Last update: 19 August 2013
Let we wish to show that
Let be arbitrary, and denote the coefficient of in any element by Then using the dual basis property. Hence for all and the nondegeneracy of tr implies that is
Hence is a projection.
Furthermore, is an Indeed, by the lemma from last time, commutes with hence with Therefore, for any and hence is an homomorphism mapping onto
Finally, we assert that Any may be written hence To show the sum is direct, note first that since fixes pointwise. Suppose so we may write for some Then as and hence which completes the proof of
Next, prove By the previous case, we may also assume that the all finite dimensional modules are completely decomposable.
Let be the hideously denoted left regular representation of Since we may write where is some (finite) index set.
Note that the representation is injective (sometimes called faithful) as for all Thus as algebras. The number are called the multiplicity of the irreducible representation in
For convenience, write Note that an arbitrary element of is of the form for some where there are occurrences of each in the block diagonal decomposition of The product in the algebra is componentwise, hence the map given by deleting duplicate copies of irreducibles, i.e. is an (onto) algebra homomorphism.
It is not hard to see that these algebras are isomorphic. Let be a basis for and let be the corresponding element to in the copy of in The set is a basis for since an element of must have identical matrices in all blocks indexed by a given Hence has the same dimension as indeed, the homomorphism above sends to (following the same indexing scheme) and is clearly invertible.
Setting we have It remains to show that This follows from the next extremely useful lemma:
Lemma 1.49 (Schur) Suppose and are irreducible representations of of dimensions and respectively. If is a matrix such that for all then either or is equivalent to and
We have to show that we will show that contains the basis of matrix units. Note that commutes with by the lemma of last time. By Schur’s lemma (with this element must be of the form for some (possibly zero). We calculate traces; using the trace property:
On the other hand, so for and if
Continued in next lecture.
This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.