Let 
Definition 16.1. A linear representation of 
Note that Definition 16.1 does not stipulate that 
We saw a special case of Definition 16.1, where we considered the special case of a subalgebra 
Definition 16.2. If 
The set of all such linear transformations is denoted
Problem 16.1. Prove that 
Definition 16.3. A linear isomorphism 
We also encountered Definition 16.3 in a special case in Lecture 15, when we discussed the classification of subalgebras of 
Problem 16.2. Let 
Let us fix a linear representation 
Proposition 16.1. We have
Proof: We have 
-QED
Definition 16.4. A linear representation 
Problem 16.3. Prove that
Proposition 16.2. If
Proof: This is Burnside’s theorem from Lecture 15: the only subalgebra of
-QED
Continuing on with 
Corollary 16.3.
Proof: Since 
-QED
We can refine the above corollary to make the following statement about intertwining maps between two possibly distince representations
Theorem 16.4. (Schur’s Lemma) If
Proof: Suppose 
which shows that 
Now suppose 
showing that 
The final statement follows from the two arguments above: any nonzero homomorphism of irreducible representations must be an isomorphism. Equivalently, there does not exist a nonzero homomorphism between two non-isomorphic irreducible representations.
-QED
and 
As we saw in Lecture 14, the trace on 
such that 
Thus, the only linear functional satisfying 
and 
is
indicates that this is an algebra with the highest degree of noncommutativity. According to our definitions in Week One, the degree of noncommutativity of an algebra is formalized as the dimension of its center. We now prove that the center of 
we have 
and the total space 
there is a third 
by Burnside’s theorem. The orthogonal complement 
of 
and we claim it is 
and 
i.e. 
of scalar operators. 
be a central element, and let 
be an eigenvalue of 
Then, the 
-eigenspace of 
is an 
is an eigenvector of 
is any operator we have
is again in the 
where subalgebras are indexed by partitions 
of the finite set 
. The classification is very simple: the subalgebra 
of a given partition 
A nice feature of this classification is that it shows every subalgebra of 
is itself isomorphic to a function algebra, because 
The first step is to formulate the vector space version of partitions of a set.
of nonzero subspaces of 
as the blocks of 
to be the set of all operators 
is a subalgebra of 
is the linear partition of 
the subalgebra 
is a linear partition of 
be an ordered basis of 
and let 
be an ordered basis of 
Then 
is an ordered basis of 
whose matrix relative to this ordered basis has the form ![[A]_{(e_1,\dots,e_m,f_1,\dots,f_n)}=\begin{bmatrix} M_1 & {} \\ {} & M_2 \end{bmatrix},](https://s0.wp.com/latex.php?latex=%5BA%5D_%7B%28e_1%2C%5Cdots%2Ce_m%2Cf_1%2C%5Cdots%2Cf_n%29%7D%3D%5Cbegin%7Bbmatrix%7D+M_1+%26+%7B%7D+%5C%5C+%7B%7D+%26+M_2+%5Cend%7Bbmatrix%7D%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
and 
Thus, 
is not isomorphic to a linear algebra, but rather to a direct sum of linear algebras,
does there exist a linear partition 
? If 
the answer is clearly “yes” since we have 
Thus we consider the case where 
. Together with the corollary to Burnside’s theorem, we thus have a linear partition 
of 
This is almost correct. Let us say that two blocks 
are 
such that the matrix of every 
be a set parameterizing the corresponding equivalence classes of blocks, and for each 
choose a representative 
of the corresponding equivalence class. 
and let 
be an orthonormal basis.
are equal if and only if 
for all 
;
with equality if and only if 
makes sense even if the basis 
be two orthonormal bases of 
.
of 
of 
defined by 
is unitary. We have 
for the same reason as above. We have thus shown that Definition 14.3 gives the same linear functional on 
That is, every operator has an eigenvalue.
?
for all 
defined by
we have
and consider the matrix element
form an orthonormal basis of 
for its matrix elements relative 
We claim that
be the operator on the right hand side of the above equation. Then, the matrix elements of 
of the function algebra 
of the linear algebra 
are orthogonal projections spanning a commutative subalgebra of 
is a scalar multiple of the trace.
of 
, we have
such that 
for all 
Thus for any 
be a finite abelian group.
the Fourier basis of 
is an eigenbasis for the image of 
-dimensional unit cube. In this case, the convolution algebra 
is 
-dimensional and the group basis is 
ranging over all 
. The convolution product of two basis vectors is
of the 
for some 
where 
are the vectors
are adjacent if and only if 
This means that the Wedderburn transform of 
the character table is
is the usual dot product. Thus, the Fourier basis of 
can be simplified: every coordinate of 
equal to zero contributes 
and every coordinate of 
contributes 
Thus writing 
for the number of entries equal to one in 
terms equal to one and 
is equal to the number of 
, which is 
Then, the left regular representation 
be any function on 
is the linear operator on the Hilbert space 
given by the image of the Fourier basis vector 
In other words, 
and 
This says that the operators 
making up the image of the commutative algebra 
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. 
consisting of orthogonal projections, called its Fourier basis. Thus any function 
can be expanded in the group basis,
defined by 
defined by 
is an algebra isomorphism 
Thus, we have
 = \sum\limits_{h \in G} A(gh^{-1})B(h), \quad g \in G,](https://s0.wp.com/latex.php?latex=%5BA%2AB%5D%28g%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7D+A%28gh%5E%7B-1%7D%29B%28h%29%2C+%5Cquad+g+%5Cin+G%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
and on the right hand side = \widehat{A}(\lambda)\widehat{B}(\lambda), \quad \lambda \in \Lambda](https://s0.wp.com/latex.php?latex=%5B%5Cwidehat%7BA%7D%5Cwidehat%7BB%7D%5D%28%5Clambda%29+%3D+%5Cwidehat%7BA%7D%28%5Clambda%29%5Cwidehat%7BB%7D%28%5Clambda%29%2C+%5Cquad+%5Clambda+%5Cin+%5CLambda&bg=FFFFFF&fg=000&s=0&c=20201002)
is the character basis of 
-scalar product in 
corresponding to a given group element 
From the inversion formula, the Fourier transform of the indicator function of 
is 
is the elementary basis of 
consisting of indicator functions 
we have
together with the definition of the norm associated to a scalar product.
the 
case of the 
-norm, which for any 
is defined by
, which is defined by 
the sum 
is approximately equal to its largest term, and this concentration becomes exact in the 
which in terms of the character basis 
we have
Since 
was arbitrary, the above shows that
it can be cancelled from both sides of the above inequality, leaving
of a group 
defined by
is a function from the convolution algebra 
Moreover, the map 
of 
and 
on 
Since 
under 
it represents 
to be a cyclic group of order four with generator 
, so that 
is the group unit. The group basis of 
and we denote these vectors by 
for brevity. Let 
with respect to the ordered basis 
. 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 
for 
is an algebra homomorphism if and only if the function 
defined by
the unitary group of the complex numbers.
is called a character of 
.
However, for highly noncommutative groups there may not be many nontrivial characters.
To begin, recall that 
and 
and consider the group
is a cyclic group of order 
with generator 
Define the dual group of 
is the additive group of integers modulo 
writing
is a group isomorphism 
(Exercise: prove this, noting that because 
it is sufficient to show the parameterization is an injective group homomorphism).
the function 
defined by
is a principal 
th 
because the identity element 
has parameters 
Moreover, for any 
is indeed a group homomorphism 
of the convolution algebra 
We now claim that the characters form a basis of 
which is the dimension of 
we have
we have
we have
is an 
th root of unity.
since this highlights the symmetry 
The 
symmetric matrix ![X=[\chi^\lambda_\alpha]](https://s0.wp.com/latex.php?latex=X%3D%5B%5Cchi%5E%5Clambda_%5Calpha%5D&bg=FFFFFF&fg=000&s=0&c=20201002)
is called the character table of 
is a symmetric unitary matrix. Another way to say the same thing is that the rescaled character basis
, every partition of 
is called a class function if it is constant on conjugacy classes, meaning that 
for all 
for all 
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 
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 
. Thus 
let 
be the indicator function of 
so 
span the class algebra of 
constant on conjugacy classes can be written
denotes the value 
for any 
We will show that 
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 
and 
are required to belong to 
and 
respectively.
 = 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)
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)
are isomorphic.
, where 
is the elementary function corresponding to 
is also an algebra homomorphism. Indeed, by Theorem 1.2 and linearity of 
Thus,
so defined is an algebra isomorphism. Note that the Fourier transform on 
The function 
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 
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 
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,