User makhalan duff - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:25:39Z http://mathoverflow.net/feeds/user/5309 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130770/can-group-cohomology-be-interpreted-as-an-obstruction-to-lifts Can group cohomology be interpreted as an obstruction to lifts? Makhalan Duff 2013-05-15T22:58:17Z 2013-05-16T06:27:41Z <p>The standard way to view the first and second group cohomologies is this:</p> <h3>The Standard Story</h3> <p>Let $G$ be a group, and let $M$ be a commutative group with a $G$-action. Then the first cohomology has the following interpretation: $H^1(G,M)$ is bijective with sections (modulo conjugation by $M$) of the short exact sequence $$1\rightarrow M\rightarrow M\rtimes G\rightarrow G\rightarrow 1.$$ Furthermore, the group of $1$-cocycles, $Z^1(G,M)$ is bijective with the sections of this short exact sequence. (Not modulo anything.) In fact, this holds even if $M$ is non-abelian.</p> <p>The second cohomology $H^2(G,M)$ is bijective with the set of isomorphism classes of group extensions $$1\rightarrow M\rightarrow H\rightarrow G\rightarrow 1$$ for which there exists a (or equivalently for every) set theoretic section $s:G\rightarrow H$ such that $g\cdot m=s(g)m(s(g))^{-1}$. (Here $g\cdot m$ denotes the action of $g$ on $m$ coming from the $G$-module structure of $M$.)</p> <h3>Liftings</h3> <p>In a paper I have been reading, they have given an entirely different interpretation to the first cohomology. Namely:</p> <p>Let $A$ and $B$ be groups, and let $C$ be a normal abelian subgroup of $B$. Let $\bar \phi:A\rightarrow B/C$ be a homomorphism. Assume $\phi$ has a lift $\alpha:A\rightarrow B$. Then $Z^1(A,C)$ is bijective with the set of lifts of $\phi$ to homomorphisms from $A$ to $B$.</p> <p>The bijection goes like this: $\theta\in Z^1(A,C)$ goes to $\alpha\theta$.</p> <p>My question is: can one give an interpretation in terms of lifts to the second cohomology, or to the group of $2$-cocycles?</p> <p>More precisely:</p> <h3>Question</h3> <p>Let $A$ and $B$ be groups, and let $C$ be a normal abelian subgroup of $B$. Let $\bar \phi:A\rightarrow B/C$ be a homomorphism.</p> <p>Is it true that there exists a lift of $\bar \phi$ to a homomorphism from $A$ to $B$ if and only if $H^2(A,C)$ is trivial? Or is there a $2$-cocycle one can define (how would one define it?) such that there exists a lift of $\bar \phi$ if and only if it is trivial in $H^2(A,C)$? Or perhaps the right group to look at is the group of $2$-cocycles $Z^2(A,C)$ rather than the cohomology group?</p> <p>I don't know if such an interpretation exists, so this is just wishful thinking. Since this is the first time I've seen the interpretation of $Z^1(A,C)$ in terms of lifts, I was curious whether such an interpretation extends to the second cohomology.</p> http://mathoverflow.net/questions/58507/how-was-the-importance-of-the-zeta-function-discovered How was the importance of the zeta function discovered? Makhalan Duff 2011-03-15T07:11:27Z 2013-05-08T17:05:34Z <p>This question is similar to <a href="http://mathoverflow.net/questions/1880/why-do-zeta-functions-contain-so-much-information" rel="nofollow">http://mathoverflow.net/questions/1880/why-do-zeta-functions-contain-so-much-information</a> , but is distinct. If the answers to that question answer this one also, I don't understand why.</p> <p>The question is this: with the benefit of hindsight, the zeta function had become the basis of a great body of theory, leading to generalizations of CFT, and the powerful Langlands conjectures. But what made the 19th century mathematicians stumble on something so big? After all $\sum \frac{1}{n^s}$ is just one of many possible functions one can define that have to do with prime numbers. How and why did was the a priori fancifully defined function recognized as being of fundamental importance?</p> http://mathoverflow.net/questions/119439/does-the-proof-of-gaga-use-the-axiom-of-choice Does the proof of GAGA use the axiom of choice? Makhalan Duff 2013-01-21T03:05:06Z 2013-01-21T04:46:52Z <p>Serre's GAGA result roughly states the following. Let $X$ be a complex projective algebraic variety. Then the natural functor from the category of coherent sheaves over the algebraic structure sheaf of $X$ to the category of coherent sheaves over the analytic structure sheaf of $X$ is an equivalence of categories.</p> <p>This theorem always seemed to have the air of magic to me. Things that are analytic must come from algebra. I want to dust away some of this magic, and get a clearer picture. With this goal in mind, I have skimmed the proof of GAGA.</p> <p>The proof of GAGA is rather involved. It uses Cartan's theorem A for both the algebraic and analytic cases, the isomorphism of the completions of the stalks of the structure sheaf in the algebraic case and the analytic case, and a variety of technical results. After having done that for a few days, I still remain with a sense of amazement and a basic lack of understanding about what makes this work. This brings me to the precise phrasing of my question: (which will hopefully help me find the precise step where the magic happens)</p> <h3>Question</h3> <p>Does the proof of Serre's GAGA theorem use the axiom of choice? If so, at what step does this happen?</p> http://mathoverflow.net/questions/38649/online-math-history-lectures Online math history lectures Makhalan Duff 2010-09-14T05:08:09Z 2012-11-20T04:49:26Z <p>This question is somewhat similar to this: <a href="http://mathoverflow.net/questions/1714/best-online-math-videos" rel="nofollow">http://mathoverflow.net/questions/1714/best-online-math-videos</a></p> <p>I'm using the word "history" loosely here. What I'm looking for are those lectures that put various mathematical developments in perspective by explaining their origins. There's something very insightful about seeing someone talk about the origins of a concept, that makes things click. Especially if he or she partook in the inception of that development.</p> <p>So: where can I find such lectures online?</p> http://mathoverflow.net/questions/62682/centralizers-of-elements-in-free-profinite-groups Centralizers of elements in free profinite groups Makhalan Duff 2011-04-23T00:39:49Z 2012-09-20T02:12:42Z <p>I believe, although I can't say that I've given a rigorous proof, that for a free group $F_r$, and an element of it $a$, $C_{F_r}(\langle a \rangle)=$ the group generated by the elements $b \in F_r$ such that $a=b^n$ for some integer $n$ (I will say: the group of powers and roots of $a$).</p> <p>One may similarly ask (and this is my interest in this), given $a\in \hat{F_r}$ (the free profinite group on $r$ generators), what is $C_{\hat{F} _r}(\langle a \rangle)$? In particular, is it the profinite completion of the group of powers and roots of $a$?</p> http://mathoverflow.net/questions/102839/what-is-the-relationship-between-motivic-cohomology-and-the-theory-of-motives What is the relationship between motivic cohomology and the theory of motives? Makhalan Duff 2012-07-21T20:58:09Z 2012-07-22T01:24:39Z <p>I will begin by giving a rough sketch of my understanding of motives.</p> <p>In many expositions about motives (for example, www.jmilne.org/math/xnotes/MOT102.pdf), the category of motives is defined to be a category such that every Weil cohomology (viewed as a functor) factors through it. This does not define the category uniquely, nor does it imply that it exists.</p> <p>There are two concrete candidates that we can construct. The category of Chow motives, which is well-defined, is trivially a category of motives. However, it has some bad properties. For example, it is not Tannakian. The second candidate is the category of numerical motives. It too is well-defined, however it is only conjectured that it is category of motives (i.e., that every Weil cohomology factors through this category). This conjecture is closely related to (or perhaps even equivalent to?) Grothendieck's standard conjectures. That would be desirable, because the category of numerical motives is very well-behaved.</p> <p>Furthermore, the original motivation for motives is that Grothendieck has proven that if the category of numerical motives is indeed a category of motives, then the Weil conjectures are correct.</p> <p>So far, even though I a murky on many of the details, I follow the storyline.</p> <h3>Question</h3> <p>Where does "motivic cohomology" (in the sense of, for example, www.claymath.org/library/monographs/cmim02.pdf) fit into this story? </p> <p>I know that motivic cohomology has something to do with Milnor K-theory, but that is more or less where my understanding of the context of motivic cohomology ends. If motives are already an abstract object that generalizes cohomology, what does motivic cohomology signify? What is the motivation for defining it? What is the context in which it arose?</p> http://mathoverflow.net/questions/100798/what-makes-geometric-cft-easier-than-cft What makes Geometric CFT easier than CFT? Makhalan Duff 2012-06-27T18:49:47Z 2012-07-05T13:39:54Z <p>I've been reading: math.stanford.edu/~conrad/249BPage/handouts/geomcft.pdf</p> <p>in an attempt to shed some geometric light on class field theory. The last paragraph there reads:</p> <p><i> In case the ground field $k$ is perfect, the essential difficulty in the proof of class field theory – proving that the Artin map kills certain principal ideals – becomes easy to prove geometrically by means of the interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups. (More precisely, one has $J_m (k) = Cl_m (K)$ when $Br(k) = 1$, as happens when $k$ is finite but not when $k$ is a number field.) </i></p> <p>What precisely is he referring to? What particular part of class field theory ("that the Artin map kills certain principal ideals") becomes easier to prove, and why ("by means of the interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups.")?</p> http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity How does one understand geometric CFT in terms of modularity? Makhalan Duff 2012-06-29T21:38:38Z 2012-07-01T23:50:26Z <p>I have recently asked a question in a similar vein: <a href="http://mathoverflow.net/questions/100798/what-makes-geometric-cft-easier-than-cft" rel="nofollow">http://mathoverflow.net/questions/100798/what-makes-geometric-cft-easier-than-cft</a></p> <p>but I'm afraid I wasn't quite ripe to ask it yet. I have since consulted with the following sources:</p> <p><a href="http://jmilne.org/math/Books/ADTnot.pdf" rel="nofollow">http://jmilne.org/math/Books/ADTnot.pdf</a>, <a href="http://math.stanford.edu/~conrad/249BPage/handouts/geomcft.pdf" rel="nofollow">http://math.stanford.edu/~conrad/249BPage/handouts/geomcft.pdf</a> and <a href="http://arxiv.org/abs/hep-th/0512172" rel="nofollow">http://arxiv.org/abs/hep-th/0512172</a></p> <p>As well as some sources refreshing my memory on classic CFT:</p> <p><a href="http://people.maths.ox.ac.uk/gounelas/projects/bmo.pdf" rel="nofollow">http://people.maths.ox.ac.uk/gounelas/projects/bmo.pdf</a> and <a href="http://www.math.dartmouth.edu/~trs/expository-papers/tex/CFT.pdf" rel="nofollow">http://www.math.dartmouth.edu/~trs/expository-papers/tex/CFT.pdf</a> among others.</p> <p>My motivation for studying geometric class field theory was first and foremost to solidify my understanding of classic class field theory. I thought that perhaps the geometric intuition will shed some light on the rather massive apparatus of CFT.</p> <p>In order to be concrete, I will specify a version of geometric CFT:</p> <p>Theorem A: Let $C$ be a smooth projective, geometrically irreducible curve over a finite field $k$, and let $K$ be its function field. Then the (Artin reciprocity) map $\Phi_K:Div(C)\rightarrow \pi_1^{ab}(C)$ given by $p\mapsto Frob_p$ factors through $Pic(C)$, and induces an isomorphism between the profinite completion of $Pic(C)$ and $\pi_1^{ab}(C)$.</p> <p>This appears in Toth's master thesis (linked above) as Theorem 1.1.4. In order to understand this as analogous to classic CFT, one need only notice that $K^{\times}\backslash \mathbb{I}_K/ \prod _{p \mbox{ closed point in } C}\hat{O_p}$ (where $\mathbb{I}_K$ is the ideles, and $\hat{O_p}$ is the completion of the stalk at the closed point $p$) is isomorphic to $Pic(C)$.</p> <p>In other words the theorem above can be viewed as analogous to the adelic point of view of CFT (i.e. the isomorphism between the profinite completions of $K^{\times}\backslash \mathbb{I}_K/\prod_v O_v^{\times}$ with $Gal(K^{ab}/K)$, where $K$ is a number field). I am interested in understanding what the analogous picture in Geometric CFT to modular formulations (rather than adelic ones) of classic CFT.</p> <h3>Question</h3> <p>How does one understand geometric CFT (as described in the theorem above) in terms of modularity results? In other words, what is the analogous statement to the fact that (up to finitely many primes) the splitting of primes in an abelian extension of $\mathbb{Q}$ is determined by what those primes are conjugate to modulo some conductor? Is there a geometric intuition behind the analogous statement in geometric CFT?</p> <h3>EDIT</h3> <p>After reading the comments carefully, and going back to my old notes on Class Field Theory to refresh my mind about a few things that I got wrong in the comments, I have come to the conclusion that the following is the question that I really wanted to ask:</p> <p>We have:</p> <p>Theorem B: or $K$ a number field, we have that $Gal(K^{ab}/K)$ is isomorphic to the profinite completion of $K^{\times}\backslash \mathbb{I}_K/D_K$ where $D_K$ is the connected component of $1$.</p> <p>Theorem A above is not analogous to Theorem B because, as Felipe pointed out, Theorem A is only about abelian extensions that are unramified, whereas Theorem B allows ramification everywhere. My question is: what is the analogous statement to Theorem A in the Number Theory case?</p> <p>I am in particular confused by the subtlety regarding the distinction between $D_K$ and the product $\prod _{p \mbox{ closed point in } C}\hat{O_p}$. Are they the same?</p> http://mathoverflow.net/questions/79132/is-the-integrality-of-the-zeta-function-easy Is the integrality of the zeta function easy? Makhalan Duff 2011-10-25T22:39:46Z 2012-05-22T02:14:05Z <p>I'm trying to get the gist of the proof of the Weil conjectures. Let $X$ be a variety over $\mathbb{F}_{p^n}$. A priori $Z(X,t)\in \mathbb{Q}[[t]]$. Due to the Grothendieck-Lefschetz fixed point theorem, $Z(X,t)=\prod P_i(t)^{(-1)^{i+1}}$, where $P_i(t)$ is the characteristic polynomial of the Frobenius acting on $H^i(X,\mathbb{Q}_l)$ where $l$ is a fixed prime different from $p$. This implies that $Z(X,t)\in \mathbb{Q}_l(t)\cap \mathbb{Q}[[t]]$ for every prime $l$ different from $p$.</p> <p>Does this suffice to determine that it is in $\mathbb{Q}(t)$? If not, then how was it proven that it is in $\mathbb{Q}(t)$?</p> http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits Are semi-direct products categorical limits? Makhalan Duff 2012-05-05T17:39:45Z 2012-05-07T19:35:33Z <p>Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits.</p> <p>As was pointed in: <a href="http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/" rel="nofollow">http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/</a></p> <p>Bourbaki (General Topology, Prop. 27) gives a universal property:</p> <p>Let $f \colon N \to G$, $g \colon H \to G$ be two homomorphisms into a group $G$, such that $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ for all $n \in N$, $h \in H$. Then there is a unique homomorphism $k \colon N \rtimes H \to G$ extending $f$ and $g$ in the usual sense.</p> <p>However, I remain unsatisfied. The condition $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ is a condition on elements of groups, rather than a condition that says that some diagram is commutative.</p> <p>So the question remains: are semi-direct products in the category of groups categorical limits?</p> http://mathoverflow.net/questions/95938/are-all-henselian-fields-algebraic-over-complete-fields Are all henselian fields algebraic over complete fields? Makhalan Duff 2012-05-04T00:08:20Z 2012-05-04T03:19:33Z <h3>Motivations and Terminology</h3> <p>The term "henselian field" is ambiguous. What I mean when I say that $K$ is a henselian field is that there exists a henselian DVR $R$, such that $K=Frac(R)$. What I mean when I say that $L$ is a complete field is that there exists a complete DVR $S$ such that $L=Frac(S)$.</p> <p>Note that every complete field is henselian. Examples of complete fields are $\mathbb{Q}((t))$ (the field of formal Laurent series with coefficients in $\mathbb{Q}$), $\mathbb{Q}_p$ (the $p$-adics), and so forth.</p> <p>When I try to think of henselian fields that are not complete, the ones that immediately come to mind are algebraic over some complete field. For example $\mathbb{Q}((t))(\sqrt{2})$, $\mathbb{Q}_p^{un}$ (the maximal unramified extension of $\mathbb{Q}_p$), and so forth.</p> <h3>Question</h3> <p>Is it true that for every henselian field $K$ there exists a subfield $L\subset K$ such that $L$ is complete and such that $K/L$ is an algebraic extension?</p> <p>EDIT: I've re-emphasized this in the comments, but I think it is important to put this in the body of the question: both the term "henselian field" and "complete field" are used in many different contexts to mean different things.</p> <p>Note that under the definitions above $\mathbb{R}$ does not constitute as a complete field. (This is because $\mathbb{R}$ is not the fraction field of a complete DVR.)</p> <p>Also note that I do not consider $\mathbb{Q}((t^{1/n}))_{n\in\mathbb{N}}$ to be a henselian field. (This is because my definition of a henselian field requires it to be the fraction field of a henselian DVR, not a general henselian ring.)</p> http://mathoverflow.net/questions/91676/is-ramification-of-number-fields-first-order Is ramification of number fields first order? Makhalan Duff 2012-03-20T00:51:03Z 2012-03-20T18:07:07Z <p>Fix a prime number $p$. Is there a first order sentence $\phi_p$ in the language of fields such that $\phi_p$ holds in a number field $K$ if and only if the prime $p$ is unramified in the field extension $K/\mathbb{Q}$?</p> http://mathoverflow.net/questions/90100/for-which-fields-is-the-inverse-galois-problem-known For which fields is the inverse Galois problem known? Makhalan Duff 2012-03-03T02:13:07Z 2012-03-03T08:15:55Z <p>The inverse Galois problem is known for (or in Jarden's and Fried's terminology, the following fields are universally admissible) function fields over henselian fields (like $\mathbb{Q}_p(x)$); function fields over large fields (like $\mathbb{C}(x)$); and large Hilbertian fields (conjecturally $\mathbb{Q}^{ab}$, although I'm not certain that any field is known to be in this category).</p> <h2>Clarification:</h2> <p>A large field $K$ (a.k.a. an ample field) is a field such that if $V$ is a variety of dimension $\geq 1$ over $K$ with at least one smooth $K$-rational point, then it has infinitely many smooth $K$-rational points. For example any algebraically closed field is large.</p> <p>A Hilbertian field is more difficult to explain, but it suffices to say that any number field and any function field (over any field) is Hilbertian.</p> <h2>My question is:</h2> <p>Is there a proof (not a conjecture) that there exists a field $K$ which is neither a function field over a henselian field, nor a function field over a large field, nor a large Hilbertian field, such that the inverse Galois problem is true over that field? (i.e. that every finite group is realizable as a Galois group over that field)</p> http://mathoverflow.net/questions/70059/help-motivating-log-structures Help motivating log-structures Makhalan Duff 2011-07-11T21:46:26Z 2012-03-02T17:17:05Z <p>I'm currently reading a thesis that uses log-structures. I should mention that this is my first encounter with them, and the thesis (as well as my expertise) is scheme-theoretic (in fact stack-theoretic) and so the original geometric motivations are lost on me.</p> <p>Here is my meek understanding. For any scheme, we can give a log-structure. This is a sheaf , $M$, fibered in monoids, on the etale site over a scheme $S$; together with a morphism of sheaves fibered in moinoids $\alpha:M\rightarrow O_S$ such that when it is restricted to $\alpha^{-1}(O_S^{\times})$ it is an isomorphism.</p> <p>This $\alpha$ is called the exponential map, and for any $t\in O_S(U)$ (for some $U$), a preimage of it via $\alpha$ is called $log(t)$($\in M(U)$).</p> <p>I am curious about a few things, and puzzled about others. First, in terms of the notation, surely it's no coincidence that these are called exponential maps and log-structures. What is the geometric motivation for it?</p> <p>Second, these come up in the thesis I'm reading in the context of tame covers. I am puzzled about what, precisely, log-structures contribute. It seems to me, in extremely vague terms (commensurate with my understanding), that the point of log-structures in this context is that if you add this <em>extra information</em> to tame covers it somehow helps you construct <em>proper</em> moduli spaces of covers.</p> <p>On top of everything I'm also confused about the role of minimal log-structures' in all of this.</p> <p>In conclusion, if you can say anything at all about the motivations of log-structures in the geometric setting, or more importantly in the context of tame covers, I would extremely appreciate it. The plethora of notationally different texts on the subject is making it hard to understand the gist of what's going on.</p> <p>Also, if you have examples that I should have in mind when thinking about it, that would be ideal.</p> http://mathoverflow.net/questions/88793/are-there-polynomials-almost-all-of-whose-intersection-numbers-are-divisible-by Are there polynomials (almost) all of whose intersection numbers are divisible by some integer? Makhalan Duff 2012-02-18T01:21:59Z 2012-02-20T18:20:10Z <p>I've been playing around with some basic intersection theory, and I've wondered the following:</p> <p>For every two integers $n$ and $m$, and complex numbers $a_1,...,a_n$, are there polynomials $f_1(x),...,f_n(x)$ with coefficients in $\mathbb{C}$ such that the following holds:</p> <ol> <li>$f_i(0)=a_i$.</li> <li>For every complex number $b$, $v_{(x-b)}(f_i(x)-f_j(x))$ is divisible by $m$ (in other words all of the intersection numbers away from $0$ are divisible by $m$).</li> <li>$f_i\neq f_j$ for $i\neq j$.</li> </ol> <p>(The $a_i$'s needn't be different from one another)</p> <p>This is clearly true if $n\leq 2$ and every $m$ and $a_1,a_2$, but I can't think of a general way to do it for every $n$. Is it impossible?</p> http://mathoverflow.net/questions/80717/why-should-the-anabelian-geometry-conjectures-be-true Why should the anabelian geometry conjectures be true? Makhalan Duff 2011-11-11T22:56:01Z 2012-02-04T01:57:46Z <p>I had probed friends of mine about Grothendieck's motivation for making the anabelian geometry conjectures, and they gave me the following explanation:</p> <p>If $X$ is a hyperbolic curve over some field $K$ (think projective and of genus $\geq 2$), then, intuitively, its universal cover is the upper half plane. This means that to distinguish between any two hyperbolic curves, it suffices to distinguish between the actions on the upper-half plane that induce those two hyperbolic curves. In some vague way, this should be the same as distinguishing between their fundamental groups.</p> <p>This seems a little tenuous to me. Is there a modification of the above argument that gives a moral reason for why anabelian geometry should be correct? Is there a completely different moral reason for anabelian geometry? If so, what is it? What intuitive reason should I have to believe anabelian geometry (<em>beside</em> the mounting evidence that it is indeed true)?</p> http://mathoverflow.net/questions/86874/is-every-finite-group-a-quotient-of-the-grothendieck-teichmuller-group Is every finite group a quotient of the Grothendieck-Teichmuller group? Makhalan Duff 2012-01-28T01:33:23Z 2012-01-28T10:50:19Z <p>The Grothendieck-Teichmuller conjecture asserts that the absolute Galois group $Gal(\mathbb{Q})$ is isomorphic to the Grothendieck-Teichmuller group. I was wondering, would this conjecture imply the Inverse Galois Problem? I.e. is every finite group a quotient of the Grothendieck-Teichmuller group?</p> http://mathoverflow.net/questions/85705/is-the-other-extreme-of-hilbert-irreducibility-true Is the other extreme of Hilbert Irreducibility true? Makhalan Duff 2012-01-15T02:15:34Z 2012-01-15T02:48:07Z <p>Let $K$ be a number field (or perhaps more generally a Hilbertian field). Let $X_K\rightarrow \mathbb{P}^1_K$ be a regular (i.e. without extension of scalars) $G$-Galois branched cover. Hilbert's Irreducibility implies that there are infinitely many $K$-rational points on $\mathbb{P}^1_K$ such that the fiber product $Spec(K)\times_{\mathbb{P}^1_K}X_K$ is connected. (This is frequently stated as: there are infinitely many $K$-rational points on $\mathbb{P}^1_K$ such that specializing to them gives a $G$-Galois extension of fields over $K$.)</p> <p>My question is whether the other extreme of this is true. I.e., is there a $K$-rational point on $\mathbb{P}^1_K$ such that $Spec(K)\times_{\mathbb{P}^1_K}X_K$ has $|G|$ connected components (each one isomorphic to $Spec(K)$)? In other words, is there a $K$-rational point on $\mathbb{P}^1_K$ that "splits completely"?</p> <p>This reminds me of the statement that in a Galois extension of number fields there are infinitely many primes that split. This is a far from trivial statement that comes from Class Field Theory. However the analogy isn't perfect, so I don't immediately see how the same methods can be used here.</p> http://mathoverflow.net/questions/82533/a-question-related-to-hilberts-irreducibility-theorem A question related to Hilbert's Irreducibility Theorem Makhalan Duff 2011-12-03T03:54:18Z 2011-12-03T04:22:12Z <p>My question is whether for every extension of number fields $L\subset K$, and for every $f_0(x),...,f_n(x)$ in $K[x]$, there is some $\alpha\in L$ such that $$f_n(\alpha)T^n+...+f_1(\alpha)T+f_0(\alpha)$$ is irreducible as a polynomial in $K[T]$.</p> <p>If $L=K$ this is known from Hilbert's Irreducibility Theorem. I find it hard to believe that there is a counter-example to this, but on the other hand I can't seem to conjure up a proof.</p> http://mathoverflow.net/questions/58901/what-are-the-pillars-of-langlands What are the pillars of Langlands? Makhalan Duff 2011-03-19T03:14:35Z 2011-11-30T04:38:05Z <p>I had previously asked: <a href="http://mathoverflow.net/questions/47943/narratives-in-modular-curves" rel="nofollow">http://mathoverflow.net/questions/47943/narratives-in-modular-curves</a></p> <p>Since then, I've read quite a bit more (but not nearly enough) and I have a few follow up questions about the big picture. As you will soon see, I'm confused about how to think about things, and seeing the big picture will help me a lot in learning the specifics (learning in the dark is difficult!).</p> <p>As I understand it, the story goes like this. First, one defined for every number field $\zeta_K(s)=\sum_{\mathfrak{a}} \frac{1}{(N\mathfrak{a})^s}$. One then defines a Dirichlet character, and for any such one defines $L(\chi,s)$. Further, for any $1$-dimensional Galois representation, $\rho: Gal(K/\mathbb{Q}) \rightarrow \mathbb{C}$, one defines $L(\rho,s)$. Now, in the $1$-dimensional case, the main two theorems that comprise class field theory are: if $K$ is abelian over $\mathbb{Q}$ with group $G$, then $\zeta_K(s)=\prod_{\rho \in \hat{G}} L(\rho,s)$; and for any such $\rho$ there exists a unique primitive Dirichlet character $\chi$ such that $L(\rho,s)=L(\chi,s)$. So far I follow the story perfectly.</p> <p>There is also the issue of what if the base field is not $\mathbb{Q}$, which, admittedly, I don't fully have down.</p> <p>Already in dimension $2$ I have a hard time figuring out what generalizes what corresponding thing from dimension $1$. For Galois representations, one continues to define $L(\rho,s)$ in a similar manner: as the product over $p$ of the characteristic polynomials of the action of the corresponding Frobenius (whenever defined for that $p$! It is still a little murky to me what happens at the bad primes). But now we have modular forms coming in to the picture, and the whole theory of modular curves. So how does this fit in as a generalization of the 1-dimensional case? Here's my best guess, you can tell me if I'm right. For a modular form $f$, one defines the $L$-function for it by the $q$-expansion of $f$: If $f(z)=\sum a(n)e(nz)$ then $L(f,s)=\sum \frac{a(n)}{n^s}$. Then various things that I do not fully understand come into play, claiming things like: $L(s,f)=\prod_{q|N}(1-a(q)q^{-s})^{-1} \prod_{p\not |N} (1-a(p)p^{-s}+f(p)p^{k-1-2s})^{-1}$ (probably just for $f$'s with some property, akin to being primitive). It seems (is this true?) that Hecke theory implies that these $L$'s are `nice'' in the sense that they generalize Dirichlet $L$ functions. Is this the right way to see it? How? What is the $1$-dimensional analogue of modular functions, and modular curves?</p> <p>Then I imagine that one has the modularity theorem, one of whose versions is(?) that for every $2$-dimensional Galois representation there's a modular function for which $L(\rho,s)=L(f,s)$.</p> <p>You will notice that at no point did I talk about the adelic aspect. This is because I don't know where to put it. Is the adelic side easily equivalent to the (Dirichlet characters)-(modular functions) side?(are these two even on the same side?) Is it another pillar with which equivalence is far from trivial with both the Galois representations side AND the (Dirichlet characters)-(modular functions) side? In short -- I'm not sure what the pillars of Langlands are!</p> <p>Further, let us assume that we have some version of Langlands. Is there a conjectural equivalent form of $\zeta_K(s)=\prod_{\rho \in \hat{G}} L(\rho,s)$ for $K/\mathbb{Q}$ not abelian?</p> http://mathoverflow.net/questions/21122/intuition-behind-existence-of-moduli-space-of-stable-curves Intuition behind existence of moduli space of stable curves Makhalan Duff 2010-04-12T16:39:03Z 2011-11-15T21:56:38Z <p>I'm not entirely sure that the title is what I'm looking for. What I'm really asking is for intuition as to why $\bar{\mathcal{M}_g}$ is the compactification of $\mathcal{M}_g$. I'm sure this is covered in the more classic papers (like Deligne and Mumford), but I still find those hard to penetrate.</p> http://mathoverflow.net/questions/80840/what-is-the-intuition-behind-the-proof-of-the-algebraic-version-of-cartans-theor What is the intuition behind the proof of the algebraic version of Cartan's theorem A? Makhalan Duff 2011-11-13T18:46:04Z 2011-11-15T02:31:52Z <p>I am trying to understand the idea behind the proof of GAGA. A crucial step is the following:</p> <p>Theorem: Let $X=\mathbb{P}^r_{\mathbb{C}}$ (either as a variety or as an analytic space), and let $\mathcal{M}$ be a coherent sheaf on $X$. Then for $n>>0$, the twisted sheaf $\mathcal{M}(n)$ is generated by finitely many global sections.</p> <p>In the algebraic case, this is Theorem 5.17 in Hartshorne chapter II. If one tries to read the proof of Theorem 5.17, one sees that it depends on Lemma 5.14, which in turn is a generalization of Lemma 5.3. Lemma 5.3 seems to me to be a completely algebraic lemma with no geometric intuition. It will disrupt the flow of the question to state it here, so I will put it at the end. My point is that I don't see any intuition in this statement</p> <p>In the analytic case this is equivalent to <a href="http://en.wikipedia.org/wiki/Cartan_theorem_A" rel="nofollow">Cartan's Theorem A</a>. To quote the wikipedia page: "Naively, they imply that a holomorphic function on a closed complex submanifold $Z$ of a Stein manifold $X$ can be extended to a holomorphic function on all of $X$". I must confess that I have not read a proof of Cartan's Theorem A itself. But I would like to get some intuition about why it is true, and how it translates to the nilpotent proof found in Hartshorne...</p> <h3>Appendix</h3> <p>Lemma 5.3 in Hartshorne chapter II: Let $X=Spec(A)$ be an affine scheme, let $f\in A$, let $D(f)\subset X$ be the corresponding open set, and let $\mathcal{F}$ be a quasi-coherent sheaf on $X$.</p> <p>a. If $s\in \Gamma(X,\mathcal{F})$ is such that its restriction to $D(f)$ is $0$, then for some $n>0$, $f^ns=0$.</p> <p>b. Given a section $t\in \mathcal{F}(D(f))$ of $\mathcal{F}$ over the open set $D(f)$, then for some $n>0$, $f^nt$ extends to a global section of $\mathcal{F}$ over $X$.</p> <p>The reason I call this a "nilpotent method" is because in complex algerbaic geometry $f^ns=0$ would imply that either $f=0$ or $s=0$.</p> http://mathoverflow.net/questions/80770/reference-request-riemanns-existence-theorem Reference Request: Riemann's Existence Theorem Makhalan Duff 2011-11-12T19:40:36Z 2011-11-13T03:46:20Z <p>By Riemann's existence theorem I mean this:</p> <p>Let $X$ be some variety defined over $\mathbb{C}$, and let $Y$ be a <em>topological</em> covering space of $X$. Then $Y$ can be given the structure of a variety over $\mathbb{C}$, and furthermore this can be done so that the covering map will be algebraic.</p> <p>It is frequently said that this theorem is not constructive. That is to say, that it is impossible to predict the polynomials that define $Y$ and the polynomial map that defines $Y\rightarrow X$. I want to read the proof, and understand for myself why this is true. Where can I find a good, preferably succinct, proof of this theorem in English?</p> http://mathoverflow.net/questions/80784/does-grothendieck-teichmuller-tell-us-something-about-galois-actions-or-just-abo Does Grothendieck-Teichmuller tell us something about Galois actions, or just about Gal(Q)? Makhalan Duff 2011-11-12T23:22:44Z 2011-11-12T23:22:44Z <p>Taking the approach in: <a href="http://www.msri.org/realvideo/ln/msri/1999/vonneumann/schneps/1/" rel="nofollow">http://www.msri.org/realvideo/ln/msri/1999/vonneumann/schneps/1/</a></p> <p>I view the Grothendieck-Teichmuller conjecture as saying that $Gal(\mathbb{Q})$ is isomorphic to a well understood object. That is, it is isomorphic to $Out^*$ of the fundamental group of the Teichmuller lego.</p> <p>This seems to indeed be informative about $Gal(\mathbb{Q})$! My question is whether the Grothendieck-Teichmuller philosophy has predictions about how $Gal(\mathbb{Q})$ acts on varieties defined over $\mathbb{Q}$ (for example $\mathbb{P}^1_{\mathbb{Q}}\smallsetminus${$0,1,\infty$}). From the way that I formulated the conjecture, it is not obvious to me that it does; but I think I am missing the greater picture.</p> http://mathoverflow.net/questions/79575/how-do-brauer-groups-relate-to-zeta-functions How do Brauer groups relate to zeta functions? Makhalan Duff 2011-10-31T02:30:32Z 2011-10-31T06:16:23Z <p>There are two approaches to class field theory that I was taught. The first, is the theory of $L$-functions, Dirichlet characters and so forth (which I described succintly in the question <a href="http://mathoverflow.net/questions/58901/what-are-the-pillars-of-langlands" rel="nofollow">http://mathoverflow.net/questions/58901/what-are-the-pillars-of-langlands</a> that I asked a long time ago), and the other is through Brauer groups.</p> <p>To be precise, I was taught that the following sketch is class field theory:</p> <p>Let $K$ be number field. Let $v$ be any non-archimedean place. Then there is some explicit isomorphism $inv_v:Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}$. If $v$ is an archimedean place, then there is an explicit isomorphism $inv_v:Br(K_v)\rightarrow \mathbb{Z}/2\mathbb{Z}$ if $K_v$ is $\mathbb{R}$, and $0$ if $K_v$ is $\mathbb{C}$.</p> <p>Furthermore, the following sequence is exact:</p> <p>$1\rightarrow Br(K)\rightarrow \bigoplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 1$</p> <p>where the first morphism is obvious, and the last morphism is $\sum_v inv_v$.</p> <p>I take both of these approaches have a lot of content, but it is not clear how people think of them as <em>equivalent</em>. How does one build a dictionary between the one approach and the other?</p> http://mathoverflow.net/questions/79554/what-is-the-general-statement-of-hilbert-90 What is the general statement of Hilbert 90? Makhalan Duff 2011-10-30T22:56:52Z 2011-10-30T23:06:24Z <p>I know two generalizations of Hilbert 90, but I don't if there is a statement that contains both:</p> <h3>The first statement</h3> <p>Let $K$ be a field, then $H^1(Gal(K), GL_n(K^{sep}))=0$.</p> <h3>The second statement</h3> <p>Let $X$ be a scheme, then $H^1(X,\mathbb{G}_m)=Pic(X)$.</p> <h3>Question</h3> <p>Is there a nice characterization of $H^1(X,GL_n)$ (where $GL_n$ is treated as a sheaf in the etale topology) for a general scheme $X$?</p> http://mathoverflow.net/questions/79011/the-etale-fundamental-group-and-etale-cohomology-with-compact-support The etale fundamental group and etale cohomology with compact support Makhalan Duff 2011-10-24T19:00:20Z 2011-10-25T00:02:36Z <p>Before me, the following was asked: <a href="http://mathoverflow.net/questions/16566/etale-fundamental-group-and-etale-cohomology-of-curves" rel="nofollow">http://mathoverflow.net/questions/16566/etale-fundamental-group-and-etale-cohomology-of-curves</a></p> <p>However, that question dealt only with projective curves.</p> <h3>Question</h3> <p>Let $X$ be any scheme (or if you prefer something more concrete, a variety over some field), and let $l$ be some prime different from the characteristics of the residue fields of $X$ (respectively, the characteristic of the field over which the variety is defined), then is there an isomorphism $Hom_{cont}(\pi_1^{et}(X),\mathbb{Q}_l)\cong H^1_c(X,\mathbb{Q}_l)$?</p> http://mathoverflow.net/questions/74600/is-the-set-of-all-curves-that-have-a-galois-map-to-the-projective-line-zariski-cl Is the set of all curves that have a Galois map to the projective line Zariski closed in M_g? Makhalan Duff 2011-09-05T17:26:27Z 2011-09-11T20:42:33Z <p>Not all Riemann surfaces have branched Galois maps to the Riemann sphere. One way to see this is that if $C\rightarrow\mathbb{P}^1$ is Galois, this implies that $C$ is defined over its field of moduli (as a curve).</p> <p>What is the shape of all genus $g$ Riemann surfaces that have a Galois map to $\mathbb{P}^1$? Is it Zariski closed in the coarse moduli space of genus $g$ curves $M_g$?</p> http://mathoverflow.net/questions/74815/is-there-for-every-variety-x-an-abelian-variety-a-such-that-their-1st-l-adic-coho Is there for every variety X an abelian variety A such that their 1st l-adic cohomologies are isomorphic? Makhalan Duff 2011-09-08T01:03:43Z 2011-09-09T12:05:39Z <p>This question is somewhat inspired by Kevin Buzzard's answer to <a href="http://mathoverflow.net/questions/74776/what-is-the-interpretation-of-complex-multiplication-in-terms-of-langlands" rel="nofollow">http://mathoverflow.net/questions/74776/what-is-the-interpretation-of-complex-multiplication-in-terms-of-langlands</a> and somewhat from my own curiosity about such topics.</p> <p>Let $X$ be a variety over $\mathbb{Q}$. This variety induces a pure motive of weight $1$ (it induces pure motives of other weights, but I will focus on the one with weight $1$). I understand that weight $1$ motives come (conjecturally, of course) from weight $2$ newforms. </p> <p>Okay, now let's trace it back. Let's start with a weight $2$ newform. Then, as James D. Taylor alluded to (and from what I know from <a href="http://staff.science.uva.nl/~bmoonen/MTGps.pdf" rel="nofollow">http://staff.science.uva.nl/~bmoonen/MTGps.pdf</a>), the corresponding newform must be the pure motive of weight 1 that is induced by an abelian variety (this is special to the weight $2$ newforms).</p> <p>If so, then it seems that this proves that any pure motive of weight 1 is equal to the pure motive of weight 1 of a motive coming from some abelian variety.</p> <p>Put back in words that are not conjectural: Is it true that for every variety $X$ over $\mathbb{Q}$ there is an abelian variety over $\mathbb{Q}$, $A$, such that $H^1_{et}(X,\mathbb{Q}_l) \cong H^1 _{et} (A,\mathbb{Q}_l)$ as $Gal(\mathbb{Q})$-representations?</p> <p>Or perhaps is the following weaker statement true (if I somehow managed to get something wrong in the above): For every variety $X$ over $\mathbb{Q}$, $L(X,s)$ (coming from the action on the pure motive of weight 1 -- which is well defined even without motives, since one can create it using $l$-adic cohomology) = $\prod_i L(A_i,s)$ where the $A_i$'s are (finitely many) abelian varieties over $\mathbb{Q}$ and the $L$'s are coming from their pure motives of weight $1$.</p> <p>I would very much like to know, if the above is wrong, where exactly the fallacy was. But if everything above is right, then is this known without assuming crazy conjectures like the standard conjectures or forms of Langlands?</p> http://mathoverflow.net/questions/63964/given-a-branched-cover-with-branch-cycle-description-g-1-g-r-does-g-i Given a branched cover with branch cycle description $(g_1,...,g_r)$, does $g_i$ generate some decomposition group? Makhalan Duff 2011-05-05T01:27:55Z 2011-08-25T20:22:12Z <p>Classically: Let $a_1,...,a_r$ be points in $\mathbb{P}^1_{\mathbb{C}}$, and let $\alpha_1,...,\alpha_r$ be simple loops around the $a_i$, all counterclockwise, and none touching (so $\alpha_1...\alpha_r=1$ in the fundamental group of the projective line minus those points). An (pointed, to be pedantic) unramified $G$-cover (meaning a normal covering space with deck transformations$=G$) of $\mathbb{P}^1_{\mathbb{C}}-a_1,...,a_r$ is given by a surjection $\pi_1(\mathbb{P}^1_{\mathbb{C}}-a_1,...,a_r) \rightarrow G$. Let $g_i$ be the image of $\alpha_i$. We say that this $G$-Galois branched cover has branch cycle description $(g_1,...,g_r)$ (note that this depends on our choice of the $\alpha_i$'s). This covering map of curves can be extended to a map of (smooth) projective groups. It can then be shown by a simple topological argument that $g_i$ generates the inertia group (=decomposition group in this case) of some point above $a_i$.</p> <p>My question is whether (and if so, how?) this is also true for the $\overline{\mathbb{F}_p}$ case.</p> <p>Let me be precise. It is known via Grothendieck that $\pi_1^{(p)}(\mathbb{P}^1_{\overline{\mathbb{F}_p}}-a_1,...,a_r)=\widehat{\langle \alpha_1,...,\alpha_r|\prod \alpha_i =1 \rangle}^{(p)}$ (the $^{(p)}$ indicates that we're taking the inverse limit of all prime-to-$p$ finite quotients). Since these $\alpha_i$'s are given in SGA1 through a rather mysterious method, I wonder if the phenomenon described in the first paragraph is still true.</p> <p>My question, therefore, is: let $G$ be a prime-to-$p$ group, and let $X\rightarrow \mathbb{P}^1_{\overline{\mathbb{F}_p}}$ be a (pointed, to be pedantic) branched $G$-cover with branch points $a_1,...,a_r$. Let $\alpha_1,...,\alpha_r$ be such that $\pi_1^{(p)}(\mathbb{P}^1_{\overline{\mathbb{F}_p}}-a_1,...,a_r)=\widehat{\langle \alpha_1,...,\alpha_r|\prod \alpha_i =1 \rangle}^{(p)}$ (I'm almost positive that what I'm about to say is false if you're allowed to choose any such $\alpha_i$'s, so let's assume that we're taking the ones from Grothendieck's construction. If you see a better way of saying what the condition should be on the $\alpha_i$'s I would be very interested in that). Let the branch cycle description of this cover be $(g_1,...,g_r)$ (with respect to these $\alpha_i$'s). Is it true that $g_i$ generates the inertia group (=decomposition group in this case) of some point of $X$ above $a_i$?</p> <p>The topological argument that we were able to use for the $\mathbb{C}$ case seems to no longer apply...</p> http://mathoverflow.net/questions/130770/can-group-cohomology-be-interpreted-as-an-obstruction-to-lifts/130792#130792 Comment by Makhalan Duff Makhalan Duff 2013-05-16T14:13:06Z 2013-05-16T14:13:06Z Thanks! Makes perfect sense. http://mathoverflow.net/questions/119439/does-the-proof-of-gaga-use-the-axiom-of-choice Comment by Makhalan Duff Makhalan Duff 2013-01-21T04:47:25Z 2013-01-21T04:47:25Z Mariano, it is! Think for example on the case of curves. There it does reduce to saying &quot;topological&quot;. http://mathoverflow.net/questions/119439/does-the-proof-of-gaga-use-the-axiom-of-choice/119441#119441 Comment by Makhalan Duff Makhalan Duff 2013-01-21T04:46:23Z 2013-01-21T04:46:23Z I'm not so sure that it's as straightforward as you say. If I give you an analytic coherent sheaf, would you be able to give me the algebraic coherent sheaf that induces it? http://mathoverflow.net/questions/102839/what-is-the-relationship-between-motivic-cohomology-and-the-theory-of-motives/102842#102842 Comment by Makhalan Duff Makhalan Duff 2012-07-22T01:26:05Z 2012-07-22T01:26:05Z It does indeed help! http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity Comment by Makhalan Duff Makhalan Duff 2012-07-02T02:19:09Z 2012-07-02T02:19:09Z $\mathbb{I}_{\mathbb{Q}}/\mathbb{Q}^{\times}$ by $D_{\mathbb{Q}}$, which is the product of the infinite places, and by $\prod \mathbb{Z}_p^{\times}$ (to account for not allow ramification anywhere). But I wonder where you're using something special about $\mathbb{Q}$, because this isn't true for all number fields, is it? http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity Comment by Makhalan Duff Makhalan Duff 2012-07-02T02:19:02Z 2012-07-02T02:19:02Z @Dror: that's very helpful. Let me ask you a few questions. 1. Is $D_K$ always the product over all infinite places? 2. Let $L$ be the maximal abelian extension of $K$ that is unramified over a specific set of primes $S$ of $O_K$. Are you saying that $L^{\times}N_{L/K}(\mathbb{I}_L)$ is equal to the product $\prod_{\mathfrak{p} \not \in S} O_{\mathfrak{p}} ^{\times}$? I guess what I'm trying to ask is: it seems that you're saying that the &quot;reason&quot; that there are no abelian unramified extensions of $\mathbb{Q}$ is that you get $Gal(\mathbb{Q}^{un,ab}/\mathbb{Q})$ by quotienting http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity Comment by Makhalan Duff Makhalan Duff 2012-06-30T16:03:14Z 2012-06-30T16:03:14Z $K^{\times}\backslash \mathbb{I}_K/\prod_{v\not \in S} O^{\times}_v$ is isomorphic to $Gal(K^{ab}/K)$. What is the correct way to think about this? http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity Comment by Makhalan Duff Makhalan Duff 2012-06-30T16:02:15Z 2012-06-30T16:02:15Z @Felipe: sorry for asking so many questions, but I'm confused again. Serre's book would be easier to read with a little motivation, so hopefully you'll indulge me. My perception was that if $S$ is a finite set of places, then $K^{\times}\backslash \mathbb{I}_K/\prod_{v\not \in S} O^{\times}_v$ would have a profinite completion which is isomorphic to $Gal(L/K)$ where $L$ is the maximal abelian cover of $K$ that is ramified only in $S$. But it seems that what you're saying is that if S is the set of infinite places, then http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity Comment by Makhalan Duff Makhalan Duff 2012-06-30T13:06:40Z 2012-06-30T13:06:40Z I'm sorry, my last comment was silly. You're right that $\pi_1^{ab}(C)\cong Gal(K^{ab,un}/K)$. My question is, why is the analogous statement for $K$ a number field that $K^{\times}\backslash \mathbb{I}_K/\prod_v O_v^{\times}$ is isomorphic (after taking profinite completions) to $Gal(K^{ab}/K)$ rather than $Gal(K^{ab,un}/K)$? Or am I wrong about that? For example $\mathbb{Q}^{\times}\backslash \mathbb{I}_{\mathbb{Q}}/\prod_p \mathbb{Z}_p^{\times}$ is far from trivial, isn't it? I guess I'm just finding it hard to draw the analogy. http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity Comment by Makhalan Duff Makhalan Duff 2012-06-30T01:24:31Z 2012-06-30T01:24:31Z In what sense? For example it is not true that every abelian cover of $\mathbb{P}^1_{\mathbb{F}_p}$ is unramified! (E.g., $y^2=x$ will define an abelian cover ramified over $0$ and $\infty$.) I feel like I'm missing the crucial point you're trying to get across... Can you tell me what I'm missing? http://mathoverflow.net/questions/100977/how-does-one-understand-geometric-cft-in-terms-of-modularity Comment by Makhalan Duff Makhalan Duff 2012-06-29T22:56:32Z 2012-06-29T22:56:32Z @Felipe: I am confused by your statement. The analogous theorem to the theorem I cited in the number theory case describes $Gal(K^{ab}/K)$ in terms of (the profinite completion of) a double quotient of the ideles of $K$. If the theorem I cited is only unramified class field theory, wouldn't it describe $Gal(K^{ab,un}/K)$ instead? http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits Comment by Makhalan Duff Makhalan Duff 2012-05-05T19:27:03Z 2012-05-05T19:27:03Z Mark, I'm still trying to figure out whether it's true even in the context of that question. Yiftach Barnea didn't explain why that isomorphism is true. http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits Comment by Makhalan Duff Makhalan Duff 2012-05-05T19:26:20Z 2012-05-05T19:26:20Z Let's put it in the simplest possible terms: is it possible to show that for $G$ and $H$ finitely generated, with an action $\phi_1:H\rightarrow Aut(G)$ which extends (by assumption) to an action $\phi_2:\hat{H}\rightarrow Aut(\hat{G})$ (where hat denotes profinite completion), it is true that $\hat{G}\rtimes \hat{H}$ is a profinite group? http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits Comment by Makhalan Duff Makhalan Duff 2012-05-05T19:11:46Z 2012-05-05T19:11:46Z Hmm... I'm looking at the question: <a href="http://mathoverflow.net/questions/60750/profinite-completion-of-a-semidirect-product" rel="nofollow" title="profinite completion of a semidirect product">mathoverflow.net/questions/60750/&hellip;</a> in which you also participated. In Yiftach Barnea's answer, I believe that he assumes that inverse limits do commute with semi-direct products when he says that $\hat{G}\rtimes \hat{H}\cong \varprojlim (G\rtimes H)/(G_n\rtimes N)$. Would you agree with his usage, or do you think he is wrong? http://mathoverflow.net/questions/96078/are-semi-direct-products-categorical-limits Comment by Makhalan Duff Makhalan Duff 2012-05-05T18:58:13Z 2012-05-05T18:58:13Z Mark, it is indeed! Is it true? Do you have a reference, or can give a reason for it to be true?