Lecture 5

Lectures in Representation Theory

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

Last update: 19 August 2013

The previous results yield a great amount of information concerning semisimple algebras. How can one tell if an algebra is semisimple?

Theorem 1.41 Let $A$ be a finite dimensional algebra. The following are equivalent:

1.	The trace $tr$ of the regular representation of $A$ is nondegenerate.
2.	Every representation $V$ of $A$ is completely decomposable.
3.	$A ≃ ⨁_{d_{λ} \in \hat{A}} M_{d_{λ}} (ℂ) .$
4.	$\sqrt{A} = 0$

We will build the proof (and the new definitions) of this theorem over the next few lectures/pages.

Our previous definition of nondegenerate trace used the hypothesis that the algebra $A$ was semisimple. The following is an alternate definition of nondegeneracy which does not assume semisimplicity.

Definition 1.42 Let $\vec{t}$ be a trace on $A,$ a finite dimensional algebra. We say that $\vec{t}$ is nondegenerate if for every nonzero $a \in A$ there exists $b \in A$ such that $\vec{t} (b a) \neq 0 .$

Example 1.43 Let $\vec{t}$ be the ordinary trace $tr$ on $M_{d} (ℂ),$ and let $a \in M_{d} (ℂ)$ be the matrix unit $E_{i j} .$ Then for $b = E_{j i},$ one has that $\vec{t} (b a) = tr (E_{j j}) \neq 0 .$

Definition 1.44 Given an algebra $A$ with basis ${g_{1}, \dots, g_{d}},$ and let $\vec{t}$ be a trace on $A .$ The Gram matrix is the matrix $M = (⟨ g_{i}, g_{j} ⟩) = (\vec{t} (g_{i} g_{j})) .$

Proposition 1.45 Given an algebra $A$ with basis $G = {g_{1}, \dots, g_{d}},$ and let $\vec{t}$ be a trace on $A .$ The following are equivalent:

1.	$\vec{t}$ is a nondegenerate trace.
2.	The Gram matrix is invertible.
3.	$A$ has a dual basis with respect to (the inner product generated by) $\vec{t} .$

Proof.

$(3) \leftrightarrow (2) .$ Suppose $G^{*} = {g_{1}^{*}, \dots, g_{d}^{*}}$ is a dual basis for $A$ with respect to $\vec{t} .$ Let $C = (c_{j k})$ be the transition matrix between the bases $G$ and $G^{*}$ (that is, ${g_{j}}^{*} = \sum c_{j k} g_{k}).$ Then by duality, $δ_{i j} = ⟨ g_{i}, {g_{j}}^{*} ⟩ = ⟨ g_{i}, \sum_{k} c_{j k} g_{k} ⟩ = \sum_{k} c_{j k} ⟨ g_{i}, g_{k} ⟩ .$ This yields the matrix equation $C M^{t} = I_{d} = C M$ (since $\vec{t}$ is symmetric, $M^{t} = M).$ Thus the existence of a dual basis implies the invertibility of the Gram matrix.

Conversely, if the Gram matrix is invertible, let $C$ be its inverse, and define the $g_{i}^{*}$ as above. The invertibility of $C$ guarantees that the $g_{i}^{*}$ will form a basis, while the above calculation will show that the basis is dual to the original one.

$(3) \leftrightarrow (2)$ $□$

$(1) \leftrightarrow (2) .$ Let $a$ be a nonzero element of $A,$ and suppose that $\vec{t}$ is degenerate at $a;$ i.e. for every $b$ in $A$ one has $\vec{t} (b a) = 0 .$ In particular, let $b$ run through the basis $G,$ so that $\vec{t} (g_{i} a) = 0$ for all $i .$

Consider $a = \sum a_{j} g_{j},$ , for $a_{j} \in ℂ .$ Now we have that $0 = \vec{t} (g_{i} a) = \sum_{j} \vec{t} (g_{i} a_{j} g_{j}) = \sum_{j} ⟨ g_{i}, g_{j} ⟩ a_{j} = 0$

This yields the matrix equation $M (\begin{matrix} a \\ ⋮ \\ a_{d} \end{matrix}) = 0,$ whence that the Gram matrix $M$ is not invertible.

Conversely, if $M$ is not invertible, one can produce $\vec{a}$ as above in the kernel of $M;$ the trace $\vec{t}$ is degenerate at $a = \sum a_{j} g_{j} .$

$□$

Definition 1.46 We can think of the $C -algebra$ $A$ as a vector space, denoted $\vec{A},$ and consider the action of $A$ on $\vec{A}$ given by $a \cdot \vec{b} = \vec{a b} .$ This is the (left) regular action, and we will say that $\vec{A}$ is the (left) regular representation of $A .$

Let $\vec{t}$ be the trace of the regular representation of $V .$ Then for $a \in A,$ and $G = {\vec{g_{1}}, \dots, \vec{g_{d}}}$ a basis for $\vec{A},$ one has $\vec{t} (a) = \sum_{\vec{g_{i}} \in G} a \cdot \vec{g_{i}} |_{\vec{g_{i}}},$ where this last $a \cdot \vec{g_{i}} |_{\vec{g_{i}}}$ is meant to denote the coefficient of $\vec{g_{i}}$ in the image of $a \cdot \vec{g_{i}} .$ In the case that we have a dual basis, one can replace $a \cdot \vec{g_{i}} |_{\vec{g_{i}}}$ with $⟨ a g_{i}, g_{i}^{*} ⟩,$ and the above formula becomes $\vec{t} (a) = \sum_{\vec{g_{i}} \in G} a \cdot \vec{g_{i}} |_{\vec{g_{i}}} = \sum_{\vec{g_{i}} \in G} ⟨ a g_{i}, g_{i}^{*} ⟩ .$

Lemma 1.47 Given an algebra $A$ with basis $G = {g_{1}, \dots, g_{d}},$ and $\vec{t}$ a non- degenerate trace on $A .$ (Note that by the previous proposition, $G$ has a dual basis $G^{*} .)$ Let $W_{1} : A \to M_{d_{1}} (ℂ)$ and $W_{2} : A \to M_{d_{2}} (ℂ)$ be representations of $A .$ Let $C$ be any $d_{1} \times d_{2}$ matrix, and let $[C] = \sum W_{1} (g_{i}^{*}) C W_{2} (g_{i}) .$ (“symmetrize $C ”)$ Then for all $a$ in $A,$ $W_{1} (a) [C] = [C] W_{2} (a),$ so $[C]$ is an algebra homomorphism.

		Proof.
	$\begin{matrix} W_{1} (a) [C] & = & W_{1} (a) \sum_{g_{i} \in G} W_{1} (g_{i}^{}) C W_{2} (g_{i}) \\ = & \sum_{g_{i}} W_{1} (a g_{i}^{}) C W_{2} (g_{i}) & since W_{1} is an alg hom \\ = & \sum_{g_{i}} W_{1} (\sum_{h_{k}} ⟨ a g_{i}^{}, h_{k} ⟩ h_{k}^{}) C W_{2} (g_{i}) & expanding g_{i}^{} in terms of a basis h_{k}^{} \\ = & \sum_{h_{k}} W_{1} (h_{k}^{}) C W_{2} (\sum_{g_{i}} ⟨ a g_{i}^{}, h_{k} ⟩ g_{i}) & change order of summation, and move scalars around \\ = & \sum_{h_{k}} W_{1} (h_{k}^{}) C W_{2} (\sum_{g_{i}} ⟨ h_{k}, g_{i}^{} ⟩ g_{i}) & since \vec{t} (a g_{i}^{} h_{k}) = \vec{t} (h_{k} a g_{i}^{}) \\ = & \sum_{h_{k}} W_{1} (h_{k}^{*}) C W_{2} (h_{k} a) = [C] W_{2} (a) \end{matrix}$ $□$

Theorem 1.48 Let $A$ be a finite dimensional algebra. Suppose that $tr,$ the trace of the regular representation, is nondegenerate. Then every representation $V$ of $A$ is completely decomposable.

Proof.

Let $V$ be a representation of $A .$

case 1: $V$ is irreducible.

Done.

case 1: $V$ is not irreducible.

Since $V$ is not irreducible, there exists a submodule $V_{1}$ in $V,$ and without loss of generality we may presume that $V_{1}$ is irreducible. We want to produce an $A -submodule$ $V_{2}$ such that $V ≃ V_{1} \oplus V_{2} .$

Let $P : V \to V$ be a projection onto $V_{1}$ (i.e. $P$ is a $C -space$ morphism such that $P v_{1} = v_{1}$ for all $v_{1} \in V_{1}$ and $P V \subseteq V_{1}).$ One may produce examples of projections as follows: pick a basis for $V_{1},$ and extend to a basis for $V .$ Define $P$ by mapping the basis elements of $V_{1}$ to themselves, and the other basis elements of $V$ to any arbitrary vectors in $V_{1} .$

Idea of proof: Let $V_{2} = (I - P) V .$ Then $V = P V + (I - P) V = V_{1} + V_{2} .$ This doesn’t quite work, since there is no reason for this $V_{2}$ to be an $A -module.$ The $[P]$ will play a useful role here.

Let $P_{1} : V \to V$ be the map given by $P_{1} = [P] = \sum_{g_{i}} V (g_{i}^{*}) P V (g_{i}) .$

Claim 1: $P_{1}$ is a projection onto $V_{1} .$

Proof.

$P_{1} V \subseteq V_{1} :$ Note that for each $i,$ $P V (g_{i}) V \subseteq P V \subset V_{1},$ and that $V (g_{i}^{*}) V_{1} \subseteq V_{1} .$ This latter containment follows since $V_{1}$ is $A -stable.$

$P_{1} v_{1} = v_{1} \forall v_{1} \in V_{1} :$ This part only works if we are using the regular trace; we’ll see how that goes in the claim below.

\begin{matrix} P_{1} v_{1} & = & \sum_{g_{i}} V (g_{i}^{*}) P V (g_{i}) v_{1} \\ = & \sum_{g_{i}} V (g_{i}^{*}) V (g_{i}) v_{1} & since P is a projection \\ = & \sum_{g_{i}} V (g_{i}^{*} g_{i}) v_{1} \end{matrix}

Claim: $\sum_{g_{i}} g_{i}^{*} g_{i} = 1$ if $g_{i}^{*}$ is constructed from tr, the trace of the regular representation.

		Proof.
	Continued next lecture.

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

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