*** Problems in this lecture due March 15 ***
*** Please complete a course evaluation ***
Let 
be the category of finite-dimensional unitary representations 
of a finite group 
.
Problem 23.1. One-dimensional unitary representations of are the same thing as multiplicative characters of 
Explain this sentence carefully.
We also discussed a way to associate a scalar-valued function on to every unitary representation of . Namely, the character of is the function on defined by
After objects come morphisms. We have been working on our understanding of the hom-spaces in 
which are vector spaces of linear transformations. Namely, for any two unitary representations and 
of , the corresponding set of morphisms 
is the subspace of 
consisting of those linear transformations which satisfy
Transformations 
are said to intertwine the actions 
and 
Equivalently, is the space of -equivariant linear maps from 
to 
The first thing we did was prove a dimension formula, which we restate now for ease of reference.
Theorem 23.1. We have
This is at least reminiscent of our main result about multiplicative characters of a finite abelian group: they form an orthogonal basis of 
So, we want to see some sort of orthogonality emerging for characters of unitary representations which, once again, are higher-dimensional generalizations of multiplicative characters. We saw that the more general version of character orthogonality we are trying to develop is related to the concept of irreducibility.
Problem 23.2. Prove that if is abelian, all its irreducible representations are one-dimensional.
Last lecture, we proved the following.
Theorem 23.2 (Schur’s Lemma). If and are irreducible unitary representations of 
then every nonzero map is an isomorphism.
As a consequence of Theorems 23.1 and 23.2 we have the following: if and are irreducible and non-isomorphic, their characters are orthogonal: we have
In particular, this generalizes character orthogonality as we know it for finite abelian groups. This leads to the following.
Conjecture 23.3. If is an irreducible unitary representation of then
The goal of this lecture is to prove Conjecture 23.3. First of all, because of Theorem 23.1, for any unitary representation of we have
So Conjecture 23.3 is equivalent to the following.
Conjecture 23.4. If is an irreducible unitary representation of , then 
is one-dimensional.
First of all, let us clarify that 
is the space of all linear operators 
such that
As we have seen, this is in fact a subalgebra of 
, and e know that and every algebra has a unique one-dimensional subalgebra. So, Conjecture 23.4 is equivalent to the following.
Theorem 23.5. If is an irreducible unitary representation of then 
.
Proof: Let 
and let 
be an eigenvalue of 
Then, 
is again in the algebra 
but at the same time 
is a nontrivial -stable subspace of . Since is irreducible, 
, which is the same as 
The above argument relies on something we all take for granted. For any algebra 
and any element 
, the spectrum of 
is the set
Theorem 23.6 (Fundamental Theorem of Algebra). Every has nonempty spectrum 
Problem 23.3. Assuming the above, prove that the spectrum of any element of any finite-dimensional von Neumann algebra is nonempty.
The following generalizes Theorem 23.6.
Theorem 23.7 (Burnside’s Theorem). The only subalgebra 
which admits no proper non-trivial stable subspace 
is 
Let be any unitary representation of and consider the corresponding linear representation of 
defined by
Problem 23.4. Prove that is irreducible if and only if 
is irreducible.
Combining this problem with Burnside’s theorem, we have that if and hence is irreducible, then the image of in is 
Consequently, is the center of and a second proof of Theorem 23.5 follows if we prove directly that the center of an endomorphism algebra is minimal.
Theorem 23.8. The center of is 
Proof: Let 
be an orthonormal basis and let 
be the corresponding elementary basis of Suppose that 
commutes with all elementary operators. Then, for any 
we have
and likewise
so that
Furthermore, we have
and also
which forces 
.
