bio | website | jmilne.org/math |
---|---|---|
location | Ann Arbor, MI, USA, and New Zealand. | |
age | ||
visits | member for | 4 years, 6 months |
seen | Oct 18 '10 at 13:35 | |
stats | profile views | 6,138 |
Arithmetic geometry (especially Shimura varieties and abelian varieties).
Apr 17 |
comment |
Why are modular forms (usually) defined only for congruence subgroups?
Certainly not canonical. For example, the curve for Gamma_N classifies elliptic curves with a level N structure. Such objects have a discrete invariant, namely, an Nth root of 1 given by applying the Weil pairing to the basis. To get a connected family, you need to specify the Nth root, so this only makes sense over a field where you have one. Of course, for nonconnected curves, everything is defined over Q. |
Apr 17 |
revised |
Why are modular forms (usually) defined only for congruence subgroups?
Minor fixes. |
Apr 17 |
revised |
What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?
Minor fixes. |
Apr 16 |
answered | Why are modular forms (usually) defined only for congruence subgroups? |
Apr 15 |
awarded | Nice Answer |
Apr 15 |
comment |
What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?
Thanks --- that looks like the correct way to interpret Iwahori's statements (and, yes, I should have assumed k to be algebraically closed). |
Apr 15 |
answered | What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra? |
Apr 15 |
comment |
What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra?
There seems to be an assumption implicit in the question that when you start with an algebraic Lie algebra g, the affine group scheme G you get has Lie algebra g. As @unknown noted, this is far from true (except for semisimple Lie algebras in characteristic zero). So before trying to understand the affine group schemes you get from nonalgebraic Lie algebras, perhaps you should try to understand those you get from algebraic Lie algebras (e.g., the one-dimensional Lie algebra). |
Apr 14 |
awarded | Nice Answer |
Apr 14 |
answered | 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.? |
Apr 12 |
revised |
Why does the naive definition of compactly supported étale cohomology give the wrong answer?
edited body; added 251 characters in body |
Apr 12 |
answered | Why does the naive definition of compactly supported étale cohomology give the wrong answer? |
Mar 16 |
comment |
The algebraicity of Hodge structure map
The maps $\mathbb{S}\to G_{\mathbb{R}}$ are defined over $\mathbb{R}$,not $\mathbb{Q}$. However, one of the axioms (2.1.1.1 in Deligne's Corvallis articles; SV1 in my articles) for Shimura varieties implies that the weight homomorphism $w$ of each $h$ factors through the centre of $G$ (hence through connected centre, which is a torus). As the $h$'s are all conjugate, this implies that the $w$'s are all equal, and that the $w$ is defined over the algebraic closure of $\mathbb{Q}$ (it is a homomorphism between tori defined over $\mathbb{Q}$). |
Mar 16 |
revised |
The algebraicity of Hodge structure map
added 316 characters in body |
Mar 16 |
answered | The algebraicity of Hodge structure map |
Mar 11 |
awarded | Nice Answer |
Mar 11 |
comment |
Books on reductive groups using scheme theory
@Brian: I agree that if you try to directly transfer the proof in the smooth case to the nonsmooth case, you can sometimes run into some very heavy scheme theory, but there are also elementary proofs using Hopf algebras. I learnt this from Waterhouse's book. As Serre pointed out, Hopf algebra proofs don't illuminate, but my strategy is to sketch the geometric argument and write out the Hopf algebra argument (when necessary). I'm only doing things over fields (or rings, when it's just as easy). |
Mar 11 |
answered | Books on reductive groups using scheme theory |
Mar 4 |
awarded | Enlightened |
Mar 4 |
revised |
Why do people think that abelian varieties are the hardest case for the Hodge conjecture?
added 377 characters in body |