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

$$\dim {\mathrm{Hom}}_G(S_i^{\ast },S_i)=\{\begin{array}{l}1,\text{mbox}if{S}_i\text{mbox}isself-dual,\\ 0,\text{mbox}if{S}_i\text{mbox}isnotself-dual.\end{array}$$

Note that $(S_i\otimes S_i{)}^G=({\mathrm{S}}^2{S}_i{)}^G\oplus ({\Lambda }^2{S}_i{)}^G$. Hence, if $S_i$ is self-dual, then $\dim (S_i\otimes S_i{)}^G=1$ and one of the following holds:

• $\dim ({\mathrm{S}}^2{S}_i{)}^G=1$ and $\dim ({\Lambda }^2{S}_i{)}^G=0$, which will correspond to symmetric type of $S_i$;
• $\dim ({\mathrm{S}}^2{S}_i{)}^G=0$ and $\dim ({\Lambda }^2{S}_i{)}^G=1$, which will correspond to skew-symmetric type of $S_i$.

Lemma

Let $V$ be $G$-module, then

• ${\chi }_{{\mathrm{S}}^{2}V}(g)=\frac{1}{2}({\chi }_V(g{)}^2+{\chi }_V(g^2))$;
• ${\chi }_{{\Lambda }^{2}V}(g)=\frac{1}{2}({\chi }_V(g{)}^2-{\chi }_V(g^2))$;
• ${\chi }_{V\otimes V}(g)={\chi }_V(g{)}^2={\chi }_{{\mathrm{S}}^{2}V}(g)+{\chi }_{{\Lambda }^{2}V}(g)$.

\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

$${\chi }_{{\mathrm{S}}^{2}V}(g)=\sum _{1⩽i⩽j⩽n}{\lambda }_i{\lambda }_j=\frac{1}{2}\left(\sum _{i=1}^n{\lambda }_i^2+(\sum _{i=1}^n{\lambda }_i{)}^2\right)=\frac{1}{2}(\chi (g{)}^2+{\chi }_V(g^2)).$$

Similarly, there is

$${\chi }_{{\Lambda }^{2}V}(g)=\frac{1}{2}({\chi }_V(g{)}^2-{\chi }_V(g^2))$$

and now we finish the proof. \end{proof}

Now we are ready to prove ^df16b3.

\begin{proof} Note that

$$\begin{array}{rl}FS({\chi }_i)& =\frac{1}{|G|}\sum _{g\in G}{\chi }_i(g^2)=\frac{1}{|G|}\sum _{g\in G}({\chi }_{{\mathrm{S}}^2{S}_i}(g)-{\chi }_{{\Lambda }^2{S}_i}(g))\\ & =⟨{\chi }_{{\mathrm{S}}^2{S}_i},1⟩-⟨{\chi }_{{\Lambda }^2{S}_i},1⟩=\dim ({\mathrm{S}}^2{S}_i{)}^G-\dim ({\Lambda }^2{S}_i{)}^G\\ & =\{\begin{array}{ll}1& \text{mbox}if\chi \text{mbox}isreal-valuedand{S}_i\text{mbox}isofsymmetrictype\\ 0& \text{mbox}if\chi \text{mbox}isnotreal-valued\\ -1& \text{mbox}if\chi \text{mbox}isreal-valuedand{S}_i\text{mbox}isofskewsymmetrictype\end{array}\end{array}$$

      by here. \end{proof}

Exercise. Let $V$ be vector space, then $Bil(V)$ is a vector space of bilinear forms on $V$ Show that $V^{\ast }\otimes V^{\ast }\simeq Bil(V)$. (I think it is easy, with $f\otimes f^{\prime }\mapsto B(v,u):=f(v)f^{\prime }(u)$.)

Now consider $P:V^{\ast }\otimes V^{\ast }\to V^{\ast }\otimes V^{\ast },v^{\ast }\otimes u^{\ast }\to u^{\ast }\otimes v^{\ast }$. Since $Bil(V)\simeq V^{\ast }\otimes V^{\ast }$, we have that ${\mathrm{S}}^{2}V$ is the space of symmetric bilinear forms on $V$ and ${\Lambda }^{2}V$ is the space of skew-symmetric bilinear forms on $V$. If $V$ is $G$-module, then $V^{\ast }\otimes V^{\ast }$ 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 $(V^{\ast }\otimes V^{\ast }{)}^G$. Recall that

$$(V^{\ast }\otimes V^{\ast }{)}^G=({\mathrm{S}}^2{V}^{\ast }{)}^G\oplus ({\Lambda }^2{V}^{\ast }{)}^G,$$

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 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 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 ${\alpha }_1{u}_1+\cdots +{\alpha }_n{u}_n$ with ${\alpha }_i\in \mathbb{C}$.

Observation. if $U$ is an irreducible $\mathbb{R}$-module, $U_{\mathbb{C}}$ is not always irreducible. For example, $C_3=⟨\left[\begin{array}{cc}0& -1\\ 1& -1\end{array}\right]⟩$ is irreducible over $\mathbb{R}$ but reducible over $\mathbb{C}$.

Proposition

If $U_{\mathbb{C}}$ is simple, then it is of symmetric type.