User anon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:34:55Z http://mathoverflow.net/feeds/user/16556 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74073/the-first-complete-proof-of-the-kronecker-weber-theorem/74112#74112 Answer by anon for The first complete proof of the Kronecker-Weber theorem anon 2011-08-31T00:45:16Z 2011-08-31T11:54:32Z <p>Weber gave the first complete proof, based partly on ideas of Kronecker. It's true that there are errors in Weber's proofs, but nothing that he couldn't have fixed if they had been pointed out to him. Kronecker and Weber had some of the most original and magnificently beautiful ideas in mathematics --- let the lesser mathematicians fuss over the details.</p> http://mathoverflow.net/questions/73879/canonical-liftings-of-endomorphisms-of-ordinary-abelian-varieties/73963#73963 Answer by anon for Canonical liftings of endomorphisms of ordinary abelian varieties anon 2011-08-29T12:18:26Z 2011-08-29T12:18:26Z <p>The canonical reference is Messing, LNM 264 1972, Chapter V, 3.4, p174.</p> http://mathoverflow.net/questions/72993/on-sufficient-conditions-on-an-analytic-map-to-be-algebraicregular/73016#73016 Answer by anon for On sufficient conditions on an analytic map to be algebraic(=regular) anon 2011-08-16T22:00:07Z 2011-08-22T20:39:19Z <p>Borel (1972, J. Diffl. Geometry) proved that $f$ is always algebraic if $Y$ is the quotient of a bounded symmetric domain by a torsion-free arithmetic subgroup. This is a super-generalization of your example 3 (the quotient of the complex upper half plane by $\Gamma(2)$ is isomorphic to the projective line minus three points). The proof uses a generalization of work of Kwack plus the resolution of singularities. </p> <p>Added: Kwack (1969) generalized the big Picard theorem by proving that any holomorphic map from the punctured unit disk into a hyperbolic complex space can be extended holomorphically to the whole unit disk. [A reduced complex space is said to be hyperbolic if the Kobayashi pseudodistance is a distance (Kobayashi 1967).]</p> <p>Borel 1972 replaced the punctured disk in Kwack's theorem with a product of punctured disks and disks. </p> <p>Resolution of singularities allows you to realize a smooth algebraic variety as an open subvariety of a smooth projective variety in such a way that the boundary is a divisor with normal crossings (hence analytically a product of punctured disks and disks).</p> <p>These statements sometimes allow you to extend your map to an analytic map of projective varieties, where you can apply Chow's theorem to prove that it is regular.</p> <p>References:</p> <p>Borel, Armand. Some metric properties of arithmetic quotients of symmetric spaces and an extension theorem. J. Differential Geometry 6 (1972), 543--560.</p> <p>Kwack, Myung H., Generalization of the big Picard theorem. Ann. of Math. (2) 90 1969 9--22. </p> <p>Kobayashi, Shoshichi, Invariant distances on complex manifolds and holomorphic mappings. J. Math. Soc. Japan 19 1967 460--480. </p> http://mathoverflow.net/questions/73006/tensor-product-in-k-linear-categories-how-canonical/73021#73021 Answer by anon for tensor product in k-linear categories how canonical anon 2011-08-16T23:26:17Z 2011-08-16T23:26:17Z <p>You can make $A\otimes V$ canonical by defining it to be a family of objects indexed by the bases of $V$. But that's a big mess, so you probably don't want to do it. If all you want is to make $V\mapsto A\otimes V$ into a genuine functor, you can do that by choosing a suitable adjoint to the inclusion functor from vector spaces of the form $k^n$ to the category of all vector spaces. </p> http://mathoverflow.net/questions/72874/how-would-a-motivic-proof-of-the-riemann-hypothesis-over-finite-fields-go/72879#72879 Answer by anon for How would a motivic proof of the Riemann hypothesis over finite fields go? anon 2011-08-14T18:26:03Z 2011-08-14T20:16:19Z <p>The standard conjectures imply directly that the category of motives over the finite field $\mathbb{F}_q$ is a polarizable (hence semisimple) Tannakian category. Using only that, we have the following result. </p> <p>PROPOSITION: Let $X$ be a motive of weight $m$ over $\mathbb{F}_q$, and let $\alpha\mapsto\alpha^t$ be the involution of $\mathrm{End}(X)$ defined by a Weil form $\varphi$. The following statements hold for the Frobenius endomorphism $\pi =\pi_X$ of $X$:</p> <p>(a) $\pi\cdot\pi^t=q^m$; hence $\mathbb{Q}[\pi ]$ is stable under the involution $\alpha\mapsto\alpha^t$;</p> <p>(b) $\mathbb{Q}[\pi ]\subset\mathrm{End}(X)$ is a product of fields;</p> <p>(c) for every homomorphism $\rho\:\mathbb{Q}[\pi ]\rightarrow \mathbb{C}$, $\rho (\pi^t)=\iota (\rho\pi )$, and $|\rho\pi |=q^{m/2}$. ($\iota$ is complex conjugation)</p> <p>PROOF: (a) By definition, $\varphi$ is a morphism $X\otimes X\to T^{\otimes (-m)}$ ($T$ is the Tate object). It is invariant under $\pi$, and so $$\varphi (\pi x,\pi y)=\pi (\varphi (x,y))=q^m\varphi (x,y)=\varphi (x,q^my).$$ But $\varphi (\pi x,\pi y)=\varphi (x,\pi^t\pi y)$, and because $\varphi$ is nondegenerate, this implies that $\pi^t\cdot\pi =q^m$. Therefore $\mathbb{Q}[\pi ]$ is stable under $\alpha\mapsto\alpha^t$, and we obtain (a). </p> <p>(b) Let $R$ be a commutative subalgebra of $\mathrm{End}(X)$ stable under $\alpha\mapsto\alpha^t$, and let $r$ be a nonzero element of $R$. Then $s=rr^t\neq 0$ because $\mathrm{Tr}(rr^t)>0$. As $s^t=s$, $\mathrm{Tr}(s^2)=\mathrm{Tr}(ss^t)>0$, and so $s^2\neq 0$. Similarly $s^4\neq 0$, and so on, which implies that $s$ is not nilpotent, and so neither is $r$. Thus $R$ is a finite-dimensional commutative $\mathbb{Q}$-algebra without nonzero nilpotents, and the only such algebras are products of fields. </p> <p>(c) In an abuse of notation, we set $\mathbb{R}[\pi ]=\mathbb{R}\otimes_{ \mathbb{Q}}\mathbb{Q}[\pi ]$. As in (b), this is a product of fields stable under $\alpha\mapsto\alpha^t$. This involution permutes the maximal ideals of $\mathbb{R}[\pi ]$ and, correspondingly, the factors of $\mathbb{R}[\pi ]$. If the permutation were not the identity, then $\alpha\mapsto\alpha^ t$ would not be a positive involution. Therefore each factor of $\mathbb{R}[\pi ]$ is stable under the involution. The only involution of $\mathbb{R}$ is the identity map (= complex conjugation), and the only positive involution of $\mathbb{C}$ is complex conjugation. Therefore we obtain the first statement of (c), and the second then follows from (a). </p> <p>This (conjectural) proof of the Riemann hypothesis for motives is very close to Weil's original proof for abelian varieties (Weil 1940). </p> http://mathoverflow.net/questions/71660/why-would-the-category-of-motives-be-tannakian/71693#71693 Answer by anon for Why would the category of Motives be Tannakian? anon 2011-07-30T23:29:08Z 2011-07-30T23:34:47Z <p>Actually, the category of motives isn't equivalent to the category of representations of an affine group scheme except in characteristic zero, and even there the equivalence depends on the choice of a fibre functor.</p> <p>The functor to motives is supposed to be a universal cohomology theory. Certainly, one would like the target of a cohomology theory to be at least tannakian.</p> <p>If you assume the Hodge conjecture, then the affine group scheme attached to the category of abelian motives over $\mathbb{C}$ (that generated by abelian varieties) is more-or-less known --- at least its algebraic quotients are classified.</p> <p>If you assume the Tate conjecture, then the affine groupoid attached to the category of motives over an algebraic closure of $\mathbb{F}_p$ is more-or-less known.</p> <p>In the general case nothing is known except that the group is VERY BIG --- for example, over $\mathbb{C}$ it has uncountably many distinct quotients isomorphic to PGL(2).</p> http://mathoverflow.net/questions/70607/exterior-powers-in-tensor-categories/70728#70728 Answer by anon for Exterior powers in tensor categories anon 2011-07-19T12:04:37Z 2011-07-19T12:04:37Z <p>Deligne (Categories Tannakiennes, 1990, p165) defines it to be the image of the antisymmetrisation $a=\sum(-1)^{\epsilon(\sigma)}\sigma\colon X^{\otimes n}\rightarrow X^{\otimes n}$.</p> http://mathoverflow.net/questions/70655/discrete-compact-duality-for-nonabelian-groups/70691#70691 Answer by anon for Discrete-compact duality for nonabelian groups anon 2011-07-19T00:14:01Z 2011-07-19T00:14:01Z <p>There is a duality between compact groups and neutral Tannakian categories equipped with a symmetric polarization --- see Deligne and Milne, Tannakian Categories 4.27, 2.33.</p> http://mathoverflow.net/questions/74853/analogue-of-shafarevich-oggs-theorem-over-complex-numbers Comment by anon anon 2011-09-08T19:08:55Z 2011-09-08T19:08:55Z Over a smooth complex algebraic variety, $R^1f_*$ is an equivalence from abelian schemes to polarizable variations of integral Hodge structures of type {(-1,0),(-1,0)}. Now there are lots of theorems about extending variations of Hodge structures in terms of monodromy. http://mathoverflow.net/questions/74073/the-first-complete-proof-of-the-kronecker-weber-theorem/74112#74112 Comment by anon anon 2011-08-31T19:31:45Z 2011-08-31T19:31:45Z We are talking about Weber's proof, not Kronecker's. I'm strongly objecting to Gray's statement &quot;it is also not true that the first proof is due to Weber&quot; (and also to the sort of mentality that leads to such statements). I also object to his statement that &quot;Weber's mistake was not found until 1979&quot;. How would he know? Not everyone writes a paper every time they find a mistake. http://mathoverflow.net/questions/74073/the-first-complete-proof-of-the-kronecker-weber-theorem/74112#74112 Comment by anon anon 2011-08-31T15:14:37Z 2011-08-31T15:14:37Z Well, they are questioning whether Weber proved the theorem, and even whether it should be called the Kronecker-Weber theorem. If you look hard enough, you can find errors in a great many proofs: the giants prove the great theorems; the pygmies fix their proofs. http://mathoverflow.net/questions/73017/what-is-the-relationship-between-the-finiteness-of-the-tate-shafarevich-group-and Comment by anon anon 2011-08-16T23:34:38Z 2011-08-16T23:34:38Z See Tate's Bourbaki talk: On the conjectures of Birch and Swinnerton-Dyer and a geometric analog. S&#233;minaire Bourbaki, Vol. 9, Exp. No. 306, 415--440. Available at numdam.org. http://mathoverflow.net/questions/71435/which-conjectures-only-need-the-grand-riemann-hypothesis-to-become-genuine-theore/71472#71472 Comment by anon anon 2011-08-12T12:42:11Z 2011-08-12T12:42:11Z Yes, I think they are using fancy words they don't fully understand. Probably they mean: A resource for both the enthusiastic amateur and the professional. http://mathoverflow.net/questions/72288/proofs-of-mordell-weil-theorem Comment by anon anon 2011-08-08T17:01:00Z 2011-08-08T17:01:00Z For elliptic curves over $\mathbb{Q}$, the proof is quite elementary, especially if you assume that the points of order 2 are rational. For elliptic curves over a number field, you need to know the finiteness of the class number and the finite generation of the group of units (basic facts in algebraic number theory). For abelian varieties, you need to know rather a lot of algebraic geometry. You should be able to find what you want in online lecture notes. http://mathoverflow.net/questions/72288/proofs-of-mordell-weil-theorem/72292#72292 Comment by anon anon 2011-08-08T00:34:04Z 2011-08-08T00:34:04Z The standard proof of the weak Mordell theorem in Silverman unnecessarily translates the &quot;putative finiteness of $E(L)/nE(L)$ into a statement about certain field extensions of $L$&quot;. There are simpler proofs that directly relate the finiteness to the finiteness of the class number of $L$ and the finite generation of its units. http://mathoverflow.net/questions/71937/principal-bundles-in-the-etale-and-zariski-topology-and-extensions-of-the-structu Comment by anon anon 2011-08-03T12:58:11Z 2011-08-03T12:58:11Z @Jason, sorry, I don't understand your question. The argument Angelo gives is essentially that in my comment, but with the cohomology removed. http://mathoverflow.net/questions/71937/principal-bundles-in-the-etale-and-zariski-topology-and-extensions-of-the-structu Comment by anon anon 2011-08-02T23:51:43Z 2011-08-02T23:51:43Z Let $f\colon X_{et}\to X_{zar}$ be the identity map. For a special group $G$, $R^1f_*G=0$ and so the Leray spectral sequence shows that the map from $H^2_{zar}$ to $H^2_{et}$ is injective in the commutative case, which is what you want in order to apply the 5-lemma. It is surely also true in the noncommutative case, but writing down a proof will be more complicated. http://mathoverflow.net/questions/71660/why-would-the-category-of-motives-be-tannakian/71693#71693 Comment by anon anon 2011-07-31T00:39:58Z 2011-07-31T00:39:58Z The concept of a motive emerged slowly in Grothendieck's mind in the 1960's. Later he had a student Saavedra develop the theory of Tannakian categories for his thesis. Although there are many naturally occurring examples of Tannakian categories in algebraic geometry, I think for Grothendieck the principal (conjectural) example was motives. To some extent Tannakian categories were developed to give the correct formalism for motives. The idea that the category of motives should be describable by some monster affine group scheme (or groupoid scheme) was probably an early part of G's thinking.