41,103 reputation
3127218
bio website math.uchicago.edu/~emerton
location
age 42
visits member for 4 years, 3 months
seen 2 days ago

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
Oct
7
awarded  Caucus
Oct
7
awarded  Constituent
Sep
6
comment What is the “right” universal property of the completion of a metric space?
@PeteL.Clark : Dear Pete, You're welcome. Cheers, Matt
Aug
28
awarded  Nice Answer
Aug
23
awarded  Nice Answer
Aug
12
comment Iwasawa Decomposition for Matrices
Dear Vishal, This is the Gram--Schmidt process for turning a basis of $\mathbb R^n$ into an orthonormal basis: think of the columns of a matrix in $GL_n(\mathbb R)$ as a basis of $\mathbb R^n$, apply Gram--Schmidt, and then reinterpret it in terms of matrix multiplications. (This is basically what Paul Garrett's answer does.) Regards,
Aug
12
comment Atiyah-MacDonald, exercise 7.19 - “decomposition using irreducible ideals”
@Efim: Dear Efim, I don't have a copy of A&M, so I can' answer your question about the relationship b/w 8.6 and 7.19. As for the $I_j$, I don't really understand your question. The ideal $I_j$ is defined as in my answer, i.e. we fix an $i$, and then for each $j$ the ideal $I_j$ is defined by a certain formula in terms of the corresponding ideal $\mathfrak c_j$. So there is no particular "choice" of $\mathfrak c_j$; for each $\mathfrak c_j$ there is a corresponding $I_j$ (while the index $i$ is fixed throughout the discussion). Regards,