Complete Reducibility

## Complete reducibility

(Maschke's theorem). Let $A$ be a finite dimensional algebra over a field $𝔽$ such that the trace $\mathrm{tr}$ of the regular representation of $A$ is nondegenerate. Then every representation of $A$ is completly decomposable.

 Proof. Let $B$ be a basis of $A$ and let ${B}^{*}$ be the dual basis of $A$ with respect to the form $⟨\phantom{\rule{.5em}{0ex}},\phantom{\rule{.5em}{0ex}}⟩:A×A\to \stackrel{‾}{𝔽}$ defined by $a 1 a 2 = tr a 2 a 1 , for all a 1 a 2 ∈ A .$ The dual basis ${B}^{*}$ exists because the trace $\mathrm{tr}$ is nondegenerate. Let $M$ be an $A$-module. If $M$ is reducible then the result is vacuously true, so we may assume that $M$ has a proper submodule $N$. Let $p\in \mathrm{End}\left(M\right)$ be a projection onto $N$, i.e. $pM=N$ and ${p}^{2}=p$. Let $p = ∑ b ∈ B b p b * , and e = ∑ b ∈ B b b * . 5$ For all $a\in A$, $tr e a = ∑ b ∈ B tr b b * a = ∑ b ∈ B a b b * = ∑ b ∈ B a b b = tr a .$ So $\mathrm{tr}\left(\left(e-1\right)a\right)=0$, for all $a\in A$. Thus, since $\mathrm{tr}$ is nondegenerate, $e=1$. Let $m\in M$. Then $p{b}^{*}m\in N$ for all $b\in B$, and so $\left[p\right]m\in N$. So $\left[p\right]M\subseteq N$. Let $n\in N$. Then $p{b}^{*}n={b}^{*}n$ for all $b\in B$, and so $\left[p\right]n=en=1·n=1$. So $\left[p\right]M=N$ and ${\left[p\right]}^{2}=\left[p\right]$ as elements of $\mathrm{End}\left(M\right)$. Note that $\left[1-p\right]=\left[1\right]-\left[p\right]=e-\left[p\right]=1-\left[p\right]$. So $M = p M ⊕ 1 - p M = N ⊕ 1 - p M ,$ and by proposition (1.2(b), Algebras) [ AGAIN THIS IS A LINK TO AN OUTSIDE PAGE WHICH WILL DIE IF THE OTHER PAGE IS UPDATED. HOW SHOULD WE FORMAT THESE LINKS? ] $\left[1-p\right]M$ is an $A$-submodule of $M$ which is complementary to $M$ [ SHOULD THIS BE 'N'? ]. By induction on the dimension of $M$, $N$ and $\left[1-p\right]M$ are completely decomposable, and therefore $M$ is completely decomposable. $\square$

1. Every representation of $A$ is completely decomposable.
2. The regular representation of $A$ is completely decomposable.
3. $A\cong \underset{\lambda \in \stackrel{ˆ}{A}}{⨁}{M}_{{d}_{\lambda }}\left(\stackrel{‾}{𝔽}\right)$ for some finite index set $\stackrel{ˆ}{A}$, and some ${d}_{\lambda }\in {ℤ}_{>0}$.
4. The trace of the regular representation of $A$ is nondegenerate.
5. $\mathrm{det}\left(\mathrm{tr}\left({b}_{i}{b}_{j}\right)\right)\ne 0$.

 Proof. The result follows from theorem 1.1, theorem 1.2 on the Regular Representations page and theorem 1.3 on the Regular Representations page. $\square$

Remark. Let $R$ be an integral domain, and let ${A}_{R}$ be an algebra over $R$ with basis $\left\{{b}_{1},\dots ,{b}_{d}\right\}$. Then $\mathrm{det}\left(\mathrm{tr}\left({b}_{i}{b}_{j}\right)\right)$ is an element of $R$ and $\mathrm{det}\left(\mathrm{tr}\left({b}_{i}{b}_{j}\right)\right)\ne 0$ in $\stackrel{‾}{𝔽}$ if and only if $\mathrm{det}\left(\mathrm{tr}\left({b}_{i}{b}_{j}\right)\right)\ne 0$ in $R$. In paricular, if $R=ℂ\left[x\right]$, then $\mathrm{det}\left(\mathrm{tr}\left({b}_{i}{b}_{j}\right)\right)$ is a polynomial. Since a polynomial has only a finite number of roots, $\mathrm{det}\left(\mathrm{tr}\left({b}_{i}{b}_{j}\right)\right)\left(n\right)=0$ for only a finite number of values in $n\in ℂ$.

## Reference

[HA] T. Halverson and A. Ram, Partition algebras, European Journal of Combinatorics 26, (2005), 869-921; arXiv:math/040131v2.