bio | website | |
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visits | member for | 3 years, 3 months |
seen | Dec 18 '12 at 0:11 | |
stats | profile views | 124 |
Mar 9 |
awarded | Nice Answer |
Mar 22 |
awarded | Yearling |
Dec 17 |
comment |
recover trace of l-adic sheaves defined over an extension
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? |
Dec 17 |
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Hodge numbers of reduction mod $p$
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). |
Dec 14 |
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Mathematician trying to learn string theory
Candelas once told me that, when he asked Atiyah how to learn algebraic geometry, Atiyah responded: "You can't". 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. |
Dec 13 |
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Elementary proof of Mordell's theorem
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. |
Dec 10 |
answered | complex multiplication |
Dec 10 |
answered | $Pic(X)/l=0$ in terms of $H^*_{et}(X,\mu_{l^n})$? |
Dec 2 |
comment |
algebraic groups and their Lie algebras
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 $\mathbb{G}_a$). |
Dec 2 |
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How does “modern” number theory contribute to further understanding of $\mathbb{N}$?
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. |
Dec 1 |
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Extremely messy proofs
Messy? yes. 62 pages? no. Hilbert's proof only takes about 12 pages of the article. |
Nov 28 |
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Is it known if the absolute Galois group is “divisible”?
Quotients of divisible groups are divisible, which is certainly not true of the absolute Galois group of Q. |
Nov 27 |
answered | Fundamental motivation for several complex variables |
Nov 27 |
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Fundamental motivation for several complex variables
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. |
Nov 25 |
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Examples of injective morphisms which are not universally injective
"Universally injective" is equivalent to "injective and all the maps on the residue fields are radicial" EGA I, 3.7.1. |
Nov 24 |
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Langlands Paper on representations of abelian algebraic groups
There are several expositions of Langlands paper in the literature, for example, Labesse, J.-P. Cohomologie, $L$-groupes et fonctorialité. (French) [Cohomology, $L$-groups and functoriality] Compositio Math. 55 (1985), no. 2, 163--184. |
Nov 19 |
comment |
Is there an algebraic geometry analogue of the closed graph theorem?
Instead of assuming that the projection from $\Gamma$ to $X$ be proper, doesn't it suffice to assume that $\Gamma$ be irreducible? |
Nov 17 |
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Groups becoming algebraic groups
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. |
Nov 16 |
awarded | Commentator |
Nov 16 |
comment |
Is there something interesting in the uniqueness condition for a sheaf?
Well, all "natural" 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. |