bio | website | math.uchicago.edu/~emerton |
---|---|---|
location | ||
age | 43 | |
visits | member for | 4 years, 11 months |
seen | Jun 25 at 18:53 | |
stats | profile views | 26,274 |
Nov 21 |
awarded | Nice Answer |
Nov 12 |
awarded | Good Answer |
Oct 26 |
awarded | Enlightened |
Oct 26 |
awarded | Nice Answer |
Oct 15 |
awarded | Notable Question |
Oct 8 |
awarded | Populist |
Aug 4 |
awarded | Nice Answer |
Jun 2 |
awarded | Good Answer |
May 23 |
comment |
When are dual modules free?
@GrahamLeuschke: Dear Graham, I think that the first half is okay provided that $M$ is assumed a priori to have finite projective dimension. (One can proceed by induction on the projective dimension.) Regards, |
Apr 30 |
comment |
Current Status on Langlands Program
@plusepsilon.de: Dear plusepsilon.de, Yes, that's right. But to actually construct/study Galois reps. attached to $GL(n)$ automorphic form, you have to work with unitary groups, for which stable conjugacy and conjugacy are distinct. For example, the proof of Sato--Tate is prima facie about proving cases of symmetric power functoriality from $GL(2)$ to $GL(n)$, and so one could imagine that it doesn't involve stabilization or the fundamental lemma. But in fact, the proof for elliptic curves with integral $j$-invariant depended on the fundamental lemma for unitary groups. Regards, |
Apr 30 |
comment |
Explicit calculation of Weil Deligne representations
Dear Hiro, The case of a Tate curve is particularly easy. The $\ell$-adic Tate module is the Kummer extension of $\mathbb Q_{\ell}$ by $\mathbb Q_{\ell}(1)$ corresponding to the Tate parameter $q$ defining the curve. (This follows directly from the description of the points as $\overline{\mathbb Q}_p^{\times}/q^{\mathbb Z}$.) Now just apply the standard recipe to get the Weil--Deligne representation. Regards, |
Apr 30 |
comment |
Current Status on Langlands Program
Dear Joel, This is an impressive summary of the current status! Cheers, |
Apr 30 |
comment |
Current Status on Langlands Program
@plusepsilon.de: Dear plusepsilon.de, Since many results for $GL(n)$ are proved via transfer to unitary groups, stabilization is in fact an issue, even in the theory for $GL(n)$. Regards, |
Apr 14 |
awarded | Enlightened |
Apr 14 |
awarded | Nice Answer |
Apr 14 |
awarded | Nice Answer |
Feb 9 |
comment |
Why are modular forms (usually) defined only for congruence subgroups?
@MarcPalm: Dear Marc, Your statement is not literally true as written. What is true is that there is a vector on which $\Gamma_0(p_v^N)$ acts through some nebentypus character (i.e. some character of the lower right-hand entry, taken modulo $p_v^N$). In classical terms, this means that one can always work on $\Gamma_1(K)$ for $K$ large. This last statement is easy to see directly: the usual matrix $(0 \quad 1, N \quad 0)$ will conjugate $\Gamma(N)$ into $\Gamma_1(N^2)$. Regards, |
Jan 20 |
revised |
Algebraic stacks: limit preserving versus locally of finite presentation
edited tags |
Jan 20 |
revised |
Finite-type Artin Stack over $\mathbb C$
edited tags |
Jan 11 |
comment |
Understanding iterated integrals
Dear Sasha, It is "Le groupe fondamental de la droite projective moins trois points"; this link used to work. (I'm not sure if its broken now, or if there's just a problem with my connection.) Regards, Matthew |