40

Here's some piece of the bigger picture. Maass forms and holomorphic modular forms are both automorphic representations for $GL(2)$ over the rationals. An automorphic representation is a typically huge representation $\pi$ of an adele group (in this case $GL(2,\mathbf{A})$, with $\mathbf{A}$ the adeles of $\mathbf{Q}$). Because the adeles is the product of ...

answered Jun 6 '12 at 6:40

Kevin Buzzard

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26

Will Sawin's answer is perfectly correct, but I wanted to add some further perspective.
What Peter Scholze has proven is a profound generalization of (one half) of Class Field Theory. The particular example you state is simply a special case of a much more general result.
To set the stage, let me recall some of the set-up from class field theory. Let $F$ ...

nt.number-theory galois-representations langlands-conjectures arithmetic-groups quadratic-reciprocity

22

A brief update: in his amazing talk yesterday at MSRI (available here), Laurent Fargues explained (building on work of Peter Scholze, see his phenomenal talk two weeks ago here and his historic Berkeley lectures here) a conjectural picture for the local Langlands correspondence as a geometric Langlands correspondence over the "Fargues-Fontaine curve".
[...

21

Well, ${\rm GL}(2,q)$ has Abelian Sylow $p$-subgroups for every odd prime $p.$ The symmetric group $S_{n}$ has non-Abelian Sylow $p$-subgroups for each prime $p$ such that $p^2 \leq n.$ Hence thesymmetric group $S_{25}$ is not a subgroup of any ${\rm GL}(2,q)$ since it has non-Abelian Sylow $3$-subgroups and non-Abelian Sylow $5$-subgroups ( even the ...

21

I am waking up an old and already well-answered question, to offer another point of view,
The Artin's conjecture appears very naturally in the context of Chebotarev's density theorem.
In fact, we can see Cheobtarev's contribution as a clever trick to circumvent the Artin conjecture by reducing the proof to cases where it is known (by works of Dirichlet and ...

20

Here is Ribet's proof (expanding on Ulrich's comment):
Let $G_K:=Gal(\bar{K} / K)$ and $V_l:=T_l(E)\otimes \mathbb{Q}_l$.
The image $\rho_{l,E}(G)$ is a closed subgroup of the $l$-adic Lie group $\text{Aut}(V_l(E)) \cong \text{GL}_{2}(\mathbb{Q}_l)$ and is therefore a Lie subgroup of $\text{Aut}(V_l(E))$. Its Lie algebra $\mathfrak{g}_l$ is a subalgebra of ...

18

It is true if $V$ is of dimension 1, essentially by class field theory (as you are considering only representations of $G_{\mathbb Q}$). Otherwise, it is still largely open for the following reasons.
If $n\geq 2$, the best tool we have to study $G_{\mathbb Q}$-representations are so called modularity lifting theorems. These theorems take as hypotheses the ...

18

The classification of 1-dimensional representations, i.e. characters $G_K \to E^\times$, is a bit more complicated than you imply in your question. Any such character lands in $O_E^\times$ by compactness; and since $O_E^\times$ is profinite, and class field theory identifies the abelianisation of $G_K$ with the profinite completion of $K^\times$, we conclude ...

nt.number-theory rt.representation-theory arithmetic-geometry galois-representations p-adic-hodge-theory

answered Jun 30 '19 at 10:52

David Loeffler

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17

We can generalize Ralph's answer to find the abelian groups that are $p$-groups for odd $p$ that cannot be subgroups of $GL_2$.
if $p|(q^2-1)(q^2-q)$ then exactly one of $p|(q+1)$, $p|(q-1)$ and $p|q$. If $p|q$ then $q=p^n$ and upper-triangular unipotent matrices provide enough, so they're a Sylow subgroup, $(\mathbb Z/p)^n$. If $p|(q-1)$ then the group of ...

17

I am not sure this answer will satisfy you totally because I am not sure what you mean exactly by $p$-adic modular forms. But at least, for a $p$-adic modular form $f$ in the sense of
Serre, which is an eigenform for almost all the Hecke operators $T_\ell$ (with eigenvalue $a_\ell)$ there always exists a semi-simple Galois representation
$r:G_{\mathbb Q} \...

17

This is more subtle than it looks. I asked exactly the same question some years back (see here); but I'm not going to flag this question as duplicate, because the answer that was given to my question at the time, which was identical to @eric's comment above, was wrong.
The correct answer is this. Let $f$ be a normalized non-CM newform. Associated to $f$ ...

17

The correct statement is: any 3-dimensional selfdual Galois representation is isomorphic to a quadratic twist of the adjoint of some 2-dimensional representation. (The quadratic twist is really necessary, as one can see from a variant of David Speyer's example.)
If $\rho: G_F \to GL_3(\overline{\mathbf{Q}}_\ell)$ is a Galois representation, where $F$ is a ...

16

The short answer to the question is that all currently known proofs of the local Langlands correspondence (and I'm just referring to GL(n) here) are "global" in the sense that they involve embedding the local problem into a global one. That is, the local field in question is realized as the completion of a global field at one of its places. Then the theory ...

16

Definitely no. To see why, consider just the case where $W=\mathbb Q_\ell$ is the trivial representation (this is not really a loss of generality, because extensions of $W$ by $V$ are "the same thing" as extensions of $\mathbb Q_\ell$ by $W^\ast \otimes V$). Then extensions (as $G$-representations) of $\mathbb Q_\ell$ by $V$ are parametrized by $H^1(G,V)$. ...

15

The first question (applied to $\mathrm{GL}(2)$-abelian varieties over $\mathbf{Q}$) seems to include the following problem: what totally real fields $F$ occur as the field of coefficients of a classical weight $2$ modular form? This seems a totally impossible question to answer.
For example, it includes the question of which Hilbert modular surfaces $X_F$ ...

15

You're asking if $Ext^1_G(\alpha,\beta)$ is one-dimensional. The short answer is no,
yet there are many cases where the answer is yes. Actually, the dimension of this group is not known in all cases, and when it is known, in general it is by a deep theorem,
not a trivial computation.
To be more precise, we have $Ext^1_G(\alpha,\beta) = Ext^1_G(1,\alpha^{-1} ...

15

The de Rham characters are the same as the Hodge-Tate ones. The semistable ones are the same as the crystalline ones, and in your notation they are the de Rham ones for which $(\chi \cdot \chi_p^{-k})(I_K)$ is trivial (and not merely finite). This can be found for example in Fontaine and Mazur's paper.

15

Actually, this is an exercise in Serre's Lectures on Mordell--Weil Theorem:
$K(A[n])$ always contains $\mu_n$ if $char(K)$ does not divide $n$ and $A$ is an abelian variety of positive dimension over $K$. (You don't need to assume that $K$ is a number field)
Here is a solution. First, it suffices to check the case when $n=\ell^m$ is a power of a prime $\...

15

I did the computation in Sage, and there is no such form $f$. There are 21 Galois orbits of newforms of level $\Gamma_0(1776)$ and trivial character, of which the largest has size 3, and none of the reductions modulo any of the primes above 7 in the coefficient fields is congruent to the form associated to $\bar\rho$.
Looking again at your question, the ...

15

Good question. I don't understand fully what's happening, but here is an idea.
Let $f=\sum a_n q^n$ be a modular form of weight $k+1/2$, nebentypus $\chi$. Assuming $k \geq 1$, the Shimura correspondence attaches to $f$ a
modular form of integral weight $2k$, $g = \sum b_n q^n$ such that
$$L(g,s) = L(\chi',s-k+1) \sum \frac{a_{n^2}}{n^s},$$
where $\chi'(n)=\...

14

Claim 1: No group
$$A_{a,b,,c} := \mathbb{Z}/2^a \times \mathbb{Z}/2^b \times \mathbb{Z}/2^c$$
with $a,b,c > 1$ is a subgroup of some $GL(2,q)$.
Claim 2: If $q \neq 2,3$, then the quotients of $GL(2,q)$ are exactly the subgroups of the cyclic group $\mathbb{F}^\times_q$ and the groups $GL(2,q)/D$ where $D \le \mathbb{F}^\times_q \cdot I$.
In case $GL(...

14

Abelian varieties over the rationals are modular if and only if they are of "$GL_2$"-type, which is a notion introduced by Ribet who proved that this statement is a consequence of Serre's conjecture which, as you know, has since been proved. Here is a link to Ribet's paper:
http://math.berkeley.edu/~ribet/Articles/korea.pdf
Generalizing the statement of ...

answered May 23 '14 at 14:53

Felipe Voloch

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13

I don't know that this is written down anywhere, but it's possible. It is known in general that $GL(T_{\ell}(E))$ is contained in the normalizer of $R_{\ell}^{\times}$, where $R_{\ell} = \mathbb{Z}[i] \otimes \mathbb{Z}_{\ell}$. (This follows for example from Corollary 2 on page 502 of Serre and Tate's ''Good reduction of abelian varieties'' Annals paper.) ...

13

Peter Scholze explains the same example in more detail in this talk https://youtu.be/cFdm0B9KLcQ?t=48m and gives even more detail here https://youtu.be/gI3Ta73yVuo?t=23m55s.
In brief, the relationship is that for every prime $\mathfrak p$ of $\mathbb Z[i]$ not dividing $183$, the polynomial has a root mod $\mathfrak p$ if and only if the eigenvalue of the $\...

nt.number-theory galois-representations langlands-conjectures arithmetic-groups quadratic-reciprocity

13

For a prime $p$ and an elliptic curve $E/\mathbb{Q}$, we have the exact sequence
$$\displaystyle 0 \rightarrow E(\mathbb{Q})/p E(\mathbb{Q}) \rightarrow S_p(E) \rightarrow \text{Sha}_E[p]\rightarrow 0,$$
where $S_p(E)$ is the $p$-Selmer group of $E$ and $\text{Sha}_E$ is the Tate-Shafarevich group of $E$ and $\text{Sha}_E[p]$ is the $p$-part of it. This ...

answered Mar 14 '18 at 0:22

Stanley Yao Xiao

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12

Here's an example, if I'm not mistaken. Let $E / K$ be an elliptic curve and $x \in E$ a non-torsion $K$-point. Then the image of the divisor $\{x\} - \{\infty\}$ under the etale cycle class map is a nontrivial class in $H^1(K, H^1(E_{\bar K}, \mathbf{Q}_\ell)(1))$ and thus corresponds to a non-split extension of $H^1(E_{\bar K}, \mathbf{Q}_\ell)(1)$ by $\...

12

There's a little more geometry here that should be accounted for in characteristic $p$. Namely, the curve $X_0(Np)_{\mathbb{F}_p}$ is reducible -- its two components are isomorphic to $X_0(N)_{\mathbb{F}_p}$ and they intersect transversally at the supersingular points (see e.g. Ribet and Stein's online notes). So the components are disjoint on the ordinary ...

12

The object (conjecturally) associated to an Artin representation by Langlands is not a classical modular form except in a very limited number of situations (odd two dimensional representations). Let $K$ be the splitting field of some polynomial over $\mathbf{Q}$ with Galois group $S_5$. If $V$ is any finite dimensional representation of dimension $n$ of $S_5$...

12

A natural generalization of the geometric modularity conjecture which is compatible with your formulation
Do you expect some form of modularity to correspond to the existence of a map from some space (like a Shimura variety) to the variety in question?
is to ask whether a variety always appears as a quotient of the Picard variety of (the smooth ...

12

Galois representations have to come from somewhere.
If you are not interested in learning about modular forms and automorphic forms at this point, the other best source of representations are elliptic curves. You can find a gentle introduction in Silverman's book "The Arithmetic of Elliptic Curves", particularly the chapter on Tate modules.
As an ...

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