7 Character Table, Frobenius-Schur Indicator & Complexification

Proposition

For a character table,

• the rows as “weighted” orthogonal by Schur’s orthogonality relations;
• for columns $j$ and $j^{\prime }$,

$$\sum _{i=1}^r{\chi }_i(C_j)\overline{{\chi }_i(C_j^{\prime })}=\frac{|G|}{|C_j|}{\delta }_{j{j}^{\prime }};$$

• the matrix $U=\left[\sqrt{\frac{|C_j|}{|G|}}\right]{\chi }_i(C_j)$ satisfies $U{U}^{\ast }=\mathrm{I}$, where $U^{\ast }={\overline{U}}^T$;
• the complex conjugate of a row is a row of the character table by ${\overline{\chi }}_s={\chi }_{s^{\ast }}$ (possible the same);
• the complex conjugate of a column is also a column by $\overline{{\chi }_s(g)}={\chi }_s(g^{-1})$ (possible the same);
• (twisting) the pointwise product of a row starting with $1$ with any other row is again a row;
• the pointwise product of any two rows is a linear combination of rows of the character table with non-negative integer coefficients.

Example. Different groups with the same character table: $D_8$ and $Q_8$.

Frobenius-Schur Indicator

Definition

Frobenius-Schur indicator is defined as $FS(\chi )=\frac{1}{|G|}{\sum }_{g\in G}\chi (g^2)$, where $\chi$ is an irreducible character.

Remark. For compact group and its irreducible character $\chi$, define $FS(\chi )={\int }_{g\in G}\chi (g^2)d\mu$ where $\mu$ is Haar measure with $\mu (G)=1$ in Gordon Liebeck, real representation.

Theorem

Let $\chi$ be an irreducible character. Then

$$FS(\chi )=\{\begin{array}{ll}1& \text{mbox}if\chi \text{mbox}isreal-valuedand{S}_i\text{mbox}isofsymmetrictype\\ 0& \text{mbox}if\chi \text{mbox}isnotreal-valued\\ -1& \text{mbox}if\chi \text{mbox}isreal-valuedand{S}_i\text{mbox}isofskewsymmetrictype\end{array}$$

where $S_i$ is called of symmetric (rep. skew-symmetric) type if it has a non-zero symmetric (rep. skew-symmetric) $G$-invariant form.

Definition

We say that $S_i$ is symmetric (rep. skew-symmetric) type if $S_i$ admits non-zero symmetric (rep. skew-symmetric) $G$-invariant bilinear form. A bilinear map $B:V\times V\to \mathbb{C}$ is called a bilinear form, and is called

• $G$-invariant if $B(g{v}_1,g{v}_2)=B(v_1,v_2)$;
• symmetric if $B(v_1,v_2)=B(v_2,v_1)$;
• skew-symmetric if $B(v_1,v_2)=-B(v_2,v_1)$.

Remark. If $S_i$ has character ${\chi }_i$, then ${\overline{\chi }}_i$ is a character of $S_i^{\ast }$. Then:

• if $S_i$ is irreducible, then $S_i^{\ast }$ is irreducible as well;
• ${\chi }_i$ is real-valued iff ${\chi }_i={\overline{\chi }}_i$ iff $S_i$ is self-dual.

Tensor Square

If $V$ is a vector space, then $V\otimes V={\mathrm{S}}^{2}V\otimes {\Lambda }^{2}V$. If $v_1,\cdots ,v_n$ is basis of $V$, then $v_i\otimes v_j$ is basis of $V\otimes V$ and

$$\begin{array}{rl}{\mathrm{S}}^{2}V& =⟨v_i\otimes v_j+v_j\otimes v_i:1⩽i⩽j⩽n⟩,\\ {\Lambda }^{2}V& =⟨v_i\otimes v_j-v_j\otimes v_i:1⩽i<j⩽n⟩.\end{array}$$

Define $P:V\otimes V\to V\otimes V,v_i\otimes v_j\to v_j\otimes v_i$, then $P$ preserves ${\mathrm{S}}^{2}V$ and ${\Lambda }^{2}V$. Note that ${\mathrm{S}}^{2}V$ (rep. ${\Lambda }^{2}V$) is the eigenspace of $P$ corresponding to eigenvalue $1$ (rep. $-1$). It yields that ${\mathrm{S}}^{2}V=\ker (P-I)$ and ${\Lambda }^{2}V=\ker (P+I)$. Since $P^2=\mathrm{id}$, we can check that $\dim {\mathrm{S}}^{2}V=n(n+1)/2$, $\dim {\Lambda }^{2}V=n(n-1)/2$ and $V\otimes V={\mathrm{S}}^{2}V\oplus {\Lambda }^{2}V$.

If $V=S_i$, then

$$(S_i\otimes S_i{)}^G\simeq {\left(\mathrm{Hom}(S_i^{\ast },S_i)\right)}^G\simeq {\mathrm{Hom}}_G(S_i^{\ast },S_i)$$

and so

1,\mbox{ if }S_i\mbox{ is self-dual,} \\ 0,\mbox{ if }S_i\mbox{ is not self-dual.} \end{cases}$$ Note that $(S_i\otimes S_i)^G=(\mathrm S^2 S_i)^G\oplus(\Lambda^2S_i)^G$. Hence, if $S_i$ is self-dual, then $\dim(S_i\otimes S_i)^G=1$ and one of the following holds: ^efck4j - $\dim (\mathrm S^2S_i)^G=1$ and $\dim(\Lambda^2S_i)^G=0$, which will correspond to symmetric type of $S_i$; - $\dim (\mathrm S^2S_i)^G=0$ and $\dim(\Lambda^2S_i)^G=1$, which will correspond to skew-symmetric type of $S_i$. > [!lemma] > > Let $V$ be $G$-module, then > - $\chi_{\mathrm S^2V}(g)=\frac{1}{2}(\chi_V(g)^2+\chi_V(g^2))$; > - $\chi_{\Lambda^2V}(g)=\frac{1}{2}(\chi_V(g)^2-\chi_V(g^2))$; >- $\chi_{V\otimes V}(g)=\chi_V(g)^2=\chi_{\mathrm S^2V}(g)+\chi_{\Lambda^2V}(g)$. `\begin{proof}` Since $G$ is finite, $\rho(g)$ has finite order and so $\rho(g)$ is diagonalizable for all $g\in G$. For a fixed $G$ and vector space $V$, we have basis of eigenvectors $v_1,\cdots,v_n$ of $\rho(g)$ such that $\lambda_1,\cdots,\lambda_n$ are corresponding eigenvalues. It follows that

\chi_{\mathrm{S}^2V}(g)=\sum_{1\leqslant i\leqslant j\leqslant n}\lambda_i\lambda_j=\frac{1}{2}\left(\sum_{i=1}^n\lambda_i^2+(\sum_{i=1}^n\lambda_i)^2\right)=\frac{1}{2}(\chi(g)^2+\chi_V(g^2)).

$$Similarly,thereis$$

\chi_{\Lambda^2V}(g)=\frac{1}{2}(\chi_V(g)^2-\chi_V(g^2))

and now we finish the proof. `\end{proof}` Now we are ready to prove [[#^df16b3|^df16b3]]. `\begin{proof}` Note that

\begin{aligned} FS(\chi_i)&=\frac{1}{|G|}\sum_{g\in G}\chi_i(g^2)=\frac{1}{|G|}\sum_{g\in G}(\chi_{\mathrm{S}^2S_i}(g)-\chi_{\Lambda^2S_i}(g))\ &=\left\langle\chi_{\mathrm{S}^2S_i},1\right\rangle-\left\langle\chi_{\Lambda^2S_i},1\right\rangle=\dim (\mathrm{S}^2S_i)^G-\dim(\Lambda^2S_i)^G\ &=\begin{cases} 1 & \mbox{if }\chi\mbox{ is real-valued and }S_i\mbox{ is of symmetric type} \ 0 & \mbox{if }\chi\mbox{ is not real-valued}\ -1 & \mbox{if }\chi\mbox{ is real-valued and }S_i\mbox{ is of skew symmetric type} \end{cases} \end{aligned}

      by [[7 Character Table, Frobenius-Schur Indicator & Complexification#^efck4j|here]]. `\end{proof}` **Exercise.** Let $V$ be vector space, then $Bil(V)$ is a vector space of bilinear forms on $V$ Show that $V^*\otimes V^*\simeq Bil(V)$. (I think it is easy, with $f\otimes f'\mapsto B(v,u):=f(v)f'(u)$.) Now consider $P:V^*\otimes V^*\to V^*\otimes V^*,v^*\otimes u^*\to u^*\otimes v^*$. Since $Bil(V)\simeq V^*\otimes V^*$, we have that $\mathrm{S}^2V$ is the space of symmetric bilinear forms on $V$ and $\Lambda^2V$ is the space of skew-symmetric bilinear forms on $V$. If $V$ is $G$-module, then $V^*\otimes V^*$ is also $G$-module as well, by $g(f\otimes h)(u\otimes v)=f(g^{-1}u)\otimes h(g^{-1}v)$. If $B\in Bil(V)$ is $G$-invariant, then $(gB)(u,v)=B(g^{-1}u,g^{-1}v)=B(u,v)$ and so the space of $G$-invariant bilinear form is $(V^*\otimes V^*)^G$. Recall that

(V^\otimes V^)^G=(\mathrm S^2V^)^G\oplus(\Lambda^2 V^)^G,

and it yields that the space of $G$-invariant bilinear form can be decomposed as the space of symmetric $G$-invariant bilinear forms and the space of skew $G$-invariant bilinear forms. In fact, for each $f(u,v)\in Bil(V)$, it can be written as $f=f_1+f_2$ where $f_1(u,v)=f(u,v)+f(v,u)$ and $f_2(v,u)=f(u,v)-f(v,u)$. The decomposition above coincides with [[7 Character Table, Frobenius-Schur Indicator & Complexification#^efck4j|it]] and we also have names *real*, *complex* and *quaternionic* type for symmetric, complex and skew-symmetric. **Remark.** From linear algebra, odd dimensional space cannot have a non-degenerate skew symmetric bilinear form. Hence, if $S_i$ is self-dual and of odd dimension, then $S_i$ is of symmetric type. **Example.** Check $FS$ for characters of $D_8$ and $Q_8$. - Note that these two groups have the same character table. - When $G=D_8$, $\{g^2:g\in G\}=\{1,1,1,1,1,1,a^2,a^2\}$ and $FS(\chi_5)=\frac{1}{8}(2\times 6+(-2)\times 2)=1$ and $FS(\chi_i)=1$ for $i=1,2,3,4$. - When $G=Q_8$, $\{g^2:g\in G\}=\{1,1,-1,-1,-1,-1,-1,-1\}$ and $FS(\chi_5)=\frac{1}{8}(2\times 2+(-2)\times 6)=-1$ and $FS(\chi_i)=1$ for $i=1,2,3,4$, where $\chi_i$ is defined [[List 2#^tsqbbt|here]]. # Complexification Note that $2$-dimensional $D_4$ complex representation is just a complexification of $2$-dimensional real representation. If $U$ is a $\mathbb{R}G$-module, then $\mathbb{C}\otimes_{\mathbb{R}}U$ is a $\mathbb{C}G$-module. That is, if $\{u_1,\cdots,u_n\}$ is an $\mathbb{R}$-basis of $U$, then $\{1\otimes u_1,\cdots,1\otimes u_n\}$ is a $\mathbb{C}$-basis of $\mathbb{C}\otimes_{\mathbb{R}}U$ and usually we write that $\alpha_1 u_1+\cdots+\alpha_nu_n$ with $\alpha_i\in \mathbb{C}$. **Observation.** if $U$ is an irreducible $\mathbb{R}$-module, $U_{\mathbb{C}}$ is not always irreducible. For example, $C_3=\left\langle \begin{bmatrix}0 &-1\\1& -1\end{bmatrix}\right\rangle$ is irreducible over $\mathbb{R}$ but reducible over $\mathbb{C}$. > [!proposition] > > If $U_\mathbb{C}$ is simple, then it is of symmetric type.