This lecture concerns characters of FINITE groups over the field of complex
numbers (complex representations). This notion of "character" persists throughout
representation theory and is a key link to (algebraic) combinatorics. Each time
(Lie groups, Lie algebras,...), one is looking to achieve the same niceness that
occurs for characters of finite groups, so we start here.
Basic question 1: Suppose V is a G-module. How do we decompose V into irreducibles
(up to isomorphism)?
Basic question 2: Suppose V and W are G-modules. How can we decide if they are
the same representation (up to isomorphism)?
Answer: Compute characters!
Definition: Let A be an nxn matrix. Then tr(A)=sum of the diagonal entries.
``Coordinate-free'' definition:
Suppose T:V->V is an endomorphism (=homomorphism of an object to itself;
in our case a linear transformation).
\lambda is an eigenvalue of T if there exists a nonzero v such that
T(v)=\lambda v. We call v the eigenvector associated to lambda.
Facts (Math 416):
* If you have a basis of eigenvectors you're diagonalizable
* In general, you might not have such a basis of eigenvectors, but if
you are over an **algebraically closed field** (such as complex
numbers), there is always a basis of generalized eigenvectors -> Jordan
canonical form.
Definition: tr(T:V->V) is the sum of the (generalized eigenvalues)
counted with multiplicity.
These definitions are natural/compatible in the following sense:
* Define an equivalence relation on nxn matrices by declaring A~B if
A=gBg^{-1} for some g in GL_n.
* A~B => Tr(A)=Tr(B)
* "~" exactly corresponds to change of basis.
* Thus the second definition defines Tr of a matrix by the trace of the
canonical representative (the Jordan form) of the matrix.
Properties of trace
--------------------
* Tr(A)=Tr(B) if A~B
* Tr(A+B)=Tr(A)+Tr(B)
* Tr(AB)=Tr(BA) even though AB=BA is usually not true.
* Tr(A @ B)=Tr(A)Tr(B) where "@" written usually \otimes is Kronecker (ahem,
tensor) product of two matrices, defined as follows:
If A is an m x n matrix and B is a p x q matrix then
[a11 B ... a1n B]
A@B = [ ............. ]
[am1 B ... amn B]
where
aB means multiply every entry of B by a.
-------------------
Having reviewed some properties of trace, we can now define:
Definition: Suppose f:G->GL(V) is a linear representation over a field F
then
\chi_f:G->F
defined by \chi_f(g)=Tr(f(g)) is the *character* of f.
Prop: If X is the character of a represention f of degree n,
(1) X(1)=n
(2) Finite group reps/C is EQUIVALENT to a representation described by
by a unitary matrix U [i.e., UU^*=I].
Consequently,
X(s^{-1})=complex conjugate(\chi(s))
(3) X(tst^{-1})=\chi(s)
Proof: (1) is obvious since f(id)=the n x n identity matrix.
(2) Over V=C^n one has the dot product
(v,w) = \sum_i v_i b_i^*
However this scalar product might not be G-invariant, i.e.,
(f_s x, f_s y)=(x,y) for all s in G.
If so, then define a NEW scalar product by averaging:
(x,y)' := \sum_{t\in G} (f_s x, f_s y)
(In class discussion: this is a scalar product. This is
**Weyl's unitary trick**.)
Now, by Gram-Schmidt orthogonalization process, one can always find an
orthonormal basis with respect to this G-invariant scalar product (-,-)'.
This MEANS f_s written with respect to this orthonormal basis is a
unitary linear transformation for each s. This MEANS the matrix of
f_s under this basis is unitary.
That is, every finite group rep is unitarizable: we can write a
homomorphism f:G->U(n); this is called a *unitary* representation.
This is cool because we can now use:
Facts from linear algebra: Every unitary matrix is diagonalizable.
The eigenvalues all have modulus 1.
Therefore
X(s)^* = Tr(f_s)^*
= \sum_i {\lambda_i}^*
= \sum_i \lambda_i^{-1}
= X(s^{-1}).
(3) This is a crucial one, it follows since trace is preserved
under ~. QED
-------
Remark: Something whose theory we'll skirt around is Lie groups
(like GL_n). These are by definition differentiable manifolds such that
the multiplication and inverses are smooth maps. We're skirting around
this because I want to stay (for the purposes of our rigorous discussion,
and time limits) within the realm of ALGEBRA. Fortunately, the
representation theory we care about (Complex simple Lie groups -- the
"nice case") is controlled by Lie algebra representations anyway.
However, you might read a bit more about all this so let me say a
few more orienting words. In the theory one reduces to Compact Lie
groups. This is the generalized Weyl Unitary trick (reduction to
"maximal compact subgroup"). One point is that representations (we
care about) are continuous. So if a group G is compact then its
image is compact. Well, if f:G->GL(V) is a linear represenation, that
means f is unitarizable, analogous to the finite group case.
--------
Definition: Define an equivalence relation ~ on G by conjugation,
that is q~p if q=gpg^{-1} for some g in G. The equivalence classes
under ~ are called *conjugacy classes* of G.
Definition: A *class function* on G is a function f:G->F that is
constant on conjugacy classes. That is f(q)=f(p) if q=gpg^{-1} for
some g in G.
Remark: Notice that class functions of G form a vector space over
F. **Later we will see that the characters form a orthonormal
basis!**
Prop 2: Let f_1:G->GL(V_1) and and f_2:G->GL(V_2) be two linear
representations of G. Let X_1 and X_2 be their characters. Then
character X of the representation V_1 \oplus V_2 satisfies
X=X_1+X_2.
Proof: G acts on V = V_1 \oplus V_2 like the block matrix
[f_1(g) 0 ]
[ 0 f_2(g) ]
and the result follows.
We arrive at the VERY IMPORTANT:
Schur's lemma:
Let f_1:G->GL(V_1) and f_2:G->GL(V_2) be irreducible representations of G.
Let p:V_1->V_2 be a G-equivariant linear transformation (a.k.a.
"intertwining map"), that is
p(g.v)= g.p(v)
or in terms of f_1, f_2, equivalently
p o f_1(g) = f_2(g) o p
THEN
(1) p=0 unless V_1 and V_2 are isomorphic as G-modules (a.k.a. f_1
and f_2 are isomorphic as representations).
(2) If V_1 = V_2 and f_1=f_2 then f = lambda Identity.
[That is, if V is a G-module, it admits no G-equivariant
endomorphisms except multiples of the identity.]
Proof: (1) Suppose f:V_1->V_2 is an intertwining map but V_1 and
V_2 are not isomorphic as G-modules.
Now the image of f is a G-module in V_2. But if im(f)=0 we're done.
Since V_2 is irreducible that means im(f)=V_2.
Also, ker(f) is a G-submodule of V_1. If ker(f)=0 that means f is
injective. Hence f is an isomorphism =><=. Thus ker(f)=V_1, which
means f=0.
(2) Let lambda be an eigenvalue of f. At least one exists if we
work over C. Let f'=f-\lambda Id. Thus ker(f') is not just 0. But
f' is also a G-equivariant map from V_1 to V_2(=V_1). Since ker(f')
is G-stable we conclude ker(f')=V_1. That is f'=0 or f=lambda I.
QED
Schur's lemma is used to prove "orthogonality relations" for characters.
It has many variations depending on the category of objects you're dealing with,
but the main message is that by studying
dim Hom_G(V,W)
you can decide "how many times V appears in W".
Application of Schur's lemma (2): Let G be an abelian group and f:G->GL(V) be
an irreducible linear representation. Then dim V =1.
Proof: Fix any g and look at f(g) in GL(V). This is (by definition) an
isomorphism of V with itself (as vector spaces). Notice that f(g) is also a
G-equivariant isomorphism. Indeed, if h in G then
f(g)[f(h)v]=f(gh)v=f(h)[f(g)v].
Thus since f(g) is not the 0 map, by Schur (2), f(g)=\lambda Id_V (lambda \neq
0. This MEANS for every g in G and every v in V, g.v=\lambda v. Thus, EVERY 1
dimensional subspace of V is G-stable. So V cannot be irreducible unless the
dimension is 1. QED
Remark: In particular this applies to G=C^* (=GL_1(C)) and G= complex numbers of
modulus 1. In either case, G is abelian, so any irrep is of the form f:G->GL_1.
Now suppose the map is polynomial, or rational. You're asking for z->p(z) with
the property that p(1)=1 and p(z)^2=p(z^2). This only gives z->z^m as
possibilities (why? Imagine z->z^4 + z^3 were a representation. Notice that
the RHS has a nonzero root which implies it's not a map to C^* after all.)
----
We've already discussed that class functions of G are a vector space,
and characters of G are vectors in this space.
In fact there's an inner product on the space Class(G) of class functions
of G.
Definition (inner product on Class(G)) Let p, q in Class(G)
(p|q):=1/|G| \sum_{t\in G} p(t) complex-conjugate[q(t)]
=1/|G| \sum_{t\in G} p(t) q(t^{-1})
Prop: This is an inner product.
Proof: By definition an inner product on a vector space V satisfies
Conjugate symmetry (x|y)= complex-conj (y|x)
Linearity in the first argument
Positive definiteness: if x is non-zero, (x|x) > 0.
This holds in our case. QED
Computations derived by Schur's lemma implies the following
Theorem 3 (of Serre):
(i) If X is the character of an irreducible rep, (X|X)=1.
I.e., irreps have norm 1 (in fact the converse is true, as we will see later).
(ii) If X, X' are characters of two nonisomorphic irreps,
(X|X')=0 (X and X' are orthogonal).
Proof: See Serre Theorem 3 and the calculations in Section 2.2. QED
Remark: In Serre Section 2.2 he defines a similar notation to (-|-), namely
<-,-> defined for *any* functions X,Y on G (not necessarily class functions,
or characters) by
= (1/|G|) \sum_{t\in G} A(t^{-1})B(t)
= (1/|G|) \sum_{t\in G} A(t)B(t^{-1})
If A and B are characters then the two notations agree since we prove
(Prop 1 of Serre) that B(t^{-1})=B(t)^* in that case.
----------
Theorem 4 (of Serre): Let V be a rep of G with character X. Suppose
V decomposes into irreps as
V=W_1 \oplus W_2 \cdots \oplus W_k
If W is an irrep with character Y then the number of W_i isomorphic to
W is =.
Proof: Since characters respect direct sum we have
X=X_1+X_2+...+X_k
Hence (X|Y)=(X_1+...+X_k|Y)
=(X_1|Y)+...+(X_k|Y)
But (X_i|Y)=1 if X_i is iso Y and is 0 otherwise. QED
Cor 1 (of Serre chapter 2) The number of X_i isomorphic to Y above does not
depend on the decomposition. (Recall the trivial rep and V decomposing into
lines.)
Proof: (X|Y) does not depend on the decomposition. QED
KNOWING THE CHARACTER MEANS YOU KNOW THE REPRESENTATION UP TO
ISOMORPHISM!:
Cor 2 (of Serre Chapter 2) Two representations with the same character are
isomorphic.
Proof: Suppose V, V' are two G-modules with the same character. By
cor 1, each irrep appears the same number of times in V and V'. QED
-------
VERY USEFUL! CHARACTERS LET YOU COMPUTATIONALLY CHECK IF A REPRESENTATION
IS IRREDUCIBLE!
Theorem 5 (of Serre): If X is the character of V then (X|X) is a positive
integer and (X|X)=1 if and only if V is irreducible.
Proof: Suppose
V=W_1 \oplus...\oplus W_k
is the decomposition of V into irreducibles W_i with character X_i.
Then by orthogonality of characters
(X|X)=\sum_j m_j^2 (*)
where m_j is the number of times some fixed irrep appears. Since at
least one irrep appears in V's decomposition, the positive integer
claim follows.
For the remaining claim, we've already shown/claimed that if X is
irreducible, (X|X)=1. Conversely, suppose X is reducible, then (*)
shows (X|X) >=2. QED
-----
Important Example: (The regular representation) Recall the regular
representation has V=C[G], the vector space of formal linear
combinations of elements of G. Thus dim V = |G|. Let f:G->V be the
associated linear representation map. Now, if g is not the identity
map f(g)h is not h for all h in G. That is tr f(g)=0. On the other
hand, if g=id then tr f(id) = |G|. We have just proved
Prop 5 (of Serre Chapter 2) The character X_reg of the regular
representation is given by the rule
X_reg(id) =|G|
X_ref(g) = 0 if g is not the identity map
Moreover,
Cor 1 (of Section 2.4 of Serre):
Every irrep W_i of G is contained in the regular representation
with multiplicity equal to its degree.
Proof: The multiplicty of W_i is
= (1/|G|) \sum_{g\ in G} X_{reg}(g^{-1})X_{W_i}(g)
= (1/|G|) (|G| * X_{W_i}(id)) [as by Prop 5]
= X_{W_i}(id)
= degree of W_i. QED
Cor 2 (of Section 2.4 of Serre):
(a) The degrees n_i of W_i in the regular representation satisfy
\sum_{i=1}^h n_i^2 = g.
(b) If g in G is not the identity then
\sum_{i=1}^h n_i X_{W_i}(s) = 0
Proof: By Cor 1 we can decompose V=regular rep as
V= \sum_i n_i W_i
Since characters are additive on direct sum
X_reg(g) = \sum_i n_i X_{W_i}(g).
Now evaluate g=id to get (a) and g neq id to get (b), using that
X_{reg}(g)=0 in that case. QED
Remark: The regular representation is interesting since "inside it" you can
find every possible irrep of G.
----------
Recall that a function f on G is a *class function* if it is constant
on conjugacy classes, i.e.,
f(tst^{-1})=f(s) for all s,t in G.
Let H=Class(G) be the set of class functions on G.
Theorem 6 (of Serre) The irreducible characters X_1,...,X_h of G form an
orthonormal basis of H.
Proof: Serre first proves this technical sounding Proposition
Prop 6 (of Serre)
Let f in Class(G) and let p:G->GL(V) be a linear representation of G.
Let p_f in End(V) (=linear maps from V -> V) defined by:
p_f = \sum_{t\in G} f(t)p_t
If (p,V) is irreducible of degree n and has character X then p_f is a
"homothety" (p_f=\lambda I) with ratio \lambda given by
\lambda=(1/n)\sum_{t \in G} f(t)X(t) = (g/n) (f,X^*)
where, recall,
X^*(g)=complex conjugate of X(g) = X(g^{-1})
and
(a|b)=(1/|G|)\sum_{t\in G} a(t)b(t)^*
Proof of Prop 6: We wish to apply Schur's lemma on homothety by
showing p_f is a G-equivariant endomorphism of V->V. That is,
we want p_f p_s = p_s p_f for all s in G.
Compute
p_s^{-1} p_f p_s = \sum_{t \in G} f(t) p_s^{-1} p_t p_s
= \sum_{t \in G} f(t) p_{s^{-1}ts}
[by homomorphism]
= \sum_{u in G} f(sus^{-1})p_u
[reindexing u=s^{-1}ts]
= \sum_{u in G} f(u) p_u [f is a Class fn.]
= p_f [by definition]
as desired. Hence Schur's lemma (2) says
p_f = \lambda I
What is \lambda?
On the one hand,
Tr(p_f) = Tr(\lambda I) = n\lambda (*)
On the other hand
Tr(p_f) = Tr(\sum_{t in G} f(t) p_t)
= \sum_{t in G} f(t) Tr(p_t)
= \sum_{t in G} f(t) X(t)
= g(f|X^*)
Now equate with (*) and divide out n. QED
Proof of Theorem 6: We already know (stated in these notes, you
read the proof details) that the irreducible characters X_1,...,X_h
form an orthonormal set in Class(G). Indeed, this is equivalent to
saying X_i^* (defined by X_i^*(g) = conjugate X_i(g) form an
orthonormal set)
GENERAL LINEAR ALGEBRA: An orthonormal set of a vector space V (an
inner product space over R,C) can always extend to an orthonormal
basis. This follows from the existence of the Gram-Schmidt
orthogonalization process.
Therefore to show X_1,...,X_h is a basis, it suffices to show
that if f in Class(G) then
(f|X_i^*)= 0 for all i => f=0.
Let f in Class(G) be such a function.
Let p:G->GL(V) be any linear representation and define, as in the above
prop,
p_f := \sum_{t in G} f(t) p_t
By the above prop,
p_f is identically 0 if p is irreducible.
However, even in p is reducible, it is a direct sum of irreducibles
and it follows from the definition that p_f is identically 0. (why?)
Now apply this to the special case that p is the regular representation
of G. Recall V has basis e_t for each t in G and G acts by
g.e_t = e_{gt}.
Compute
p_f(e_{id}) = \sum_{t in G} f(t) p_t e_{id}
= \sum_{t in G} f(t) e_t (*)
But we already saw p_f = 0 (identically). Hence (*) says one has
a linear combination:
0 = \sum_{t in G} f(t) e_t
Since {e_t} is a basis, this means f(t)=0 for all t in G.
That is, f is identically the 0-function in Class(G), which is what
we wanted. QED
---------
Theorem 7 (of Serre) The number of irreducible representations of G
(up to isomorphism) equals the number of conjugacy classes of G.
Proof: A function f in H=Class(G) is determined by its values on the conjugacy
classes C_1,...,C_k. But there's no other restriction on f in H (f(C_i) may be
arbitrarily assigned). Hence the dimension of H is k. On the other hand, Theorem 6
says the dimension is the number h of irreps. Hence k=h. QED
--------
Example: Do example on page 20 of Serre: computing the character
table of S_3: you know the trivial 1 dimensional rep and sign rep
are irreducible (being one dimensional).
Let 1=identity perm, t=(12)(3) and c=(123). These represent the
conjugacy classes of S_3. Since characters are class functions, to
write down the *character table* it suffices to know the reps on
these permutations.
Since S_3 has three conjugacy classes means we're missing one
irrep, and we even know its degree (2) b/c 1^2+1^2+?^2=3!=6 => ?=2.
Thus we have a partial character table of S_3:
1 t c
trivial 1 1 1
sgn 1 -1 1
?????? 2 ?? ??
for the unknown irrep we know the char value at 1 is the degree (2) but
don't know the rest of the table (yet).
But, we can figure out the missing values in at least two ways.
Way 1: Expand the character table
1 (12) (23) (13) (123) (132)
trivial 1 1 1 1 1 1
sgn 1 -1 -1 -1 1 1
?????? 2 a a a b b
We know that the characters are orthogonal with respect to the
scalar product
So we get
2 +3a^* +2b^*=0
2 -3a^* +2b^*=0
where a^* is a-conjugate.
=> -4=2b^* => b=-1 => a=0
Thus the table is
1 t c
trivial 1 1 1
sgn 1 -1 1
?????? 2 0 -1
Way 2: trivial+sgn+2???? is the regular representation, and we can
compute the character of that directly to deduce a,b.
What is ?????: This is the standard representation: the action of S_3
on the hyperplane {(x,y,z) in C^3:x+y+z=0} (why?).
Remarks:
* Notice that although in theory the character values of S_3 are
complex numbers, they are in fact integers. This is always true,
and a combinatorial rule for the value is given by the
Murnaghan-Nakayama rule AND by the Fomin-Greene rule. People
actually want to be able to compute these values efficiently and no
such efficient method exists. In fact, as far as I know, no
computational comparison of the Murnaghan-Nakayama rule vs
Fomin-Greene rule has been done (based on Barcelo-Ram, which is old
-- so if one wants to look into this, one should do a literature
search).
* There are old problems associated to the S_n character table. One of
them is this. Notice that even though the character table has signs, if
you fix a row of the (compressed) character table and sum over the columns
(i.e., conjugacy classes), the number is nonnegative! This has a rep
theory explanation, but:
Problem 12 of https://math.mit.edu/~rstan/papers/problems.pdf
asks for a combinatorial interpretation of this nonnegative integer.
* Refer to Alexander Miller's papers
* Besides the pure math reasons to care about rep theory of the symmetric
group, check out Diaconis' viewpoint on applications to statistics:
https://jdc.math.uwo.ca/M9140a-2012-summer/Diaconis.pdf
https://projecteuclid.org/ebooks/institute-of-mathematical-statistics-lecture-notes-monograph-series/Group-representations-in-probability-and-statistics/toc/10.1214/lnms/1215467407
---
Next look at the expanded table
1 (12) (23) (13) (123) (132)
trivial 1 1 1 1 1 1
sgn 1 -1 -1 -1 1 1
standard 2 0 0 0 -1 -1
The pointwise product of two rows is a class function, say for example
standard^2 = 4 0 0 0 1 1
Notice that by the fact that the irreducibles form a basis of class functions,
this vector is expressible as a linear combination of the rows. It is
1.trivial+1.sgn+1.standard = standard^2
The "1,1,1" are "tensor product multiplicities" and (magically)
are all positive (see however the actual explanation in the next lecture).
Famous open problem: Give a combinatorial rule for these numbers in general.
See problem 10 of http://math.mit.edu/~rstan/papers/problems.pdf
as well as update notes
https://mathoverflow.net/questions/349406/updates-to-stanleys-1999-survey-of-positivity-problems-in-algebraic-combinatori
---------
References for further reading:
Serre's book on finite group rep theory
http://www.math.tau.ac.il/~borovoi/courses/ReprFG/Hatzagot.pdf
Karen Smith's notes
http://www.math.lsa.umich.edu/~kesmith/rep.pdf