7* Notes for Frobenius-Schur Indicator

Equivalent definition of real type/complex type

To the extent that we are interested in the action of a group $G$ on real rather than complex vector spaces, the problem we face is to say which of the complex representations of $G$ we have studied are in fact real.

Definition

If a group $G$ acts on a real vector space $V_0$, then we say the corresponding complex representation of $V=V_0{\otimes }_{\mathbb{R}}\mathbb{C}$ is real.

Remark. In the other word, if $\rho (g)$ is a matrix over $\mathbb{R}$ for all $g\in G$, then the corresponding $V$ is real and so $V$ can be written as $V=V_0{\otimes }_{\mathbb{R}}\mathbb{C}$.

Recall that the character tables of $Q_8$ and $D_8$ are same, and so all characters of $Q_8$ are real. But the $2$-dimensional representation of $Q_8$, which is defined here, can not defined over $\mathbb{R}$.

Proposition

The $2$-dimensional representation of $Q_8$ can not be defined over $\mathbb{R}$.

\begin{proof} Let $C=\{\pm 1,\pm i\}$ be the subgroup of $Q_8$ and let $V$ be a real vector space of dimension $2$ where $i$ acts by $M:=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]$. Assume that the $2$-dimensional representation $\rho$ of $Q_8$ can be defined over $\mathbb{R}$, then the restriction of $\rho$ to $C$ is isomorphic to $V$. Assume that $\rho (j)=N$, then

$$MN=N{M}^3\text{mbox}and{M}^2=N^2.$$

By solving it directly, $M$ does not exist. \end{proof}

Motivated by ^12152a, we have a classification of representations. An irreducible representation $V$ is one of the following:

• $V$ is real;
• $V$ is not real but $\chi$ is real;
• $\chi$ is not real.

Proposition

Assume that $\rho$ is an irreducible representation, then the followings are equivalent:

• $\rho$ is self-dual;
• there exists non-degenerated $G$-invariant bilinear form.

\begin{proof} Assume that $B$ is a non-degenerated $G$-invariant bilinear form, then define

$$\mathcal{F}:V\to V^{\ast },x\mapsto B(x,\cdot ).$$

It follows that $V\simeq V^{\ast }$ and so $\rho ,{\rho }^{\ast }$ are equivalent.

Assume that $\rho$ and ${\rho }^{\ast }$ are equivalent, then there exists an isomorphism $\phi :V\to V^{\ast }$ as modules, satisfying $\phi (gv)=g\phi (v)$. Define $B(x,y):=\phi (x)(y)$. Note that

$$\begin{array}{rl}B(gx,gy)& =\phi (gx)(gy)=(g\phi (x))(gy)\\ & =\phi (x)(g^{-1}(gy))=\phi (x)(y)\\ & =B(x,y)\end{array}$$

Therefore, $B$ is non-degenerated and $G$-invariant. \end{proof}

Remark. Notice that $B$ is determined by $\phi$. If $B_1,B_2$ are two different $G$-invariant bilinear forms, then there exists distinct isomorphisms of modules ${\phi }_1,{\phi }_2:V\to V^{\ast }$. It follows that ${\phi }_1^{-1}\circ {\phi }_2:V\to V$ is an isomorphism. By Schur’s lemma, ${\phi }_1^{-1}\circ ({\phi }_2(v))=\lambda v$ for some $\lambda \in \mathbb{C}$ and so ${\phi }_1(\lambda v)={\phi }_2(v)$ and so $\lambda B_1=B_2$. Therefore, for a self-dual representation, there exists the unique non-degenerated $G$-invariant bilinear form up to scalar.

Proposition

An irreducible representation V of $G$ is real if and only if there is a non-degenerate symmetric bilinear form B on V preserved by $G$.

\begin{proof} See here. It is just a rewritten version of this lemma. \end{proof}

Real Irreducible Representation

Let $V_{\mathbb{R}}$ be a real vector space and let $e_1,\cdots ,e_n$ be a set of basis. The complexification of $V_{\mathbb{R}}$ is defined to be $V_{\mathbb{C}}:=V_{\mathbb{R}}{\otimes }_{\mathbb{R}}\mathbb{C}$.

Note that

$$V=\left\{\sum _{k=1}^n{\lambda }_k{e}_k:{\lambda }_k\in \mathbb{C}\right\}=\{u+iv:u,v\in V_{\mathbb{R}}\}$$

and $V_{\mathbb{C}}=\{v_1\otimes 1+v_2\otimes i:v_1,v_2\in V_{\mathbb{R}}\}$ can be identified by $v_1\otimes 1+v_2\otimes i=v_1+i{v}_2$. Define $\sigma :V\to V,u+iv\mapsto u-iv$. Then for a complex vector space $V$, $U:=\{u:\sigma (u)=u\}$ is decomplexification of $V$.

The following proposition comes from here.

Proposition

Let $V_0$ be a real vector space on which $G$ acts irreducibly, $V=V_0\otimes \mathbb{C}$ the corresponding real representation of $G$. Show that if $V$ is not irreducible, then it has exactly two irreducible factors, and they are conjugate complex representations of $G$.

\begin{proof} Since $V$ is not irreducible, there exists an irreducible submodule $W\subseteq V$. Define $U=\sigma (W)$, then $U$ is also an irreducible submodule. It follows that either $W\cap U=\{0\}$ or $W\cap U=W$. If $W\cap U=W$, then $W_0:=\{v\in W:\sigma (v)=v\}$ is a submodule of $V_0$, which is impossible. Therefore, $W\cap U=\{0\}$.

Define $V_1:=W+U\subseteq V$, then $V_1$ is a submodule of $V$ and $\sigma (V_1)=V_1$. Define $V_0^{\prime }=\{v\in V_1:\sigma (v)=v\}$, then $V_0^{\prime }$ is a submodule of $V_0$ and so $V_0^{\prime }=V_0$. It deduces that

$$V_1=V_0^{\prime }\otimes \mathbb{C}=V_0\otimes \mathbb{C}=V$$

and so $V=W\oplus U=W\oplus \sigma (W)$. Now we finish the proof. \end{proof}

Applications

For a finite group $G$, define $\vartheta (g)=\left|\left\{h\in G:h^2=g\right\}\right|$. Then

$$\begin{array}{rl}⟨\vartheta ,{\chi }_{\rho }⟩& =\frac{1}{|G|}\sum _{g\in G}\vartheta (g){\chi }_{\rho }\left(g^{-1}\right)\\ & =\frac{1}{|G|}\sum _{g\in G}\sum _{h^2=g}{\chi }_{\rho }\left(h^{-2}\right)\\ & =\frac{1}{|G|}\sum _{h\in G}{\chi }_{\rho }\left(h^{-2}\right)\\ & =FS(\rho ).\end{array}$$

It follows that $\vartheta ={\sum }_{\rho \in \mathrm{Irr}(G)}FS(\rho ){\chi }_{\rho }$ and so the number of involutions of $G$ is equal to ${\sum }_{\rho }FS(\rho )\mathrm{deg}\rho -1$.

If $|G|$ is even, then $G$ has at least one involutions and so there exists an irreducible $\rho$ such that $FS(\rho )=1$, i.e. $\rho$ is of real type.

If $|G|$ is odd, then $FS(\chi )=\frac{1}{|G|}{\sum }_{g\in G}\chi (g^2)=\frac{1}{|G|}{\sum }_{g\in G}\chi (g)\psi (g^{-1})=⟨\chi ,\psi ⟩=0$ for all non-trivial irreducible representation. Therefore, for any non-trivial representation $\rho$, $\rho$ is not self-dual and for each $n\ne 1$, the number of representations of dimension $n$ is even. Furthermore, for each non-trivial representation $\rho$ of $G$, $\dim \rho$ is odd.