# Unitary representations of finite groups over finite fields

I would like to learn the basic theory of unitary representations of finite groups over finite fields. Here, the unitary group $$\operatorname{GU}(n,\mathbb{F}_{q^2})$$ consists of all invertible transformations of $$\mathbb{F}_{q^2}^n$$ that preserve the Hermitian form $$\langle x, y \rangle = \sum_{i \in [n]} x_i y_i^q$$, and "unitary representation" means a group homomorphism $$\rho \colon G \to \operatorname{GU}(n,\mathbb{F}_{q^2})$$. This is a special case of the usual notion of a representation $$\rho \colon G \to \operatorname{GL}(n,\mathbb{F}_{q^2})$$.

Over the complex numbers, every representation $$\rho \colon G \to \operatorname{GL}(n,\mathbb{C})$$ of a finite group $$G$$ is similar to a unitary representation $$\rho' \colon G \to \operatorname{GU}(n,\mathbb{C})$$, in the sense that there is an invertible operator $$M$$ such that $$\rho'(g) = M\rho(g) M^{-1}$$ for every $$g \in G$$. In this sense and others, the theory of unitary representations over $$\mathbb{C}$$ is essentially the same as that of ordinary representations.

However, over finite fields the notions are distinct. If $$G$$ is a finite group and $$\rho \colon G \to \operatorname{GL}(n,\mathbb{F}_{q^2})$$ is a representation, there might not be an invertible operator $$M$$ such that $$M \rho(g) M^{-1} \in \operatorname{GU}(n,\mathbb{F}_{q^2})$$ for every $$g \in G$$. For example, $$\mathbb{Z}_5$$ has a faithful 2-dimensional representation over $$\mathbb{F}_{3^2}$$ that is not similar to any unitary representation, since 5 divides $$|\operatorname{GL}(2,\mathbb{F}_{3^2})|$$ but not $$|\operatorname{GU}(2,\mathbb{F}_{3^2})|$$.

Question: Have unitary representations of finite groups over finite fields been systematically studied, and if so where can I learn the basics?

Here is one example of what I want to learn to do:

1. Describe all the unitary representations of the dihedral group of order 8 when $$q=11$$.

At the moment I do not even know how to:

1. Describe all the unitary representations of $$\mathbb{Z}_2 \times \mathbb{Z}_2$$ when $$q=3$$.

Some other things I want to learn include:

1. Where Maschke's Theorem holds (i.e. $$(|G|,q) = 1$$ so that $$\mathbb{F}_{q^2}[G]$$ is semisimple), does every unitary representation decompose as an orthogonal direct sum of irreducible unitary subrepresentations?

2. Again where Maschke's Theorem holds, is there any analogue of the Peter-Weyl Theorem to give the space $$L^2(G)$$ of functions $$f \colon G \to \mathbb{F}_{q^2}$$ an orthogonal basis consisting of matrix elements for irreducible unitary representations?

3. What are some conditions for a modular representation to be similar to a unitary representation? (i.e. which subgroups of $$\operatorname{GL}(n,\mathbb{F}_{q^2})$$ are conjugate with subgroups of $$\operatorname{GU}(n,\mathbb{F}_{q^2})$$?)

Bonus for answers understandable to a humble analyst.

• @LSpice I see we need to be careful about what "irreducible" means for a unitary representation. Probably the right thing is to talk about unitary representations not just on $\mathbb{F}_{q^2}^n$ but on general unitary spaces, i.e. vector spaces over $\mathbb{F}_{q^2}$ equipped with nondegenerate sesquilinear forms. Then a "unitary subrepresentation" should mean an invariant subspace on which the form restricts nondegenerately (e.g. isotropic subspaces are out), and an "irreducible unitary representation" has no proper unitary subrepresentations. Apr 12, 2019 at 23:39
• @LSpice I guess the orthogonal complement of a unitary subspace is again unitary, and that they make an algebraic direct sum together. Then your argument shows that every unitary representation is the orthogonal direct sum of irreducible unitary subrepresentations. Great! Apr 12, 2019 at 23:41
• @LSpice One caveat is that an "irreducible unitary representation" is not necessarily irreducible as a $G$-module. For example, choose $G$ and $q$ so that $\mathbb{F}_{q^2}[G]$ is not semisimple. The regular representation is unitary when we identify $\mathbb{F}_{q^2}[G]$ with what I want to call $L^2(G)$. As a unitary representation it decomposes as a direct sum of irreducibles, but as a $G$-module it does not. Hence some of the irreducible unitary representations in the decomposition of $L^2(G)$ are not irreducible as $G$-modules. Apr 12, 2019 at 23:42
• @LSpice If we use this definition of irreducibility, then the embedding of $V \otimes V^*$ into $L^2(G)$ that you mentioned might not be injective. For instance, if $W \subset V$ is an invariant subspace on which the form is degenerate, then we can choose $u \in W \setminus \{0\}$ and $v \in (W \cap W^\perp) \setminus \{0\}$, and the map you described sends $u \otimes v^* \mapsto 0$ (if I understand you correctly) since $\langle \rho(g) u, v \rangle = 0$ for every $g \in G$. Apr 12, 2019 at 23:43
• I agree that my comments are not appropriate for the definition of irreducibility you have in mind. Sorry; I'm not used to thinking about modular representations. (Perhaps, in case others share my confusion, it is a good idea to edit the definition into the question?) Apr 13, 2019 at 2:43

We had to deal with this problem when classifying maximal subgroups of the finite classical groups, which is the aim of our book:

The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, by John N. Bray, Derek F. Holt, Colva M. Roney-Dougal.

The most difficult maximal subgroups to classify, are those in the so-called Aschbacher class $${\mathscr S}$$, consisting of absolutely irreducible subgroups that are almost simple mod scalars. Many of these arise as reductions of complex representations over finite fields. Tables of complex representations of groups that are close to simple are available up to dimension about $$250$$, but we needed to know which classical group the reduction lies in, which means identifying the fixed form.

We generally relied on Lemma 4.4.1 of the book, which says:

For a given absolutely irreducible representation over $${\mathbb F}_{q^2}$$ of a group $$G$$, with Frobenius-Schur indicator $$\circ$$, the image of $$G$$ under the representation consists of isometries of a unitary form if and only if the action of the field automorphism $$\sigma :x \to x^q$$ on the Brauer character is the same as complex conjugation.

In many cases, such as when $$q$$ is coprime to the group order, the Brauer character is just the ordinary complex character.

As an example, the reduction of the complex representation of degree $$3$$ of the $$3$$-fold cover $$3.A_6$$ of $$A_6$$ lies in $${\rm PSL}(3,p)$$ for primes $$p \equiv 1,4 \bmod 15$$, in $${\rm PSU}(3,p)$$ (as a subgroup of $${\rm PSL}(3,p^2)$$) when $$p \equiv 11,14 \bmod 15$$ (or when $$p=5$$), and in $${\rm PSL}(3,p^2)$$ without preserving a unitary form when $$p \equiv 2,3 \bmod 5$$.

• Unless you think there's no risk of confusion, it may be worth it, for people like me who read "representation of a group over $\mathbb F_{q^2}$" as "representation of (a group $G$ over $\mathbb F_{q^2}$)", seeming to consider representations of algebraic groups, to re-phrase as "representation over $\mathbb F_{q^2}$ of a group $G$", so that it's clear we're specifying the field of definition of the representation and not of the group (the latter meaningless anyway in the abstract setting considered). Apr 12, 2019 at 0:50
• OK, I have reworded it, but I was quoting the lemma directly from the book! Apr 12, 2019 at 2:03
• A collision of cultures! :) Feb 5 at 23:58