User js milne - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:10:41Z http://mathoverflow.net/feeds/user/930 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21415/what-algebraic-group-does-tannaka-krein-reconstruct-when-fed-the-category-of-modu/21429#21429 Answer by JS Milne for What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra? JS Milne 2010-04-15T07:47:52Z 2013-03-08T16:55:08Z <p>Rather than an answer, this is more of an anti-answer: I'll try to persuade you that you probably don't want to know the answer to your question.</p> <p>Instead of some exotic nonalgebraic Lie algebra, let's start with the one-dimensional Lie algebra $\mathfrak{g}$ over a field $k$. A representation of $\mathfrak{g}$ is just a finite-dimensional vector space + an endomorphism. I don't know what the affine group scheme attached to this Tannakian category is, but thanks to a 1954 paper of Iwahori, I can tell you that its Lie algebra can be identified with the set of pairs $(\mathfrak{g},c)$ where $\mathfrak{g}$ is a homomorphism <em>of abelian groups</em> $k\to k$ and $c$ is an element of $k$. So if $k$ is big, this is huge; in particular, you don't get an algebraic group. (Added: $k$ is algebraically closed.)</p> <p>By contrast, the affine group scheme attached to the category of representations of a semisimple Lie algebra $\mathfrak{g}$ in characteristic zero is the simply connected algebraic group with Lie algebra $\mathfrak{g}$. </p> <p>In summary: this game works beautifully for semisimple Lie algebras (in characteristic zero), but otherwise appears to be a big mess. See arXiv:0705.1348 for a few more details. </p> http://mathoverflow.net/questions/15595/math-keyboard-does-it-exist/39865#39865 Answer by JS Milne for Math keyboard: does it exist ? JS Milne 2010-09-24T15:01:38Z 2010-09-24T15:01:38Z <p>One almost existed. The program I use to write TeX has several "keyboards". For example, typing Ctrl-g switches to the Greek "keyboard" and then "a" puts "\alpha" in your file; typing Ctrl-s and then "a" puts "\angle" in your file. At one time, the owners of the program were planning to produce a keyboard with the property that the symbol on the keyboard would change when you switched keyboards. So when you typed Ctrl-g, you would see $\alpha$ on the "a" key (I think they were planning to use LEDs). It was certainly an interesting idea, but, no, I wouldn't buy such a thing --- like most people here, I touch-type, and prefer to enter everything with a sequence of normal characters.</p> http://mathoverflow.net/questions/38168/is-the-category-of-commutative-group-schemes-abelian/38181#38181 Answer by JS Milne for Is the category of commutative group schemes abelian? JS Milne 2010-09-09T14:42:23Z 2010-09-10T12:26:40Z <p>The category of commutative affine algebraic group schemes over any field is abelian. More generally, the standard isomorphism theorems in abstract group theory hold in the category of affine algebraic group schemes over a field. See, for example, Chapt 1 of my online notes. Over other bases, this need not be true. If you are foolish enough to work with reduced algebraic group schemes (Borel, Humphreys, Springer, et al) then you run into all sorts of problems.</p> <p>Added: In 1962, Cartier gave a conference talk in which he noted that the standard isomorphism theorems etc. fail with the usual definition of algebraic groups (no nilpotents), but then observed that the "preceding difficulties vanish" when one works with schemes. As far as I know, this is the first statement in print. Of course, it is all covered in the 1963-64 Grothendieck-Demazure seminar (SGA3). </p> <p>Over arbitrary bases, you can still form kernels, but quotients are a problem because the subgroup need not be flat. As noted, there are also problems when you drop the condition "affine and finite type".</p> <p>When you don't allow nilpotents (i.e., when you work in the category of reduced group schemes), the Frobenius map $\mathbb{A}^1\to \mathbb{A}^1$ (this is a homomorphism when $\mathbb{A}^1$ is regarded as the additive group scheme) is mono and epi but not an isomorphism (no inverse). In the full category of affine algebraic group schemes, it has a kernel, namely, the finite group scheme $\alpha_p$, so there is no problem. </p> http://mathoverflow.net/questions/25774/where-can-we-find-delignes-paper-theorie-de-hodge-i/36441#36441 Answer by JS Milne for Where can we find Deligne's paper " Theorie de Hodge I"? JS Milne 2010-08-23T12:22:27Z 2010-08-23T12:22:27Z <p>Amazingly, ALL the ICM talks since the beginning of time can now be found online at <a href="http://mathunion.org/ICM/" rel="nofollow">http://mathunion.org/ICM/</a></p> http://mathoverflow.net/questions/30042/jordan-decomposition-in-a-classical-group/30876#30876 Answer by JS Milne for Jordan decomposition in a classical group JS Milne 2010-07-07T11:43:20Z 2010-07-07T11:43:20Z <p>The proof of the Jordan decomposition for algebraic groups over perfect fields has two parts:</p> <p>(a) Linear algebra: An automorphism of a vector space has a unique multiplicative Jordan decomposition, which is compatible with maps and tensor products...</p> <p>(b) Some baby Tannakian stuff.</p> <p>Most proofs in the literature mix the two parts, making the proof seem more difficult than it is. If you accept the baby Tannakian stuff, which everyone should know anyway, one is left with some easy linear algebra (see, for example, I Section 9 of my online notes).</p> http://mathoverflow.net/questions/26062/automorphism-of-algebraic-group-preserving-a-hyperspecial-maximal-compact/26078#26078 Answer by JS Milne for Automorphism of algebraic group preserving a hyperspecial maximal compact JS Milne 2010-05-26T23:16:38Z 2010-05-27T03:42:50Z <p>As noted, there seem to be some "reductive"s missing from the question. Here's what is known: let $R$ be a Henselian discrete valuation ring with field of fractions $K$, and let $R^{\prime}$ be the integral closure of $R$ in the maximal unramified extension $K^{\prime}$ of $K$; a smooth affine scheme $X$ over $R$ defines a scheme $X_{K}$ over $K$ and a subset $X(R^{\prime})$ of $X_{K}(K^{\prime})$; the functor $X\rightsquigarrow(X_{K},X(R^{\prime}))$ is fully faithful.</p> <p>The proof of this is fairly easy (cf. 1.7.3 of .Bruhat, F., and Tits, J., Groupes reductifs sur un corps local II, Publ. Math. IHES, 60, 1984). </p> <p>As stated this fails without "affine and smooth".</p> <p>In general you can't replace $X(R^{\prime})$ with $X(R)$ (because $X(R)$ may be empty). Perhaps if $X_{k}(k)$ is Zariski dense in $X_{k}$ it's OK ($k$=residue field).</p> <p>Added: As blt points out, Lemma 6.2 of Snowden and Wiles, arXiv:0908.1991v3, states that, when $K$ is a finite extension of $\mathbb{Q}_{\ell}$, $X$ is a simply connected semisimple group, and the map on the generic fibre is an automorphism, if $X(R)$ maps into $X(R)$ then $X(R^{\prime})$ maps into $X(R^{\prime})$. Thus, for simply connected semisimple groups, the answer is YES, and as BCnrd points out, that implies that the answer is YES for all reductive groups.</p> http://mathoverflow.net/questions/25354/constructing-coherent-sheaves-on-shimura-varieties/25368#25368 Answer by JS Milne for Constructing coherent sheaves on Shimura varieties. JS Milne 2010-05-20T14:24:25Z 2010-05-20T15:24:05Z <p>As others have pointed out, the word you are looking for is automorphic vector bundle. The holomorphic automorphic forms are exactly the sections of these bundles. To define them, you start with a homogeneous bundle on the (compact) dual hermitian symmetric space. Initially, the construction of the automorphic vector bundle attached to the homogeneous bundle is analytic, but it can be made algebraic by the introduction of the standard principal bundle. It is known that the standard principle bundle has a canonical model over the reflex field, so this all works arithmetically. [The definition and proof of the existence of a canonical model of the standard principal bundle is in my 1988 Inventiones paper, after earlier work of Shimura, Harris, and others; there is a discussion of such things in Chapter III of my article in the Proceedings of the 1988 Ann Arbor conference, which are available on my website.]</p> http://mathoverflow.net/questions/24585/if-erds-is-published-as-erdos-in-a-paper-which-do-i-cite/24646#24646 Answer by JS Milne for If Erdős is published as Erdös in a paper, which do I cite? JS Milne 2010-05-14T18:02:18Z 2010-05-14T18:02:18Z <p>My rule of thumb is to spell names as MR spells them. I just looked up Erdos in MR, and it turn out that there is a Paul Erdös as well as a Paul Erdős (different people). Since you mean the second, you should spell it correctly, or else put (sic) in your bibliography. </p> http://mathoverflow.net/questions/23624/lifting-etale-morphisms/23636#23636 Answer by JS Milne for Lifting Etale Morphisms JS Milne 2010-05-05T22:02:32Z 2010-05-05T22:02:32Z <p>SGA 1, IX, 1.10: Let $Y$ be a scheme proper over a complete local noetherian ring $R$, and let $Y_0$ be the closed fibre of $Y/R$. Then the functor $X\mapsto X_0$ from finite etale coverings of $Y$ to finite etale coverings of $Y_0$ is an equivalence of categories.</p> http://mathoverflow.net/questions/21555/why-are-modular-forms-usually-defined-only-for-congruence-subgroups/21611#21611 Answer by JS Milne for Why are modular forms (usually) defined only for congruence subgroups? JS Milne 2010-04-16T20:01:04Z 2010-04-17T06:34:04Z <p>Actually, modular curves for congruence groups have canonical models over number fields, not Q (there exist congruence subgroups other than $\Gamma _0(N)$!). They even have reasonably nice integral models. Moreover, the modular forms are sections of a sheaf defined on the canonical model, and the sheaf extends to the integral model. This has the following consequence: if a modular form has Fourier coefficients $a_n$ in a number field $K$ (so the form is a section of the sheaf over $K$), then the $a_{n}$ have bounded denominators, i.e., lie in $d^{-1}\mathcal{O}_{K}$ for some $d$.</p> <p>Modular curves for arithmetic groups also have models over number fields, but the last statement definitely fails for forms that don't come from congruence groups. So something goes wrong with the beautiful picture we have for congruence modular curves, but I've never understood exactly what. However, this is another indication that the link to arithmetic is more tenuous in the noncongruence case, and helps explain why number theorists are mainly interested in the congruence case.</p> <p>[This was written as a comment on Buzzard's answer, but the site wouldn't let me post it (too long).]</p> http://mathoverflow.net/questions/21265/how-many-groups-of-size-at-most-n-are-there-what-is-the-asymptotic-growth-rate/21283#21283 Answer by JS Milne for How many groups of size at most n are there? What is the asymptotic growth rate? And what of rings, fields, graphs, partial orders, etc.? JS Milne 2010-04-14T01:37:38Z 2010-04-14T01:37:38Z <p>Roughly speaking, the more high powers of primes divide $n$, the more groups of order $n$ there should be. In fact, if $f(n)$ is the number of isomorphism classes of groups of order $n$, then $$f(n)\leq n^{(\frac{2}{27}+o(1))e(n)^{2}}%$$ where $e(n)$ is the largest exponent of a prime dividing $n$ and $o(1)\rightarrow0$ as $e(n)\rightarrow\infty$ (see Pyber, L. Enumerating finite groups of given order. Ann. of Math. (2) 137 (1993), no. 1, 203--220. MR1200081).</p> <p>From my Group Theory notes page 12.</p> http://mathoverflow.net/questions/21051/why-does-the-naive-definition-of-compactly-supported-etale-cohomology-give-the-wr/21056#21056 Answer by JS Milne for Why does the naive definition of compactly supported étale cohomology give the wrong answer? JS Milne 2010-04-12T01:21:20Z 2010-04-12T01:28:17Z <p>It is important in etale cohomology, as it is topology, to define cohomology groups with compact support --- we saw this already in the case of curves in Section 14. They should be dual to the ordinary cohomology groups.</p> <p>The traditional definition (Greenberg 1967, p162) is that, for a manifold $U$, $H_{c}^{r}(U,\mathbb{Z})=dlim_{Z}H_{Z}^{r}(U,\mathbb{Z})$ where $Z$ runs over the compact subsets of $U$. More generally (Iversen 1986, III.1) when $\mathcal{F}$ is a sheaf on a locally compact topological space $U$, define $\Gamma_{c}(U,\mathcal{F})=dlim_{Z}\Gamma_{Z}(U,\mathcal{F})$ where $Z$ again runs over the compact subsets of $U$, and let $H_{c}% ^{r}(U,-)=R^{r}\Gamma_{c}(U,-)$.</p> <p>For an algebraic variety $U$ and a sheaf $\mathcal{F}$ on $U_{\mathrm{et}}$, this suggests defining $\Gamma_{c}(U,\mathcal{F})=dlim_{Z}\Gamma_{Z}(U,\mathcal{F}),$ where $Z$ runs over the complete subvarieties $Z$ of $U$, and setting $H_{c}^{r}(U,-)=R^{r}\Gamma_{c}(U,-)$. However, this definition leads to anomolous groups. For example, if $U$ is an affine variety over an algebraically closed field, then the only complete subvarieties of $U$ are the finite subvarieties, and for a finite subvariety $Z\subset U$, $H_{Z}^{r}(U,\mathcal{F})=\oplus_{z\in Z}H_{z}^{r}(U,\mathcal{F}).$ Therefore, if $U$ is smooth of dimension $m$ and $\Lambda$ is the constant sheaf $\mathbb{Z}/n\mathbb{Z}$, then $H_{c}^{r}(U,\Lambda)=dlim H_{Z}^{r}(U,\Lambda)=\oplus_{z\in U}H_{z}% ^{r}(U,\Lambda)=\oplus_{z\in U}\Lambda(-m)$ if $r=2m$, and it is 0 otherwise These groups are not even finite. We need a different definition...</p> <p>If $j\colon\ U\rightarrow X$ is a homeomorphism of the topological space $U$ onto an open subset of a locally compact space $X$, then $H_{c}^{r}(U,\mathcal{F})=H^{r}(X,j_{!}\mathcal{F})$ (Iversen 1986, p184). We make this our definition.</p> <p>From Section 18 of my notes: Lectures on etale cohomology.</p> http://mathoverflow.net/questions/18341/the-algebraicity-of-hodge-structure-map/18343#18343 Answer by JS Milne for The algebraicity of Hodge structure map JS Milne 2010-03-16T08:22:40Z 2010-03-16T09:27:50Z <p>It looks false to me. Let $V=\mathbb{Q}^{2}$, and let $V(\mathbb{R})=V^{0}\oplus V^{2}$ where $V^{0}$ is the line defined by $y=ex$ and $V^{2}$ is the line defined by $y=\pi x$. Give $V^{0}$ the unique Hodge structure of type $(0,0)$ and $V^{2}$ the unique Hodge structure of type $(1,1)$. To say that $w$ is defined over the subfield $\mathbb{Q}^{\mathrm{al}}$ of $\mathbb{C}$ means that the gradation $V(\mathbb{R})=V^{0}\oplus V^{2}$ arises from a gradation of $V(\mathbb{Q}{}^{\mathrm{al}})$ by tensoring up, but this isn't true. Perhaps the all the "resources" have additional conditions, or perhaps they are all ...</p> <p>Added: When you are defining a Shimura variety, the weight homomorphism w factors through a Q-subtorus of GL(V), and then it is true that w is defined over the algebraic closure of Q (because, for tori T,T', the group Hom(T,T') doesn't change when you pass from one algebraically closed field to a larger field).</p> http://mathoverflow.net/questions/17662/books-on-reductive-groups-using-scheme-theory/17793#17793 Answer by JS Milne for Books on reductive groups using scheme theory JS Milne 2010-03-11T01:15:18Z 2010-03-11T01:15:18Z <p>Personally, I find the "classical" books (Borel, Humphreys, Springer) unpleasant to read because they work in the wrong category, namely, that of reduced algebraic group schemes rather than all algebraic group schemes. In that category, the isomorphism theorems in group theory fail, so you never know what is true. For example, the map $H/H\cap N\rightarrow HN/N$ needn't be an isomorphism (take $G=GL_{p}$, $H=SL_{p}$, $N=\mathbb{G}_{m}$ embedded diagonally). Moreover, since the terminology they use goes back to Weil's Foundations, there are strange statements like "the kernel of a homomorphism of algebraic groups defined over $k$ need not be defined over $k$". Also I don't agree with Brian that if you don't know descent theory, EGA, etc. then you don't "know scheme theory well enough to be asking for a scheme-theoretic treatment'. </p> <p>Which explains why I've been working on a book whose goal is to allow people to learn the theory of algebraic group schemes (including the structure of reductive algebraic group schemes) without first reading the classical books and with only the minimum of prerequisites (for what's currently available, see my website under course notes). In a sense, my aim is to complete what Waterhouse started with his book.</p> <p>So my answer to the question is, no, there is no such book, but I'm working on it....</p> http://mathoverflow.net/questions/17020/why-do-people-think-that-abelian-varieties-are-the-hardest-case-for-the-hodge-con/17025#17025 Answer by JS Milne for Why do people think that abelian varieties are the hardest case for the Hodge conjecture? JS Milne 2010-03-03T23:21:43Z 2010-03-04T08:08:37Z <p>I would say the answer to both questions is no. In fact, abelian varieties should be an "easy" case. For example, it is known that for abelian varieties (but not other varieties), the variational Hodge conjecture implies the Hodge conjecture. It is disconcerting that we can't prove the Hodge conjecture even for abelian varieties, even for abelian varieties of CM-type, and we can't even prove that the Hodge classes Weil described are algebraic. So if the Hodge conjecture was proved in one interesting case, e.g., abelian varieties, that would be a big boost.</p> <p>Added: As follow up to Matt Emerton's answer, a proof that the Hodge conjecture for abelian varieties implies the Hodge conjecture for all varieties would (surely) also show that Deligne's theorem (that Hodge classes on abelian varieties are absolutely Hodge) implies the same statement for all varieties. But no such result is known (and would be <em>extremely</em> interesting).</p> http://mathoverflow.net/questions/1720/can-algebraic-varieties-be-rigidified-by-finite-sets-of-points Can algebraic varieties be rigidified by finite sets of points? JS Milne 2009-10-21T21:02:26Z 2010-02-25T16:48:12Z <p>For an algebraic variety X over an algebraically closed field, does there always exist a finite set of (closed) points on X such that the only automorphism of X fixing each of the points is the identity map? If Aut(X) is finite, the answer is obviously yes (so yes for varieties of logarithmic general type in characteristic zero by Iitaka, Algebraic Geometry, 11.12, p340). For abelian varieties, one can take the set of points of order 3 [added: not so, only for polarized abelian varieties]. For P^1 one can take 3 points. Beyond that, I have no idea.</p> <p>The reason I ask is that, for such varieties, descent theory becomes very easy (see Chapter 16 of the notes on algebraic geometry on my website).</p> http://mathoverflow.net/questions/16261/quotient-of-a-reductive-group-by-a-non-smooth-central-finite-subgroup/16348#16348 Answer by JS Milne for Quotient of a reductive group by a non-smooth central finite subgroup JS Milne 2010-02-25T02:18:15Z 2010-02-25T02:18:15Z <p>The standard isomorphism theorems in abstract group theory all hold for group schemes of finite type over a field. This is implicit in SGA (a key point is the statement mentioned by Ekedahl) and is explicit in the notes on algebraic groups,... appearing on my website (Section 7 of Chapter I). This makes a lot of things obvious (including your questions).</p> <p>[The isomorphism theorems fail when you don't allow nilpotents, which is why the standard expositions on algebraic groups are so complicated.]</p> http://mathoverflow.net/questions/15979/motivation-for-the-etale-topology-over-other-possibilities/15980#15980 Answer by JS Milne for Motivation for the étale topology over other possibilities JS Milne 2010-02-21T18:38:43Z 2010-02-21T18:38:43Z <p>As observed, the Zariski topology has too few open subsets to compute cohomology with constant coefficients. It had already been observed by Serre that etale covers were enough to trivialize principal bundles for many algebraic groups. That suggested using etale covers. The etale "topology" is the coarsest for which the inverse function theorem holds. The flat topology also gives Weil cohomologies (the same ones as the etale topology), but why use flat covers when etale covers are enough.</p> http://mathoverflow.net/questions/15370/tools-for-the-langlands-program/15394#15394 Answer by JS Milne for Tools for the Langlands Program? JS Milne 2010-02-16T00:44:11Z 2010-02-16T00:44:11Z <p>Hey! We're making progress. It used to be called the Langlands philosopy. [Oops, this was meant to be a comment on fpqc's comment.]</p> http://mathoverflow.net/questions/14607/when-to-start-reviewing/14674#14674 Answer by JS Milne for When to start reviewing JS Milne 2010-02-08T18:44:48Z 2010-02-08T18:44:48Z <p>Actually, all reviews are written by research mathematicians, since MR/Zentralblatt typically only invite those who have published papers. In fact, it is not a great burden. As everyone knows, the best way to understand some mathematics is to explain it others. If you get a paper to review that you have read, or were planning to read, then writing a review doesn't take long. If you get a paper that you weren't planning to read, then reviewing it broadens your understanding. So my advice is to begin reviewing when invited, although no harm will be done if you prefer to wait until you are more experienced. Since all of us benefit from reviews, all of us have an ethical obligation to contribute for at least part of our careers.</p> http://mathoverflow.net/questions/14278/haar-measure-on-a-quotient-references-for/14305#14305 Answer by JS Milne for Haar measure on a quotient, References for. JS Milne 2010-02-05T18:45:14Z 2010-02-05T18:45:14Z <p>The book I always look at for such things is Nachbin, The Haar Integral, which is short, and has a whole chapter on Integration on Locally Compact Homogeneous Spaces. </p> <p>And a plus: he gives you a choice of reading the proof of the existence and uniqueness of the Haar integral according to Weil or according to Henri Cartan.</p> http://mathoverflow.net/questions/13619/is-a-free-alternative-to-mathscinet-possible/14087#14087 Answer by JS Milne for Is a free alternative to MathSciNet possible? JS Milne 2010-02-04T01:08:29Z 2010-02-05T18:27:46Z <p>Everyone I know in the AMS would like to make MR/MathSci free, but the problem is that it costs millions of dollars to produce and maintain (it requires a large staff in Ann Arbor and elsewhere, including many mathematicians), and no one has managed to find any other way to pay for it*. This is certainly something the mathematicians in the AMS are aware of and have thought about. The AMS attempts to make it as widely available as possible given the constraint that it has to be paid for. As far as I know, the revenue from MR/MathSci only pays to support it, not any of the AMS's other activities [not so; see below.]</p> <p>Posters don't seem to realize the huge effort that goes into maintaining a project like this (for example, every article has to be assigned to a reviewer, and every review has to edited). Certainly, I don't believe a free alternative would be able to come anywhere near the quality MathSci maintains, so my answer is no, a free alternative to MR/MathSci is not possible.</p> <p>Perhaps free supplements to MathSci could be useful, but anything that drew potential reviewers away from MR/MathSci would harm, not help, what is an extremely valuable resource.</p> <p>*Of course, the intelligent thing would be for the funding agencies in the wealthy countries (US,EU,Japan,...) to pay for MR/MathSci directly, so that it could be distributed freely, but getting them to do this seems to be hopeless.</p> <p><b>Added:</b> In response to Anton Petrunin's comment, here are some numbers. The AMS employs 15 mathematical editors (i.e., mathematicians) and a total staff of over 70 at Mathematical Reviews (in Ann Arbor). The total direct cost of producing MR/MathSci in 2008 was 6,569,000USD. However, contrary to what I thought, the AMS does use the revenue from MR/MathSci and its other publications to support a large part of its other activities (Member and professional services, general and administrative expenses). In 2008, about 24% of total publication revenue was used in this way. See <a href="http://www.ams.org.proxy.lib.umich.edu/notices/200908/rtx090801006p.pdf" rel="nofollow">2008 report</a> and <a href="http://www.ams.org/notices/200706/tx070600770p.pdf" rel="nofollow">ad</a>.</p> http://mathoverflow.net/questions/13451/rational-hilbert-modular-surfaces/13473#13473 Answer by JS Milne for Rational Hilbert modular surfaces JS Milne 2010-01-30T13:47:11Z 2010-01-31T16:23:56Z <p>The answer is (probably) yes.</p> <p>A theorem of Margulis et al. shows that an irreducible lattice in a Lie group is arithmetic unless the group is isogenous to SO(1,n)x(compact) or SU(1,n)x(compact).</p> <p>A theorem of Mumford shows that the quotient of a hermitian symmetric domain by a neat arithmetic subgroup is of logarithmic general type in the sense of Iitaka (hence not rational). (An arithmetic subgroup is neat if the subgroup generated by the eigenvalues of any element of the subgroup is torsion-free. Every sufficiently small subgroup is neat).</p> <p>For a discussion of the first theorem, see Section 5B of Witte Morris, Introduction to Arithmetic Groups, <a href="http://front.math.ucdavis.edu/0106.5063" rel="nofollow">http://front.math.ucdavis.edu/0106.5063</a></p> <p>For the second theorem, see: Mumford, D. Hirzebruch's proportionality theorem in the noncompact case. Invent. Math. 42 (1977), 239--272. MR0471627 </p> <p>Added: As @moonface points out, this doesn't prove it.</p> <p>The congruence subgroup problem is known for $SL_{2}$ over totally real fields $F$ other than $Q$. Let $\Gamma$ be an irreducible lattice in $SL_{2}(\mathbb{R})\times SL_{2}(\mathbb{R})$. By Margulis, it is arithmetic, hence congruence. Moreover, after conjugating we may suppose that $\Gamma\subset\Gamma(1)=SL_{2}(O)$. Now $\Gamma\backslash\mathbb{H}\times\mathbb{H}$ covers $\Gamma(1)\backslash \mathbb{H}\times\mathbb{H}$, and so if the first is rational, so is the second (Zariski 1958; requires dimension 2). Thus, we have a finite list of the possible totally real quadratic extensions $K$ to work with. For some $N$, $\Gamma\supset\Gamma(N)$, and we may suppose that $\Gamma(N)$ is neat. Since $\Gamma(1)/\Gamma(N)$ is finite, for any integer $N$ we get (in sum) only finitely many rational Hilbert modular surfaces of level $N$.</p> <p>The problem remains to show that, for each totally real quadratic field on our list, every rational Hilbert modular surface is of level $N$ for some fixed $N$. Apart from looking case by case, I don't see how to do this (but it is surely true).</p> http://mathoverflow.net/questions/12009/is-there-a-slick-proof-of-the-classification-of-finitely-generated-abelian-groups/12135#12135 Answer by JS Milne for Is there a slick proof of the classification of finitely generated abelian groups? JS Milne 2010-01-17T19:31:11Z 2010-01-17T19:31:11Z <p>The slickest (nonconstructive!) proof I know of is the one I put in my Group Theory <a href="http://www.jmilne.org/math/CourseNotes/GT.pdf" rel="nofollow">notes</a>, p22. You choose a generating set $x_1,\ldots,x_n$ for the group such that $x_1$ has the minimum possible order, and then prove that the group is the direct sum of the subgroups generated by $x_1$ and by $x_2,\ldots,x_n$. Now apply induction on $n$ to see that the group is a direct sum of cyclic groups.</p> http://mathoverflow.net/questions/11087/a-telegram-by-grothendieck-to-serre/11172#11172 Answer by JS Milne for A telegram by Grothendieck to Serre JS Milne 2010-01-08T18:41:17Z 2010-01-08T22:37:13Z <p>I think I heard Grothendieck's talk at the 1958 ICM described as a "a telegram by Grothendieck to Serre". But this would have been in a conversation (with neither Serre nor Grothendieck). I don't know whether anyone said it in a lecture, much less wrote it down.</p> <p>MR0130879 (24 #A733) Grothendieck, Alexander The cohomology theory of abstract algebraic varieties. 1960 Proc. Internat. Congress Math. (Edinburgh, 1958) pp. 103--118 Cambridge Univ. Press, New York </p> <p>Added: I don't know who first described Grothendieck's 1958 ICM talk as a "telegram to Serre", but I probably heard it from Lubin. Since Lubin and the article's author spent their careers at Brown, this fits.</p> <p>Grothendieck's talk is available <a href="http://www.math.jussieu.fr/~leila/grothendieckcircle/CohomologyVarieties.pdf" rel="nofollow">here</a>. In it he lays out his plans for schemes and their cohomology. In 1958, not many people would have understood it.</p> http://mathoverflow.net/questions/10335/about-hilbert-and-siegel-modular-varieties-forms/10338#10338 Answer by JS Milne for about Hilbert and Siegel modular varieties (forms) JS Milne 2010-01-01T00:43:53Z 2010-01-01T00:43:53Z <p>The Shimura datum for a Hilbert modular variety is the pair (G,X) where G is the group over Q obtained from GL(2) over a totally real number field F by (Weil) restriction of scalars and X is a product of copies of C minus the real axis indexed by the real embeddings of F.</p> <p>In the definition of Shimura varieties we work only over Q by convention: we could work over number fields, but this gives nothing new (one can always take Weil restriction). However, in some cases, e.g., Hilbert modular varieties, it seems (to me to be) more natural to work over a number field other than Q.</p> http://mathoverflow.net/questions/9778/complete-discrete-valuation-rings-with-residue-field-p/9783#9783 Answer by JS Milne for Complete discrete valuation rings with residue field ℤ/p JS Milne 2009-12-26T06:32:58Z 2009-12-26T06:32:58Z <p>I don't know what <em>all</em> the cdvr's with residue field $F_p$ are, but local class field theory gives you a huge family. Those that are finite extensions of $Z_p$ are the rings of integers in the totally ramified extensions of $Q_p$. For each prime element $\pi$ of $Z_p$, there is a unique totally ramified extension $K_{\pi}$ of $Q_p$ such that (a) $\pi$ is a norm from every finite subextension of $K_{\pi}$ and (b) the maximal abelian extension of $Q_p$ is obtained by composing $K_{\pi}$ with the unramified extensions of $Q_{p}$. For example, if $\pi=p$, then $K_{\pi}$ is obtained by adjoining the $p^{n}$th roots of $1$ to $Q_{p}$ for all $n$. In general, $K_{\pi}$ is given quite explicitly by Lubin-Tate theory. Different $\pi$ give different field extensions, so the $K_{\pi}$'s are parametrized by the units in $Z_p$. See, for example, Chapter I of my <a href="http://www.jmilne.org/math/CourseNotes/cft.html" rel="nofollow">notes</a></p> http://mathoverflow.net/questions/8809/is-every-subgroup-of-an-algebraic-group-a-stabilizer-for-some-action/8831#8831 Answer by JS Milne for Is every subgroup of an algebraic group a stabilizer for some action? JS Milne 2009-12-14T03:12:13Z 2009-12-14T03:12:13Z <p>In his book "Linear algebraic groups", 6.8, p98, Borel shows that the quotient of an affine algebraic group over a field by an algebraic subgroup exists as an algebraic variety, and he notes p.105 that Weil proved a similar result for arbitrary algebraic groups.</p> http://mathoverflow.net/questions/8599/tips-on-cohomology-for-number-theory/8644#8644 Answer by JS Milne for tips on cohomology for number theory JS Milne 2009-12-12T02:51:47Z 2009-12-12T02:51:47Z <p>Two-cocycles turned up, and were used in the study of central simple algebras before group cohomology was defined. The use of group cohomology greatly simplifies the classification of of central simple algebras over $\mathbb{Q}$, for example. Moreover, once you restate the classification (the Albert-Brauer-Hasse-Noether theorem) more abstractly in terms of group cohomology, you can apply it elsewhere.</p> <p>Homology/cohomology is just like anything else in mathematics. At first it may seem strange, but once you understand it, it becomes familiar.</p> <p>In algebraic number theory and class field theory, you mainly need the cohomology of finite groups, which in fact is a very good place to start, so my advice would be to first study thoroughly the cohomology of finite groups, as for example, in Part 3 of Serre's book Local Fields (Corps Locaux). From there, it is not difficult to understand the cohomology of profinite groups.</p> <p>Loosely speaking, etale cohomology unites the usual cohomology of manifolds with the cohomology of the Galois groups of fields, and is of fundamental importance in arithmetic geometry.</p> http://mathoverflow.net/questions/8324/what-does-linearly-disjoint-mean-for-abstract-field-extensions/8328#8328 Answer by JS Milne for What does "linearly disjoint" mean for abstract field extensions? JS Milne 2009-12-09T07:43:18Z 2009-12-09T18:06:24Z <p>Regard all the fields as subfields of the algebraic closure of K (or K(X)). More precisely, choose an algebraic closure of K and form $k^{1/p^{\infty}}$ inside of it.</p> <p>Added: Linear disjointness is <em>only</em> defined for subfields of some big field. If you choose a different algebraic closure of K, then an isomorphism from it to the first will carry $k^{1/p^{\infty}}$ onto $k^{1/p^{\infty}}$ (with my definition), so you get an isomorphic situation. A similar remark applies to the second example (take $\bar k$ to be the algebraic closure of k in an algebraic closure of K(X)). This is why some authors don't bother to make this explicit (judging by this discussion, they should). </p> <p>There is no ambiguity.</p> http://mathoverflow.net/questions/42272/are-there-any-good-computer-programs-for-drawing-algebraic-curves Comment by JS Milne JS Milne 2010-10-18T13:36:46Z 2010-10-18T13:36:46Z If you have access to Mathematica and it produces the diagram you want, fine, but for more control use gnuplot+tikz. http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts Comment by JS Milne JS Milne 2010-10-17T20:07:13Z 2010-10-17T20:07:13Z Brauer groups and cohomology are certainly overkill for Wedderburn's theorem: if $D$ is a finite division algebra and $L$ is a maximal subfield, then the Noether-Skolem theorem shows that the multiplicative group of $D$ is a union of conjugates of that of $L$; hence $D$=$L$. http://mathoverflow.net/questions/42126/chapters-1-4-of-the-artin-tate-notes-on-class-field-theory Comment by JS Milne JS Milne 2010-10-14T14:45:47Z 2010-10-14T14:45:47Z As far as I know, the notes for the first part of the seminar were never written up. Lang missed this part of the seminar because he started as a philosophy student. http://mathoverflow.net/questions/41253/who-fixed-the-topology-on-ideles/41332#41332 Comment by JS Milne JS Milne 2010-10-07T00:30:00Z 2010-10-07T00:30:00Z Yes, I think the answer is Weil. In his short 1936 paper &quot;Remarques sur des resultats recent de C. Chevalley&quot; he complains about Chevalley's topology, and writes down other constructions and topologies (and mentions Grossencharacters) but doesn't write down anything immediately recognizable (to me) as the ideles. In his 1951 paper (J. Math. Soc. Japan) he defines without comment the ideles with the natural (modern) topology. That is also where he wrote &quot;La recherche d'une interpretation pour C_k ... me semble constituer l'un des problemes fondamentaux...&quot; See also his Commentaries in his CW. http://mathoverflow.net/questions/4422/is-a-torsion-free-abelian-group-finitely-generated-if-all-of-its-localizations-a/40805#40805 Comment by JS Milne JS Milne 2010-10-03T00:10:22Z 2010-10-03T00:10:22Z Have you seen the erratum <a href="http://www.jmilne.org/math/CourseNotes/errata.html#AV" rel="nofollow">jmilne.org/math/CourseNotes/errata.html#AV</a> ? http://mathoverflow.net/questions/39485/are-there-motives-which-do-not-or-should-not-show-up-in-the-cohomology-of-any-s Comment by JS Milne JS Milne 2010-09-21T13:09:16Z 2010-09-21T13:09:16Z Blasius has pointed out that the naive generalization of the modularity conjecture fails --- there exist elliptic curves over number fields that are not quotients of the albanese of any Shimura variety --- but I don't know of any reason why the more general version (4) can't be true. (Blasius 2004 MR2058605). http://mathoverflow.net/questions/38356/what-heuristic-evidence-is-there-concerning-the-unboundedness-or-boundedness-of-m Comment by JS Milne JS Milne 2010-09-11T13:13:44Z 2010-09-11T13:13:44Z Cassels said it best (JLMS 1966, p.257):... it has been widely conjectured [on the basis of calculations] that there is an upper bound for the rank depending only on the groundfield. This seems to me implausible because the theory makes it clear that an abelian variety can only have high rank if it is defined by equations with very large coefficients. . (For there must be a lot of alternative factorizations to be possible in the arguments of &#167;24.) http://mathoverflow.net/questions/38168/is-the-category-of-commutative-group-schemes-abelian/38181#38181 Comment by JS Milne JS Milne 2010-09-10T12:28:10Z 2010-09-10T12:28:10Z Thanks - fixed. http://mathoverflow.net/questions/37246/is-there-a-classification-of-embeddings-of-sl-2-into-sp-6-as-algebraic-groups-o/37256#37256 Comment by JS Milne JS Milne 2010-09-01T13:08:53Z 2010-09-01T13:08:53Z Assume the base field has characteristic zero. The category of representations of a semisimple Lie algebra is Tannakian, and the algebraic group attached to this category is the simply connected algebraic group with the given Lie algebra. A homomorphism of Lie algebras defines a tensor functor of the Tannakian categories, and hence a homomorphism of the corresponding simply connected algebraic groups. http://mathoverflow.net/questions/37214/why-arent-there-more-classifying-spaces-in-number-theory/37259#37259 Comment by JS Milne JS Milne 2010-08-31T14:29:12Z 2010-08-31T14:29:12Z Continued: It was then found that $H^{1}$ coincided with the group of crossed homomorphisms modulo principal crossed homomorphisms, and $H^{2}$ with the group of equivalence classes of &quot;factor sets&quot;, which had been introduced much earlier (e.g., I. Schur, \&quot;{U}ber die Darstellung der endlichen$\ldots$ , 1904; O. Schreier, \&quot;{U}ber die Erweiterungen von Gruppen, 1926; R. Brauer, \&quot;{U}ber Zusammenh\&quot;{a}nge$\ldots$ , 1926). For more on the history, see MacLane 1978 (Origins of the cohomology of groups. Enseign. Math. (2) 24 (1978), no. 1-2, 1--29. MR0497280). http://mathoverflow.net/questions/37214/why-arent-there-more-classifying-spaces-in-number-theory/37259#37259 Comment by JS Milne JS Milne 2010-08-31T14:27:06Z 2010-08-31T14:27:06Z From my Class Field Theory Notes p86. In the mid-1930s, Hurewicz showed that the homology groups of an &quot;aspherical space&quot; $X$ depend only on the fundamental group $\pi$ of the space. Thus one could think of the homology groups $H_{r}(X,\mathbb{Z})$ of the space as being the homology groups $H_{r}(\pi,\mathbb{Z})$ of the group $\pi$. It was only in the mid-1940s that Hopf, Eckmann, Eilenberg, MacLane, Freudenthal and others gave purely algebraic definitions of the homology and cohomology groups of a group $G$. http://mathoverflow.net/questions/35746/inaccessible-cardinals-and-andrew-wiless-proof Comment by JS Milne JS Milne 2010-08-16T18:48:35Z 2010-08-16T18:48:35Z The Stack Project develops a huge amount of Grothendieck style mathematics, including a lot of etale cohomology, using only ZFC (specifically, NOT using universes). If anyone has any doubt that this can be done, I suggest that they look at it. http://mathoverflow.net/questions/35746/inaccessible-cardinals-and-andrew-wiless-proof Comment by JS Milne JS Milne 2010-08-16T13:41:10Z 2010-08-16T13:41:10Z Actually, Cosmonut misquoted the article by leaving out the rest of the statement: &quot;But there is a general consensus among mathematicians that this was just a convenient short cut rather than a logical necessity. With a little work, Wiles's proof should be translatable into Peano arithmetic or some slight extension of it.&quot; http://mathoverflow.net/questions/33808/how-connected-are-you Comment by JS Milne JS Milne 2010-07-29T18:10:46Z 2010-07-29T18:10:46Z MR collaboration numbers aren't accurate. For example, MR claims my collaboration number with Gauss is an improbable 6. They get that number by claiming Landau and Riemann are coauthors and Riemann and Gauss are coauthors because works by them were included together in reprint collections. Similarly, they incorrectly give my Erdos number as 3 because they count me as a co-author of Gerardin when we only published articles in the same collection. http://mathoverflow.net/questions/30453/does-the-tannaka-krein-theorem-come-from-an-equivalence-of-2-categories Comment by JS Milne JS Milne 2010-07-07T03:38:19Z 2010-07-07T03:38:19Z @Theo et al. --- well, you could try looking up the definition in Cat&#233;gories Tannakiennes (Saavedra 1972, Deligne 1990) or Tannakian Categories (Deligne and ... 1982, Breen 1994) or .. A Tannakian category over a field $k$ is neutral if it <i>admits</i> a fibre functor over $k$. In general, it only admits a fibre functor over an extension of $k$. There are various expressions of Tannaka duality in 2-category terms in Saavedra, e.g., III 2.3.2, p180.