The convolution and class algebras and
of a group
are apparently quite different from the other types of algebras we have seen, namely function algebras
and linear algebras
However, they are in fact closely related.
Theorem 9.1. The convolution algebra is isomorphic to a subalgebra of a linear algebra.
Proof: Since has a natural scalar product, the
-scalar product
it is a Hilbert space as well as an algebra. For each , consider the function
defined by
Observe that is a linear operator on
i.e.
Thus, is a function from the convolution algebra
to the linear algebra
Moreover, the map
itself is a linear transformation, i.e.
We claim that is an injective linear transformation. Since
is a linear transformation of
, it is completely determined by its action on the group basis
of
. Thus, it suffices to show that the linear operators
and
on
are distinct unless
Since
and
are linear operators on
they are uniquely determined by their actions on the group basis
of
. We have
hence
From this we can conclude that the image of in
under
is a subspace isomorphic to
. It remains to show that
is not just a linear transformation, but an algebra homomorphism, and we leave this as an exercise (that you really should do).
-QED
The proof of Theorem 8.1 is the linear version of Cayley’s theorem from group theory: instead of representing as a subgroup of the group of permutations of
it represents
as a subalgebra of the algebra of linear operators on
This is called the left regular representation of
Corollary 9.2. The class algebra is isomorphic to a function algebra.
Proof: Since is a subalgebra of
, and since
is isomorphic to a subalgebra of a linear algebra,
is isomorphic to a subalgebra of a linear algebra. Thus,
is isomorphic to a commutative subgalgebra of a linear algebra, and we have previously shown that all commutative subalgebras of linear algebras are isomorphic to function algebras.
-QED
For an example illustrating how the proof of Theorem 8.1 works, let us take to be a cyclic group of order four with generator
, so that
is the group unit. The group basis of
is thus
and we denote these vectors by
for brevity. Let
be any function on
and let
be its expansion in the group basis, so that
Problem 9.1. Show by direct calculation that the matrix of with respect to the ordered basis
of
is
In the case where is an abelian group, as in the example above, we have that
. Moreover, it is possible to establish Corollary 8.1 directly, without appealing to Theorem 8.1. This is done constructively, by finding an explicit basis of orthogonal projections in the commutative convolution algebra
called its Fourier basis. The advantage of this direct approach is that it also gives us an explicit description of the spectrum of all matrices in the left regular representation of
This is very useful in applications – in particular, matrices in the left regular representation of a cyclical group are called circulant matrices and they are important in engineering.
To see begin to see where a projection basis of might come from, recall that we previously showed the non-existence of an algebra homomorphism
for
a Hilbert space of dimension at least two. This reflects the fact that linear algebras
are maximally noncommuative. But we have also seen that for
a group with at least two elements, the dimension of
is at least two, so convolution algebras are always at least one degree more commutative than linear algebras, and therefore might admit homomorphisms to the complex numbers.
Theorem 9.3. A linear transformation is an algebra homomorphism if and only if the function
defined by
is a group homomorphism taking values in the unitary group of the complex numbers.
Problem 9.2. Prove Theorem 8.3.
Definition 9.1. A group homomorphism is called a character of
The set of all characters of
is denoted
.
Observe first of all that is a set of functions on
is a subset of
the space of all
-valued functions on
It is this special subset of functions tethered to the group law in
that we look to for orthogonal projections. No matter what the group
is, the set
is nonempty: it contains at least the trivial character defined by
However, for highly noncommutative groups there may not be many nontrivial characters.
Problem 9.3. Determine the Fourier dual of the symmetric group
For abelian groups the situation is much better and in fact we always have To begin, recall that every finite abelian group is isomorphic to a product of cyclic groups. We thus fix a positive integer
and
positive integers
and consider the group
where is a cyclic group of order
with generator
Define the dual group of
to be
That is,
where is the additive group of integers modulo
. We can parameterize
by the points of
writing
Indeed, the parameterization is a group isomorphism
(Exercise: prove this, noting that because
it is sufficient to show the parameterization is an injective group homomorphism).
Theorem 9.4. For every the function
defined by
where is a principal
th root of unity, is a character of
and every character of
is of this form.
Proof: For any it is clear that
because the identity element
has parameters
Moreover, for any
we have
so is indeed a group homomorphism
The fact that every homomorphism
is
for some
is left as an exercise.
-QED
We now have a special subset of the convolution algebra
of the finite abelian group
, namely the set
of all homomorphisms to the unitary group
We now claim that the characters form a basis of
Since the number of characters is
which is the dimension of
, it is sufficient to show that
is a linearly independent set in
Theorem 9.5. The set is orthogonal with respect to the
-scalar product on
– we have
Proof: For any we have
where
Thus if we have
and if we have
where the denominator of each fraction is nonzero and the numerator is zero, because is an
th root of unity.
-QED
The orthogonal basis of
is called its character basis. It is convenient to write
since this highlights the symmetry
The
symmetric matrix
is called the character table of
and Theorem 8.3 says that
is a symmetric unitary matrix. Another way to say the same thing is that the rescaled character basis
is an orthonormal basis of the convolution algebra In fact, the further scaling
is even better, for the following reason.
Theorem 9.6. The elements of the basis are orthogonal projections in
Proof: For any we have
For any we have
The internal sum is
where the final equality is Theorem 8.3. Thus
-QED
Since we know that any algebra with a basis of orthogonal projections is isomorphic to the function algebra (Lecture 3), Theorem 8.4 gives the promised second proof of the fact that the convolution algebra of an abelian group
is isomorphic to a function algebra. In this particular case, the basis
is known as the Fourier basis of
