How to test if a $\mathbb{R} G$-module is irreducible?

Question

Let $V$ be a $\mathbb{C} G$-module with character $\chi$. We know that $V$ is irreducible if and only if the inner product $\left<\chi,\chi\right>=1$.

But what if $V$ is a $\mathbb{R} G$-module?

Answer 1

Over $\Bbb R G$ there are three types of irreducible modules. The first is got by considering an irreducible $\Bbb CG$-module with non-real character $\rho$ and restricting to $\Bbb RG$. This has character $\chi= \rho+\overline\rho$ and has $\left<\chi,\chi\right>=2$.

The second type are "orthogonal" representations. These are modules $M$ with $\left<\chi,\chi\right>=1$ with the property that $M\otimes\Bbb C$ is an irreducible module over $\Bbb C$ with real-valued character.

The third type are "symplectic" representations. An example is the representation of the quaternion group on $\Bbb R^4$ (as identified with $\Bbb H$). These come from restricting certain irreducible modules over $\Bbb CG$ to $\Bbb RG$. The character over $\Bbb R$ will be $\chi=2\rho$ where $\rho$ is the character over $\Bbb C$, and then $\left<\chi,\chi\right>=4$.

Answer 2

A $\mathbb{C}G$-module $V$ is irreducible if and only if its endomorphism ring $\operatorname{End}_{\mathbb{C}G}(V)$ consists of scalars, i.e. has dimension $1$. Since $\langle \chi_V, \chi_W \rangle = \dim \operatorname{Hom}_{\mathbb{C}G}(V, W)$, we have that $V$ is irreducible if and only if $\langle \chi_V, \chi_V \rangle = 1$.

For $\mathbb{R}G$-modules $V$ and $W$, we still have that $ \langle \chi_V, \chi_W \rangle = \dim \operatorname{Hom}_{\mathbb{R}G}(V, W)$, however now there are three possibilities for $\dim \operatorname{End}_{\mathbb{R}G}(V)$ when $V$ is irreducible. $\operatorname{End}_{\mathbb{R}G}(V)$ must be a finite-dimensional real division algebra, and so isomorphic to $\mathbb{R}$, $\mathbb{C}$, or the quaternions $\mathbb{H}$, which have real dimension $1$, $2$, and $4$ respectively. So if $\langle \chi_V, \chi_V \rangle > 4$, then $V$ cannot be irreducible, however if $\langle \chi_V, \chi_V \rangle \leq 4$, we need some extra information.

However, we have another characterisation of irreducibility for $\mathbb{R}G$-modules: $V$ is irreducible if and only if it admits a unique (up to scalar) invariant symmetric bilinear form. Invariant bilinear forms are elements of $(V^* \otimes V^*)^G$, and symmetric invariant bilinear forms are elements of $(\operatorname{Sym}^2 V^*)^G$, where $\operatorname{Sym}^2 V^*$ denotes the symmetric tensors in $V^* \otimes V^*$. So $V$ is irreducible if and only if $\dim (\operatorname{Sym}^2 V^*)^G = 1$.

We can write this last condition in terms of characters. Since $(\operatorname{Sym}^2 V^*)^G \cong \operatorname{Hom}_{\mathbb{R}G}(1, \operatorname{Sym}^2 V^*)$ as $\mathbb{R}$-vector spaces, we have $$ \dim (\operatorname{Sym}^2 V^*)^G = \dim \operatorname{Hom}_{\mathbb{R}G}(1, \operatorname{Sym}^2 V^*) = \langle \chi_1, \chi_{\operatorname{Sym}^2 V^*} \rangle$$ and since $$\chi_{\operatorname{Sym}^2 V^*}(g) = \frac{1}{2}\left( \chi_V(g^2) + \chi_V(g)^2 \right)$$ we finally get $$ \dim (\operatorname{Sym}^2 V^*)^G = \frac{1}{2|G|}\sum_{g \in G} \chi_V(g^2) + \chi_V(g)^2$$ and hence $V$ is irreducible if and only if this last expression is $1$.

If we define the Frobenius-Schur indicator $\nu_V = \frac{1}{|G|}\sum_{g \in g} \chi_V(g^2)$, then we can say that $V$ is irreducible if and only if $\langle \chi_V, \chi_V \rangle + \nu_V = 2$. For an irreducible $\mathbb{R}G$-module $V$, the Frobenius-Schur indicator $\nu_V$ also takes different values depending on whether $V$ is of real, complex or quaternionic type. Here is a short table of these numbers for $V$ an irreducible $\mathbb{R}G$-module:

$$\begin{matrix} & \langle \chi_V, \chi_V \rangle & \nu_V \\ V \text{ real} & 1 & 1 \\ V \text{ complex} & 2 & 0 \\ V \text{ quaternionic} & 4 & -2 \end{matrix}$$