User james d. taylor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:20:48Z http://mathoverflow.net/feeds/user/5756 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126430/is-there-a-nice-criterion-for-when-the-splitting-fields-of-two-irreducible-polyno Is there a nice criterion for when the splitting fields of two irreducible polynomials are equal? James D. Taylor 2013-04-03T19:02:58Z 2013-04-04T12:50:05Z <p>This question is a bit vague, but I was wondering if someone might have an insightful answer.</p> <p>Let $f_1$ and $f_2$ be irreducible polynomials in $\mathbb{Q}[x]$. Is there an easy criterion for knowing when the splitting fields of $f_1$ and $f_2$ yield the same field extensions of $\mathbb{Q}$?</p> <p>Here is a related question. Let $L/\mathbb{Q}$ be a finite field extension. Assume both $f_1$ and $f_2$ remain irreducible in $L$. Given such an $L$, is there a way to determine when the splitting fields of $f_1$ and $f_2$ over $L$ are the same? (It is possible that the splitting fields of $f_1$ and $f_2$ over $\mathbb{Q}$ are different, but their splitting fields over $L$ are the same.)</p> http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture Philosophy behind Mochizuki's work on the ABC conjecture James D. Taylor 2012-09-07T01:06:48Z 2013-02-22T18:21:30Z <p>Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?</p> http://mathoverflow.net/questions/51395/a-soft-introduction-to-physics-for-mathematicians-who-dont-know-the-first-thing A soft introduction to physics for mathematicians who don't know the first thing about physics James D. Taylor 2011-01-07T14:39:17Z 2012-12-12T14:17:31Z <p>There have been similar questions on mathoverflow, but the answers always gave some advanced introduction to the mathematics of quantum field theory, or string theory and so forth. While those may be good introduction to the mathematics of those subjects, what I require is different: what provides a soft and readable introduction to the (many) concepts and theories out there, such that the mathematics involved in it is in comfortable generality. What makes this is a "for mathematicians" question, is that a standard soft introduction will also assume that the reader is uncomfortable with the word "manifold" or certainly "sheaf" and "Lie algebra". So I'm looking for the benefit of scope and narrative, together with a presumption of mathematical maturity.</p> <p>N.B. If your roadmap is several books, that is also very welcome.</p> http://mathoverflow.net/questions/108831/why-is-there-no-regular-etale-fundamental-group Why is there no "regular etale fundamental group"? James D. Taylor 2012-10-04T16:03:59Z 2012-10-04T18:04:11Z <p>Let $K$ be a number field, and let $X_K$ be a $K$-variety.</p> <p>The etale fundamental group of $X_K$, as defined in SGA1, classifies the automorphisms of finite etale covers of $X_K$. Some of these etale covers are not geometric. For example, if $L$ is a finite field extension of $K$, then $X_L\rightarrow X_K$ is a finite etale cover. A cover $Y$ of $X_K$ is called regular (think: geometric) if it doesn't have an extension of scalars (to be precise, if $K$ is algebraically closed in the function field of every connected component of $Y$).</p> <p>I know, however, that there is no "regular etale fundamental group" for $X_K$, that classifies the automorphisms of just the regular finite etale covers of $X_K$. I vaguely recall hearing in a conference that this is because the regular etale covers don't form a Galois category. Is that correct? What fails for them to form a Galois category? This question has been in the back of my mind ever since I started working with the etale fundamental group.</p> http://mathoverflow.net/questions/97051/what-is-the-intuition-for-mathbbqab-having-cohomological-dimension-1 What is the intuition for $\mathbb{Q}^{ab}$ having cohomological dimension $1$? James D. Taylor 2012-05-15T20:57:46Z 2012-05-15T20:57:46Z <p>I frequently talk to people who think of finite fields as arithmetic analogs of punctured discs. This makes some sense: the absolute Galois group of a finite field is the profinite completion of $\mathbb{Z}$. Since the absolute Galois group of a field is its algebraic fundamental group, this gives the feel of punctured disc. </p> <p>The cohomological dimension of a field is the cohomological dimension of its absolute Galois group. Therefore the cohomological dimension of a finite field is $1$. This agrees with the intuition that finite fields are "like" punctured discs. (Given a projective curve $C$ over $\mathbb{C}$ and a point $P$ on $C$, a small punctured analytic neighborhood of $P$ has dimension $1$ in the sense that it is a neighborhood of a curve.)</p> <p>The picture gets murky when we get to $\mathbb{Q}^{ab}$. The absolute Galois group of $\mathbb{Q}^{ab}$ is not known, but is conjectured to be profinite free. It is known that the cohomological dimension of $\mathbb{Q}^{ab}$ is $1$. Is there some geometric intuition associated with $\mathbb{Q}^{ab}$? It is surely much more complex than a punctured disc, because its algebraic fundamental group (absolute Galois group) is more complicated. </p> http://mathoverflow.net/questions/68647/understanding-the-wiki-page-on-verdier-duality Understanding (the wiki page on) Verdier duality James D. Taylor 2011-06-23T19:48:20Z 2012-04-19T01:11:04Z <p>My familiarity with concepts related to derived categories is only tangential, and little by little I intend to get more comfortable with them. I was playing around with Caldararu's introduction to the topic, and looking up various things on the web.</p> <p>Here is my question (that I have every confidence is trivial for experts):</p> <p>On the wiki page on Verdier duality <a href="http://en.wikipedia.org/wiki/Verdier_duality" rel="nofollow">http://en.wikipedia.org/wiki/Verdier_duality</a> it says the following: Let $F$ be a field, and $X$ a finite dimensional (dimension is defined here cohomologically, but for our purposes a finite dimensional manifold will do) locally compact space.</p> <p>In the part about Poincare duality, it says: $H^k(X,F)=[F,X[k]]$. What is the interpretation of this notation? As I see it, $[F,X[k]]$ means $Hom(F,X[k])$ in the derived category. But this means that $X$ is seen as a complex. How? And why would $Hom(F,X[k])$ equal $H^k(X,F)$?</p> http://mathoverflow.net/questions/91009/which-groups-are-quotients-of-symmetric-groups Which groups are quotients of symmetric groups? James D. Taylor 2012-03-12T19:05:12Z 2012-03-12T19:27:38Z <p>By Cayley's embedding theorem every group $G$ embeds into the symmetric group $S_{|G|}$. But which groups $G$ have the property that there exists some $n$ such that $G$ is a quotient of $S_n$? My intuition is that it couldn't possibly be all finite groups. Is there a nice characterization of the groups that do have this property?</p> http://mathoverflow.net/questions/88289/are-there-n-polynomials-for-which-all-intersection-multiplicities-are-at-least-m Are there n polynomials for which all intersection multiplicities are at least m? James D. Taylor 2012-02-12T17:56:04Z 2012-02-12T23:47:10Z <p>I don't know whether this is known or not, but I was thinking of the following problem.</p> <p>Let $n$ and $m$ be natural numbers. Are there $n$ polynomial $f_1,...,f_n\in \mathbb{C}[x]$, such that all of their intersection numbers are at least $m$?</p> <p>It is also possible to state an even harder question in a more natural (albeit less basic) way: Are there $n$ homogeneous polynomials in 2 variables for which all of their intersection numbers in $\mathbb{P}^1_{\mathbb{C}}$ are at least $m$?</p> <p>EDIT: Sorry, the comments made me realize I wanted a slightly different condition: that for every $j$ there's a unique $k\neq j$ such that $f_j(0)=f_k(0)$. (In particular, $n$ is assumed to be even.) I will allow the intersection at $x=0$ to not be of multiplicity $\geq m$, but I will ask that over $x\neq 0$ the intersection multiplicity will be $\geq m$.</p> <h3>Clarification</h3> <p>The question was posed in a way that algebraic geometers would understand, because I suspect they are most likely to come up with a solution to this question. I wanted to emphasize, however, that intersection multiplicity is something that every high-school student can understand: If $f_1$ and $f_2$ are polynomials with coefficients in $x$, and they meet at say $x=3$, then their intersection multiplicity at $x=3$ is the greatest natural number $l$ such that $(x-3)^l$ divides the polynomial $f_1-f_2$.</p> http://mathoverflow.net/questions/80717/why-should-the-anabelian-geometry-conjectures-be-true/87502#87502 Answer by James D. Taylor for Why should the anabelian geometry conjectures be true? James D. Taylor 2012-02-04T01:57:46Z 2012-02-04T01:57:46Z <p>I think you should take a look at: <a href="http://www.renyi.hu/~szamuely/heid.pdf" rel="nofollow">http://www.renyi.hu/~szamuely/heid.pdf</a></p> <p>The section there about anabelian geometry gives several reasons why one might believe it is true.</p> http://mathoverflow.net/questions/86670/what-is-the-obstruction-for-a-local-set-of-models-of-a-curve-to-come-from-a-globa What is the obstruction for a local set of models of a curve to come from a global model? James D. Taylor 2012-01-25T23:48:35Z 2012-01-31T09:41:09Z <p>If $X_{\mathbb{Q}}$ is a curve over $\mathbb{Q}$, we get a curve $X_{\mathbb{Q}_p}$ over $\mathbb{Q}_p$ for every prime $p$.</p> <p>My question is about the reverse process. Say we are given curves $X_{\mathbb{Q} _ p}$ for every $p$ such that $X_ { \bar {\mathbb{Q}} _ p} \cong X_ {\bar{ \mathbb{Q}} _ q}$ for any two primes $p$ and $q$ (where the isomorphism is as schemes; alternatively, they are isomorphic when base-changed to an algebraically closed field that contains both $\mathbb{Q}_p$ and $\mathbb{Q} _ q$). Then is there a nice way to describe the obstruction for these models to come from a global curve $X_{\mathbb{Q}}$ over $\mathbb{Q}$?</p> http://mathoverflow.net/questions/51136/what-are-galois-categories-used-for What are Galois Categories used for? James D. Taylor 2011-01-04T17:01:59Z 2012-01-12T21:35:57Z <p>Galois categories are introduced (for the first time?) in SGA1, but here's an English introduction that's available online: <a href="http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Lynn.pdf" rel="nofollow">http://www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Lynn.pdf</a></p> <p>It seems that Galois Categories are a way of axiomatizing all the Galois correspondences in the various fields: Galois theory for fields, Galois theory for covers, Galois theory tame covers and so forth.</p> <p>What is the benefit, if at all, of this formalism? Is it just to outline the commonalities of these seemingly different topics, or is there some applicable virtue to this language?</p> http://mathoverflow.net/questions/83140/where-was-riemann-existence-first-proven Where was Riemann Existence first proven? James D. Taylor 2011-12-10T19:48:36Z 2011-12-12T19:35:15Z <p>Nowadays the standard reference for Riemann's Existence theorem is SGA1, where the proof heavily relies on Serre's GAGA. I imagine that the theorem is much older, as its name suggests, and that its original proof is quite different. I thought it would be instructive for me to look at how this theorem was viewed in the pre-SGA era.</p> <h3>Question</h3> <p>Where does a proof of Riemann's Existence Theorem original appear? Or better yet, where is a readable summary of it in English (or somewhat less preferably in French)?</p> http://mathoverflow.net/questions/81759/general-cohomology-groups-and-motives General cohomology groups and motives James D. Taylor 2011-11-23T22:51:17Z 2011-11-23T22:51:17Z <p>Let $X$ be a variety over $\mathbb{Q}$. Let $\mathcal{F}$ be a sheaf on $X$. Then we have an action of $Gal(\mathbb{Q})$ on $H_{et}^i(X,\mathcal{F})$. In certain cases we can say a lot about this action. For example if $\mathcal{F}$ is the constant sheaf $\mathbb{Q}_ p$ for some prime $p$, or the constant sheaf $\mathbb{C}$. In those cases $H_{et}^*(\ \underline{},\mathcal{F})$ is a Weil cohomology; and so we conjecture that these representations come from motives, and in particular we have the Langlands conjectures about how these representations are nice'' (i.e. automorphic).</p> <p>My question is: what can we say about the representation $H_{et}^i(X,\mathcal{F})$ for a general $\mathcal{F}$? What if $\mathcal{F}$ is, for example, not constant? Do we have an equivalent conjecture to the Langlands conjectures? (i.e. is there a generalization for a general $\mathcal{F}$ to the statement that $H_{et}^i(X,\mathcal{F})$ should be automorphic?)</p> <p>My humility when it comes to the Langlands conjectures behooved me to put a community wiki stamp on this question on the off chance my question strikes experts as silly and/or vague.</p> http://mathoverflow.net/questions/75335/what-is-the-reason-for-modularity-results What is the "reason" for modularity results? James D. Taylor 2011-09-13T17:18:54Z 2011-11-14T21:30:48Z <p>The question is a little wishy-washy, but I take my cues from other popular questions that relate to the philosophy behind the mathematics as <a href="http://mathoverflow.net/questions/2551/why-do-groups-and-abelian-groups-feel-so-different" rel="nofollow">http://mathoverflow.net/questions/2551/why-do-groups-and-abelian-groups-feel-so-different</a> .</p> <p>I am aware of the statements of class field theory and the modularity theorem, as well as far-reaching generalizations that have to do with the conjectural Langlands group and motives. But on a basic level, I just don't understand why such statements <em>should</em> be true, other than that there is a lot of evidence that they are.</p> <p>What is the philosophical impetus behind modularity results?</p> <p>When I read about number theory, I can very easily understand the intuition behind ramification of primes (because the intuition is geometric), but as soon as we start talking about splitting of primes, and are therefore in the realm of modularity results, I lose all intuition of <em>why</em> things should be true (even though I can read and understand the results as an undergraduate can -- agreeing line by line).</p> <p>An example of an answer for CFT can be the following thing I've heard, but was somewhat unsatisfied with because I didn't fully understand it: that it grew out of generalizations of Fourier analysis. (if you also think of this as the <em>reason</em> it's true, and can expatiate -- do!)</p> http://mathoverflow.net/questions/80786/did-grothendieck-have-a-plan-for-proving-riemann-existence-algebraically Did Grothendieck have a plan for proving Riemann Existence algebraically? James D. Taylor 2011-11-13T00:27:41Z 2011-11-13T17:35:19Z <p>A <a href="http://mathoverflow.net/questions/80770/reference-request-riemanns-existence-theorem" rel="nofollow">recent question</a> reminded me of a question I've had in the back of my mind for a long time. It is said that Grothendieck wanted the center-piece of SGA1 to be a completely algebraic proof (without topology) of the following theorem: $\pi_1^{et}(\mathbb{P}^1_{\mathbb{C}}\smallsetminus a_1,...,a_r)\cong$ the profinite completion of $\langle \alpha_1,...,\alpha_r|\alpha_1...\alpha_r=1\rangle$.</p> <p>As you may know, he did not succeed.</p> <p>From my experience with Grothendieck's ideas, he often has a proposed proof in mind that would take years (if it all) to be realized. Did Grothendieck have an idea of how to prove this fact algebraically? If so, what was the missing element in his proposed proof?</p> http://mathoverflow.net/questions/79566/does-a-curve-have-infinitely-many-k-rational-points-under-these-hypotheses Does a curve have infinitely many $K$-rational points under these hypotheses? James D. Taylor 2011-10-31T00:15:13Z 2011-10-31T02:35:27Z <p>The question was a bit long for the title, so let me explain what I mean here. Let $k$ be some field (ideally, a number field). Let $X$ be a curve that is defined over $k$, such that it has a $k$-point. Let $K$ be an algebraic extension of $k$ which is <em>not</em> finitely generated. Does this imply that $X$ has infinitely many $K$-points?</p> <h3>Motivation</h3> <p>The canonical example I have of a field which is infinitely generated over its prime field, together with a curve, such that the curve doesn't have infinitely many rational points is: $x^2+y^2=-1$ where the field is $\mathbb{R}$. However in this case, $\mathbb{R}$ is not an algebraic extension of a field $k$ for which this curve does have a rational point. (Let alone is a not-finitely-generated extension of such a $k$.)</p> http://mathoverflow.net/questions/79115/in-what-way-do-the-weil-conjectures-pertain-to-langlands In what way do the Weil Conjectures pertain to Langlands? James D. Taylor 2011-10-25T19:43:09Z 2011-10-25T21:42:49Z <p>For a relative variety $X$ over a ring of integers $O_K$, we can define a zeta function. This zeta function is defined as the product of the zeta functions of the variety specialized to $O_K/\mathfrak{p}$ as $\mathfrak{p}$ runs over $Spec(O_K)$. In turn, zeta functions of varieties over finite fields are easy to define using the counting of rational points. As Grothendieck proved, these zeta functions can be expressed as a product of $L$-functions indexed by $i$, where the $i^{th}$ $L$-function is related to the $i^{th}$ (Weil) cohomology of $X_{O_K/\mathfrak{p}}$. The $i^{th}$ $L$-function of $X$ over $O_K$ is defined to be the product over $\mathfrak{p}$ of the $i^{th}$ $L$-function of $X_{O_K/\mathfrak{p}}$.</p> <p>The Weil conjectures give us a lot of information about the zeta functions of varieties over finite fields, and in fact about their $L$-functions.</p> <p>The Langlands program is about properties of $L$-functions of $X$ over $O_K$.</p> <p>Is it possible to interpret the Weil conjectures as telling us something meaningful about the Langlands program?</p> http://mathoverflow.net/questions/78928/what-are-non-abelian-l-functions What are non-abelian $L$-functions? James D. Taylor 2011-10-23T21:31:53Z 2011-10-23T21:42:09Z <p>I have heard people discussing the utility of $L$-functions, claiming that since they are essentially cohomological entities, they are "abelian" and therefore lack force.</p> <p>From looking around on the web, I see that this idea has a base of believers, and that there is some notion of non-abelian $L$-functions.</p> <h3>Question</h3> <p>What is the definition of non-abelian $L$-functions? Does it have to do with replacing cohomology with homotopy in some way? How does it relate to the original definition of $L$-function (in particular, what is the analogue of the characteristic polynomial?)? What is the context in which it arises?</p> http://mathoverflow.net/questions/78696/is-there-an-intuitive-reason-for-zariskis-main-theorem Is there an intuitive reason for Zariski's main theorem? James D. Taylor 2011-10-20T19:21:01Z 2011-10-21T22:57:16Z <p>Zariski's main theorem has many guises, and so I will give you the freedom to pick the one that you find to be most intuitive. For the sake of completeness, I will put here one version:</p> <p>Zariski's main theorem: Let $f:X\rightarrow Y$ be a quasi-finite separated birational map of varieties, where $Y$ is normal. Then $f$ is an open immersion.</p> <p>There is no reason to pick this particular formulation. In fact, every formulation seems to me like a technical lemma rather than a theorem with geometrically intuitive content.</p> <h3>Question</h3> <p>Is there a formulation of Zariski's main theorem that has an intuitive/pictorial reason'' for it? Or is Zariski's main theorem in its core a technical result with no geometric reason?</p> http://mathoverflow.net/questions/51531/theorems-that-are-obvious-but-hard-to-prove/76821#76821 Answer by James D. Taylor for Theorems that are 'obvious' but hard to prove James D. Taylor 2011-09-30T02:02:59Z 2011-09-30T02:02:59Z <p>I think Godel's completeness theorem is very intuitive. For example, can you imagine a first order theorem that would be true for all groups, that you wouldn't be able to prove (by Godel's definition of prove'). Of course not! But the proof of the completeness theorem is hard.</p> http://mathoverflow.net/questions/76349/is-there-a-connected-non-affine-scheme-s-such-that-it-is-the-union-of-rings-of Is there a connected non-affine scheme $S$ such that it is the union of rings of integers of number fields? James D. Taylor 2011-09-25T17:41:21Z 2011-09-26T05:01:44Z <p>I was woolgathering about the notion of a scheme, and it occurred to me that I know of no non-affine scheme $S$ that is the union of $Spec(O_K)$'s of some number field $K$ (I allow $K$ to vary - so that $S$ might be $Spec(O_K)\cup Spec(O_L)$ for example).</p> <p>It would an interesting notion if one could patch rings of integers together to form some non-affine $1$-dimensional normal scheme $S$. The fact that I've never seen an example makes me think it's impossible.</p> <h3>Question</h3> <p>Is there a connected non-affine scheme $S$ such that it is the union of open subschemes of it that are $Spec$'s of rings of integers of number fields?</p> <p>More pointedly, if $Spec(O_K)$ (the ring of integers of some number field $K$) is an open subscheme of a normal scheme $S$ then is it equal to it?</p> http://mathoverflow.net/questions/74454/are-the-l-functions-of-x-0n-automorphic Are the $L$-functions of $X_0(N)$ automorphic? James D. Taylor 2011-09-03T18:23:23Z 2011-09-19T20:15:18Z <p>This question, like all of my previous questions regarding Langlands, is very naive.</p> <p>All $g\geq 1$ curves come from quotients of the upper half plane. The curves $X_0(N)$ come from quotients of special subgroups of the group of automorphisms of the upper half plane. This might imply that they are easier to work with.</p> <p>$Gal(\mathbb{Q})$ acts on the Tate module of $X_0(N)$, which leads to a motivic $L$-function. Can one prove that $L$-functions arising from these $X_0(N)$'s are $L$-functions coming from automorphic forms?</p> <p>Furthermore, is this the motivation for these curves to begin with? If this is true, is this the reason that the modularity theorem (Taniyama-Shimura) often phrased in terms of parametrizing elliptic curves via $X_0(N)$'s? If not, then why do these curves come up in the formulation of Taniyama-Shimura?</p> http://mathoverflow.net/questions/75709/what-is-the-non-motivic-motivation-behind-automorphic-representations What is the non-motivic motivation behind automorphic representations? James D. Taylor 2011-09-17T23:58:35Z 2011-09-18T04:47:13Z <p>In one of my last questions: <a href="http://mathoverflow.net/questions/75335/what-is-the-reason-for-modularity-results" rel="nofollow">http://mathoverflow.net/questions/75335/what-is-the-reason-for-modularity-results</a></p> <p>it was pointed out to me that "the notion of automorphic representation developed independently of any concern with modularity or class field theory. As I said, it has its own history, arising ultimately from the theory of elliptic integrals." Since automorphic representations never made intuitive sense to me, and since they are usually presented nowadays in relation to the Langlands program, I wondered what their original motivation was. In particular, can you recommend a readable and introductory source where I can learn about the non-motivic motivation behind automorphic representations? Why did they come onto the scene? What problem did they solve?</p> http://mathoverflow.net/questions/74776/what-is-the-interpretation-of-complex-multiplication-in-terms-of-langlands What is the interpretation of complex multiplication in terms of Langlands? James D. Taylor 2011-09-07T18:05:42Z 2011-09-07T20:48:46Z <p>I'm trying to interpret things in the following terminology:</p> <p>Assume the standard conjectures, the existence of the conjectural Langlands group, and anything else you wish.</p> <p>I assume the following statement: the category of isomorphism classes of irreducible continuous algebraic (I added the word algebraic' after it was pointed out that it was needed) representations $\mathcal{L}_{\mathbb{Q}}$ (the Langlands group) $\rightarrow GL_n(\mathbb{C})$ is equivalent to the category of $\mathbb{Q}$-motives with coefficients in $\overline{\mathbb{Q}}$.</p> <p>In the above terminology, there are three numbers to keep track of: $n$ (of $GL_n$), dimension of the scheme that the motive comes from, and the degree of purity of the motive within that scheme (by this I mean that if we denote it by $i$, then in any Weil cohomology it would be realize as the $i^{th}$ cohomology of this scheme).</p> <p>I will give two examples:</p> <p>i. Classic CFT comes from $n=1$, the dimension of the scheme that the motive comes from is $0$ (because we're dealing with fields), and the degree of purity of the motive is $0$.</p> <p>I imagine that when people talk about CFT in higher dimensions they mean the $n=1$ case where you allow the other two numbers to be different.</p> <p>ii. Taniyama Shimura: this is the $n=2$ case, where the dimension of the scheme is $1$ and the degree of purity of the motive is $1$. Here on the Langlands side, this must correspond to newforms of weight $2$ with rational Hecke eigenvalues.</p> <p>In general, one could take $n=2$, and on the Langlands side look at a general weight $k$ newform. Let $F$ be the field generated by the Hecke eigenvalues. Then the corresponding motive is coming from a $[F:\mathbb{Q}]$-dimensional variety, and the purity degree of the motive is $k-1$. If $k=2$ one can prove that the variety from which the motive is coming is an abelian variety via an argument involving Hodge structures.</p> <p>Now I wish to understand complex multiplication in this context. It seems that the goal is to classify abelian extensions of quadratic fields. This seems to imply that we are in the $n=1$ case. But what are the other two numbers? Are we really looking at motives coming from $0$-dimensional schemes? This seems weird, because in the theory of complex multiplication elliptic curves are implicated.</p> <p>Can one give a similar context to complex multiplication as I did to CFT and Taniyama Shimura? How would that work? What would the three numbers I described be?</p> http://mathoverflow.net/questions/71939/what-is-the-strongest-most-natural-conjectural-form-of-langlands What is the strongest, most natural, conjectural form of Langlands? James D. Taylor 2011-08-02T23:33:25Z 2011-09-07T03:43:13Z <p>This is inspired by my previous question: <a href="http://mathoverflow.net/questions/71743/what-is-the-precise-relationship-between-langlands-and-tannakian-formalism" rel="nofollow">http://mathoverflow.net/questions/71743/what-is-the-precise-relationship-between-langlands-and-tannakian-formalism</a></p> <p>As well as the excellent link that Tom Leinster put in a comment to that thread: <a href="http://golem.ph.utexas.edu/category/2010/08/what_is_the_langlands_programm.html" rel="nofollow">http://golem.ph.utexas.edu/category/2010/08/what_is_the_langlands_programm.html</a></p> <p>It seems that people are reluctant to say a form of Langlands that is too strong, but as consequence the statement is less natural, and more convoluted. So here I prefer that the statement be natural and bold rather than unnatural (for example, I consider the statement that each $L$ function coming from Galois representations is the $L$ function of some automorphic form to be unnatural).</p> <h3>Question</h3> <p>What is the strongest, most natural statement of Langlands? It would be nice if you can give a short definition of the words you use, but I am mostly interested in the narrative (each this has a blah, to each blah is a this, this is associated to this category by blah, and this is conjectured to be an equivalent category to blah, and so forth)</p> <p>Words like: stack, motive, Tannakian, motivic Galois group, L-packets are encouraged. (of this list $L$-packets are by far the thing I know the least about)</p> <p>This is subjective, so community wiki it is.</p> http://mathoverflow.net/questions/71939/what-is-the-strongest-most-natural-conjectural-form-of-langlands/74706#74706 Answer by James D. Taylor for What is the strongest, most natural, conjectural form of Langlands? James D. Taylor 2011-09-07T03:43:13Z 2011-09-07T03:43:13Z <p>Now that I've been exposed for a little longer to literature about Langlands (this answer is a month after I asked this question), let me submit my own non-expert answer to this question. I did enjoy the generality and naturality of the formulation of Langlands described in a newer question of mine:</p> <p><a href="http://mathoverflow.net/questions/74698/how-does-the-conjectural-langlands-group-fit-into-the-tannakian-point-of-view" rel="nofollow">http://mathoverflow.net/questions/74698/how-does-the-conjectural-langlands-group-fit-into-the-tannakian-point-of-view</a></p> <p>As you can see from that question, I'm still learning about this. But I humbly suggest that this formulation might be what I was looking for, and so for the sake of completeness I add this as an answer.</p> http://mathoverflow.net/questions/74698/how-does-the-conjectural-langlands-group-fit-into-the-tannakian-point-of-view How does the conjectural Langlands group fit into the Tannakian point of view? James D. Taylor 2011-09-07T00:48:26Z 2011-09-07T03:39:58Z <p>I've read that one way to formulate the Langlands program is the following:</p> <p>Let $\mathcal{L}_ {\mathbb{Q}}$ be the conjectural Langlands group. Then the category of semi-simple (continuous) representations of $\mathcal{L}_{\mathbb{Q}}$ that are algebraic(!) is equivalent to the category of motives over $\mathbb{Q}$ with coefficients in $\overline{\mathbb{Q}}$.</p> <p>This is peculiar to me, since the category of motives is Tannakian, and so is equivalent to the category of (all!) representations of some affine group scheme.</p> <p>How does one think of the condition that the representations must be algebraic? Does this mean that the Langlands group is not meant to be the motivic Galois group (the group guaranteed by Tannakian formalism applied to the category of motives, using <em>some</em> fiber functor)? Is there a way to reconcile the two approached in an insightful way?</p> http://mathoverflow.net/questions/71743/what-is-the-precise-relationship-between-langlands-and-tannakian-formalism What is the precise relationship between Langlands and Tannakian formalism? James D. Taylor 2011-07-31T17:38:37Z 2011-08-18T15:25:09Z <p>As anyone who's been reading the forums closely can see, I've been averaging a question a day about Tannakian formalism for the past few days. It's quite an interesting concept!</p> <p>In any case, I wish now to relate it to yet another topic of which I have only a tenuous grasp: the Langlands program. As I understood more and more about Tannakian formalism, it seemed more and more like it has something to do with Langlands. A google search confirms this. There are several sources that group these two things together. Here is a sample:</p> <p><a href="http://www.claymath.org/cw/arthur/pdf/automorphic-langlands-group.pdf" rel="nofollow">http://www.claymath.org/cw/arthur/pdf/automorphic-langlands-group.pdf</a></p> <p><a href="http://www.institut.math.jussieu.fr/~harris/Takagi.pdf" rel="nofollow">http://www.institut.math.jussieu.fr/~harris/Takagi.pdf</a></p> <p>But my understanding of Langlands is weak. I'm certainly familiar with Class Field Theory, and to some limited extent with Taniyama-Shimura. I always found Langlands difficult to penetrate. But now that I know that there is a relationship between Langlands and Tannakian formalism, I am hopeful that this will give me a bird's eye view of Langlands.</p> <p>So the question is: Does Tannakian formalism simplify the statement of Langlands, or at least motivate it? Does it have to do with the motivic Galois group (defined to be the group predicted from Tannakian formalism on the category of numerical motives)? How precisely is Tannakian formalism used in Langlands?</p> <p>In light of these ideas, I ask a secondary question: is there a relationship between the standard conjectures and Langlands? (does one imply the other?)</p> http://mathoverflow.net/questions/72874/how-would-a-motivic-proof-of-the-riemann-hypothesis-over-finite-fields-go How would a motivic proof of the Riemann hypothesis over finite fields go? James D. Taylor 2011-08-14T16:51:29Z 2011-08-14T20:16:19Z <p>It is well known that Grothendieck had a different idea than Deligne about how one should go about proving the Riemann hypothesis for finite fields. However, since Grothendieck's desired proof never came to fruition, I find it hard to look up what his proposed proof was.</p> <p>It is clear from texts on the subject that he wanted to use motives in some way or other, and that he wanted to prove the standard conjectures first. Given the standard conjectures, is there an easy proof the Riemann hypothesis over finite fields? What is it? Did he want other things to be true also? What is the sketch he had in mind -- can you write a proof of the Riemann hypothesis with some conjectural black boxes like the standard conjectures?</p> http://mathoverflow.net/questions/72546/is-the-l-function-of-the-complex-cohomology-of-a-motive-equal-to-the-l-function Is the "L-function of the complex cohomology" of a motive equal to the L-function of its l-adic realization? James D. Taylor 2011-08-10T04:31:19Z 2011-08-11T05:26:50Z <p>Let's say I have a motive in $\mathcal{M}_{num}(K)$ ($K$ a number field). For each prime $l$ there is a realization of this motive in terms of etale cohomology with coefficients in $\mathbb{Q}_l$. This has more structure than being a mere vector space: it is a representation of $Gal(K)$! This representation has an $L$-function. As I understand it, the $L$ function doesn't depend on $l$.</p> <p>What I wonder is whether there is something special about etale cohomology with coefficients in an $l$-adic field, or whether every Weil cohomology has an $L$-function attached to it that would equal the $L$-function of any other realization.</p> <p>The usual (singular) cohomology with the complex topology, is also a representation of $Gal(K)$ (factoring through the motivic Galois group, if this means anything to you). Is it true, then, that the $L$-function attached to this representation is the same as the $L$-function of the realizations via etale cohomology with coefficients in $\mathbb{Q}_l$?</p> <p>How does one go about proving equalities between $L$-functions of different realizations of the same motive?</p> http://mathoverflow.net/questions/126430/is-there-a-nice-criterion-for-when-the-splitting-fields-of-two-irreducible-polyno/126436#126436 Comment by James D. Taylor James D. Taylor 2013-04-03T19:30:36Z 2013-04-03T19:30:36Z I'm looking for something with more insight, I'm afraid... I suppose if there is none then there is none. But I'm still holding out hope that someone will tell me that there is some paper that I should read about it, or that it somehow has to do with cohomology, or whatever insight might come this way... (BTW, the third question was a statement, not a question: &quot;It is possible&quot; rather than &quot;Is it possible&quot;.) http://mathoverflow.net/questions/119439/does-the-proof-of-gaga-use-the-axiom-of-choice/119441#119441 Comment by James D. Taylor James D. Taylor 2013-01-21T05:35:33Z 2013-01-21T05:35:33Z *sorry, I meant: &quot;the algebraic sheaf that induces a particular analytic sheaf&quot;. @nosr, that sounds more along the lines I was thinking of, but I can't say that I myself am familiar with all the details of the proof. It most certainly is false that GAGA is constructive, and there have been many papers trying to understand the relationship between the analytic side and the algebraic side better. http://mathoverflow.net/questions/119439/does-the-proof-of-gaga-use-the-axiom-of-choice/119441#119441 Comment by James D. Taylor James D. Taylor 2013-01-21T05:12:44Z 2013-01-21T05:12:44Z Duff is right. It is generally not well understood how to find the algebraic sheaf that induces a particular algebraic sheaf. It is very misleading to say that GAGA is obvious in any way. http://mathoverflow.net/questions/108831/why-is-there-no-regular-etale-fundamental-group Comment by James D. Taylor James D. Taylor 2012-10-04T16:19:09Z 2012-10-04T16:19:09Z unknown, it sounds like you have an example in mind. http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture Comment by James D. Taylor James D. Taylor 2012-09-07T15:53:16Z 2012-09-07T15:53:16Z David, your comments are precisely the type of answer I'm looking for. It is okay that it's 20 years old. http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture/106603#106603 Comment by James D. Taylor James D. Taylor 2012-09-07T15:48:10Z 2012-09-07T15:48:10Z You are correct that I was inaccurate on that point, although I did know that it was Weil's idea. As for you argument for patience, I think you have misunderstood my question. I am not asking for a sketch of the methods, but only what those methods aim to achieve. An example of a good answer is David Speyer's comments. So saying &quot;here is the rough argument in the function fields case, and so what we want is a number theoretic analogue of ____&quot; is precisely the answer I was looking for. http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture Comment by James D. Taylor James D. Taylor 2012-09-07T02:21:01Z 2012-09-07T02:21:01Z @quid: you're being stubborn. Is it not legitimate to ask questions about mathematics that is available but difficult to read and understand? @Kevin: thanks! http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture Comment by James D. Taylor James D. Taylor 2012-09-07T02:00:17Z 2012-09-07T02:00:17Z @quid: the expositions I've seen (such as <a href="http://www.kurims.kyoto-u.ac.jp/~motizuki/2010-10-abstract.pdf" rel="nofollow">kurims.kyoto-u.ac.jp/~motizuki/&hellip;</a>) are mostly teasers to make people read more. My question is about the sketch underlying the proof of the ABC conjecture, which I don't see evident there. If you have an exposition that you would recommend, I suggest that you write it as an answer. http://mathoverflow.net/questions/106560/philosophy-behind-mochizukis-work-on-the-abc-conjecture Comment by James D. Taylor James D. Taylor 2012-09-07T01:55:39Z 2012-09-07T01:55:39Z Correction: &quot;an enthusiastic report&quot;. Sorry, Jordan! http://mathoverflow.net/questions/97051/what-is-the-intuition-for-mathbbqab-having-cohomological-dimension-1 Comment by James D. Taylor James D. Taylor 2012-05-16T16:26:43Z 2012-05-16T16:26:43Z Carnahan: can you expatiate a little more about your comment? http://mathoverflow.net/questions/97051/what-is-the-intuition-for-mathbbqab-having-cohomological-dimension-1 Comment by James D. Taylor James D. Taylor 2012-05-16T16:26:18Z 2012-05-16T16:26:18Z Alex: $\mathbb{Q}^{ab}$ is indeed the maximal abelian extension of $\mathbb{Q}$, but its absolute Galois group is $Gal(\bar{\mathbb{Q}}/\mathbb{Q}^{ab})$ not $Gal(\mathbb{Q}^{ab}/\mathbb{Q})$. http://mathoverflow.net/questions/89252/automorphism-group-of-the-algebraic-numbers Comment by James D. Taylor James D. Taylor 2012-02-23T05:17:00Z 2012-02-23T05:17:00Z <a href="http://en.wikipedia.org/wiki/Inverse_Galois_problem" rel="nofollow">en.wikipedia.org/wiki/Inverse_Galois_problem</a> But this question really belongs in: <a href="http://math.stackexchange.com/" rel="nofollow">math.stackexchange.com</a> http://mathoverflow.net/questions/88289/are-there-n-polynomials-for-which-all-intersection-multiplicities-are-at-least-m Comment by James D. Taylor James D. Taylor 2012-02-12T23:03:25Z 2012-02-12T23:03:25Z Yes, you're right! http://mathoverflow.net/questions/88289/are-there-n-polynomials-for-which-all-intersection-multiplicities-are-at-least-m Comment by James D. Taylor James D. Taylor 2012-02-12T22:57:24Z 2012-02-12T22:57:24Z @Florian: $0$ and $(x-1)^2(x+1)$ intersect with multiplicity $1$ at $x=-1$. http://mathoverflow.net/questions/88289/are-there-n-polynomials-for-which-all-intersection-multiplicities-are-at-least-m Comment by James D. Taylor James D. Taylor 2012-02-12T22:29:38Z 2012-02-12T22:29:38Z Look at the discussion above -- what you suggest will only make the condition hold at $x=a$, but not for other values of $x$. For example try multiplying $0,x,1,x+1$ (which satisfy the condition over $x=0$) by $(x-3)^2$ and see that there are values where some of these intersect with multiplicity $1$.