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This question is inspired by some interesting comments on this recent question.

Fix an integer $n \geq 1$ and a finite subgroup $G$ of $\mathrm{GL}_n(\mathbf{C})$. It is known that there are infinitely many primes $p$ such that $G$ embeds into $\mathrm{GL}_n(\mathbf{Q}_p)$. This follows from the Cassels Embedding Theorem : every finitely generated field of characteristic 0 embeds into $\mathbf{Q}_p$ for infinitely many primes $p$ (see Cassels's book Local Fields or Cassels's article An embedding theorem for fields. Bull. Austral. Math. Soc. 14 (1976), 193-198; see also Chapter 4 of Pete L. Clark's notes).

Representation theoretic arguments actually show that $G$ occurs in $\mathrm{GL}_n(\overline{\mathbf{Q}})$ and thus in $\mathrm{GL}_n(K)$ for some number field $K$. From this and Chebotarev's density theorem, we deduce that the set $S(G)$ of primes $p$ such that $G \hookrightarrow \mathrm{GL}_n(\mathbf{Q}_p)$ has positive density.

Does the set $S(G)$ have a natural density?

Is the set $S(G)$ a Galoisian set of prime numbers in the sense of this question?

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I'm glad you posted the question. I think though you should also refer the reader interested in CET to either Cassels's book Local Fields or to: J.W.S. Cassels, An embedding theorem for fields. Bull. Austral. Math. Soc. 14 (1976), 193-198. –  Pete L. Clark Apr 23 '13 at 3:25

2 Answers 2

up vote 9 down vote accepted

$\def\Gal{\mathrm{Gal}}$ $\def\Res{\mathrm{Res}}$ $\def\GL{\mathrm{GL}}$ $\def\F{\mathbf{F}}$ $\def\Q{\mathbf{Q}}$

Edited to include more details.

Let $K$ be a field whose characteristic is prime to the order of $G$. The algebra $K[G]$ is a product of matrix algebras over division rings. Given any absolutely irreducible character $\chi$ with coefficients in $L/K$, the question of whether $\chi$ may be realized over $L$ is equivalent to whether the corresponding division algebra $D/K$ splits over $L$.

Suppose that $K = k$ is a finite field. Then, by Weddeburn's theorem, there are no non-trivial division algebras and irreducible characters with values in $k$ have models over $k$.

Suppose that $K$ is a number field. For all but finitely many places $v$ of $K$, $D \otimes_K K_v$ is a matrix algebra. So, at least away from finitely many places, irreducible characters with values in $K$ have models over $K_v$.

Example: Let $G = Q_8$ be the quaternion group, and let $\chi$ denote the absolutely irreducible faithful character of degree $2$. Then $\chi$ is valued in $\Q$. However, the corresponding quaternion algebra $D$ is ramified at $2$ and $\infty$. Hence $G$ does not have a faithful representation over either $\mathbf{R}$ nor $\mathbf{Q}_2$, but it does for $\Q_p$ and all odd primes $p$, because $D \otimes \Q_p = M_2(\Q_p)$ in those cases.

In the example above, we see that the obstruction arises for finite places only at the prime $2$ which divides $|G|$. Let us show that there is no obstruction for any $v$ such that the residue characteristic is finite and prime to $|G|$. In such cases, there is a bijection between absolutely irreducible characters in characteristics zero and $p$ given by reduction modulo $p$. Let $\chi$ be an irreducible character with values in a local field $E$. It preserves a lattice $\mathcal{O}$, and gives rise to an irreducible character over the finite field $k$, which has a model over $k$ by the remarks above. Hence (because the characters are in bijection) it suffices to prove the following: any representation: $$G \rightarrow \GL_n(k)$$ with image $H$ admits a lift to $\GL_n(W(k))$, since $W(k)[1/p]$ will be necessarily be a subfield of $E$. (Not surprisingly, we see that all representations have models over unramified extensions when $p \nmid |G|$, since the characters are all valued in $\Q(\zeta_{m})$ where $m$ is the exponent of $G$.) There is a projection $\GL_n(W(k)) \rightarrow \GL_n(k)$; let $\Gamma$ denote the inverse image of $H$. Since $H$ has order prime to $p$ and the kernel of $\Gamma \rightarrow H$ is pro-$p$, by Schur-Zassenhaus there is a splitting $H \rightarrow \Gamma$ which gives the required lift.

Suppose now that $\eta$ is a general (genuine) character of $G$ over a local field $K/\Q_p$ with values generate the field $L/K$, and suppose that $p$ does not divide $|G|$. We prove that $\eta$ has a model over $L$. As noted above, $L/K$ is unramified, so in particular is Galois and $\Gal(L/K)$ is cyclic. The result is true for irreducible characters from the discussion above. Suppose that $\chi$ is an irreducible constituent of $\eta$. Since $\Gal(L/K)$ fixes $\eta$, it follows that the $[L:K]$ distinct characters $\sigma \chi$ occur inside $\eta$. Hence

$$\eta = \phi + \sum_{\Gal(L/K)} \sigma \chi,$$ where $\eta$ is a genuine character of lower dimension. Hence it suffices to note that $$\bigoplus \sigma \chi$$ is defined over $K$, because if $V/L$ is a realization of $\chi$, then the above is $\mathrm{Res}_{L/K}(V)$, where $L/K$ is thought of as a $[L:K]$-dimensional vector space in the usual way. The result follows by induction.

Example: Suppose that $p \equiv -1 \mod 4$, and let $\chi$ be a faithful character of $\mathbf{Z}/4\mathbf{Z}$. Then $K = \Q_p(\chi)$ is unramified over $\Q_p$ of degree $2$, and the representation $\chi + \sigma \chi$ is sends a generator to $i \in K$ thought of as a vector space over $\Q_p$. If one chooses the basis $\{1,i\}$ of $K/\Q_p$, this is just the matrix: $$\left( \begin{matrix} 0 & 1 \\\ -1 & 0 \end{matrix} \right).$$

Conclusion: Hence, at least for primes $p$ not dividing $|G|$, there is an injection from $G$ to $\GL_n(\Q_p)$ if and only if there is a faithful character $\eta$ with values in a field $K$ which has a prime of norm $p$. Since $K$ will be abelian and ramified only at primes dividing $|G|$, this is equivalent to asking that $K$ split completly at $p$. So the set $S(G)$ (up to primes dividing $|G|$) is the union of primes which split completely in some finite number of fields determined by the faithful characters of degree $n$. If one restricts to a fixed character $\eta$, then $S_{\eta}(G)$ is indeed Galoisian.

Providing that $\eta$ has at least one faithful character of degree $n$, then $S(G)$ has rational positive density, answering 1. As for 2, it is not strictly Galoisian according to the definition of the previous question, since that required that the set be the set of primes which split completely in a single field. For example, one can take $$G = \mathbf{Z}/12 \mathbf{Z}$$ and $n = 2$. Then $G$ is a subgroup of $\mathbf{GL}_2(\Q_p)$ for $p > 3$ if and only if $p \equiv 1,5,7 \mod 12$, but not $11 \mod 12$, and this is not the set of primes which splits completely in any field $L$. In this case, if $\chi$ is a faithful character of $G$ of degree $1$, then $\chi + \chi^5$ and $\chi + \chi^{-1}$ are faithful characters with values in $\Q(\sqrt{-1})$ and $\Q(\sqrt{-3})$ respectively.

If $G$ has a faithful character of degree $n$ with values in $\Q$, then we see that $G$ embeds into $\GL_n(\Q_p)$ for all but finitely many primes $p$. The converse is quite possibly false, however; one could imagine $G$ having faithful characters of degree $n$ with values in $\Q(\sqrt{2})$, $\Q(\sqrt{3})$, and $\Q(\sqrt{6})$ but not in $\Q$ (although I don't have an example off the top of my head.)

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Thanks for your answer! Your nice example shows that one can refine the question by looking only at those embeddings of $G$ into $\mathrm{GL}_n(\mathbf{Q}_p)$ which are compatible with a given complex faithful character of $G$. Regarding your argument, I don't see why the existence of a faithful $\mathbf{Q}_p$-valued character implies that $G$ embeds into $\mathrm{GL}_n(\mathbf{F}_p)$. Is it where you use your result about reps over finite fields? If so then why is the reduction mod $p$ still faithful? –  François Brunault Apr 23 '13 at 12:31
    
That's true, although one has to be a little careful by what one means by the reduction of a character mod $p$. If $\chi$ is faithful, there's a naive sense in which the reduction mod $p$ of $p \cdot \chi$ is trivial! So I re-wrote the argument in a slightly different way. –  Socky Apr 24 '13 at 14:56
    
Thank you very much for taking the time to explain in more details! –  François Brunault Apr 24 '13 at 18:39

The answer is yes, I think, $S(G)$ is Frobenian. (Edited: actually, as noticed by Rabelais and the OP, there is a gap if the representation $r_i$ below are not all irreducible. I leave the argument as it still gives a positive answer in the case where all faithful rep. of $G$ of dim $n$ are irreducible -- for instance, when $G$ is non-abelian and $n=2$).

Let $(r_i)$ for $i=1,\dots,a$ the list of all faithful complex representations of $G$ of dimension $n$. Let $(\rho_{i,j})_{i,j=1,\dots,b_i}$ the set of irreducible representations appearing as a component of $r_i$. Let $K_{i,j}$ be the field of rationality of $\rho_{i,j}$. There is a unique (up to $\mathbb K_{i,j}$-isomorphism) central simple algebra $D_{i,j}$ together with a map $\tilde \rho_{i,j}: G \rightarrow D_{i,j}$ which after tenderization by $\mathbb C$ gives back $\rho_{i,j}$.

Claim: for a prime $p$, the two following properties are equivalent. (1) $G$ can be embedded in $Gl_n(\mathbb Q_p)$ (2) There exists $i$ such that for all $j \in \{1,\dots,b_i\}$, there exists a prime $\mathcal P$ in $K_{i,j}$ of degree $1$ above $p$ and at which $D_{i,j}$ is split.

Proof of the claim: (2) implies (1) because if $\mathcal P$ is as in (2), $(K_{i,j})_{(\mathcal P)}=\mathbb Q_p$, and $D_{i,j} \otimes (K_{i,j})_{(\mathcal P)} = \mathbb Q_p$; thus the $\rho_{i,j}$ for that fixed $i$ are all defined over $\mathbb Q_p$, and so is their sum over $j$, that is $r_i$, and $r_i$ being faithful defined an embedding of $G$ into $Gl_n(\mathbb Q_p)$.

Proof of (1) implies (2). If $G$ can be embedded into $Gl_n(\mathbb Q_p)$ let $r$ be the embedding, that we see as an $n$-dimensional representation over $\mathbb Q_p$. Embedding $\mathbb Q_p$ to $\mathbb C$, we see $r$ as a complex representation which is thus one of the $r_i$. So $r_i$ is defined over $\mathbb Q_p$, which means that all the $\rho_{i,j}$ are, which in turns means that (2) holds y the lemma below. (end of proof of the claim)


Now it is clear that (2) defines a Frobenian set of primes.


Lemma: let $\rho: G \rightarrow Gl_n(\mathbb Q_p)$ be an absolutely irreducible representation, that we see as a complex irreducible representation by the choice of an embedding $\mathbb Q_p \rightarrow \mathbb C$. Let $K$ be its rationality field (the subfield of $\mathbb C$ generated by the $tr \rho(g))$, so that there exists a central simple algebra $D/K$ of rank $n^2$ such that $\rho$ comes from a "representation" $\tilde \rho: G \rightarrow D^\ast$. Then there is a place $\mathcal P$ of $K$ such that the completion $K_{\mathcal P}$ is $\mathbb Q_p$, and $D \otimes_K K_{\mathcal P} = M_n(K_{\mathcal P})$.

Proof: Since $K$ is generated (as a field over $\mathbb Q$) by the $tr \rho(g)$, and the $tr \rho(g)$ are in $\mathbb Q_p$, one has $K \subset \mathbb Q_p$. Let us take for $\mathcal P$ the prime corresponding to the restriction of the standard absolute value on $\mathbb Q_p$ to $K$. Then clearly $K_{\mathcal P} = \mathbb Q_p$. Next $\tilde \rho$ defines a representation $\tilde \rho: G \rightarrow (D \otimes \mathbb Q_p)^\ast$, hence a representation of algebra $\tilde \rho : \mathbb Q_p[G] \rightarrow D \otimes \mathbb Q_p$, which is surjective by absolute irreducibility, and whose kernel is $\ker tr red (\tilde \rho)$ where $tr red \tilde \rho$ is the reduced trace of $\tilde \rho$, but as one sees by going up to $\mathbb C$, is also $tr \rho$. Now the same argument applied to $\rho$ itself gives also a surjective morphism $\mathbb Q_p \rightarrow M_n(\mathbb Q_p)$ with kernel $ker\ tr \rho$. Hence $M_n(\mathbb Q_p) \simeq D\otimes \mathbb Q_p$, QED.


PS: I think my presentation in the argument is in fact unnecessarily complicated, because I stick to the language of representation, while everything would be much clearer in the language of pseudo-characters. Indeed, there are four representations in the short argument above (over $\mathbb C$, over $\mathbb Q_p$, over a central simple algebra K, over a central simple algebra over $K_{\mathcal P}$, but all those are incarnation of one single object, the pseudo-character $T: G \rightarrow \mathbb Q_p \subset \mathbb C$ defined by $T=tr \rho$. Now the proof of the lemma is simply that (1) $K$ which is defined as the field generated by $T(G)$ is by definition inside $\mathbb Q_p$ (2) Taylor-Rouquier's theorem: Over a given field (here $\mathbb Q_p$), an absolutely irreducible pseudo-character has one and only one (we use the only one part here, which is the easy part of the result) realization as the (reduced) trace of a representation over a central simple algebra. We apply this to the field $\mathbb Q_p$, giving us that our two realization, in $M_n(\mathbb Q _p)$ and in $D \otimes \mathbb Q_p$ have to be the same.

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Thanks! I have a little difficulty in understanding the step (1) implies (2) : by $r_i$ is defined over $\mathbf{Q}_p$, you mean that it has a model over $\mathbf{Q}_p$ up to isomorphism, then why are the subrepresentations $\rho_{i,j}$ also defined over $\mathbf{Q}_p$? (sorry if this turns out to be obvious) –  François Brunault Apr 22 '13 at 15:54
    
Dear François, I have added some details. –  Joël Apr 22 '13 at 17:16
    
$\mathbf{Z}/3\mathbf{Z}$ has a faithful $2$-dimensional representation over $\mathbf{Q}_p$ for all primes $p$, but this is not true of the sub-representations which are only defined over $\mathbf{Q}_p(\zeta_3)$. –  Socky Apr 22 '13 at 18:20
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Pourquoi François, Joël et Rabelais sont-ils en train de discuter des maths en anglais ? –  Chandan Singh Dalawat Apr 23 '13 at 3:11
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J'ai du mal à comprendre l'ancien français de Rabelais. –  Joël Apr 23 '13 at 13:15

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