In this lecture we describe the lattice of subalgebras of 
Theorem 4.1. A subalgebra 
Proof: One direction is clear: if 
Now suppose we are given a separating subalgebra 
Now observe that
Since each of the factors 
-QED
We are now ready to classify the subalgebras of
Definition 4.1. A partition of 
The set 
Now observe that associated to every partition 
Problem 4.1. Prove that 
We are now ready for the classification of subalgebras of the function algebra
Theorem 4.2. If 
Proof: The proof is by induction on the cardinality of 
For the induction step, let 
-QED
Problem 4.2. Prove that the number of non-isomorphic subalgebras of 
of a finite-dimensional Hilbert space. In this lecture, we introduce a third family of algebras associated to finite groups. In Math 202B, all groups are finite, so going forward “group” always means “finite group.”
be a group with group law written multiplicatively: for any elements 
, their product is 
. Since 
of functions 
. As explained in 
so from the point of view of this construction the fact that 
whose elements are functions 
Thus for any two functions
viewed as elements of 
, we define it using the the natural involution in 
, where 
is the group unit, as opposed to being the constant function equal to 
on every 
, which is the multiplicative unit in 
 = \sum\limits_{h \in G}A(gh^{-1})B(h)](https://s0.wp.com/latex.php?latex=%5BAB%5D%28g%29+%3D+%5Csum%5Climits_%7Bh+%5Cin+G%7DA%28gh%5E%7B-1%7D%29B%28h%29&bg=FFFFFF&fg=000&s=0&c=20201002)
 = \overline{A(g^{-1})}.](https://s0.wp.com/latex.php?latex=%5BA%5E%2A%5D%28g%29+%3D+%5Coverline%7BA%28g%5E%7B-1%7D%29%7D.&bg=FFFFFF&fg=000&s=0&c=20201002)
are a basis of 
are isomorphic groups then their convolution algebras 
are isomorphic algebras (the converse is false, bonus points if you give a counterexample). Also show that 
of 
.
is a subalgebra of 
is the constant function
is selfadjoint (make sure you understand why), and also
just permutes the points of the group (make sure you understand why). Therefore, 
and we conclude that the average of all elementary functions 
is a projection in 
and 
.
are linearly independent (make sure you understand why), we have located a two-dimensional subalgebra inside 
must also have odd order, and consequently the subalgebra 
cannot be the algebra of functions vanishing outside some subgroup 
cannot be even.
and computed their characters. Proceeding by bare hands becomes difficult quickly since the cardinality of 
grows very rapidly (indeed, superexponentially) with 
.
dimensional. This isn’t much – the number of irreducible representations of 
which according to the 
of 
is a Hilbert space with orthonormal basis 
, 
and 
is the group homomorphism defined by
is the number of fixed points (i.e. one-cycles) of the permutation 
Therefore,
is the set of fixed points of 
. To understand this sum, let us begin with the simpler version without the squares. By 
is the number of orbits of 
is one, i.e. 
so
To get the exponent of 
into the equation, let 
by 
Then, we have 
. Applying Burnside’s Lemma we thus have 
Indeed, consider any point 
Either 
or 
. In the first case, the orbit of 
under 
which gives the number of fixed points in a uniformly random permutation is equal to two. Since ![\mathbb{E}[Z]=1](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BE%7D%5BZ%5D%3D1&bg=FFFFFF&fg=000&s=0&c=20201002)
and ![\mathbb{E}[Z^2]=2](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BE%7D%5BZ%5E2%5D%3D2&bg=FFFFFF&fg=000&s=0&c=20201002)
you might be tempted to assume that the 
th moment of ![\mathbb{E}[Z^k]=k,](https://s0.wp.com/latex.php?latex=%5Cmathbb%7BE%7D%5BZ%5Ek%5D%3Dk%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
but this is wrong and the correct answer is quite a bit more interesting. 
be the random variable defined by 
of the permutation representation of 
be the one-dimensional subspace of 
which is an irreducible representation of 
be its orthogonal complement, which is the standard representation of 
we have 
and therefore
Indeed, 
and for 
its dimension is strictly larger than one. 
follows from its irreducibility. Likewise, since 
the standard representation of the symmetric group is irreducible. 
is a subgroup of 
that have 
be a set indexing isomorphism classes of irreducible unitary representations of 
or equivalently irreducible linear representations of its convolution algebra 
For each 
let 
be a representative of the corresponding isomorphism class. We know 
in 
under 
is all of 
by Burnside’s theorem. On the other hand, viewing 
as a subalgebra of 
the image of 
is a unitary representation of 
but not necessarily an irreducible one. So viewed as a representation of 
of partitions of 
and 
is the multiplicity of 
in 
be arbitrary but fixed, and let 
be a representative of the corresponding isomorphism class of irreducible unitary representations of the symmetric group 
where 
as vectors 
with 
entries, which are weakly decreasing positive integers whose sum is 
-dimensional. An important point is that there is no canonical way to say which of the abstract irreducible representations 
The main result in the representation theory of the symmetric groups is that it is possible to give a concrete construction of each 
such that 
can be obtained from 
by adding a single cell.
of 
is viewed as a representation of 
as a fixed point, its decomposition into irreducible representations of 
If we instead start with 
then as a representation of 
called 
These combinatorial objects can either be viewed as increasing paths in the Hasse graph of Young’s lattice from the bottom unicellular diagram up to 
in the cells of 
, 
A related question is to calculate the character of this representation, i.e. the function on 
of the class algebra 
where we recall that 
of the class basis 
. This is an equivalent problem because
is the value of 
on any permutation 
For any Young diagram 
define the content 
of this cell to be its column index minus its row index.
let 
be the corresponding irreducible character, i.e. function 
defined by
of the class algebra 
satisfies the orthogonality relations
be the set of conjugacy classes in 
let 
be the indicator function of the class 
The class basis 
satisfies the orthogonality relations
character orthogonality takes the form
are conjugate for every 
If 
we have
as the column index and 
![[c_{\alpha\beta\gamma}]_{\alpha,\beta,\gamma \in \Lambda}](https://s0.wp.com/latex.php?latex=%5Bc_%7B%5Calpha%5Cbeta%5Cgamma%7D%5D_%7B%5Calpha%2C%5Cbeta%2C%5Cgamma+%5Cin+%5CLambda%7D&bg=FFFFFF&fg=000&s=0&c=20201002)
such that
we have
On one hand, we have
is the identity operator. On the other hand, 
multiply both sides by 
and sum over 
consists of orthogonal projections. In particular, the span of 
we have
leaving
In the case of an ambivalent group, the Fourier basis is simply a rescaling of the character basis by nonzero (indeed, positive) factors. In the general case we have the following.
, we have
of the convolution algebra 
Recall also that the class algebra 
if and only if 
then 
is an algebra homomorphism, we have 
to be in 
is all of 
Since the center of 
, with 
and 
the identity operator in 
for every 
we have 
for some 
Thus, we have a map
of the operator 
This map is called the central character of 
is an algebra homomorphism 
and that
is a central function on 
i.e.
be the subspace of 
. We will do this by showing that 
In other words, the only central function on 
be the central character of 
and let
to be the zero operator in 
, and hence 
be the indicator function of 
i.e.
denotes the value of 
for any group element 
Thus, the functions 
and 
are the same set if 
and disjoint if 
is orthogonal in 
extending our result from the case of abelian groups, where 
be a representative of the isomorphism class of irreps corresponding to 
i.e. that the number of irreducible unitary representations of 
This is a subalgebra of the linear algebra 
of 
may thus be regarded as a unitary representation 
of 
is the restriction of 
to the subspace 
Since 
is the number of blocks of the linear partition 
and
we have
is the multiplicity of 
of 
Let 
implies 
For the converse, suppose it is the case that 
so that 
forces
has exactly one term equal to one and all other terms equal to zero, so that the isotypic decomposition of 
for some 
in which 
-scalar product and 
 = \sum\limits_{k \in G}E_g(k)A(k^{-1}h) = A(g^{-1}h), \quad h \in G.](https://s0.wp.com/latex.php?latex=%5BE_gA%5D%28h%29+%3D+%5Csum%5Climits_%7Bk+%5Cin+G%7DE_g%28k%29A%28k%5E%7B-1%7Dh%29+%3D+A%28g%5E%7B-1%7Dh%29%2C+%5Cquad+h+%5Cin+G.&bg=FFFFFF&fg=000&s=0&c=20201002)
acts on the group basis of 
be their characters, i.e. the functions 
are defined using the trace, they are central functions on 
Thus, 
Indeed, if these representations are isomorphic, then (by definition) there is a unitary vector space isomorphism 
which has the additional property that the equation
for every 
whence 
and in particular we do not know whether or not it is finite. For each 
are irreducible representations, Schur’s Lemma tells us that 
We will see next lecture that in fact equality holds: up to isomorphism, the number of irreducible unitary representations of 
. This suggests that the commutativity index of 
we have 
is an 
or 
of scalar operators.
be a central element, and let 
be an eigenvalue of 
By definition, this means that 
is not invertible in 
is a nonzero subspace of 
-eigenspace of 
we have
is again in the 
so that 
for every 
are commuting normal operators, then there exists an orthonormal basis 
such that 
is isomorphic to 
where 
If 
of 
, and therefore 
, 
induces a partition 
of 
Letting 
be the coarsest partition of 
, the lattice of 
and 
there is a third 
by Burnside’s theorem. The orthogonal complement 
and 
Then, for any 
i.e. 
as the blocks of 
and write
to be the set of all operators 
such that every block 
is the linear partition of 
the subalgebra 
is a linear partition of 
. Let 
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,
? 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 
Unfortunately, this is not quite correct. For example, consider the case where 
Let 
be an othornormal basis of 
, where 
and 
and this is a linear partition of 
is not 
of 
and 
such that, for every 
the matrix of the restriction of 
;
;
with equality if and only if 
for all 
The expansion of any vector 
with respect to an orthonormal basis is
is 
together with multiplication defined by composition, = A(B(v)),](https://s0.wp.com/latex.php?latex=%5BAB%5D%28v%29+%3D+A%28B%28v%29%29%2C&bg=FFFFFF&fg=000&s=0&c=20201002)
which is characterized by the condition 
The matrix elements of 
in 
for all 
compute the matrix elements of the product 
in terms of the matrix elements of its factors.
are defined by
elementary operators in 
and 
, 
are orthogonal projections in 
and 
Thus, 
is a linearly independent set in 
We call 
we say that 
of elementary operators spans 
be its matrix elements relative to the orthonormal basis 
by
Indeed, the matrix elements of 
;
with equality if and only if 
be two orthonormal bases of 
.
of 
of 
defined by 
is unitary. We have 
of numbers 
such that 
is not invertible. Elements of 
being non-invertible is equivalent to 
and because 
is a polynomial in 
the 
is an eigenvalue of an operator 
is called the 
and any nonzero vector 
is said to be an eigenvector of 
for all 
form an orthonormal basis of 
we have
is a linear functional which satisfies 
for all 
, then 
is a scalar multiple of the trace.
of 
such that 
for all 
Thus for any
