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visits | member for | 3 years, 1 month |
seen | Jun 6 '12 at 12:55 | |
stats | profile views | 175 |
Sep 23 |
awarded | Yearling |
Jun 21 |
awarded | Nice Answer |
Apr 22 |
comment |
Analytic isomorphisms above two etale maps
For proofs, see: Grauert and Remmert, Math. Ann. (1958), 245--318; SGA 4, XI.4.3; and SGA 1, XII. If you are willing to use resolution of singularities, the proof is not too hard (see SGA 1). |
Mar 25 |
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Any local algebraic group is birationally equivalent to an algebraic group
It's not a textbook, but the theorem is proved in the setting of schemes in an expose of Artin in SGA 3. Artin also gives a very sketchy proof of the theorem in his first article in Arithmetic Geometry 1986 (Cornell/Silverman) --- I think I would look there. |
Mar 23 |
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Derived Functors Versus Spectral Sequences
"This is usually proved using the Leray spectral sequence" I hope not (Example 1). The hypotheses imply that $f_*$ takes an injective resolution of $F$ to an acyclic resolution of $f_*F$, which can be used to compute its cohomology. This gives the result. |
Mar 18 |
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Why do twists of an algebraic group over k correspond to k-torsors over G
Look at Hom from G to the twist (as a functor). This is a G-torsor. |
Mar 10 |
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Soft(?) algebraic groups question
Well first you need to do it over Q. If H and G are semisimple and G is split over Q, you can take the split form of H. If G is not split, it's a twist of the split form, so you need the cocycle to come from one on H. Looks dubious to me. |
Mar 7 |
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Decomposition of semisimple Lie group into almost simple factors
A semisimple Lie group is a covering of a semisimple algebraic group .... Alternatively, use that its Lie algebra is a product of simple Lie algebras. |
Feb 18 |
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Lie algebras with abelian Cartan subalgebras.
Even simpler, an abelian Lie algebra is a Cartan subalgebra of itself. |
Feb 15 |
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Role of fiber functor monoidal structure in Tannakian bialgebra reconstruction
My guess is the answer is no. Specifically, I'd guess that there exist really different fibre functors that become isomorphic when you forget their monoidal structures. For example, two fibre functors send an object to vector spaces of the same dimension, so they become equal on objects when you replace the category of vector spaces with its skeleton. Perhaps this can be pushed further to show that sometimes (often? always?) two fibre functors will become isomorphic when you forget their monoidal structures. |
Jan 19 |
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Should I write to the referee?
It shouldn't "restart the review procedure" --- the editor will just send the new version of the article to his current referee. |
Jan 13 |
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The historical development of automorphic geometry
"Langlands did quite a good job of suggesting that the Jugendtraum was some sort of wrong turning." Where did he do that? |
Jan 5 |
revised |
Permission to use Online Notes
added 351 characters in body |
Jan 5 |
answered | Permission to use Online Notes |
Jan 5 |
awarded | Commentator |
Jan 5 |
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finite non-commutative local group schemes
Over a field of characteristic $p$, there is an obvious action of $\mathbb{G}_m$ on $\alpha_p$, and hence an action of $\mu_p$ on $\alpha_p$. The semi-direct product is a noncommutative connected group scheme of order $p^2$. |
Jan 3 |
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Topological examples of profinite groups
"I would like to exclude Galois groups". Actually, all profinite groups are Galois groups, so you may be in trouble. More seriously, I agree with KConrad: for most us of a profinite group is a projective limit of finite groups, so it's better to start with that as the definition. |
Dec 20 |
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understanding Milne's article “Duality in the flat cohomology of a surface”
Have you tried looking at the exposition of the theorem in: Berthelot, P. Le théorème de dualité plate pour les surfaces (d'après J. S. Milne). (French) [The flat duality theorem for surfaces (according to J. S. Milne)] Algebraic surfaces (Orsay, 1976--78), pp. 203--237, Lecture Notes in Math., 868, Springer, Berlin-New York, 1981. MR0638601? |
Dec 16 |
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Why is the definition of l-adic sheaves so complicated?
With the naive definition, $H^{1}(X,\mathbb{Z}_{\ell))=\Hom_{\text{continuous}}(\pi_{1}(X),\mathbb{Z}_{\ell})$ with the discrete topology on $\mathbb{Z}_{\ell}$. This is generally zero, because (for nice schemes) $\pi_{1}(X)$ is profinite. With the nonnaive definition, it is Hom with the natural $\ell$-adic topology on $\mathbb{Z}_{\ell}$, which is what you want. |
Nov 22 |
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Constructive proof of algebraic elements forming a subfield
Take the product of all polynomials $X-a'b'$ where $a'$ and $b'$ range over the conjugates of a and b, and use the symmetric function theorem to show that its coefficient lie in $E$. |