5 Semi-Simple and Characters

Semi-Simple

Example. For any algebra $A$, $A^{op}$ is the opposite algebra of $A$, where $a{\circ }_{A^{op}}b=ba$. Let $A={\mathrm{Mat}}_n(\mathbb{C})$, then $A^{op}=({\mathrm{Mat}}_n(\mathbb{C}),A{\circ }_{A^{op}}B=BA)$. Define $\phi :{\mathrm{Mat}}_n(\mathbb{C})\to {\mathrm{Mat}}_n^{op}(\mathbb{C}),M\mapsto M^T$. It follows that $A\simeq A^{op}$.

Definition

An algebra $A$ is called semi-simple if $\mathrm{rad}A=0$.

Theorem

Let $A$ be a finite-dimensional associative algebra. Then the followings are equivalent:

• $A$ is semi-simple.
• ${\sum }_{i=1}^r\dim V_i^2=\dim A$, where $V_1,\cdots ,V_r$ are all pairwise non-isomorphic irreducible $A$-modules.
• $A={\oplus }_{i=1}^r{\mathrm{Mat}}_{d_i}(\mathbb{C})$
• Any finite dimensional representation of $A$ is completely reducible, that is, the representation is a direct sum of irreducible representation.
• $A$ is completely reducible as a $A$-module.

\begin{proof}

(i)←>(ii). Note that $A$ is semi-simple iff $A\simeq {\oplus }_{i=1}^r\mathrm{End}(V_i)$ by ^b8akty iff $\dim A={\sum }_{i=1}^r\dim V_i^2$ by $\mathrm{End}(V_i)\simeq n_i{V}_i$.

(i)→(iii). Note that $A\simeq {\oplus }_{i=1}^r\mathrm{End}(V_i)={\oplus }_{i=1}^r{M}_{d_i}(\mathbb{C})$.

(iii)→(i). Any representation is a direct sum of copies of irreducible modules $V_i$, where $V_i$ is unique for $M_{d_i}(\mathbb{C})$. Thus $\mathrm{rad}A=0$ and so $A$ is semi-simple.

(iii)→(iv). By ^i6fgf3, any module is direct sum of irreducible. Then we get (iv).

(iv)→(v) is trivial.

(v)→(iii). Let $A={\oplus }_{i=1}^r{n}_i{V}_i$, where $V_i$ are pairwise non-isomorphic. By Schur lemma, ${\mathrm{End}}_A(V_i)\simeq \mathbb{C}$. It follows that ${\mathrm{End}}_A(n_i{V}_i)=M_{n_i}(\mathbb{C})$ and ${\mathrm{End}}_A(A)={\oplus }_{i=1}^r{M}_{n_i}(\mathbb{C})$. Take $f\in {\mathrm{End}}_A(A)$, then $f$ satisfies that $f(av)=af(v)$ for any $a,v\in A$. Since $f(a\cdot 1_A)=af(1_A)$, then $f$ is a multiplication by $f(1_A)$ and so ${\mathrm{End}}_A(A)\simeq A^{op}$. Therefore, $A\simeq {\oplus }_{i=1}^r{M}_{n_i}(\mathbb{C})$ and we get (iii). \end{proof}

Remark.

• $G$ and $\mathbb{C}G$ have the same representation. Let $G$ be a finite group. Then $\mathbb{C}G$ is semi-simple $⟺$ By ^7b841b, any representation for $\mathbb{C}G$, or for $G$, is completely reducible $⟺$ $|G|=\dim \mathbb{C}G=\sum (\dim V_i{)}^2$.
• Completely reducible is composable, but not vice versa. See here.

Characters

Definition

For any group representation $(V,\rho )$, its character is mapping ${\chi }_V:G\to \mathbb{C},g\mapsto \mathrm{tr}\rho (g)$.

Remarks.

• If $\dim V=1$, then ${\chi }_V(g)$ is just scalar $\rho (g)$.
• ${\chi }_V$ is in fact in $G^V$ (dual group)
• $\mathrm{Tr}(AB)\ne \mathrm{Tr}(A)\mathrm{Tr}(B)$, so ${\chi }_V$ is not a group homomorphism.
• ${\chi }_V(g)$ is the same on conjugacy class of $g$ in G.
• If $V\simeq W$ as $G$-modules, then ${\chi }_V={\chi }_W$ as ${\rho }_W(g)={\phi }^{-1}({\rho }_V(\phi (g)))$.

Definition

Define the vector space of class function on $G$ to be the set of functions from $G$ to $\mathbb{C}$ which are constant on conjugacy classes, denoted by $F_C(G,\mathbb{C})$. In particular, ${\chi }_V\in F_C(G,\mathbb{C})$.

Lemma

Let $V=V_1\oplus \cdots \oplus V_r$ be a $G$-module. Then ${\chi }_V={\chi }_{V_1}+\cdots +{\chi }_{V_r}$.

Example. Consider the permutation representation of $S_3$. Then ${\chi }_{{\mathbb{C}}^3}(g)=\{\text{mbox}numberoffixedpointsbyg\}$. Note that ${\chi }_{{\mathbb{C}}^3}={\chi }_S+{\chi }_W$ where $S=\{(x,y,z)\in {\mathbb{C}}^3:x+y+z=0\}$ and $W=\{(x,x,x):x\in \mathbb{C}\}$. Then ${\chi }_W=1$ and ${\chi }_S={\chi }_{{\mathbb{C}}^3}-{\chi }_W$.

Definition

The irreducible characters are characters of irreducible $G$-modules.

Lemma

Suppose that $f:M_n(\mathbb{C})\to \mathbb{C}$ is a linear map with $f(AB)=f(BA)$ for all $A,B\in M_n(\mathbb{C})$. Then $f=c\cdot \mathrm{Tr},c\in \mathbb{C}$.

\begin{proof} For any $i\ne j$, we have that $f(E_{ij})=f(E_{ii}{E}_{ij})=f(E_{ij}{E}_{ii})=0$ and $f(E_{ii})=f(E_{ij}{E}_{ji})=f(E_{ji}{E}_{ij})=f(E_{jj})$. Therefore, $f=c\cdot \mathrm{Tr}$ for some $c\in \mathbb{C}$. \end{proof}

Theorem

Irreducible characters form a basis of $F_C(G,\mathbb{C})$.

\begin{proof} Note that $F_C(G,\mathbb{C})=\{f\in \mathrm{Hom}(\mathbb{C}G,\mathbb{C}):f(ab)=f(ba),\forall a,b\in \mathbb{C}G\}$. Since $\mathbb{C}G$ is semi-simple, by ^7b841b, $\mathbb{C}G={\mathrm{Mat}}_{n_1}(\mathbb{C})\oplus \cdots \oplus {\mathrm{Mat}}_{n_r}(\mathbb{C})$. By ^230d2b, any $f\in F_C(G,\mathbb{C})$ can be written as

$$f=c_1{\mathrm{tr}}_{n_1}+\cdots +c_r{\mathrm{tr}}_{n_r}=c_1{\chi }_{V_1}+\cdots +c_r{\chi }_{V_r}.$$

Thus ${\chi }_{V_1},\cdots ,{\chi }_{V_r}$ is a set of generators of $F_C(G,\mathbb{C})$.

Assume that $c_1{\mathrm{tr}}_{n_1}+\cdots +c_r{\mathrm{tr}}_{n_r}=0$, where $c_i\in \mathbb{C}$. Note that

$$c_1=(c_1{\mathrm{tr}}_{n_1}+\cdots +c_r{\mathrm{tr}}_{n_r})(E_{11})=0$$

and similarly each $c_i=0$. Therefore, ${\chi }_{V_1},\cdots ,{\chi }_{V_r}$ are linear independent, and so irreducible characters form a basis of $F_C(G,\mathbb{C})$. \end{proof}

Corollary

The number of irreducible characters is equal to the number of conjugacy classes of $G$.

\begin{proof} Since $\dim F_C(G,\mathbb{C})=\#\text{mbox}conjugacyclasses$, then by ^c62c01 we have that

$$\#\text{mbox}conjugacyclasses=\#\text{mbox}irreducibleG\text{mbox}-modules.$$

Now we finish the proof. \end{proof}

Proposition

$V\simeq W$ iff ${\chi }_V={\chi }_W$.

\begin{proof} Assume that $V=m_1{V}_1\oplus \cdots \oplus m_r{V}_r$ where $V_i$ are pairwise non-isomorphic irreducible modules and $W=n_1{V}_1\oplus \cdots \oplus n_r{V}_r$. Then ${\chi }_V=m_1{\chi }_{V_1}+\cdots +m_r{\chi }_{V_r}$ and ${\chi }_W=n_1{\chi }_{V_1}+\cdots +n_r{\chi }_{V_r}$. Therefore, ${\chi }_V={\chi }_W$ iff $m_i=n_i$ iff $V\simeq W$. \end{proof}