Representations of $\operatorname{Sp}(2g,\mathbb{Z}_3)$

Question

Let $V$ be a $2g$-dimensional vector space over $\mathbb{Z}_3 := \mathbb{Z}/3\mathbb{Z}$. First, $\operatorname{Sp}(2g,\mathbb{Z}_3)$ acts on $\Lambda^2(V)$, and this decomposition is reducible, as the kernel of the symplectic form is a subrepresentation. Whenever $3$ does not divide $g$, this representation is decomposable, but it seems to me that it is not the case when $3$ divides $g$, as i cannot find a stable supplementary to the kernel of the symplectic form. Am I right ? Is there a way to prove/disprove what I am claiming ?

I can give more details, without certainty.

1. In the case where $3$ does not divide $g$, if we denote $\omega$ the symplectic form, and $(a_1,a_2,...a_g,b_1,...b_g)$ a symplectic basis, then we have $\Lambda^2(V) = \operatorname{Ker}(\omega) \oplus \operatorname{Span}(\xi)$, where $\xi := \sum a_i \wedge b_i$. The vector $a_1 \wedge a_2$ generates $\operatorname{Ker}(\omega)$, just like a highest weight vector in the complex case. How to be sure that this is irreducible ?

Edit : In this case the Maschke theorem applies, and the representations are completely reducible.

2. If $3$ divides $g$, then the element $\xi$ belongs to the kernel of $\omega$, hence $\operatorname{Ker}(\omega)$ is not irreducible. Also I don't know if it has a supplementary stable by the action.

Edit : I was able to check the following : any element in the kernel of $\omega$ which is not a multiple of $\xi$ generates the kernel. Hence $\operatorname{Span}(\xi)$ has no stable supplementary in $\operatorname{Ker}(\omega)$, and the kernel is not completely reducible. Also, $\operatorname{Ker}(\omega)$ itself has no stable supplementary.

3. Of course i could use the known generators of $\operatorname{Sp}(2g,\mathbb{Z}_3)$ to answer the second question by checking if any other element in $\Lambda_2(V)$ is invariant by the action. But i am not so sure about the generators of $\operatorname{Sp}(2g,\mathbb{Z}_3)$, are they the "same" as in the $\mathbb{Z}$-case ? Someone is saying it is here : Symplectic group over integers and finite fields, but i am not so sure of what he means and why this is true.

Edit : The generators of $\operatorname{Sp}(2g,\mathbb{Z}_3)$ are indeed the "obvious" lifts of the generators of $\operatorname{Sp}(2g,\mathbb{Z})$.

More generally, where can I find information about the representation theory of $\operatorname{Sp}(2g,\mathbb{Z}_3)$ ? I am lost when not working with a field of characteristic $0$. I need to study a quotient of $\operatorname{Sp}(2g,\mathbb{Z})$-modules $Q$ that is both a $\mathbb{Z}_3$-module and a $\operatorname{Sp}(2g,\mathbb{Z})$-module. If that can help, it is known that the reduction arrow from $\operatorname{Sp}(2g,\mathbb{Z})$ to $\operatorname{Sp}(2g,\mathbb{Z}_3)$ is surjective (a result by Morris Newman i think).

Edit : I was partially able to answer my questions, but i am not satisfied with the method i used : case by case computations and tricks to reduce the problem. Hence i am wondering : for such a finite group, how do we study its representations ? Should we just use the fact that it is finite, or is there a way to use the Lie structure ?

Thanks for helping. Also I am new to MathOverflow so feel free to comment on the form of my question.

Answer 1

I only just saw this question, and I know I'm late to the game, but let me attempt an answer. Let $p$ be an odd prime and $V$ be a $2g$ dimensional vector space over $k=\mathbb{Z}/p\mathbb{Z}$, carrying a non-degenerate symplectic form. Then I claim that as a module for $G=\mathop{\rm Sp}(V)$, the exterior square $\Lambda^2(V)$ has a trivial summand if and only if $g$ is not divisible by $p$.

Here is one way to see this. The symplectic form induces a $kG$-module isomorphism between $V$ and the dual space $V^*$. So $V \otimes V^*$ decomposes as a direct sum, $V\otimes V^* \cong S^2(V)\oplus\Lambda^2(V)$. Then $\mathop{\rm Hom}_{kG}(k,V\otimes V^*)\cong\mathop{\rm Hom}_{kG}(V,V)\cong\mathop{\rm Hom}_{kG}(V\otimes V^*,k)$ is one dimensional by Schur's lemma. The symplectic form gives a $G$-invariant element of $\Lambda^2(V)$, so actually $\mathop{\rm Hom}_{kG}(k,\Lambda^2(V))\cong \mathop{\rm Hom}_{kG}(\Lambda^2(V),k)$ is one dimensional, while $\mathop{\rm Hom}_{kG}(k,S^2(V))\cong\mathop{\rm Hom}_{kG}(S^2(V),k)=0$.

Now let us compute the composite of the homomorphism $k\to V\otimes V^*$ and the homomorphism $V\otimes V^* \to k$, in each case adjoint to the identity. Thinking of $V\otimes V^*$ as the space of matrices, the first has image the scalar matrices and the second is the trace. The composite is multiplication by $2g$, and is therefore non-zero if and only if $g$ is not divisible by $p$. Finally, $k$ is a direct summand of $\Lambda^2(V)$ if and only if the composite $k\to \Lambda^2(V)\to k$ is non-zero.

Since $\mathop{\rm Ext}^1_{kG}(k,k)=0$, the structure of $\Lambda^2(V)$ is as follows. If $p$ does not divide $g$, it is $k\oplus M$, where $M$ is an irreducible $kG$-module of dimension $2g^2-g-1$. If $p$ does divide $g$, it is uniserial $[k,M,k]$, where $M$ is an irreducible $kG$-module of dimension $2g^2-g-2$.

Answer 2

My guess would be that it is indecomposable and I would approach it as follows: consider first the exact sequence induced by $\omega$ over the rationals:

Inside $\bigwedge^2 V_Q$ you will have a lattice $\bigwedge^2 V_Z$ compute its image under $\omega$, and its intersection with the kernel of $\omega$. You will get an exact sequence of $Z$ modules. Reduce it modulo 3.