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bio website math.uchicago.edu/~emerton
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age 42
visits member for 4 years, 8 months
seen Jun 25 at 18:53

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
Jan
9
awarded  Great Answer
Dec
29
awarded  Yearling
Dec
16
awarded  Guru
Dec
4
awarded  Nice Answer
Dec
2
comment Is there an intuitive reason for Zariski's main theorem?
Dear Roy, This is a really nice right up. Best wishes, Matt
Nov
30
awarded  Good Answer