Representation Theory of Sn

11 Pages·2009·0.169 MB·English

by  Evan Jenkins

REPRESENTATION THEORY OF S
n
EVANJENKINS
Abstract. These are notes from three lectures given in MATH 26700, In-
troduction to Representation Theory of Finite Groups, at the University of
Chicago in November 2009. Most of the material comes from chapter 7 of
[Ful97],someofitverbatim.
1. Introduction
Most of this course has focused on the general theory for compact groups, but
our examples have focused on a particular class of compact groups, namely, the
compact Lie groups. The usual constructions of representations of compact Lie
groups relies heavily upon special structures that exist in the world of Lie groups,
namely maximal tori and Borel subgroups.
Finite groups, while in some ways much simpler than compact Lie groups, lack
a nice unified structure theory. As a result, we can’t expect that there will be any
very powerful general methods for constructing representations of finite groups.
The symmetric groups, however, were one of the first classes of groups studied
via the lens of representation theory. The character theory of S for arbitrary n
n
was worked out by Frobenius in 1900. Since then, many explicit constructions of
the representations of S were developed, some of them very closely connected to
n
the study of Lie groups.
2. Representation Theory of Finite Groups
All of our results for compact groups hold in particular for finite groups, which
canbethoughtofascompactgroupswiththediscretetopology. Here, thedescrip-
tion of Haar measure is very simple; it is just the normalized counting measure.
Explicitly, if G is a group, and f function on G, then
(cid:90) 1 (cid:88)
f(g)dg = f(g).
|G|
G g∈G
We’lldenotebyCGthevectorspaceofcomplex-valuedfunctionsonG. Notethat
dimCG=|G|. Thisvectorspaceappearsinmanydifferentguises,soit’simportant
to be able to distinguish how we’re thinking about it at any given time. First, we
notethatwecanmakeCGintoanalgebraintwonaturalways,namely,bypointwise
multiplicationoffunctionsorbyconvolution. Infact,eventheconvolutionproduct
has two conventions, and there are good reasons to use either one, but we’ll take
our convolution product to be
(cid:90)
(cid:88)
(f ∗f )(g)=|G| f (h)f (h−1g)dh= f (h)f (h−1g).
1 2 1 2 1 2
G h∈G
1
2 EVANJENKINS
Under convolution, this ring is called the group algebra of G and will play a very
important role in the representation theory of finite groups.
There are three natural actions of G on CG, the left and right regular represen-
tations L and R and the conjugation action A . These are defined by
G G G
L (g)f(x)=f(g−1x),
G
R (g)f(x)=f(xg),
G
A (g)f(x)=f(g−1xg).
G
An element of CG that is invariant under the conjugation action is called a class
function. The class functions form a subspace of CG, which will be denoted by
CGG.
Exercise 2.1. Prove that CGG is a subalgebra of CG under both the pointwise
product and the convolution product.
Let’snowseehowourresultsfromthegeneralcompactcaselookinthissetting.
Recall that if (T,V) is a representation of G, its character χ ∈CGG is defined as
V
χ (g)=trT(g). We can give CG (and hence CGG) the L2 inner product,
V
(cid:90)
(cid:104)f ,f (cid:105)= f (g)f (g)dg.
1 2 1 2
G
Under this inner product, we proved the following orthogonality relation for char-
acters.
Theorem 2.2. Let T and T be irreducible representations with characters χ and
1 2 1
χ . Then
2
(cid:40)
1 if T ∼T ,
(cid:104)χ ,χ (cid:105)= 1 2
1 2
0 if T (cid:54)∼T .
1 2
Another big result of the compact theory is the Peter-Weyl theorem.
Theorem 2.3 (Peter-Weyl). The right regular representation CG decomposes as
(cid:77)
CG= d V,
V
V irred.
where d is the dimension of V.
V
Corollary 2.4. (cid:80) d2 =|G|.
V irred. V
AconsequenceofPeter-Weylthatwehaven’tseenyetsaysthatthecharactersof
representations of G form a uniformly dense subspace of continuous class functions
on G. In the finite setting, this gives the following result.
Theorem 2.5. CGG is equal to the span of the characters of G. In particular, the
characters of the irreducible representations form an orthonormal basis.
This result, combined with the following lemma, tells us that the number of
(isomorphism classes of) irreducible representations of a finite group G is equal to
the number of conjugacy classes of G.
Lemma 2.6. dimCGG is equal to the number of conjugacy classes of G.
REPRESENTATION THEORY OF Sn 3
Proof. A class function is precisely a function on G that is constant on conjugacy
classes. We can thus identify CGG with the algebra of functions on the set of
conjugacy classes of G. The dimension of this algebra is the number of conjugacy
classes. (cid:3)
Next, we will develop some theory that is particular to the case of finite groups.
Theorem 2.7. A representation of G is the same as a (finite dimensional) left
module over the group algebra CG.
Proof. Recallthatthegroupalgebrareferstotheconvolutionproductstructureon
CG. Toseethis,wefirstreinterpretthegroupalgebraCG. AfunctiononGcanbe
thought of as a formal C-linear combination of elements of G via (f : G → C) (cid:55)→
(cid:80)
f(g)g.
g∈G
Exercise 2.8. If we view an element of CG as (cid:80) a g, where a ∈C, then the
g∈G g g
convolution product becomes
 (cid:32) (cid:33)
(cid:88) (cid:88) (cid:88) (cid:88)
 agg bhh = agbhk.
g∈G h∈G k∈Ggh=k
What does it mean to give a left CG-module? First, we note that since CG
is a C-algebra, a left CG-module is in particular a C-vector space V, which we
are assuming to be finite dimensional. To specify a CG-module structure on V,
it is enough to say how a basis acts. We have a particularly nice basis, namely,
the elements g ∈ CG. Each g must act C-linearly, as C lies in the center of
CG. The element e acts trivially, as it is the identity for the convolution product.
Furthermore,sinceg·(h·v)=(gh)·v,itfollowsthattheactionofghisthecomposite
oftheactionsofgandh. Finally,theactionofeachgisinvertible,becausegg−1 =e
inCG. WemaythusconcludethatgivingaCG-modulestructureonV isthesame
as giving a homomorphism G → GL(V), i.e., a giving a representation of G on
V. (cid:3)
Whatdoesthispointofviewbuyus? Nowwecanapplythetheoryofringsand
modules to the study of representations of finite groups. Recall that a module is
simpleifithasnopropernontrivialsubmodules. Thus,aCG-moduleisirreducible
if and only if its corresponding module is simple.
3. The Symmetric Group
We know from the above discussion how many representations we need to look
for, namely, one for every conjugacy class. However, in general, there is no known
explicit one-to-one correspondence between the conjugacy classes and the irre-
ducible representations. For S , however, we can take advantage of the fact that
n
the conjugacy classes admit a very nice combinatorial description to construct ir-
reducible representations corresponding to them.
RecallthatanyelementofS canbewrittenasaproductofdisjointcycles, and
n
twoelementsofS areconjugateifandonlyiftheyhavethesamenumberofcycles
n
of any given length. Thus, a conjugacy class of S can be specified by giving an
n
unordered partition of the number n.
Wewillwriteapartitionofnasatupleλ=(λ ,λ ,...,λ ),where(cid:80)k λ =n,
1 2 k i=1 i
and λ ≥λ ≥···≥λ >0. We can also express such a partition pictorially using
1 2 k
4 EVANJENKINS
aYoung diagram,whichisaseriesofrowsofboxes,inwhichtheithrowcontains
λ boxes.
i
For example, the partition (5,4,1,1) of 11 can be represented by the following
Young diagram.
Before we find a general construction for all n, let’s first see how we can use the
theory we’ve developed above to study the character theory of S and S .
3 4
Let’s start with a few explicit examples of irreducible representations of S . In
n
fact, aside from a couple of small-n exceptions, these are the smallest irreducible
representations of S .
n
For every n ≥ 2, there are two one-dimensional representations of S . One is
n
the trivial representation V , in which every σ ∈ S acts trivially. The other is
triv n
the sign representation V . Recall that every element σ = S can be written
sgn n
as a product of transpositions, the number of which is unique modulo 2. We the
define sgnσ ∈ Z/2 to be the number of transpositions. In the sign representation,
σ acts as multiplication by (−1)sgnσ.
The third representation is a bit more complicated to define. Let V be an
perm
n-dimensional vector space with chosen basis (e ,...,e ). Then S acts on V
1 n n perm
bythepermutation representation,inwhichσ ∈S actsbysendinge toe .
n i σ(i)
However,thisrepresentationisreducible,becausethesubspacespannedby(cid:80)n e
i=1 i
is invariant.
Proposition 3.1. The (n−1)-dimensional representation V = V /C(e +
stan perm 1
...+e ) is irreducible.
n
Proof. Let A ∈ End (V). Then A = (a ) must commute with all permutation
G ij
matrices. Inotherwords,switchinganytworowsofAmustbethesameasswitching
any two columns. This implies that a =a for all i and j, and that a =a for
ii jj ij k(cid:96)
all i(cid:54)=j, k (cid:54)=(cid:96). So the space of such A is two-dimensional, which implies that V is
thedirectsumoftwoirreduciblerepresentations. Thus,W mustbeirreducible. (cid:3)
We call this irreducible representation the standard representation.
ThegroupS hasthreeconjugacyclasses, correspondingtotheYoungdiagrams
3
, , .
Thus,thethreeirreduciblerepresentationsdefinedabovemustbetheonlythree.
We can now construct the character table of S . The character table shows the
3
value of each irreducible character on each conjugacy class.
S
3
V 1 1 1
triv
V 1 -1 1
sgn
V 2 0 -1
stan
Exercise 3.2. Show that χ (σ) is equal to the number of fixed points of σ ∈S
perm n
acting as permutations of n points.
REPRESENTATION THEORY OF Sn 5
Since characters are additive with respect to direct sums, we can now compute
χ =χ −χ .
stan perm triv
The character table of S is a bit more complicated. We know three lines of the
4
table.
S
4
V 1 1 1 1 1
triv
V 1 -1 1 -1 1
sgn
V 3 1 0 -1 -1
stan
We know that the sum of the squares of the dimensions of the representations
mustbe4!=24. Sofar,wehave12+12+32 =11,sowearemissingarepresentation
of dimension 2 and one of dimension 3.
Wecangetonemorerepresentationofdimension3bytakingV ⊗V . Since
stan sgn
characters are multiplicative with respect to tensor product, we get the following.
S
4
V 1 1 1 1 1
triv
V 1 -1 1 -1 1
sgn
V 3 1 0 -1 -1
stan
V ⊗V 3 -1 0 1 -1
stan sgn
To get the last irreducible character χ, we can use orthogonality. We know it
must be two-dimensional, so χ(1)=2.
S
4
V 1 1 1 1 1
triv
V 1 -1 1 -1 1
sgn
V 3 1 0 -1 -1
stan
V ⊗V 3 -1 0 1 -1
stan sgn
V 2 a b c d
?
We now have four linear equations (cid:104)χ,χ (cid:105)=0 in four unknowns a, b, c, d, so we
i
can hopefully solve for the character. For example, the equation (cid:104)χ,χ (cid:105)=0 says
triv
that
2+6·a+8·b+6·c+3·d=0.
Note the multiplicities; we must multiply the value at each conjugacy class by the
number of elements in the conjugacy class.
Exercise 3.3. Compute the last irreducible character of S .
4
4. Specht Modules
Our general approach to the representation theory of S will be combinatorial.
n
Recall our construction of the standard representation. We got this representa-
tion by looking at the representation corresponding to a certain group action of S
n
(namely, its usual action as permutations of n elements). This permutation repre-
sentation was not irreducible, but it had an interesting irreducible representation
sitting inside of it.
6 EVANJENKINS
Ourgeneralapproachwillfollowalongthesamelines. Namely,wewillfindsome
combinatorial objects on which S acts, construct a corresponding representation,
n
and then find an interesting irreducible representation sitting inside of it.
Let λ be a partition of n. Recall that we can think of λ as a Young diagram.
A numbering of the Young diagram λ is a labeling of the boxes by the numbers
1,2,...,n. For instance, we can number the diagram by 2 5 1.
4 3
There are, of course, n! such numberings. There is an action of S on the set of
n
numberings by permuting the numbers.
A tabloid is an equivalence class of numberings of λ, where we consider two
numberings T and T(cid:48) equivalent if the entries of each row of T agree with those in
the corresponding row of T(cid:48). For instance, we have
2 5 1 1 2 5
∼ .
4 3 3 4
We denote by {T} the tabloid corresponding to T. There is an action of S on
n
the set of tabloids of shape λ by σ·{T}={σ·T}.
We now define Mλ to be the complex vector space with basis the tabloids of
shape λ, and we let S act by permuting the basis elements. We note that this is a
n
generalization of V , because if we take tabloids of shape (n−1,1), then there
perm
areexactlynofthem, oneforeachnumberthatcanappearinthesecondrow, and
the action of S is the usual action on the set {1,2,...,n}.
n
For any numbering T, we define C(T) to be the subgroup of S that leaves the
n
columnsofT invariant. Asanabstractgroup, C(T)∼=S ×S ×···×S , where
µ1 µ2 µ(cid:96)
µ = (µ ,µ ,...,µ ) is the partition we get by flipping the Young diagram for λ
1 2 (cid:96)
over the diagonal. This partition µ is called the conjugate partition to λ.
We can similarly define R(T) to be the subgroup of S that leaves the rows of
n
T invariant. As before, we have R(T)∼=S ×···×S .
λ1 λk
Let
(cid:88)
v = (−1)sgnq{q·T}.
T
q∈C(T)
Proposition 4.1. For any σ ∈S , σ·v =v .
n T σ·T
Proof. First, we note that C(σ·T)=σC(T)σ−1. We have
(cid:88)
σ·v =σ· (−1)sgnq{q·T}
T
q∈C(T)
(cid:88)
= (−1)sgnq{(σ·q)·T}
q∈C(T)
= (cid:88) (−1)sgn(σ−1q(cid:48)σ){(σ·σ−1q(cid:48)σ)·T}
q(cid:48)∈C(σ·T)
= (cid:88) (−1)sgnq(cid:48){q(cid:48)·(σ·T)}
q(cid:48)∈C(σ·T)
=v .
σ·T
(cid:3)
Corollary 4.2. Let Sλ be the subspace of Mλ spanned by the v . Then Sλ is a
T
subrepresentation.
The representation Sλ is called the Specht module associated with λ.
REPRESENTATION THEORY OF Sn 7
Exercise 4.3. Exhibit the following isomorphisms.
(1) S(n) ∼=V .
triv
(2) S(1,1,1,...,1) ∼=V .
sgn
(3) S(n−1,1) ∼=V .
stan
Constructing the Specht modules was fairly easy, but showing that they are
irreducible and pairwise nonisomorphic is a little trickier.
We will need to do a bit of combinatorics. First, we define a total ordering on
the set of all partitions called the lexicographic ordering. Given partitions λ
and µ, we say that µ<λ if µ <λ for the first i where µ (cid:54)=λ . (Here we use the
i i i i
convention that µ =0 if µ has fewer than i parts.)
i
Thesetofallpartitionshasapartialorderingcalledthedominance ordering.
We write µ(cid:69)λ, and we say that λ dominates µ, if for all i,
µ +µ +···+µ ≤λ +λ +···+λ .
1 2 i 1 2 i
We say that λ strictly dominates µ if µ(cid:69)λ and µ(cid:54)=λ.
Note that µ(cid:69)λ implies µ≤λ, but not conversely.
Lemma 4.4. Let T and T(cid:48) be numberings of λ and λ(cid:48), where λ does not strictly
dominate λ(cid:48). Then either
(1) there exist two distinct integers that occur in the same row of T(cid:48) and the
same column of T; or
(2) λ(cid:48) =λ, and there exist p(cid:48) ∈R(T(cid:48)) and q(cid:48) ∈C(T(cid:48)) such that p(cid:48)·T(cid:48) =q·T.
Proof. Suppose(1)doesnothold. ThenalloftheentriesinthefirstrowofT(cid:48) must
appear in different columns of T, so we can find a q ∈ C(T) such that the first
1
row of q ·T has all of the entries of the first row of T(cid:48). All of the entries in the
1
second row of T(cid:48) must also appear in different columns of T, and hence also those
of q ·T. So we can find a q ∈ C(T) such that the second row of q ·(q ·T) has
1 2 2 1
all the entries of the second row of T(cid:48), and the first row still has the entries of the
first row of T(cid:48). Iterating this procedure gives, for any i, q ,q ,...,q ∈ C(T) such
1 2 i
that the first i rows of q ·q ·····q ·T contain the entries of the first i rows of
i i−1 1
T(cid:48). This implies that λ(cid:48) +···+λ(cid:48) ≤λ +···+λ , and so λ(cid:48)(cid:69)λ.
1 i 1 i
Since we assumed λ did not strictly dominate λ(cid:48), it follows that λ = λ(cid:48). If k is
the number of rows of λ, and q =q ·q ·····q , it follows that q·T and T(cid:48) have
k k−1 1
the same entries in each row, by construction. Thus, there exists p(cid:48) ∈ R(T(cid:48)) such
that p(cid:48)·T(cid:48) =q·T. (cid:3)
We define b =(cid:80) (−1)sgn(q)q ∈CG. Then v =b ·{T}.
T q∈C(T) T T
Exercise 4.5. (1) If q ∈C(T), then b ·q =(−1)sgnq·b .
T T
(2) b ·b =|C(T)|b .
T T T
Lemma 4.6. Let T and T(cid:48) be numberings of shapes λ and λ(cid:48), and assume that λ
does not strictly dominate λ(cid:48). If there is a pair of integers in the same row of T(cid:48)
and the same column of T, then b ·{T(cid:48)}=0. Otherwise, b ·{T(cid:48)}=±v .
T T T
Proof. Suppose such a pair exists. Let t ∈ C(T) ∩ R(T(cid:48)) be the transposition
permuting them. Then
(cid:88)
b t= (−1)sgn(q)qt
T
q∈C(T)
8 EVANJENKINS
= (cid:88) (−1)sgn(q(cid:48)t−1)q(cid:48)
q(cid:48)∈C(T)
=− (cid:88) (−1)sgn(q(cid:48))q(cid:48)
q(cid:48)∈C(T)
=−b .
T
But t∈R(T(cid:48)) implies t·{T(cid:48)}={T(cid:48)}, so
b ·{T(cid:48)}=b ·(t·{T(cid:48)})=(b ·t)·{T(cid:48)}=−b ·{T(cid:48)},
T T T T
so b ·{T(cid:48)}=0.
T
On the other hand, if there is no such pair, Lemma 4.4 gives us p(cid:48) ∈ R(T(cid:48)),
q ∈C(T) such that p(cid:48)·{T(cid:48)}=q·{T}. We may then conclude
b ·{T(cid:48)}=b ·{p(cid:48)·T(cid:48)}
T T
=b ·{q·T}
T
=b ·q·{T}
T
=(−1)sgn(q)b ·{T}
T
=(−1)sgn(q)·v .
T
(cid:3)
Theorem 4.7. The Specht module Sλ is irreducible, and Sλ (cid:54)∼=Sλ(cid:48) for λ(cid:54)=λ(cid:48).
Proof. Let T be any numbering of λ. Then by Lemma 4.6 and Exercise 4.5,
b ·Sλ =b ·Mλ =C·v .
T T T
On the other hand, if λ < λ(cid:48) (in the lexicographic ordering), then λ does not
dominate λ(cid:48), so by Lemmas 4.4 and 4.6, it follows that
b ·Sλ(cid:48) =b ·Mλ(cid:48) =0.
T T
Suppose Sλ =V ⊕W. Then
C·v =b ·Sλ =b ·V ⊕b ·W.
T T T T
ThisimpliesthateitherV orW containsv ,butv generatesSλ asaCS -module,
T T n
so Sλ will be equal to whichever of V and W contains v . Thus, Sλ is irreducible.
λ
Suppose λ (cid:54)= λ(cid:48). Without loss of generality, we may assume λ < λ(cid:48). Then
b ·Sλ(cid:48) =0, but b ·Sλ (cid:54)=0, so we conclude that Sλ and Sλ(cid:48) are distinct. (cid:3)
T T
We have placed the Specht modules in one-to-one correspondence with the con-
jugacy classes of S , so we have indeed constructed all irreducible representations
n
of S .
n
Ofcourse,wearefarfromfinishedstudyingtherepresentationtheoryofS . We
n
wouldlike,forinstance,aniceformulaforthecharacterofSλ. Theusualprocedure
for producing such a formula is somewhat involved, but we will instead exhibit a
simple algorithm.
Thestartingpointforthisalgorithmistonotethatifλ<λ(cid:48),thenMλ(cid:48) doesnot
containanycopiesofSλ;indeed,thisfollowsdirectlyfromthefactthatb ·Mλ(cid:48) =0
T
if T is a numbering of λ. We therefore know that Mλ is already irreducible when
λ is highest in the lexicographic ordering (i.e., λ=(n)), and that for any lower λ,
Mλ only contains Specht modules corresponding to higher partitions λ(cid:48).
REPRESENTATION THEORY OF Sn 9
We can thus proceed inductively. If we know the characters of the Mλ, and the
characters of the Sλ for all λ > µ, then we can deduce the character for Sµ by
subtracting an appropriate number of copies of the characters of Sλ.
Exercise 4.8. Investigate the characters of the Mλ.
As an example, we recompute the character table for S . First, we write down
4
the characters for the Mλ in reverse lexicographic order.
S
4
M 1 1 1 1 1
M 4 2 1 0 0
M 6 2 0 0 2
M 12 2 0 0 0
M 24 0 0 0 0
As we said earlier, M is irreducible, so M =S .
S
4
S 1 1 1 1 1
M 4 2 1 0 0
M 6 2 0 0 2
M 12 2 0 0 0
M 24 0 0 0 0
Taking the inner product of the first two characters gives 1, so we know that
there is one copy of S in M . So we subtract the character of S
from that of M to the character of S .
10 EVANJENKINS
S
4
S 1 1 1 1 1
S 3 1 0 -1 -1
M 6 2 0 0 2
M 12 2 0 0 0
M 24 0 0 0 0
We continue in this fashion. M contains one copy of S , and one copy
of S , so we subtract each character once.
S
4
S 1 1 1 1 1
S 3 1 0 -1 -1
S 2 0 -1 0 2
M 12 2 0 0 0
M 24 0 0 0 0
S
4
S 1 1 1 1 1
S 3 1 0 -1 -1
S 2 0 -1 0 2
S 3 -1 0 1 -1
M 24 0 0 0 0