4 Division algebra, density theorem and radical

Division Algebra

Definition

An algebra $\mathcal{D}$ is called a division algebra over field $k$ if any non-zero element has an inverse.

Example. $\mathbb{C}$ is a division algebra over $\mathbb{R}$.

Theorem

If $\mathcal{D}$ is finite-dimensional division algebra over algebraically closed field $k$, then $\mathcal{D}=k\cdot 1_{\mathcal{D}}$.

\begin{proof} Note that ${\mathcal{D}}^{\ast }=\mathcal{D}\setminus \{0\}$ is a multiplicative group of $\mathcal{D}$. Then $\mathcal{D}$ is a representation of ${\mathcal{D}}^{\ast }$, defined as $\rho :{\mathcal{D}}^{\ast }\to \mathcal{D},u\mapsto u$. Since $\rho (u_1{u}^{-1})u=u_1$ for any $u_1\in \mathcal{D}$, we have that $\rho$ is a irreducible representation. For any $c\in \mathcal{D}$, $r_c:\mathcal{D}\to \mathcal{D},v\mapsto vc$ is a homomorphism of irreducible ${\mathcal{D}}^{\ast }$-modules. Then by Schur’s lemma, $r_c=\lambda \mathrm{id}$. If $v=1_{\mathcal{D}}$, then $c=\lambda 1_{\mathcal{D}}$. It induces a map $f:\mathcal{D}\to k,c\mapsto \lambda$. Note that $f(c)=f(d)$ iff $c=d$, so $f$ is injective. On the other hand, for any $\mu \in k^{\ast }$, there is $f(\mu {\lambda }^{-1}c)=\mu$ and $f$ is surjective. Therefore, $\mathcal{D}=k\cdot 1_{\mathcal{D}}$. \end{proof}

Density Theorem

Proposition

Let $V_i(1⩽i⩽m)$ be irreducible finite-dimensional pairwise non-isomorphic representations of $A$, and let $V={\oplus }_{i=1}^m{n}_i{V}_i$, where $n_i$ is a number of copies of $V_i$ and $W$ is a sub-representation of $V$. Then $W={\oplus }_{i=1}^m{r}_i{V}_i$, $0⩽r_i⩽n_i$.

\begin{proof} See here. \end{proof}

To prove the density theorem, we need the following two lemmas.

Lemma

Let $V$ be an irreducible finite dimensional representation of algebra $A$, and let $v_1,\dots ,v_n\in V$ be any linearly independent vectors. Then for any $w_1,\dots ,w_n\in V$ there exists an element $a\in A$ such that $a{v}_i=w_i$.

\begin{proof} It is a exercise in the List 1. See here. \end{proof}

Lemma

Let $V$ be an irreducible representation of $A$ on dimension $n$. Then $\mathrm{End}(V)\simeq nV$ as $A$-modules.

\begin{proof} Let $Y=\mathrm{End}(V)$, and choose basis $\{v_1,\cdots ,v_n\}$ for $V$. Define $\phi :Y\to nV$ as $\phi (x)=(x{v}_1,\cdots ,x{v}_n)$ for any $x\in Y=\mathrm{End}(V)$. We claim that $\phi$ is an isomorphism of $A$-modules, that is, $\mathrm{End}(V)\simeq nV$. Since $a(x{v}_1,\cdots ,x{v}_n)=(ax{v}_1,\cdots ,ax{v}_n)=\phi (ax)$ we have that $a\phi (x)=\phi (ax)$ and $\phi$ is a homomorphism. With fixed $v_1,\cdots ,v_n$, any $T\in Y$ becomes $[T]$ with respect to basis $v_1,\cdots ,v_n$. Then $\phi :[T]\mapsto (T{v}_1,\cdots ,T{v}_n)$ is a bijection. \end{proof}

Density theorem

i) Let $V$ be irreducible finite-dimensional representation of $A$. Then the map $\rho :A\to \mathrm{End}(V)$ is surjective.

ii) Let $V=V_1\oplus \cdots \oplus V_n$, where $V_i$ are irreducible finite-dimensional pairwise non-isomorphic representations of $A$. Then $\rho =\oplus {\rho }_i:A\to \oplus \mathrm{End}(V_i)$ is surjective as well.

\begin{proof} (i) Let $B=\rho (A)\subseteq \mathrm{End}(V)$. Take any $c\in \mathrm{End}(V)$ and let $v_1,\cdots ,v_n$ be a basis of $V$. Let $w_i=c{v}_i$. But it follows that there exists $a\in A$ such that $w_i=a{v}_i$ by ^f1e9c8. So $\rho (a)=c$ and so $\rho$ is surjective.

(ii) Let $B_i={\rho }_i(A)=\mathrm{End}(V_i)\simeq d_i{V}_i$ by ^67e432. Let $B=\rho (A)\subseteq \oplus \mathrm{End}(V_i)\simeq \oplus d_i{V}_i$ where $d_i=\dim V_i$. Then $B=\mathrm{Im}\rho$ is a submodule in ${\oplus }_{i=1}^k{d}_i{V}_i$. Conversely, $B=\rho (A)\supseteq {\rho }_i(A)\simeq d_i{V}_i$ for all $i$. Therefore, we have that $B=\oplus \mathrm{End}(V_i)$. \end{proof}

Theorem

Let $A={\oplus }_{i=1}^r{M}_{d_i}(k)$ where $M_{d_i}(k)$ is the algebra generated by $d_i\times d_i$ matrices with entries in $k$. Then the irreducible representations of $A$ are $V_1=k^{d_1},\cdots ,V_r=k^{d_r}$ and any finite-dimensional representation of $A$ is direct sum of $V_i$.

\begin{proof} First, the given representations are clearly irreducible, as for any $v\ne 0,w\in$ $V_i$, there exists $a\in A$ such that $av=w$. Next, let $X$ be an $n$-dimensional representation of $A$. Then, $X^{\ast }$ is an $n$-dimensional representation of $A^{\mathrm{op}}$. But ${\left({\mathrm{Mat}}_{d_i}(k)\right)}^{\mathrm{op}}\cong {\mathrm{Mat}}_{d_i}(k)$ with isomorphism $\phi (X)=X^T$, as $(BC{)}^T=C^T{B}^T$. Thus, $A\cong A^{\text{op }}$ and $X^{\ast }$ may be viewed as an $n$-dimensional representation of $A$. Define

$$\varphi :\underset{n\text{ copies }}{\underbrace{A\oplus \cdots \oplus A}}⟶X^{\ast },(a_1,\cdots ,a_n)\mapsto a_1{y}_1+\cdots +a_n{y}_n$$

where $\left\{y_i\right\}$ is a basis of $X^{\ast }$. Note that $\varphi$ is clearly surjective, as $k\subset A$. Thus, the dual map ${\varphi }^{\ast }:X⟶A^{n\ast }$ is injective. But $A^{n\ast }\cong A^n$ as representations of $A$, which is easy to check. Hence, $\mathrm{Im}{\varphi }^{\ast }\cong X$ is a sub-representation of $A^n$. Next, ${\mathrm{Mat}}_{d_i}(k)=d_i{V}_i$, so $A={\oplus }_{i=1}^r{d}_i{V}_i,A^n={\oplus }_{i=1}^{r}n{d}_i{V}_i$, as a representation of $A$. Hence by ^ey40pv, $X={\oplus }_{i=1}^r{m}_i{V}_i$, as desired. \end{proof}

Remark. An alternating proof: see Exercise 16 and Etingof.

Radical of Algebra

Definition

We call $\mathrm{rad}A$ the set of elements of finite-dimensional algebra $A$ which acts trivially on any vector of irreducible $A$-module $V$.

Lemma

$\mathrm{rad}A$ is an ideal in $A$.

Theorem

A finite-dimensional algebra $A$ has only finitely many irreducible representations up to isomorphism. Suppose that $V_1,\cdots ,V_m$ is the complete set of irreducible pairwise non-isomorphic modules. Then $V_i$ are finite-dimensional and $A/\mathrm{rad}A\simeq {\oplus }_{i=1}^m\mathrm{End}{V}_i$.

\begin{proof} Let $V$ be irreducible. Take non-zero $v\in V$. Then $Av$ is a non-zero $A$-submodule in $V$ and $Av$ is finite-dimensional. Since $V$ is irreducible, then $Av=V$ is finite-dimensional. Let $V_1,\cdots ,V_r$ be any set of pairwise non-isomorphic irreducible modules. Then by ^ca6f8a, $\rho :A\to {\oplus }_{i=1}^r\mathrm{End}{V}_i$ is surjective. Since $\dim A⩾\dim {\oplus }_{i=1}^r\mathrm{End}{V}_r⩾r$, then $r$ is finite by $\dim A$ finite. Consider $\{V_1,\cdots ,V_m\}$ is the complete set of irreducible pairwise non-isomorphic modules. Then the map $\rho :A\to {\oplus }_{i=1}^m\mathrm{End}{V}_i$ is surjective with $\ker \rho =\mathrm{rad}A$ and so $A/\mathrm{rad}A\simeq {\oplus }_{i=1}^m\mathrm{End}{V}_i$. \end{proof}