Let 
where 
Theorem 17.1. The linear map 
Proof: Suppose first that 
and moreover for 
where 
which shows that 
Conversely, suppose we start from the assumption that 
and also
where in both calculations we are using the hypothesis that
where we used our hypothesis that the codomain of the group homomorphism
-QED
You will recognize that Theorem 17.1 was proved previously in the special case where 
Definition 17.1. A unitary representation of 
where
By Theorem 17.1, there is a bijective correspondence between linear representations of the convolution algebra 
Definition 17.2. If 
The set of all such linear maps is denoted 
Theorem 17.2. Suppose that 
Proof: The proof is an (interesting) exercise in polar decomposition, or singular value decomposition if you prefer. -QED
To further connect up with linear representations of algebras, let 
Theorem 17.3. The linear representation
Problem 17.1. Prove Theorem 17.3. (This is straightforward and is just to get you accustomed to the definitions in play).
Theorem 17.4 (Schur’s Lemma) Let
Proof: Special case of Schur’s lemma for linear representations of algebras. -QED
Theorem 17.5. (Maschke’s Theorem) Let
of 
Proof: Special case of Maschke’s theorem for linear representations of algebras.
-QED
Our goal is to classify all unitary representations of a finite group 
Now suppose that 
be the function on
One of the two main theorems next week is the following.
Theorem 17.6. The set 
This theorem generalizes the construction of the Fourier basis in
is just a homomorphic image of 
of 
for some given Hilbert space 
Then, Burnside’s theorem says that either 
or there is a proper non-trivial subspace 
invariant under all operators 
In this situation, 
is defined by declaring 
to be the restriction of 
to the invariant subspace 
In the case where 
the pair 
is a linear representation of 
the identity homomorphism; this is called the tautological representation or defining representation of 
are representations of 
a linear transformation 
is said to be a homomorphism of representations (or an intertwining transformation) if 
is a vector subspace of the space 
of all linear transformations 
is said to be an isomorphism of representations. When such an intertwining linear isomorphism exists, we say that 
of 
of 
-basis of 
in the 
-basis of 
for every algebra element 
, which is a subalgebra of 
is the subalgebra consisting of all operators 
that commute with every operator 
if and only if 
which is exactly the condition for 
to be an element of 
-invariant subspaces of 
we have the following.
is the centralizer of 
the statement is equivalent to the fact that the center of 
We claim that 
is a 
-invariant subspace of 
then for any 
we have
is again in 
Since 
or 
and since 
is a 
-invariant subspace of 
Indeed, suppose 
i.e. 
for 
Then, for any 
remains in the image of 
as claimed.
be any function on 
Taking the left regular representation, we have
is the linear operator on the Hilbert space 
given by the image of the Fourier basis vector 
under the injective algebra homomorphism 
In other words, 
and 
This says that the operators 
making up the image of the commutative algebra 
and moreover that calculating the eigenvalues of the operator 
is the same thing as calculating the Fourier transform of the function 
Hint: show that every 
circulant matrix is the image of a function on the cyclic group of order 
in the regular representation. 
, every partition of 
is called a class function if it is constant on conjugacy classes, meaning that 
for all 
for all 
belongs to 
if and only if it is constant on conjugacy classes.
For any 
, we have = \sum\limits_{h \in G} A(gh^{-1})B(h) = \sum\limits_{h \in G} A(g(gh)^{-1}) B(gh) = \sum\limits_{h \in G} A(gh^{-1}g^{-1})B(gh),](https://s0.wp.com/latex.php?latex=%5BAB%5D%28g%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7D+A%28gh%5E%7B-1%7D%29B%28h%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7D+A%28g%28gh%29%5E%7B-1%7D%29+B%28gh%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7D+A%28gh%5E%7B-1%7Dg%5E%7B-1%7D%29B%28gh%29%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
simply permutes the terms of the sum. Continuing the calculation, we have = \sum\limits_{h \in G} A(gh^{-1}g^{-1})B(gh) =\sum\limits_{h \in G} A(h^{-1})B(gh) = \sum\limits_{h \in G} B(gh)A(h^{-1}),](https://s0.wp.com/latex.php?latex=%5BAB%5D%28g%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7D+A%28gh%5E%7B-1%7Dg%5E%7B-1%7D%29B%28gh%29+%3D%5Csum%5Climits_%7Bh+%5Cin+G%7D+A%28h%5E%7B-1%7D%29B%28gh%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7D+B%28gh%29A%28h%5E%7B-1%7D%29%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
 = \sum\limits_{h \in G} B(gh^{-1})A(h) = [BA](g),](https://s0.wp.com/latex.php?latex=%5BAB%5D%28g%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7D+B%28gh%5E%7B-1%7D%29A%28h%29+%3D+%5BBA%5D%28g%29%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
is a central function; we will prove it is constant on conjugacy classes. Since 
commutes with all functions in 
it commutes with every elementary function 
. Since = \sum\limits_{k \in G}Z(hk^{-1})E_g(k)=Z(hg^{-1}),](https://s0.wp.com/latex.php?latex=%5BZE_g%5D%28h%29+%3D+%5Csum%5Climits_%7Bk+%5Cin+G%7DZ%28hk%5E%7B-1%7D%29E_g%28k%29%3DZ%28hg%5E%7B-1%7D%29%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
 = \sum\limits_{k\in G}E_g(hk^{-1})Z(k)=Z(g^{-1}h),](https://s0.wp.com/latex.php?latex=%5BE_gZ%5D%28h%29+%3D+%5Csum%5Climits_%7Bk%5Cin+G%7DE_g%28hk%5E%7B-1%7D%29Z%28k%29%3DZ%28g%5E%7B-1%7Dh%29%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
for all 
rather than 
let 
be the indicator function of 
so 
of 
span the class algebra of 
constant on conjugacy classes can be written
denotes the value 
for any 
We will show that 
-scalar product on 
is a partition of 
is an orthogonal (but not orthonormal) set of functions in ![[c_{\alpha\beta\gamma}]](https://s0.wp.com/latex.php?latex=%5Bc_%7B%5Calpha%5Cbeta%5Cgamma%7D%5D&bg=FFFFFF&fg=000&s=0&c=20201002)
of 
i.e. 
For any 
and any 
, show that
counts solutions to the equation 
in 
is any particular point of the conjugacy class 
and 
are required to belong to 
and 
respectively.
be a finite nonempty set.
-valued functions on  = A(x)B(x), \quad x \in X,](https://s0.wp.com/latex.php?latex=%5BAB%5D%28x%29+%3D+A%28x%29B%28x%29%2C+%5Cquad+x+%5Cin+X%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
are defined 
For each 
define the corresponding elementary function 
by
where 
is the =E_x(z)E_y(z)=\delta_{xz}\delta_{yz}=\delta_{xy}E_x(z).](https://s0.wp.com/latex.php?latex=%5BE_xE_y%5D%28z%29%3DE_x%28z%29E_y%28z%29%3D%5Cdelta_%7Bxz%7D%5Cdelta_%7Byz%7D%3D%5Cdelta_%7Bxy%7DE_x%28z%29.&bg=FFFFFF&fg=000&s=0&c=20201002)
is a linearly independent set in 
can be written as
are said to be isomorphic if there exists a 
are isomorphic.
be a vector space basis of 
be the function algebra of this set, and define a linear transformation
, where 
is the elementary function corresponding to 
. This is a vector space isomorphism, since it maps a basis of 
is also an algebra homomorphism. Indeed, by Theorem 1.2 and linearity of 
Next,
rather than 
Thus,
so defined is an algebra isomorphism. Note that the Fourier transform on 
The function 
is called the Fourier transform of the algebra element 
is the product of 
, which might be convoluted and difficult to calculate. On the other hand, the pointwise product of the corresponding functions 
is very simple, both conceptually and computationally. 
, the complex number system, elements of which are called scalars and denoted by lower-case Greek letters,
and 
as usual, and we write 
for the 
whose values are denoted by concatenating its arguments: 
. Associativity means that the symbol 
is unambiguous because its two possible meanings coincide:
denote the zero vector in 
for all 
such that
is distinct from the additive unit 
In fact, there is only one multiplicative unit. 
be multiplicative units in 
.
for the unique multiplicative unit. Later on, we may omit the subscript and simply write 
such that 
.
are such that 
. Prove that 
.
is the inverse of 
This is written 
and 
.
range over ![[\gamma_{xyz}]](https://s0.wp.com/latex.php?latex=%5B%5Cgamma_%7Bxyz%7D%5D&bg=FFFFFF&fg=000&s=0&c=20201002)
of complex numbers called the multiplication tensor of 
The elements of this three-tensor are called the connection coefficients of 
This set of 
numbers completely determines multiplication in 
whose values are denoted by a superscript asterisk: 
. Antilinearity means that conjugation interacts with the vector space operations according to the rule
![[\eta_{xy}]](https://s0.wp.com/latex.php?latex=%5B%5Ceta_%7Bxy%7D%5D&bg=FFFFFF&fg=000&s=0&c=20201002)
relative to the the basis 
of invertible elements in an algebra 
is invertible, and in fact 
.
.
.
.
.
of all selfadjoint elements in an algebra 
with 
selfadjoint. We say that 
is called a projection. Projections 
are said to be orthogonal if 
are such that 
where 
we have
we must have 
for each 
of all unitary elements in 
be its decomposition into real and imaginary parts. Then 
be any two selfadjoint elements, and set 
commute even if they are not selfadjoint. Then we can write 
and 
where 
are selfadjoint and thus commute with one another. Thus, 
is said to be an algebra homomorphism if
to indicate that 
together with an associative, bilinear, unital multiplication and a linear, involutive conjugation.
We write 
as 
and 
we declare
for some 
and 
be real elements of 
is also a real element of 
and 
are selfadjoint elements of 
we have that the real part of 
and its imaginary part is 
Since 
commute, 
of 
is the smallest subalgebra of 
are called central elements.
be any two central elements and 
, we have
.
(maximally commutative, minimally noncommutative). Any algebra 
have the lowest commutativity index; such maximally noncommutative algebras are called 
is the set of all elements in 
.
is indeed a subalgebra of 
are subalgebras of 
then 
and whose “largest” element is 
, a relation on 
of 
to denote the membership of 
in 
For example, you are familiar with the notion of an 
with the following properties :
for all 
;
iff 
;
then 
.
for all 
and 
then 
;
and 
then 
.
consisting of a set together with a partial order on it is called a 
by definition means 
. Importantly, note the use of the adjective “partial” – Definition 2.3 allows for the possibility that there are pairs of elements 
for which neither 
by inclusion:
, then 
subsets of 
is the largest subset of 
is the smallest subset of 
set is the only subset of 
. Similarly, the whole set 
is the only set satisfying 
to be the set of all vector subspaces of 
is an 
as intersection, because the intersection of two subspaces is again a subspace. However, the union of two subspaces is not, and we have to modify the definition to 
, as the empty subset of 
to be the set of all subalgebras of 
is an induced subposet of 
of subalgebras of 
into a lattice, because the span of the union 
of two subalgebras of 
, define 
to be the intersection of all subalgebras of 
be the set of all commutative subalgebras of 
This definition will fail if one of 
with the following property: if 
, then 
.
. If 
such that 
In this case, the algebra generated by 
is a commutative subalgebra of 
and this contradicts the maximality of 
Then for a commutative subalgebra 
we have
(note that the conclusion of Problem 2.1 was used here) .