Complete Reducibility

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 22 January 2010

Complete reducibility

(Maschke's theorem). Let $A$ be a finite dimensional algebra over a field $𝔽$ such that the trace $tr$ 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 $⟨⟩ : A \times A \to \overline{𝔽}$ defined by $⟨a_{1}, a_{2}⟩ = tr (a_{2} a_{1}), for all a_{1}, a_{2} \in A .$ The dual basis $B^{}$ exists because the trace $tr$ 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 $p \in End (M)$ be a projection onto $N$ , i.e. $p M = N$ and $p^{2} = p$ . Let $[p] = \sum_{b \in B} b p b^{}, and e = \sum_{b \in B} b b^{} . 5$ For all $a \in A$ , $\begin{array}{rcl} tr (e a) & = & \sum_{b \in B} tr (b b^{} a) \\ = & \sum_{b \in B} ⟨a b, b^{}⟩ \\ = & \sum_{b \in B} a {b\|}_{b} \\ = & tr (a) . \end{array}$ So $tr ((e - 1) a) = 0$ , for all $a \in A$ . Thus, since $tr$ is nondegenerate, $e = 1$ . Let $m \in M$ . Then $p b^{} m \in N$ for all $b \in B$ , and so $[p] m \in N$ . So $[p] M \subseteq N$ . Let $n \in N$ . Then $p b^{} n = b^{} n$ for all $b \in B$ , and so $[p] n = e n = 1 \cdot n = 1$ . So $[p] M = N$ and ${[p]}^{2} = [p]$ as elements of $End (M)$ . Note that $[1 - p] = [1] - [p] = e - [p] = 1 - [p]$ . So $M = [p] M \oplus (1 - [p]) M = N \oplus [1 - p] M,$ 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? ]* $[1 - p] M$ is an $A$ -submodule of $M$ which is complementary to $M$ [ SHOULD THIS BE 'N'? ]. By induction on the dimension of $M$ , $N$ and $[1 - p] M$ are completely decomposable, and therefore $M$ is completely decomposable. $□$

(Artin-Wedderburn theorem). Let $A$ be a finite dimensional algebra over an algebraically closed field $\overline{𝔽}$ . Let $\{b_{1}, \dots, b_{d}\}$ be a basis of $A$ and let $tr$ be the trace of the regular representation of $A$ . Then the following are equvalent

Every representation of

A

is completely decomposable.

The regular representation of

A

is completely decomposable.

A ≅ ⨁_{λ \in \hat{A}} M_{d_{λ}} (\overline{𝔽})

for some finite index set

\hat{A}

, and some

d_{λ} \in ℤ_{> 0}

The trace of the regular representation of

A

is nondegenerate.

\det (tr (b_{i} b_{j})) \neq 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. $□$

Remark. Let $R$ be an integral domain, and let $A_{R}$ be an algebra over $R$ with basis $\{b_{1}, \dots, b_{d}\}$ . Then $\det (tr (b_{i} b_{j}))$ is an element of $R$ and $\det (tr (b_{i} b_{j})) \neq 0$ in $\overline{𝔽}$ if and only if $\det (tr (b_{i} b_{j})) \neq 0$ in $R$ . In paricular, if $R = ℂ [x]$ , then $\det (tr (b_{i} b_{j}))$ is a polynomial. Since a polynomial has only a finite number of roots, $\det (tr (b_{i} b_{j})) (n) = 0$ 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.

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