# Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$

Let $p$ be a prime number, $q=p^e$ a power of $p$, and $G=SL_2(\mathbb F_q)$. Let $V$ be the adjoint representation of $G$, i.e. $V$ is the 3-dimensional $\mathbb F_q$-space of of (2,2)-matrices of trace $0$ with coefficients in $\mathbb F_q$, and $G$ acts on it by conjugation. So $V$ is an irreducible representation of $G$.

By "restriction of scalar", $V$ can also be considered as a representation of dimension $3 e$ over $\mathbb F_p$.

When is this representation $V/\mathbb F_p$ of dimension $3e$ irreducible?

(Edit: it is obviously not absolutely irreducible, as remarked by Venkataramana below).

In general, I would be interested by any pointer to a reference mentioning questions of irreducibility of representations obtained by "restriction of scalars".

• Thanks for all the three good answers to my question. I accept Jim's because it is the most general, but the others are very good too.
– Joël
Nov 9, 2014 at 16:37

There is some textbook literature which essentially covers the issues raised here, though it often deals with more general situations. (Over finite fields life is simpler, since Schur indices are 1.) See for example Curtis & Reiner (1962), Section 70, and also the book Character Theory of Finite Groups by Isaacs (reprinted by AMS), especially Chapter 9 and problem (9.6).

The basic criterion deals with an arbitrary finite group $G$ and its representations over a finite splitting field $\mathbb{F}_q$. In your situation, this applies to finite Chevalley groups by the 1963 paper of Steinberg (see $\S13$ of his 1967-68 Yale lectures here). What you need to know is that an absolutely irreducible $FG$-module $M$ over a subfield $F$ of the splitting field $\mathbb{F}_q$ is irreducible over the prime field $\mathbb{F}_p$ if and only if $F$ is the "field of definition" of $M$. Here a field of definition is one generated over the prime field by the traces of representing matrices. (It's probably hard to find an explicit statement and proof of this in the textbooks, but it's implicit.)

In your specific example, it's therefore necessary to be careful about where the traces of the representing matrices lie.

• Thanks a lot! That's a very nice criterion, and simple to remember since one direction is easy (at least in the case where $p \not \mid |G|$).
– Joël
Nov 7, 2014 at 21:40

$G$ acts irreducibly on $V$: Suppose that $W$ is an $\mathbb F_p$-subspace of $V$ invariant under $G$, of dimension $d$. So $W$ has $p^d-1$ nonzero elements. Hence there is an orbit of $G$ on these elements of length prime to $p$. Let $A\in W$ be a trace $0$ matrix in this orbit. The orbit length through $A$ is the index $[G:C_G(A)]$. If $A$ has no eigenvalue in $\mathbb F_q$, then $A$ is irreducible on $\mathbb F_q^2$, so by Schur's Lemma we get that $C_G(A)$ is a subgroup of the multiplicative group of $\mathbb F_{q^2}$. In particular, $p$ does not divide $|C_G(A)|$, so $p$ divides the orbit length through $A$, a contradiction.

Thus $A$ has eigenvalues in $\mathbb F_q$. First suppose that $p$ is odd. If the eigenvalues are equal, then they vanish by the trace $0$ condition. If they are however distinct, then $A$ is conjugate in $G$ to a diagonal matrix. So we may assume that there is a nonzero $a$ such that either $A=\begin{pmatrix}0 & a\\0 & 0\end{pmatrix}$ or $A=\begin{pmatrix}a & 0\\0 & -a\end{pmatrix}$. In either case, one quickly computes the conjugacy class $A^G$, and finds out that in both cases the $\mathbb F_p$ span of the classes are $V$. So $W=V$ because $A\in W$.

Similarly, in the case $p=2$ we obtain that we may assume $A=\begin{pmatrix}a & b\\0 & a\end{pmatrix}$ for some $a,b\in\mathbb F_q$. Again, $\mathbb F_p[A^G]=V$.

• Are you looking at the two dimensional rep of $V$? Joel looks at the three dimensional adjoint representation $V$. Nov 8, 2014 at 2:47
• @Venkataramana: So do I. The matrices $A$ are clearly elements from $V$, and as I said in the proof, an orbit of $G$ through $A$ is the conjugacy class $A^G$ ($=\{g^{-1}Ag\;|\;g\in G\}$). If $A$ lies in a submodule $W$, then $A^G\subseteq W$, and then also the $\mathbb F_p$ span of $A^G$ is contained in $W$. But the latter is $V$ for the given matrices $A$. Nov 8, 2014 at 8:54
• @Michor: thank you. After reading the ans carefully, I see that you were looing at the 3 dimensional adjoint module. Nov 8, 2014 at 14:53

It is not absolutely irreducible: over the algebraic closure, it is the direct sum of $V^{Fr ^i}$ ($0\leq i \leq e-1$), where $Fr : {\mathbb F}_q\rightarrow {\mathbb F}_q$ is the map $x\mapsto x^p$.

I think $e$ being odd or even does not matter: $V$ is irreducible over ${\mathbb F}_p$. The restriction of scalars may be viewed as follows. First, $V$ viewed as a ${\mathbb F}_p$ vector space has dimension $3e$, and one has an embedding ${\mathbb F}_q\subset End_{{\mathbb F}_p}(V)$. The algebra $M_3({\mathbb F}_q)$ is simply the commutant of this copy of ${\mathbb F}_q$ in $End _{{\mathbb F}_p}(V)$ and is spanned as a ${\mathbb F}_p$ vector space (a small check) by elements of $Ad (SL_2({\mathbb F}_q))$.

Suppose $W\subset V$ is irreducible for the adjoint action of $G=SL_2({\mathbb F}_q)$. Then $W$ is stabilised by $M_3({\mathbb F}_q)$; in particular, $W$ is an ${\mathbb F}_q$ subspace of $V$ (now viewed as a ${\mathbb F}_q$ vector space). But $SL_2({\mathbb F}_q)$ acts irreducibly on $V={\mathbb F}_q ^3$, hence $W=V$. [This irreducibility fails of course when $p=2$].

 I think I can find a proof that the ${\mathbb F}_p$ algebra $R$ generated by $Ad (SL_2){\mathbb F}_q)$ is all of $M_3({\mathbb F}_q)$. First of all, $V \otimes _{{\mathbb F}_p}{\overline {\mathbb F}_p}$ is the direct sum of $W_i=V^{Fr ^i}$ as $i$ varies from $0$ to $e-1$, and is semi-simple. Therefore, $V$ is semi-simple as an $R$-module. Hence $R$ is semi-simple (the action of $R$ being faithful on $V$). Therefore, $R$ is a product of simple rings $R_i$, each $R_i$ of the form $M_{d_i}({\mathbb F}_{q_i})$ a matrix algebra over a finite field containing ${\mathbb F}_p$.

The module $V_i={\mathbb F}_{q_i}^{d_i}$ is absolutely irreducible for $R_i$. However, $V$ over the algebraic closure, is a sum of irreducible three dimensional representations $W_i$. Therefore, each $V_j$ is some $Wi$ over the algrbraic closure, and is hence three dimensional: $d_i=3$. Moreover, since $R$ commutes with the action of $F={\mathbb F}_{q}$ on $V$, each ${\mathbb F}_{q_i}$ contains $F$. We have thus,

$$M_3({\mathbb F}_q)\supset R\supset R_i=M_3({\mathbb F}_{q_i})$$ which shows that $R=M_3({\mathbb F}_q)$ as needed.

• Oh you're right for the absolute irreducibility question, which was not thoughtful of me to ask. But what really interests me is the irreducibility (not absolute). Can you explain you argument for $e$ odd in more details?
– Joël
Nov 7, 2014 at 16:00
• Meanwhile, let me edit my question to remove the part about absolute irreducibility.
– Joël
Nov 7, 2014 at 16:01
• Thanks, but I am a little lost. It looks like for you $V$ is the natural 2-dimensional representation while I defined it as the adjoint, 3-dimensional, representation.
– Joël
Nov 7, 2014 at 16:50
• Dear Venkaramana, I understand your strategy. But I am not sure how to make the "small check" you mention. So basically you are saying that in the space $M_3(\mathbb F_q)$ the elements of $Ad(SL_2(\mathbb F_q)$ generates everything as an $\mathbb F_p$-vector space, i.e. as an abelian group. Those elements are matrices $((-a^2,-2ab,-b^2),(-ac,ad+bc,bd),(-c^2,2cd,d^2))$ for $((a,b),(c,d)) \in SL_2(\mathbb F_q)$, and it isn't obvious to me what's the $\mathbb F_p$ linear span of those matrices. It would be enough to prove that the scalar matr. are in that span, but even this I find not clear.
– Joël
Nov 7, 2014 at 18:15