We have been comparing and contrasting the function algebra of a finite group
with its convolution algebra
These objects are the same as vector spaces, but very different as algebras. In particular, we understand what is happening inside
very well, in the sense that we know all of its subalgebras, whereas our understanding of subalgebras of
is limited to those coming from subgroups
together with the class algebra
Today we will compare and contrast and
through the lens of states and traces. The starting point is the same in both cases: a very simple lemma identifying linear functionals on a function algebra
with functions in
Lemma 14.1. For any set , linear functionals on
are in linear bijection with functions in
.
Proof: For any functional , linear or not, we get a corresponding function
defined by
If is a linear functional on
, then it is uniquely determined by its values on the basis
of elementary functions, hence
is uniquely determined by
Conversely, if
is any function, we can define a linear functional on
by
It is clear that the maps and
are inverses of one another.
Recall our classification of states on , obtained using Lemma 14.1.
Theorem 14.2. The following are equivalent:
is a probability function on
is a state on
.
In fact, we know a few enhancements of Theorem 14.2: first, is a faithful state if and only if
for all $x \in X$; second,
is an algebra homomorphism if and only if
for some
In particular, all homomorphisms
are evaluation at a point, i.e. maps of the form
for a particular
As vector spaces, and
are not just isomorphic they are equal, hence Lemma 14.1 applies verbatim. However, since
and
are quite different as algebras, the classification of states on
is going to be quite different. To prepare yourself for this, solve the following problem.
Problem 14.1. Let be a probability function. Prove that the corresponding linear functional
is a state on
if and only if
is the indicator function of the group identity
Conclude that in this case
is evaluation at
but that this is not an algebra homomorphism
Now let us determine which functions correspond to states on
. The normalization condition is straightforward: since the multiplicative identity
is
we have
Now let us consider what property must have in order for
to be a nonnegative functional. For an arbitrary function
we have
so that
Hence if we pick an ordering of
, and write
this becomes
Definition 14.3. A complex-valued function on a group
is said to be nonnegative if for any
any
and any
we have
Problem 14.2. Prove is a nonnegative function on
if and only if, for all
and
the matrix
is Hermitian and has nonnegative eigenvalues. What is this matrix if
?
With the above in place, we can state our classification of states on the convolution algebra of a finite group
as follows.
Theorem 14.4. The following are equivalent:
is a nonnegative function on
;
is a state on
The classification of states on is even more straightforward.
Theorem 14.5. The following are equivalent:
is a central function on
;
is a trace on
.
Proof: Suppose is central. We know from last lecture that this is equivalent to
Thus,
Conversely, if is a trace on
then
whence is central.
We now have a complete classification of states and traces on the convolution algebra of a finite group
Namely, every linear functional on
has the form
for some function . We know that
is a state if and only if
and
is a nonnegative Hermitian matrix. We know that
is a trace if and only if
is a central function on