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Let $\text{St}_n(\mathbb{F}_q)$ be the Steinberg module (over $\mathbb{C}$) for $\text{SL}_n(\mathbb{F}_q)$.

What is the irreducible decomposition of the restriction of $\text{St}_{2n}(\mathbb{F}_q)$ to the symplectic group $\text{Sp}_{2n}(\mathbb{F}_q)$?

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    $\begingroup$ This does not seem to be an easy problem and to my knowledge it is not solved. One has to work with a nice model of the Steinberg representation. One can for instance work with the model afforded by the top homology module of the Tits building, or with the model as an alternate sum of ${\rm Ind}_P^G 1_P$, where $P$ runs over the standard parabolic subgroups. In any case it seems that one has to understand the geometric of the embedding of the Tits building of ${\rm Sp}_{2n}$ in the building of ${\rm SL}(2n)$. $\endgroup$
    – Paul Broussous
    Commented Sep 29, 2014 at 8:54
  • $\begingroup$ @Melanie: I'd add to what Paul says the suggestion to look at the smallest special cases of your question when $n=1,2, \ldots$ There seem to be no easy branching rules for characters of the finite groups of Lie type in general, but conceivably the pattern in this special situation is reasonable (if there is a pattern). $\endgroup$
    – Jim Humphreys
    Commented Sep 29, 2014 at 12:35
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    $\begingroup$ P.S. 1) Though your question seems to be directed toward ordinary characters, it might be interesting to reduce modulo $p$ (where $q$ is a power of $p$) and see how the resulting projective module St decomposes. Unfortunately, too little is yet known about the modular representations, such as dimensions of indecomposable projectives. 2) In characteristic 0, you can get a rough idea of the size of the decomposition: "most" irreducible characters (for large $p$) have degrees approximately equal to the degree $q^N$ of the Steinberg character, with $N$ the number of positive roots. $\endgroup$
    – Jim Humphreys
    Commented Sep 29, 2014 at 17:12

3 Answers 3

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While this should be a comment rather than an answer, I cannot comment yet since I've just joined, and figured that the following points could be useful at least so figured I'd post them (although you may already be familiar).

1) One thing that is known, which is related to your question, is the trivial character induced from Sp(2n,q) to SL(2n,q) is multiplicity-free. The trivial character induced from Sp(2n,q) to GL(2n,q) is also multiplicity-free and given explicitly in the paper:

N.F.J. Inglis and J. Saxl, An explicit model for the complex representations of the finite general linear groups, Arch. Math. (Basel) 57 (1991), no. 5, 424–431.

You can use that decomposition to understand the decomposition when induced up to SL(2n,q). The possible relevance here is that the Steinberg is the Alvis-Curtis dual of the trivial character, which you may be able to take advantage of.

2) A less relevant, but still possibly helpful paper, is:

An, Jianbei and Hiss, Gerhard Restricting the Steinberg character in finite symplectic groups, J. Group Theory 9 (2006), no. 2, 251–264.

In this paper, they restrict the Steinberg of Sp(2n,q) down to a specific maximal parabolic.

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This is also not a complete answer, but I think technically the right track: Paul Broussous already suggested in his answer to define the Steinberg representation by induction of parabolics and Jim Humphreys pointed out that for general inclusions of Lie-Groupe there seem to be no easy branching rules (and suggested a small example, say $n=2$). I just want to tell you only what I see special about your case:

a) To be slightly more explicit on the first point: Let $I$ be a set of simple roots for $G$ $$St:=\sum_{J\subset I} (-1)^{|J|} Ind_{P_J}^{G}(1)$$ and this is related to to the similar expression for a Weyl group character, with $W_J\subset W$ the Weyl group generated by the simple elements longest element of the parabolic subsystem generated by the simple roots $J$: $$\epsilon:=\sum_{J\subset I} (-1)^{|J|} Ind_{W_J}^{W}(1)$$ One shows (Carter "Finite Groups of Lie Type", Chp. 6.2) that $\epsilon(w)=(-1)^{length(w)}$ and $\epsilon(w)=det(w)$ in the reflection representation.

Especially for $SL_{2n}=A_{2n-1}$ all $W_J$ are of type $A_k\times A_{k'}\cdots$ and $\epsilon$ is the sign-function in the symmetric group $\mathbb{S}_{2n}$.

b) A special relation between the Weyl groups / root systems / Tits buildings of $SL_{2n}=A_{2n-1}$ and $Sp_{2n}=C_n$ is as follows:

On $SL_{2n}$ you have an outer automorphism, say explicitly $f(X)=(X^T)^{-1}$. Correspondingly on $A_{2n-1}$ you have a diagram-automorphism $f$ of order $2$ and the orbits form a root system of type $C_n$.

Hence you can

  • identify the $J'$-subsets for $Sp_{2n}$ with those $J$-subsets for $SL_{2n}$ which are $f$-stable; and especially the simple roots $\alpha_i',i\in I'$ of $Sp_{2n}$ with $f$-orbits $\{\alpha_i,\alpha_{2n-i}\}$ resp. $\{\alpha_n\}$ in $SL_{2n}$.
  • Identify the Weyl group $W'\subset W$ of $Sp_{2n}\subset SL_{2n}$ to be generated by $s_is_{2n-i}$ and $s_n$.
  • and so on...I think this is a good embedding of the Tits building in the sense of the comments above?

I am not sure this is enough to compute explicit branching rules for your question $$\chi:=Res^{SL_{2n}}_{Sp_{2n}}(St_{SL_{2n}})=\sum_{J\subset I} (-1)^{|J|} Res^{SL_{2n}}_{Sp_{2n}}(Ind_{P_J}^{SL_{2n}}(1))$$ i.e. to calculate $(\psi,\chi)$ for some irreps $\psi$ of $Sp_{2n}$. You might have a chance by starting with the principal series (maybe it's enough?), proceeding as in many proofs (e.g. $(St,St)=1$ in Carter Prop. 6.2.2), but I don't know an analogon to calculate double cosets $P_{J_1}'xP_{J_2}$ with $P_{J_1}'\subset Sp_{2n}$. At least on the level of Weyl groups computing double cosets $W_{J_1}'xW_{J_2}$ is definitely doable with the description in b), but I think this is not enough...

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In my paper "The high-dimensional cohomology of the moduli space of curves with level structures" with Neil Fullarton, we show how to decompose this representation into a large number of pieces (indexed by partitions of $g$), each of which is a specific induced representation. This is not a complete branching rule since we do not know whether these pieces are irreducible, but it is at least a start.

This paper can be downloaded from my webpage here. The result you are looking for is Theorem 4.1; see Remark 4.2 for an alternate formulation that is more representation-theoretic.

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