of a situation where keeping track of isomorphisms leads to tangible benefit?</em></p> <p>I believe this to be a serious question because it actually is oftentimes a good idea casually to identify isomorphism classes. To bring up an intermediate-level example I've alluded to often, consider the classification of topological surfaces. When I explain it to students, I do somewhat consciously write equalities as I manipulate one shape into another homeomorphic one. I even do it rather quickly to encourage intuitive associations that are likely to be useful. In any case, for arguments of that sort, it would be really tedious, and probably pointless, to write down isomorphisms with any precision.</p> <p>Meanwhile, at other times, I've also joined in the chorus of criticism that greets the conflation of equality and isomorphism.</p> <p>The problem is it's quite challenging to come up with really striking examples where this care is rewarded. Let me start off with a somewhat specialized class of examples. These come from <em>descent theory</em>. The setting is a map $$X\rightarrow Y,$$ which is usually submersive, in some sense suitable to the situation. You would like criteria for an object $V$ lying over $X$, say a fiber bundle, to arise as a pull-back of an object on $Y$. There is a range of formalism to deal with this problem, but I'll just mention two cases. One is when $Y=X/G$, the orbit space of a group action on $X$. For $V$ to be pulled-back from $Y$, we should have $g^*(V)\simeq V$ for each $g\in G$. But that's not enough. What is actually required is that there be a collection of isomorphisms $$f_g: g^*(V)\simeq V$$ that are compatible with the group structure. This means something like $$f_{gh}=f_g\circ f_h,$$ except you have to twist in an obvious way to take into account the correct domain. So you see, I have at least to introduce notation for the isomorphisms involved to formulate the right condition. In practice, when you want to <em>construct</em> something on $Y$ starting from something on $X$, you have to specify the $f_g$ rather precisely. </p> <p>Another elementary case is when $X$ is an open covering $(U_i)$ of $Y$. Then an object on $Y$ is typically equivalent to a collection $V_i$ of objects, one on each $U_i$, but with additional data. Here as well, $V_i$ and $V_j$ obviously have to agree on the intersections. But that's again not enough. Rather there should be a collection of isomorphisms $$\phi_{ji}: V_i|U_i\cap U_j\simeq V_j|U_i\cap U_j$$ that are compatible on the triple overlaps: $$\phi_{kj}\circ \phi_{ji}=\phi_{ki}.$$ Incidentally, for something like a vector bundle, since any two of the same rank are locally 'the same,' it's clear that keeping track of isomorphisms will be the key to the transition from collections of local objects to a global object. The formalism is concretely applied in situations where you can define some objects only locally, but would like to glue them together to get a global object. For a really definite example that comes immediately to mind, there is the determinant of cohomology for vector bundles on a family of varieties over a parameter space $Y$. Because a choice of resolution is involved in defining this determinant, which might exist only locally on $Y$, Knudsen and Mumford struggled quite a bit to show that the local constructions glue together. Then Grothendieck suggested the remedy of defining the determinant provisionally as a <em>signed</em> line bundle, which then allowed them to nail down the correct $\phi_{ji}$. These days, this determinant is a very widely useful tool, for example, in generating line bundles on moduli spaces.</p> <p>I apologize if this last paragraph is a bit too convoluted for non-specialists. Part of my reason for writing it down is to illustrate that my main examples for bolstering the 'keep track of isomorphisms' paradigm are a bit too advanced for most undergraduates.</p> <p>So, to conclude, I'd be quite happy to hear of better examples. As already suggested above, it would be nice to have them be accessible but substantively illuminating. If you would like to discuss, say, different bases for vector spaces, it would be good if the language of isomorphism etc. clarifies matters in a really obvious way, as opposed to a sets-and-elements exposition.</p> <p>Added: Oh, if you have advanced examples, I would certainly like to hear about them as well.</p> <p>Added: I see now there are three levels at least to distinguish:</p> <p>Regarding objects as equal vs. regarding them as isomorphic vs. paying attention to specific isomorphisms.</p> <p>I somehow conflated the two transitions in the course of asking the question. Of course I'm happy to see good examples illustrating the nature of either, but I'm especially interested in the second refinement.</p> <p>Added yet again:: I'm grateful to everyone for contributing nice examples, and to Urs Schreiber who put in some effort to instruct me over at the <a href="http://golem.ph.utexas.edu/category/2009/10/math_overflow.html#comments" rel="nofollow">n-category cafe</a>. As I mentioned to Urs there, it would be especially nice to see examples of the following sort.</p> <ol> <li><p>One usually thinks $X=Y$;</p></li> <li><p>A careful analysis encourages the view $X\simeq Y$;</p></li> <li><p>This perspective leads to genuinely new insight and benefit. </p></li> </ol> <p>Even better would be if some specific knowledge of the isomorphism in 2. is important. Of course, more than two objects might be involved. I was initially hoping for some input from combinatorics, with the emphasis on 'bijective proofs' and all that. Anything?</p> <p>Added, 14 May:</p> <p>OK, I hope this will be the last addition. Because this question flowed over to the n-category cafe, I ended up having a small discussion there as well. I thought I'd copy here my last response, in case anyone else is interested.</p> <p>n-cafe post:</p> <p>I suppose it's obvious by now that I'm using a specific request to drive home the need for 'small but striking examples' in favor of category theory.</p> <p>Last fall, Eugenia Cheng told me of a visit to some university to give a colloquium talk. The host greeted her with the observation that he doesn't regard category theory as a field of research. OK, he was probably a bit extreme, but milder versions of that view are quite common. Now, one possible response is to regard all such people as unreasonable and talk just to friends (who of course are the reasonable people!). This is not entirely bad, because that might be a way to buy time and gain enough stability to eventually prove the earth-shattering result that will show everyone! Another way is to take up the skepticism as a constructive everyday challenge. This I suppose is what everyone here is doing at some level, anyways.</p> <p>Other than the derived loop space, which is not exactly small, Urs' examples are all of the simple subtle sort that can, over time, contribute to a really important change in scientific outlook and maybe even the infrastructure of a truly glorious theory. For example, I agree wholeheartedly about the horrors of the old tensor formalism. But it's not unreasonable to ask for more striking accessible evidence of utility when it comes to the current state of category theory.</p> <p>The importance of small insights and language that gradually accumulate into the edifice of a coherent and powerful theory is the usual interpretation of Grothendieck's 'rising sea' philosophy. However, the process is hardly ever smooth along the way, especially the question of acceptance by the community. I'm not a historian, but I've studied arithmetic geometry long enough to have some sense of the changing climate surrounding etale cohomology theory, for example, over the last several decades. The full proof of the Weil conjectures took a while to come about, as you know. Acceptance came slowly with many bits and pieces sporadically giving people the sense that all those subtleties and abstractions are really worthwhile. Fortunately, the rationality of the zeta function was proved early on. However, there was a pretty concrete earlier proof of that as well using $p$-adic analysis, so I doubt it would have been the big theorem that convinced everyone. One real breakthrough came in the late sixties when Deligne used etale cohomology to show that Ramanujan's conjecture on his tau function could be reduced to the Weil conjectures. There was no way to do this without etale cohomology and the conjecture in question concerned something very precise, the growth rate of natural arithmetic functions. This could even be checked numerically, so impressed people in the same way that experimental verification of a theoretical prediction does in physics. Clearly something deep was going on. Of course there were many other indications. The construction of entirely new representations of the Galois group of $\mathbb{Q}$ with very rich properties, the unification of Galois cohomology and topological cohomology, a clean interpretation of arithmetic duality theorems that gave a re-interpretation of class field theory, and so on.</p> <p>For myself, being a fan of you folks here, I believe this kind of process is going on in category theory. But I don't think you have to be too unreasonable to doubt it. In a similar vein, I don't agree with Andrew Wiles' view that physics will be irrelevant for number theory, but also think his pessimism is perfectly sensible.</p> <p>I think I'm trying to make the obvious point that the presence of pessimists can be very helpful to the development of a theory, in so far as the optimists interact with them in constructive ways. I haven't been coming to this site much lately, because the bit of internet time I have tends to be absorbed by Math Overflow. But I did catch David's <a href="http://golem.ph.utexas.edu/category/2010/05/quinn_on_higherdimensional_alg.html#more" rel="nofollow">recent post</a> on Frank Quinn's article, which ended up as a catalyst for my MO question.</p> <p>At the Boston conference following the proof of Fermat's last theorem, I've been told Hendrik Lenstra said something like this: 'When I was young, I knew I wanted to solve Diophantine equations. I also knew I didn't want to represent functors. Now I have to represent functors to solve Diophantine equations!' So should we conclude that he was foolish to avoid representable functors for so long? I wouldn't.</p> <p><a href="http://mathoverflow.net/questions/24378/equality-vs-isomorphism-vs-specific-isomorphism/24563#24563" rel="nofollow">This response</a> to the MO question brings up the importance of knowing the specific isomorphism between some Hilbert spaces given by the Fourier transform. This is an excellent example, especially when we consider how it relates to the different realizations of the representations of the Heisenberg group and the attendant global issues, say as you vary over a family of polarizations. But I couldn't resist recalling Irving Segal's insistence that 'There's only <em>one</em> Hilbert space!' Obviously, he knew, among many other things, the different realizations of the Stone-Von-Neumann representation as well as anyone, so you can take your own guess as to the reasoning behind that proclamation. He certainly may have lost something through that kind of philosophical intransigence. But I suspect that he, and many around him, gained something as well.</p> http://mathoverflow.net/questions/119015/what-precisely-does-kleins-erlangen-program-state/119116#119116 Comment by Minhyong Kim Minhyong Kim 2013-01-29T00:05:07Z 2013-01-29T00:05:07Z @Terry Tao: I don't know if Klein would have agreed, but I'm sympathetic to the view that category theory inherits the Erlangen programme. If one allows the categorical generalization of symmetry, a version of the 'recover space from symmetries' view that is almost tautologically true is the Yoneda Lemma. In a different direction, there are a number of interesting theorems in algebraic and arithmetic geometry that allow the reconstruction of a space from an associated category. http://mathoverflow.net/questions/119015/what-precisely-does-kleins-erlangen-program-state/119116#119116 Comment by Minhyong Kim Minhyong Kim 2013-01-16T22:21:26Z 2013-01-16T22:21:26Z This answers seems to capture the rough understanding I had of the programme. I recall hearing an even more simplistic formulation to the effect that 'a geometry is determined by its symmetries.' If some idea of this sort really was articulated in the Erlangen programme, then it seems to have rested on a very restricted (one might even say old-fashioned) notion of geometry. From the Riemannian viewpoint, most geometries have no symmetries at all. That is, from a modern perspective, such a statement is clearly wrong. http://mathoverflow.net/questions/115489/why-is-algebraic-de-rham-cohomology-via-completion-independent-of-embedding Comment by Minhyong Kim Minhyong Kim 2012-12-05T11:32:08Z 2012-12-05T11:32:08Z There are a number of approaches using tubular neighborhoods. But a nice way is to use the *infinitesimal topology,' a characteristic zero version of the crystalline topology. The ideas are explained in Illusie's article in Arcata 1974. http://mathoverflow.net/questions/115430/what-is-the-geometry-of-an-undecidable-diophantine-equation Comment by Minhyong Kim Minhyong Kim 2012-12-05T00:17:13Z 2012-12-05T00:17:13Z I never checked this myself, but Mazur told a number of us many years ago that he had looked at Matiyasevich's equations, and found that they all had plenty of rational solutions. Someone should ask him about this. Alternatively, you could look at the equations yourself. This doesn't answer your question, but it's a start. http://mathoverflow.net/questions/114383/examples-where-adding-complexity-made-a-problem-simpler Comment by Minhyong Kim Minhyong Kim 2012-11-26T04:42:14Z 2012-11-26T04:42:14Z My favorite elementary example is the computation of $1+2+\cdots +n$, which is easier to do twice than once. http://mathoverflow.net/questions/111519/why-might-andre-weil-have-named-carl-ludwig-siegel-the-greatest-mathematician-of/111697#111697 Comment by Minhyong Kim Minhyong Kim 2012-11-08T14:04:30Z 2012-11-08T14:04:30Z (That should have been <i>integral</i> points.) It's easy to forget now that the significance of proving the finite generation of points on something as exotic as a Jacobian, not corresponding in any obvious way to an actual equation, must not have been obvious in the early twentieth century. Siegel's concrete theorem was what made it clear that this was a useful thing. It's also easy to forget that Siegel's theorem was essentially the best thing Diophantine geometry had to offer for many decades preceding Faltings, and probably one of the best theorems in all of number theory. http://mathoverflow.net/questions/111519/why-might-andre-weil-have-named-carl-ludwig-siegel-the-greatest-mathematician-of/111697#111697 Comment by Minhyong Kim Minhyong Kim 2012-11-08T13:58:19Z 2012-11-08T13:58:19Z Dear Paul, This is a very interesting answer, and I think I understand your caution with respect to the larger-than-life people. However, I would have to agree with quid in this case: Weil speaks highly of Siegel in a number of different places. John Coates points out to me that Siegel was the person who really thrust Weil's name into the mathematical landscape by using the Mordell-Weil theorem to prove the finiteness of points on affine hyperbolic curves. http://mathoverflow.net/questions/111519/why-might-andre-weil-have-named-carl-ludwig-siegel-the-greatest-mathematician-of/111529#111529 Comment by Minhyong Kim Minhyong Kim 2012-11-05T07:31:34Z 2012-11-05T07:31:34Z Pete's answer demonstrates very well the correctness of this expectation, but I agreed even before I saw it. http://mathoverflow.net/questions/111519/why-might-andre-weil-have-named-carl-ludwig-siegel-the-greatest-mathematician-of/111529#111529 Comment by Minhyong Kim Minhyong Kim 2012-11-05T07:31:08Z 2012-11-05T07:31:08Z This is a very nice answer and, a bit paradoxically, illustrates how the question is not very strange. On the one hand, I do understand how some people may be wary of questions that somehow feel 'gossipy.' However, I understood Jonah's question to be something like this: 'Of course Siegel is great, but there are many great mathematicians. What special circumstance made Weil name Siegel in particular? I think a good answer to this question by someone knowledgeable in number theory and some history would be quite illuminating from a mathematical point of view.' http://mathoverflow.net/questions/111329/varieties-with-infinitely-many-etale-covers-and-rational-points/111331#111331 Comment by Minhyong Kim Minhyong Kim 2012-11-03T16:06:11Z 2012-11-03T16:06:11Z Maybe I also misunderstand the question, but I don't see what the difficulty is over a number field. Take any genus one curve $X$ without a rational point, $3x^3+4y^3+5z^3=0$ over $Q$, for example. Then $X$ will have a finite-to-one map $f:X\rightarrow E$ to its Jacobian $E$, which is an elliptic curve. You can then pull back, say, any isogeny of $E$ of degree prime to that of $f$. You can generalize this easily to other curves without rational points. http://mathoverflow.net/questions/111084/why-are-derived-categories-natural-places-to-do-deformation-theory Comment by Minhyong Kim Minhyong Kim 2012-11-02T00:28:47Z 2012-11-02T00:28:47Z @Fernando Muro I hope you don't mind if I revise my statement into a slightly more presumptuous form. I know a number of great mathematicians who share your view as a matter of personal <i>conviction</i>. My impression is that they find it rather hard to follow consistently as a matter of actual practice. http://mathoverflow.net/questions/111084/why-are-derived-categories-natural-places-to-do-deformation-theory Comment by Minhyong Kim Minhyong Kim 2012-11-02T00:22:27Z 2012-11-02T00:22:27Z @Fernando Muro: That's fine of course. I know a number of great mathematicians who essentially share your view as a matter of personal practice. I was just puzzled that you were puzzled by some <i>other</i> people having the view I outlined! Clearly these different temperaments can coexist and contribute to mathematics. http://mathoverflow.net/questions/111084/why-are-derived-categories-natural-places-to-do-deformation-theory Comment by Minhyong Kim Minhyong Kim 2012-11-01T14:32:30Z 2012-11-01T14:32:30Z And then other times, some popular thing looks vaguely interesting, even if you're not motivated by too specific a need. Then you'd like to hear what all the fuss is about in a relatively painless way. http://mathoverflow.net/questions/111084/why-are-derived-categories-natural-places-to-do-deformation-theory Comment by Minhyong Kim Minhyong Kim 2012-11-01T14:28:30Z 2012-11-01T14:28:30Z Fernando: That seems to be rather a hard line to take. Clearly, many people ask questions like that when they feel like something <i>may</i> be useful to them, but they're not sure enough to devote a lot of time to studying it. In such a situation, isn't it quite helpful to hear friendly explanations of why other people find it useful? http://mathoverflow.net/questions/110931/what-can-we-do-to-raise-awareness-of-reciprocity-laws Comment by Minhyong Kim Minhyong Kim 2012-10-29T02:44:12Z 2012-10-29T02:44:12Z Dear Jonah: Possibly, the tone of your question reads a bit like a call to political action. I'm guessing this reflects your sincerity, enthusiasm, and the important activism outside of mathematics that you are currently engaged in. This is fine by me, but I think it may have induced some confusion on the part of other commentators. Maybe if it were rephrased along the lines of 'ideas for conveying some deep number theory to students in secondary school (and non-arithmeticians)' the question could be just as valuable while avoiding controversy.