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 be a finite dimensional algebra. The following are equivalent:
|1.||The trace of the regular representation of is nondegenerate.|
|2.||Every representation of is completely decomposable.|
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 was semisimple. The following is an alternate definition of nondegeneracy which does not assume semisimplicity.
Definition 1.42 Let be a trace on a finite dimensional algebra. We say that is nondegenerate if for every nonzero there exists such that
Example 1.43 Let be the ordinary trace on and let be the matrix unit Then for one has that
Definition 1.44 Given an algebra with basis and let be a trace on The Gram matrix is the matrix
Proposition 1.45 Given an algebra with basis and let be a trace on The following are equivalent:
|1.||is a nondegenerate trace.|
|2.||The Gram matrix is invertible.|
|3.||has a dual basis with respect to (the inner product generated by)|
Suppose is a dual basis for with respect to Let be the transition matrix between the bases and (that is, Then by duality, This yields the matrix equation (since is symmetric, Thus the existence of a dual basis implies the invertibility of the Gram matrix.
Conversely, if the Gram matrix is invertible, let be its inverse, and define the as above. The invertibility of guarantees that the will form a basis, while the above calculation will show that the basis is dual to the original one.
Let be a nonzero element of and suppose that is degenerate at i.e. for every in one has In particular, let run through the basis so that for all
Consider , for Now we have that
This yields the matrix equation whence that the Gram matrix is not invertible.
Conversely, if is not invertible, one can produce as above in the kernel of the trace is degenerate at
Definition 1.46 We can think of the as a vector space, denoted and consider the action of on given by This is the (left) regular action, and we will say that is the (left) regular representation of
Let be the trace of the regular representation of Then for and a basis for one has where this last is meant to denote the coefficient of in the image of In the case that we have a dual basis, one can replace with and the above formula becomes
Lemma 1.47 Given an algebra with basis and a non- degenerate trace on (Note that by the previous proposition, has a dual basis Let and be representations of Let be any matrix, and let (“symmetrize Then for all in so is an algebra homomorphism.
Theorem 1.48 Let be a finite dimensional algebra. Suppose that the trace of the regular representation, is nondegenerate. Then every representation of is completely decomposable.
Let be a representation of
case 1: is irreducible.
case 1: is not irreducible.
Since is not irreducible, there exists a submodule in and without loss of generality we may presume that is irreducible. We want to produce an such that
Let be a projection onto (i.e. is a morphism such that for all and One may produce examples of projections as follows: pick a basis for and extend to a basis for Define by mapping the basis elements of to themselves, and the other basis elements of to any arbitrary vectors in
Idea of proof: Let Then This doesn’t quite work, since there is no reason for this to be an The will play a useful role here.
Let be the map given by
Claim 1: is a projection onto
This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.