User anon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:50:28Z http://mathoverflow.net/feeds/user/22295 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116010/complex-multiplication/116018#116018 Answer by anon for complex multiplication anon 2012-12-10T21:00:12Z 2012-12-10T21:00:12Z <p>There are a number of definitions of complex multiplication in the literature.</p> <p>(a) Shimura says that an abelian variety of dimension g has complex multiplication if its endomorphism algebra $End(A)\otimes Q$ contains a field of degree 2g.</p> <p>(b) Deligne et al. say an abelian variety has complex multiplication if it is a product of abelian varieties with complex multiplication in the sense of Shimura (equivalently, but better, if its Mumford-Tate group is a torus).</p> <p>(c) Classical algebraic geometers say an abelian variety has complex multiplication if it is acted on by an order in a CM field.</p> <p>With definition (a), the field is automatically a CM field in characteristic zero, but not otherwise. (A CM field is a quadratic totally imaginary extension of a totally real field.)</p> http://mathoverflow.net/questions/115953/picx-l0-in-terms-of-h-etx-mu-ln/115957#115957 Answer by anon for $Pic(X)/l=0$ in terms of $H^*_{et}(X,\mu_{l^n})$? anon 2012-12-10T07:52:02Z 2012-12-10T07:52:02Z <p>$Pic(X)$ mod $l$ injects into $H^2(X,\mu_l)$ with cokernel the group of elements of order $l$ in the Brauer group of $X$. Depending on your $X$, the Brauer group may be known, or it may be as mysterious as a Tate-Shafarevich group. For example, for a complete smooth surface over a finite field, the Brauer group is conjectured to be finite, but this is not known (it's been proved to be equivalent to the Tate conjecture for the surface).</p> http://mathoverflow.net/questions/114626/fundamental-motivation-for-several-complex-variables/114689#114689 Answer by anon for Fundamental motivation for several complex variables anon 2012-11-27T18:46:17Z 2012-11-27T18:46:17Z <p>Siegel developed the theory of holomorphic functions of several complex variables in order to study hermitian symmetric spaces (especially the Siegel modular space).</p> http://mathoverflow.net/questions/56100/why-does-tates-conjecture-imply-semisimplicity-of-crystalline-frobenius/111644#111644 Answer by anon for Why does Tate's conjecture imply semisimplicity of crystalline Frobenius? anon 2012-11-06T14:18:19Z 2012-11-06T14:18:19Z <p>Actually, the proof of Remark 8.6 is elementary. Let $\alpha$ be an endomorphism of vector space $W$, and suppose that it doesn't act semisimply. Then there exists a vector $e$ and a scalar $a$ such that $(\alpha-a)^2e=0$ but $(\alpha-a)e\neq 0$, say $\alpha e=ae+t$ with $t\neq 0$ (after possibly extending scalars). In the situation of 8.6, we have another vector space $W$, endomorphism $\beta$, and a vector $f$ such that $\beta f=a^{-1}f$. Now one checks that $(\alpha\otimes\beta-1)^2(e\otimes f)=0$ but $(\alpha\otimes\beta-1)(e\otimes f)\neq 0$, contradicting the semisimplicity of $\alpha\otimes\beta$ at $1$.</p> http://mathoverflow.net/questions/106660/kroneckers-jugendtraum-for-real-quadratic-fields/106667#106667 Answer by anon for Kronecker's Jugendtraum for real quadratic fields? anon 2012-09-08T12:42:14Z 2012-09-08T13:47:36Z <p>For totally real quadratic fields, Stark's conjecture (still a conjecture) gives an answer. I quote:</p> <p>In the case that $k$ is totally real, Tate (1984, 3.8) determines the subfield they generate; for example, when $[K:Q]=2$, they generate the abelian closure of $k$ in $\mathbb{R}$. This has implications for Hilbert's 12th problem. To paraphrase Tate (ibid. p.95):</p> <p>"If the conjecture $\mathrm{St}(K/k,S)$ is true in this situation, then the formula $\varepsilon=\exp(-2\zeta^{\prime}(0,1))$ gives generators of abelian extensions of $k$ that are special values of transcendental functions. Finding generators of class fields of this shape is the vague form of Hilbert's 12th problem, and the Stark conjecture represents an important contribution to this problem. However, it is a totally unexpected contribution: Hilbert asked that we discover the functions that play, for an arbitrary number field, the same role as the exponential function for $\mathbb{Q}{}$ and the elliptic modular functions for a quadratic imaginary field. In contrast, Stark's conjecture, by using $L$-functions directly, bypasses the transcendental functions that Hilbert asked for. Perhaps a knowledge of these last functions will be necessary for the proof of Stark's conjecture."</p> <p>Remarkably, $St(K/k,S)$ is useful for the explicit computation of class fields, and has even been incorporated into the computer algebra system PARI/GP.</p> <p>That was the situation in 1985, but, as Lemmermeyer notes, Darmon and others have been studying these questions.</p> http://mathoverflow.net/questions/94776/analytic-isomorphisms-above-two-etale-maps/94898#94898 Answer by anon for Analytic isomorphisms above two etale maps anon 2012-04-23T01:32:13Z 2012-04-23T01:32:13Z <p>This (from SGA 1) is the proof that the functor is fully faithful: We may suppose that $X$ is connected. To give an $X$-morphism $Y\rightarrow Y^{\prime}$ is to give a section to $Y\times_{X}Y^{\prime}\rightarrow Y$, which is the same as to give a connected component $\Gamma$ of $Y\times_{X}Y^{\prime}$ such that the morphism $\Gamma\rightarrow X$ induced by the projection $Y\times_{X}Y^{\prime}\rightarrow Y$ is an isomorphism. But the connected components of $Y\times_{X}Y^{\prime}$ coincide with the connected components of $Y^{an}\times_{X^{an}}Y^{\prime an}$, and if $\Gamma$ is a connected component of $Y\times_{X}Y^{\prime}$, then the projection $\Gamma\rightarrow X$ is an isomorphism if and only if $\Gamma^{an}\rightarrow Y^{an}$ is an isomorphism.</p> http://mathoverflow.net/questions/91810/non-trivial-facts-about-primes-coming-out-of-algebraic-number-theory/91822#91822 Answer by anon for Non-trivial facts about primes coming out of Algebraic Number Theory anon 2012-03-21T13:28:24Z 2012-03-21T13:28:24Z <p>Algebraic number theory solves the ancient/long-standing problem of providing a proof of quadratic reciprocity that those of us who are not Gauss can actually remember. Let p be an odd prime, and let K be the field obtained from Q by adjoining a primitive pth root of 1. Then K contains a unique quadratic extension of Q, which one sees easily is that obtained by adjoining a square root of p or -p according as p is congruent to 1 mod 4 or not. Now let q be a second odd prime. By computing the action of the Frobenius at q on the unique quadratic subfield of K in two different ways, one obtains the main statement of quadratic reciprocity.</p> http://mathoverflow.net/questions/116625/recover-trace-of-l-adic-sheaves-defined-over-an-extension Comment by anon anon 2012-12-17T20:08:04Z 2012-12-17T20:08:04Z Since the Frobenius for $\mathbb{F}_n$ is the nth power of the Frobenius for $\mathbb{F}$, you seem to be asking: what can you tell about (the trace of) a matrix from knowing (the trace of) its nth power? http://mathoverflow.net/questions/116639/hodge-numbers-of-reduction-mod-p Comment by anon anon 2012-12-17T19:56:23Z 2012-12-17T19:56:23Z For 2.), the answer is NO! and for 3.) the answer is the smooth-proper base change theorem, which gives an isomorphism on the cohomology groups (see any book on etale cohomology). http://mathoverflow.net/questions/116251/mathematician-trying-to-learn-string-theory/116305#116305 Comment by anon anon 2012-12-14T21:27:06Z 2012-12-14T21:27:06Z Candelas once told me that, when he asked Atiyah how to learn algebraic geometry, Atiyah responded: &quot;You can't&quot;. At first Candelas thought Atiyah was making a statement about him personally, but what he was saying is that algebraic geometry is such a large subject that understanding it is a full-time occupation. I'm sure string theory is the same. Therein lies the problem, and it doesn't help that when mathematicians and physicists talk about the same object they often do so in very different ways. http://mathoverflow.net/questions/116318/elementary-proof-of-mordells-theorem Comment by anon anon 2012-12-13T21:01:37Z 2012-12-13T21:01:37Z There may be proofs of Mordell's theorem that don't use any algebraic number theory (perhaps Mordell's original proof doesn't), but I doubt whether you would want to read them. You could try looking in Mordell's book or at the article Cassels, J. W. S., Mordell's finite basis theorem revisited. Math. Proc. Cambridge Philos. Soc. 100 (1986), no. 1, 31–41. http://mathoverflow.net/questions/114974/algebraic-groups-and-their-lie-algebras Comment by anon anon 2012-12-02T18:31:20Z 2012-12-02T18:31:20Z This is not a research level question. The answer is complicated, depending on the hypotheses, but can be found in books and in online notes. Briefly, over a field of characteristic zero, the category of representations of a semisimple Lie algebra is equal to that of the associated simply connected semisimple algebraic group, but otherwise the categories usually differ (e.g., for $\mathbb{G}_a$). http://mathoverflow.net/questions/115112/how-does-modern-number-theory-contribute-to-further-understanding-of-mathbbn Comment by anon anon 2012-12-02T00:30:53Z 2012-12-02T00:30:53Z Enough brilliant mathematicians have studied the Riemann hypothesis without success that we can be reasonably confident that a proof will only come by looking at it in some broader context. What that broader context will be, alas, we don't know. http://mathoverflow.net/questions/43820/extremely-messy-proofs/60127#60127 Comment by anon anon 2012-12-01T19:11:03Z 2012-12-01T19:11:03Z Messy? yes. 62 pages? no. Hilbert's proof only takes about 12 pages of the article. http://mathoverflow.net/questions/114796/is-it-known-if-the-absolute-galois-group-is-divisible Comment by anon anon 2012-11-28T18:39:31Z 2012-11-28T18:39:31Z Quotients of divisible groups are divisible, which is certainly not true of the absolute Galois group of Q. http://mathoverflow.net/questions/114626/fundamental-motivation-for-several-complex-variables Comment by anon anon 2012-11-27T07:37:05Z 2012-11-27T07:37:05Z The theory of several complex variables is such a rich and beautiful subject I have trouble understanding how anyone can ask the question. For example, the the theory of symmetric Hermitian spaces is quite interesting. http://mathoverflow.net/questions/114400/examples-of-injective-morphisms-which-are-not-universally-injective/114409#114409 Comment by anon anon 2012-11-25T20:23:28Z 2012-11-25T20:23:28Z &quot;Universally injective&quot; is equivalent to &quot;injective and all the maps on the residue fields are radicial&quot; EGA I, 3.7.1. http://mathoverflow.net/questions/114328/langlands-paper-on-representations-of-abelian-algebraic-groups Comment by anon anon 2012-11-24T19:31:49Z 2012-11-24T19:31:49Z There are several expositions of Langlands paper in the literature, for example, Labesse, J.-P. Cohomologie, $L$-groupes et fonctorialit&#233;. (French) [Cohomology, $L$-groups and functoriality] Compositio Math. 55 (1985), no. 2, 163--184. http://mathoverflow.net/questions/113858/is-there-an-algebraic-geometry-analogue-of-the-closed-graph-theorem/113860#113860 Comment by anon anon 2012-11-19T21:08:27Z 2012-11-19T21:08:27Z Instead of assuming that the projection from $\Gamma$ to $X$ be proper, doesn't it suffice to assume that $\Gamma$ be irreducible? http://mathoverflow.net/questions/112698/groups-becoming-algebraic-groups Comment by anon anon 2012-11-17T19:36:17Z 2012-11-17T19:36:17Z What's the motivation for the question? Even if the answer were yes (which I doubt), it wouldn't make it easier to verify that a variety is an algebraic group. http://mathoverflow.net/questions/112536/is-there-something-interesting-in-the-uniqueness-condition-for-a-sheaf Comment by anon anon 2012-11-16T03:54:09Z 2012-11-16T03:54:09Z Well, all &quot;natural&quot; presheafs are presheafs of functions, for which uniqueness is automatic. However, the presheaf quotient of a sheaf by a subpresheaf need not satisfy uniqueness. For example, consider at the presheaf quotient of the sheaf of locally constant functions on a space by the subpresheaf of constant functions. http://mathoverflow.net/questions/111760/a-natural-way-of-thinking-of-the-definition-of-an-artin-l-function Comment by anon anon 2012-11-08T00:46:39Z 2012-11-08T00:46:39Z 4. It behaves correctly with respect to inducing characters.