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Representation Theory 2013 Lecture Notes & Questions

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Representation Theory

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Part III Representation Theory

Please send any comments and corrections tocn319@cam.ac.

Lectured by Stuart Martin, Lent Term
0 Preliminaries Contents
1 Semisimple algebras
2 Irreducible modules for the symmetric group
3 Standard basis of Specht modules
4 Character formula
5 The hook length formula
6 Multilinear algebra and algebraic geometry
Interlude: some reminders about affine varieties
7 Schur–Weyl duality
8 Tensor decomposition
9 Polynomial and rational representations ofGL(V)
Interlude: some reminders about characters ofGLn(C)
10 Weyl character formula
11 Introduction to invariant theory and first examples
12 First fundamental theorem of invariant theory
Example sheet
Example sheet - Last updated Thursday 21 March

0 Preliminaries Contents

In this course we will study representations of

(a) (finite) symmetric groups

(b) (infinite) general linear groups

all overC, and apply the theory to “classical” invariant theory.

We require no previous knowledge, except for some commutative algebra, ordinary representation theory and algebraic geometry, as outlined below.

Commutative algebra

References:Part III course, Atiyah–Macdonald.

It is assumed that you will know about rings, modules, homomorphisms, quotients and chain conditions (the ascending chain condition, ACC, and descending chain condition, DCC).

Recall:achain, orfiltration, of submodules of a moduleMis a sequence (Mj: 0≤j≤n) such that M=M 0 &gt; M 1 &gt;···&gt; Mn&gt; 0 Thelengthof the chain isn, the number of links. Acomposition seriesis a maximal chain; equivalently, each quotientMj− 1 /Mjisirreducible.

Results:

SupposeMhas a composition series of lengthn. Then every composition series ofMhas length n, and every chain inMcan be extended to a composition series.

2 a composition series if and only if it satisfies ACC and DCC.

Modules satisfying both ACC and DCC are calledmodules of finite length. By (1), all composition series ofMhave the same length, which we denote byℓ(M) and call thelengthofM.

(Jordan–H ̈older) If (Mi) and (Mi′) are composition series then there exists a one-to-one cor- respondence between the set of quotients (Mi− 1 /Mi) and the set of quotients (Mi′− 1 /Mi′) such that the corresponding quotients are isomorphic.

Remark a vector spaceVover a fieldK, the following are equivalent:

(i) finite dimension

(ii) finite length

(iii) ACC

(iv) DCC

and if these conditions are satisfied thenℓ(V) = dim(V).

(Ordinary) representation theory (of finite groups)

References:Part II course, James–Liebeck, Curtis–Riener Vol. 1,➜1 below.

Work overCor any fieldKwhere char(K)∤|G|. LetVbe a finite-dimensional vector space.

1 Semisimple algebras

Conventions All rings are associative and have an identity, which we denote by 1 or 1R. Modules are always leftR-modules.

(1) Definition AC-algebrais a ringRwhich is also aC-vector space, whose notions of addition coincide, and for which λ(rr′) = (λr)r′=r(λr′) for allr,r′∈Randλ∈C. There are obvious notions ofsubalgebrasandalgebra homomor- phisms.

Usually, our algebras will be finite-dimensional overC.

(1) Examples and remarks

(a)Cis aC-algebra, andMn(C) is aC-algebra. IfGis a finite group then thegroup algebraCG={

∑

gαgg:αg∈C}is aC-algebra with ❼pointwise addition:

∑

gαgg+

∑

gβgg=

∑

g(αg+βg)g ❼multiplication:

∑

(∑

hh′=gαhβh′

)

Recall the correspondence between representations ofGoverCand finite-dimensionalCG-modules.

(b) IfR,SareC-algebras then thetensor productR⊗S=R⊗CSis aC-algebra. [More on⊗can be found in Atiyah–MacDonald p. 30.]

(c) If 1R= 0RthenR={ 0 R}is thezero ring. Otherwise, forλ,μ∈C, we haveλ 1 R=μ 1 Rif and only ifλ=μ, so we can identifyλ∈Cwithλ 1 R∈R. With the identificationC֒→Rwe can viewCas a subalgebra ofR.

(d) AnR-moduleMbecomes aC-space viaλm= (λ 1 R)mforλ∈C,m∈M.

Given anyR-moduleN, HomR(M,N) is aC-subspace of HomC(M,N). In particular, it is a C-space, with structure

(λφ)(m) =λφ(m) =φ(λm) ∀λ∈C, m∈M, φ∈HomR(M,N)

(e) IfMbe anR-module. Then EndR(M) = HomR(M,M) is aC-algebra with multiplication given by composition of maps:

(φψ)(m) =φ(ψ(m)) ∀m∈M, φ,ψ∈EndR(M)

In particular, ifVis aC-space then EndC(V) is an algebra. Recall that EndC(V)∼=Mn(C), where n= dimC(V).

(f) LetRbe aC-algebra andX⊆Rbe a subset. Define thecentraliserofXinRby

cR(X) ={r∈R:rx=xr∀x∈X}

It is a subalgebra ofR. ThecentreofRiscR(R) =Z(R).

(g) LetRbeC-algebra andMbe aR-module. Then the mapαR→EndC(M) given byα(r)(m) =rm forr∈R,m∈Mis a map ofC-algebras. Now, EndR(M) =cEndC(M)(αR), and soαR⊆cEndC(M)(EndR(M)).

(h) AnR-moduleMis itself naturally an EndR(M)-module, where the action is evaluation, and the action commutes with that ofR. The inclusionαas in (g) says that elements ofRact as EndR(M)-module endomorphisms ofM. Given anR-moduleN, HomR(M,N) is also an EndR(M)-module, where the action is composition of maps.

(1) Lemma LetRbe a finite-dimensionalC-algebra. Then there are only finitely many isomor- phism classes of irreducibleR-modules. Moreover, all the irreducibleR-modules are finite-dimensional.

Proof an irreducibleR-module and pick 0 6 =x∈S. Define a mapR→Sbyr7→rx. This map has nonzero image, so its image isSby irreducibility. In particular,

dimC(S)≤dimC(R)&lt;∞

ButSmust occur in any composition series ofR, so by the Jordan–H ̈older theorem, there are only finitely many isomorphism classes of irreducible modules.

(1) Lemma(Schur’s lemma) LetRbe a finite-dimensionalC-algebra. Then

(a) IfS≇Tare nonisomorphic irreducibleR-modules then HomR(S,T) = 0.

(b) IfSis an irreducibleR-module then EndR(S)∼=C.

Proof

(a) Iff:R→Sis a map of algebras then ker(f)≤Rand im(f)≤S. By irreducibility, ker(f) = 0 orRand im(f) = 0 orS. If ker(f) = 0 thenR∼=im(f) =S, contradicting nonisomorphism; so we must have ker(f) =Rand im(f) = 0, i.e= 0.

(b) ClearlyD= EndR(S) is a division ring. SinceSis finite-dimensional, so isD. Thus, ifd∈D, then the elements 1,d,d 2 ,.. linearly dependent. Letp∈C[X] be a nonzero polynomial with p(d) = 0. SinceCis algebraically closed,pfactors as

p(X) =c(X−a 1 )···(X−an)

for some 0 6 =c∈Candai∈C. Hence

(d−a 11 D)···(d−an 1 D) = 0

But, being a division ring,Dhas no zero divisors, so one of the terms must be zero. Thus d=aj 1 D∈C 1 D. Butd∈Dwas arbitrary, soD=C 1 D∼=C.

(1) Definition AnR-module issemisimple(orcompletely reducible) if it is a direct sum of irreducible submodules. A finite-dimensionalC-algebraRis semisimple if it is so as anR-module.

(1) Proposition

(a) Every submodule of a semisimple module is semisimple, and every such submodule is a direct summand.

(b) Every quotient of a semisimple module is semisimple.

In fact,Mis semisimple if and only if every submodule ofMis a direct summand. The proofs of this and of(1)are left as an exercise.

Remarks

1 semisimpleC-algebra⇒anyR-module is semisimple.

2 finite group⇒CGis semisimple. (This is precisely Maschke’s theorem.)

(1) Definition IfRandSareC-algebras,Mis anR-module andNis anS-module, thenM⊗N has the structure of anR-module via

r(m⊗n) =rm⊗n, r∈R, m∈M, n∈N

Now EndC, and henceR, has transitively onVi−{ 0 }, thus theViare irreducibleR-modules.

IfVihas basis{ei 1 ,...,eimi}then the map

R→V︸ 1 ⊕···⊕︷︷ V 1 ︸ m 1 copies

⊕···⊕V︸k⊕···⊕︷︷ Vk︸ mkcopies

given by (f 1 ,...,fk)7→(f 1 (e 11 ),...,f 1 (e 1 m 1 ),...,fk(ek 1 ),...,fk(ekmk))

is an injective map ofR-modules, and hence is an isomorphism by counting dimensions. ThusRis semisimple.

TheViform a complete set of irreducibleR-modules (by Jordan–H ̈oldeer, as in(1)), and they are nonisomorphic: ifi 6 =jthen (0,..., 0 , 1 , 0 ,...,0) annihilatesVjbut notVi.

Step 2 LetRbe any ring. Now the natural map

α:R→EndEndR(R)(R)

sendingrto the mapx7→rxis an isomorphism. It’s injective, and ifθ∈EndEndR(R)(R) then it commutes with the endomorphismsαr∈EndR(R), whereαr(x) =xr. Now ifr∈Rthen

θ(r) =θ(1r) =θ(αr(1)) =αr(θ(1)) =θ(1)r

soθacts by left-multiplication. Henceθ=α(θ(1))∈α(R), soαis surjective.

AssumeRis a semisimpleC-algebra. Then EndR(R) is semisimple by(1), and soRis a semisimple EndR(R)-module. By(1)and the isomorphismα,Rhas the required form.

(1) Lemma IfRis semisimple andMis a finite-dimensionalR-module then the natural map

α:R→EndEndR(M)(M)

sendingrto the mapm7→rmis surjective.

Proof LetI=M◦={r∈R:rM= 0}be the annihilator ofM; then kerα=I. Now,R/Iis semisimple andMis anR/I-module, and EndR(M) = EndR/I(M). So we can replaceRbyR/Iand suppose thatMis faithful (andαis injective).

By(1), EndR(M) =

∏

SEndCHomR(S,M) and, sinceMis faithful, all the spaces HomR(S,M) are nonzero because they are precisely the simple EndR(M)-modules. By(1),Mis isomorphic (as an EndR(M)-module) to

⊕

SS⊗HomR(S,M), so it’s the direct sum of dimCScopies of the simple module HomR(S,M) for eachS.

As in(1), this implies that

dimCEndEndR(M)(M) =

∑

(dimCS) 2

But this is equal to dimCR, soαis an isomorphism.

Finally, it’s worth noting:

(1) Lemma LetRbe aC-algebra and lete∈Rbe ‘almost idempotent’ in the sense thate 2 =λe for some 0 6 =λ∈C. Then for anyR-moduleM, we have an isomorphism of EndR(M)-modules

HomR(Re,M)∼=eM

Proof eMis an EndR(M)-submodule ofM: ifφ∈EndR(M) andem∈eMthenφ(em) =eφ(m)∈ eM.

Replacingebyeλ, we may suppose thateis idempotent. Now ther exists a map of EndR(M)-modules

Hom(Re,M)→eM

given byφ7→φ(e), whose inverse sendsmto the mapr7→rm.

Chapter I: Representation theory of the symmetric group

2 Irreducible modules for the symmetric group

Recall the correspondence between representationsρ:G→GL(V) andCG-modules given by setting g·v=ρ(g)vforg∈Gandv∈V. Thetrivial representationG→C×= GL(C) is given byg7→1, and the correspondingCG-module isC, sometimes writtenCGto emphasise the context.

IfX⊆{ 1 ,...,n}, write SXfor the subgroup of Snfixing every number outside ofX.

Letε: Sn→{± 1 }be thesign representation, and write

εσ=

∏

1 ≤i&lt;j≤n

σ(i)−σ(j) i−j

Throughout this section, writeA=CSn; this is a finite-dimensional semisimpleC-algebra.

(2) Definition Forλi∈N 0 , we sayλ= (λ 1 ,λ 2 ,λ 3 ,...) is apartitionofn, and writeλ⊢n, if λ 1 ≥λ 2 ≥λ 3 ≥···and

∑

iλi=n. Write (λ

a 1 1 ,λ

a 2 2 ,...,λ

ak k) to denote

(λ 1 ,...,λ 1 ︸︷︷︸ a 1

,λ 2 ,...,λ 2 ︸︷︷︸ a 2

,...,λk,...,λk ︸︷︷︸ ak

, 0 , 0 ,...)

withλ 1 &gt; λ 2 &gt;···&gt; λk.

Partitions ofncorrespond with conjugacy classes of Sn, e.

(5, 22 ,1) ↔ (• • • • •)(• •)(•)

Ifλ,μ⊢n, writeλ &lt; μif and only if there is somei∈Nwithλj=μjfor allj &lt; iandλi&lt; μi. This is thelexicographic(ordictionary)orderon partitions. It is a total order on the set of partitions ofn. E. (5)&gt;(4,1)&gt;(3,2)&gt;(3, 12 )&gt;(2 2 ,1)&gt;(2, 13 )&gt;(1 5 )

(2) Definition Partitionsλ⊢nmay be geometrically represented by theirYoung diagram

[λ] ={(i,j) :i≥ 1 , 1 ≤j≤λi}⊆N×N

If (i,j)∈[λ] then it is called anodeof [λ]. Thekthrow(resp) of a diagram consists of those nodes whose first (resp. second) coordinate isk.

Example Ifλ= (4, 22 ,1) then

[λ] =

• • • ←λ 1
• ←λ 2
•
←λ 3

Theheight(orlength) ofλis the length of the first column of [λ], i. max{j :λj 6 = 0}, and is denoted by ht(λ).

(2) Definition Aλ-tableautλis one of then! arrays of integers obtained by replacing each node in [λ] by one of the integers 1,...,n, with no repeats. E. the following are (4, 3 ,1)-tableaux

t 1 =

1 2 4 5

3 6 7

8

, t 2 =

4 5 7 3

2 1 8

6

Equivalently, aλ-tableau can be regarded as a bijection [λ]→{ 1 , 2 ,...,n}.

(A) There exist distinct integersiandjoccurring in the same row oftλand the same column oftμ;

(B)λ=μandtμ=rctλfor somer∈Rtλandc∈Ctλ.

Proof Suppose (A) is false. Ifλ 16 =μ 1 thenλ 1 &gt; μ 1 , so [μ] has fewer columns than [λ]. Hence two of the numbers in the first row oftλare in the same column oftμ, so (A) holds, contradicting our assumption. Soλ 1 =μ 1.

Since (A) fails, somec 1 ∈Ctμforcesc 1 tμto have the same members in the first row oftλ. Now ignore the first rows oftλandc 1 tμ. The same argument implies thatλ 2 =μ 2 and there isc 2 ∈Ctμsuch that tλandc 2 c 1 tμhave the same numbers in each of their first two rows.

Continuing in this way we see thatλ=μand there existsc′∈Ctλsuch thattλandc′tμhave the same numbers in each row. Thenrtλ=c′tμfor somer∈Rtλ.

Now define c=r− 1 (c′)− 1 r∈r− 1 c′Ctμ(c′)− 1 r=r− 1 Cc′tμr=Cc′tμ=Ctλ Thentμ=rctλ.

(2) Lemma Ifσ∈Sncannot be written asrcfor anyr∈Rtλandc∈Ctλ then there exist transpositionsu∈Rtλandv∈Ctλsuch thatuσ=σv.

Proof Since(2)(B) fails fortλandσtλ, there exist integersi 6 =jin the same row oftλand the same column ofσtλ. Takeu= (i j) andv=σ− 1 uσ.

(2) Lemma Given a tableautλanda∈A, the following are equivalent:

(1)rac=εcafor allr∈Rtλandc∈Ctλ;

(2)a=kh(tλ) for somek∈C.

Proof

(2)⇒(1) Asr′runs throughRtλso doesrr′, and ascruns throughCtλso doesc′cwithεc′c=εc′εc. Calculation reveals thatrh(tλ)c=εch(tλ).

(1)⇒(2) Leta=

∑

σ∈Snaσσ. Ifσis not of the formrcthen by(2)there exist transpositionsu∈Rtλ andv∈Ctλsuch thatuσv=σ. By assumption,uav=εva, and the coefficient ofσgivesauσv=εvaσ, soaσ=−aσ, and hence aσ= 0. The coefficient ofrcin (1) givesa 1 =εcarc∈C. Thus

∑

r,c

arcrc=

∑

εca 1 rc=a 1 h(tλ)

as required

(2) Lemma Ifa∈Athenh(tλ)ah(tλ) =kh(tλ) for somek∈C.

Proof x=h(tλ)ah(tλ) satisfies(2)(1).

(2) Proposition Definefλ= dimCAhλ. Then

(a)h(tλ) 2 =

n! fλ

h(tλ)

(b) fλ|n!

(c)h(tλ)Ah(tλ) =Ch(tλ)

In particular,

fλ n!

h(tλ) is idempotent.

Proof

(a) Leth=h(tλ). We already know thath 2 =khfor somek∈C. Right-multiplication byhinduces a linear mapˆh:A→A. Fora∈A, (ah)h=k(ah), soˆk|Ahis multiplication byk. Extend a basis ofAhto a basis ofA. With respect to this basis,

ˆh∼

(

kIfλ ∗ 0 0

)

and so tr(ˆh) =kfλ. With respect to the basis SnofA,ˆhhas matrixHwhose entries are Hστ= coefficient ofσinτh ButHστ= 1, so tr(ˆh) =n!.

Hencek= n! fλ

.

(b) The coefficient of 1 inh 2 =khis

∑

εc 1 εc 2 ∈Zwhere the sum runs over allr 1 ,r 2 ∈Rtλand c 1 ,c 2 ∈Ctλsuch thatr 1 c 1 r 2 c 2 = 1. (c) By(2), the only other possibility ishAh= 0; buth 26 = 0.

(2) Proposition Ah(tλ) is an irreducibleA-module.

Proof Puth=h(tλ). ThenAh 6 = 0 andAis semisimple, so it is enough to show thatAhis indecomposable.

SupposeAh=U⊕V. ThenCh=hAh=hU⊕hV, so one ofhUandhVis nonzero; without loss of generality,hU 6 = 0. ThenhU=Ch, soAh=AhU⊆U, soU=AhandV= 0.

(2) Lemma Ifλ &gt; μare partitions andtλ,tμare tableaux thenh(tμ)Ah(tλ) = 0.

Proof By(2)there exist integers in the same row oftλand the same column oftμ. Letτ∈ Rtλ∩Ctμbe the corresponding transposition. Then h(tμ)h(tλ) =h(tμ)ττh(tλ) =−h(tμ)h(tλ) soh(tμ)h(tλ) = 0.

Applying this to (σtλ,tμ) forσ∈Sngives 0 =h(tμ)h(tλ) =h(tμ)σh(tλ)σ− 1

and soh(tμ)σh(tλ) = 0.

Thush(tμ)Ah(tλ) = 0.

(2) Lemma Ifλ 6 =μandtλ,tμare tableaux, thenAh(tλ)≇Ah(tμ).

Proof Assume without loss of generality thatλ &gt; μ. If there exists anA-module isomorphism f:Ah(tλ)∼=Ah(tμ) then

f(Ch(tλ)) ( 2. 10 ) = f(h(tλ)Ah(tλ)) =h(tλ)f(Ah(tλ)) =h(tλ)Ah(tμ) ( 2. 12 ) = 0

which is a contradiction.

(2) Theorem The left-ideals{Ahλ :λ⊢n}form a complete set of nonisomorphic irreducible A-modules.

Proof They are irreducible by(2)and nonisomorphic by(2)and, since

|{λ:λ⊢n}|=|{ccls of Sn}|=|{irreducibleCSn-modules}| they form a complete set.

Further reading Chapter 4 of James’s lecture notes (Springer) – alternative proof using ‘poly- tabloids’ and certain permutation modules.

as required.

(3) Lemma

∑

λ⊢m

Fλ 2 =m!

Proofby induction onm.

The casem= 1 is clear. Suppose the statement holds true for partitions ofm−1. Then

∑

λ⊢m

Fλ 2

( 3. 2 ) =

∑

λ⊢m,λ/τ

FλFτ

( 3. 4 ) =

∑

τ⊢m− 1

mFτ 2 =m·(m−1)! = m!

as required.

(3) Lemma Iftλ&gt; t′λare standard thenh(tλ)h(t′λ) = 0.

Proof It suffices to show that there are integersi 6 =jlying in the same row oft′λand the same column oftλ. For then the corresponding transpositionτ= (ij)∈Rt′λ∩Ctλgives the result as in the proof of(2).

It remains to find such integersiandj. Letx∈[λ] be the first place wheretλandt′λfirst differ. Let t′λ=iandy= (t′λ)− 1 (i)∈[λ]. Thenymust be below and to the left ofxsincetλis standard. In particular,xcannot lie in the first column or the last row. Letz∈[λ] be in the same row asxand the same column ofy, and letj=tλ(z) =t′λ(z) be the common value oftλandt′λatz.

It is now elementary to verify thatiandjsatisfy the above assumption.

(3) Theorem CSn=

⊕

CSnh(tλ), where the direct sum runs over all standard tableauxtλfor all partitionsλ⊢n.

Proof The sum is direct: suppose we have a nontrivial relation

∑

a(tλ)h(tλ) witha(tλ)∈CSn are not all zero. Chooseμmaximal (in the ordering on partitions) such that somea(tμ)h(tλ) 6 = 0, and for thisμpickt′μminimal (in the ordering on standardμ-tableaux) such thata(t′μ)h(t′μ) 6 = 0. Multiplying the relation on the right byh(t′μ) givesa(t′μ)h(t′μ) 2 = 0 by(3)and(2), and hence a(t′μ)h(t′μ) = 0, contradicting our assumption.

It is clear that the

⊕

CSnh(tλ)≤CSn. By Artin–Wedderburn,CSn∼=

⊕

fλCSnhλasCSn-modules, while in our direct sum there areFλcopies ofChλ. By Jordan–H ̈older,Fλ≤fλ. Butfλ≤Fλsince, by(3),

∑

Fλ 2 =n! =

∑

fλ 2. HenceFλ=fλfor eachλ⊢n, and soCSn≤

⊕

CSnh(tλ).

(3) Corollary The number of standardλ-tableaux is equal to dimC(CSnhλ).

Proof Hλ=hλas in the proof of(3).

(3) Lemma LetMbe a finite-dimensionalCSn-module. ThenM∼=

⊕

h(tλ)M, where the direct sum runs over all standard tableauxtλfor all partitionsλ⊢n.

Proof If

∑

m(tλ) = 0 is a nontrivial relation withm(tλ)∈h(tλ)Mthen chooseμminimal andt′μ maximal withm(t′μ) 6 = 0. Premultiply the relation byh(t′μ) to contradict(3)and(2). Finally,

⊕ h(tλ)M

( 1. 12 ) ∼=

⊕

HomCSn(CSnh(tλ),M)

∼=HomCSn

(⊕

CSnh(tλ),M

)

∼=HomCSn(CSn,M)∼=M

4 Character formula

Given a finite-dimensionalCG-algebraM, itscharacteris defined by

χM(g) = tr

(

M→M

m7→gm

)

andχMis aclass functionG→C, i. a function that is constant on the conjugacy classes ofG. If αis a conjugacy class inG, writeχM(α) for the common value ofχMonα.

Givenλ⊢n, denote the character of the irreducibleCSn-moduleCSnhλbyχλ. We will calculate χλ(α) and later use the formula to derive Weyl’s character formula for GLn(C).

Now Sn-classαcomprises all permutations with fixed cycle typenαn··· 2 α 21 α 1 , i. havingαnn-cycles, ...,α 2 2-cycles andα 1 1-cycles. Letnα=|α|.

(4) Lemma nα=

n! 1 α 12 α 2 ..αn·α 1 !α 2 !...αn!

Proof Any permutation inαis one of then! of the form

(∗)(∗)···(∗)

︸︷︷︸

α 1

(∗∗)(∗∗)···(∗∗)

︸︷︷︸

α 2

···(

n ︷︸︸︷ ∗···∗)(

n ︷︸︸︷ ∗···∗)···(

n ︷︸︸︷ ∗···∗) ︸︷︷︸ αn Each such permutation can be represented in 1α 1 ..αn·α 1 !...αn! ways: each of theαkk-cycles can be represented inkways by cycling their components, and each block ofαkk-cycles can be represented inαk! ways by permuting the position of the factors.

(4) Theorem(orthogonality relations) Letλ,μ⊢nand letαandβbe conjugacy classes in Sn. Then

(a)

∑

cclsα

nαχλ(α)χμ(β) =

{

n! ifλ=μ 0 ifλ 6 =μ

(b)

∑

λ⊢n

χλ(α)χλ(β) =

{

n!/nα ifα=β 0 ifα 6 =β

Proof Every element of Snis real, i. conjugate to its inverse, and so

χλ(σ) =χλ(σ− 1 ) =χλ(σ)

soχλ(σ)∈Rfor allσ∈Sn. These relations are then just the usual ones for general finite groups.

Notation Givenx= (x 1 ,...,xn)∈Cm,ℓ 1 ,...,ℓm∈Z, define

|xℓ 1 ,...,xℓm|= det(x ℓj i)

Usuallyℓj≥0, in which case this is a homogeneous polynomial of degree

∑m

ℓjinx 1 ,...,xm.

ExampleThe Vandermonde determinant is given by

V(x) =|xm− 1 ,..., 1 |

(4) Lemma V=V(x) =

∏

i&lt;j(xi−xj)

Proof Exercise.

In particular, it this term is zero unless theℓiare all distinct. So

det

(

1

1 −xiyj

)

=

∑

ℓ 1 ,...,ℓmdistinct

xℓ 11 ..ℓmm

∣

∣yℓ 1 ,...,yℓm

∣

=

∑

ℓ 1 &gt;···&gt;ℓm≥ 0

(

∑

π∈Sm

x ℓπ(1) 1 ..

ℓπ(m) m

∣

∣yℓπ(1),...,yℓπ(m)

∣

)

=

∑

ℓ 1 &gt;···&gt;ℓm≥ 0

(

∑

π∈Sm

x ℓπ(1) 1 ..

ℓπ(m) m επ

∣

∣yℓ 1 ,...,yℓm

∣

)

=

∑

ℓ 1 &gt;···&gt;ℓm≥ 0

∣

∣xℓ 1 ,...,xℓm

∣

∣yℓ 1 ,...,yℓm

∣

which is what we needed.

Notation Ifλ= (λ 1 ,...,λm)∈Zm, letℓi=λi+m−i, i.ℓ 1 =λ 1 +m−1,...,ℓm=λm. Sometimes in the literature, theℓiare denoted byβiand are thus called ‘β-numbers’. Note that

(1)λ 1 ≥···≥λmif and only ifℓ 1 &gt;···&gt; ℓm

(2) If

∑

iλi=nandλi≥0 then

∣

∣xℓ 1 ,...,xℓm

∣

|xm− 1 ,..., 1 |

is a polynomial of degreemin thexis.

Write Λ+(m,n) for the set of all partitionsλofninto≤mparts.

(4) Definition Givenx 1 ,...,xm,y 1 ,...,ym∈C, then for anyi∈N, set

si=xi 1 +···+xim, ti=y 1 i+···+tim

These are called thepower sums(sometimes referred to in the literature asNewton sums).

(4) Lemma

∑

λ∈Λ+(m,n)

∣

∣xℓ 1 ,...,xℓm

∣

|xm− 1 ,..., 1 |

∣

∣yℓ 1 ,...,yℓm

∣

|ym− 1 ,..., 1 |

=

1 ∑

cclsα

nαsα 11 ..αnntα 11 ..αnn

Note that both quotients on the left-hand side genuinely are polynomials (in fact, Schur polynomials), so they make sense even if thexi,yiare not distinct.

Proof Both sides are polynomials, so without loss of generality we may take all thexi,yito have modulus&lt;1. Then

log

∏n

i,j=

1

1 −xiyj

=

∑n

i,j=

(

xiyj 1

+

(xiyj) 2 2

+

(xiyj) 3 3

+···

)

=

s 1 t 1 1

+

s 2 t 2 2

+

s 3 t 3 3

+···

and hence

∏n

i,j=

1

1 −xiyj

= exp

(

s 1 t 1 1

+

s 2 t 2 2

+

s 3 t 3 3

+···

)

=

∑∞

1 (

s 1 t 1 1

+

s 2 t 2 2

+

s 3 t 3 3

+···

=

∑

(∗)

(

1

·

n! α 1 !α 2 !···

(

s 1 t 1 1

)α 1 ( s 2 t 2 2

)α 2 ···

)

where (∗) denotes the sum over allnand all sequencesα 1 ,α 2 ,... of nonnegative integers with only finitely many nonzero terms andα 1 +α 2 +···=n. But then

∏n

i,j=

1

1 −xiyj

=

∑

(∗)

sα 11 sα 22 ···tα 11 tα 22 ··· 1 α 12 α 2 ···α 1 !α 2 !···

By(4)(Cauchy) and(4), we have

∑

ℓ 1 &gt;ℓ 2 &gt;···&gt;ℓm≥ 0

∣

∣xℓ 1 ,...,xℓm

∣

|xm− 1 ,..., 1 |

∣

∣yℓ 1 ,...,yℓm

∣

|ym− 1 ,..., 1 |

=

∏m

i,j=

1

1 −xiyj

=

∑

(∗)

sα 11 sα 22 ···tα 11 tα 22 ··· 1 α 12 α 2 ···α 1 !α 2 !···

so we can equate terms which are of degreenin thexito obtain the desired result.

(4) Definition Forλ= (λ 1 ,...,λm)∈Zmandαa conjugacy class in Sn, letψλ(α) be the coefficient of the monomialxλ 11 ..λmminsα 11 ..αnn, i.

sα 11 ..αnn=

∑

λ∈Zm

ψλ(α)xλ 11 ..λmm

Remarks

(1)ψλis a class function for Sn.

(2)ψλ(α) = 0 if anyλiis negative or if

∑m

λi 6 =n.

(3)ψλis a symmetric function of theλis.

Notation Set ωλ(α) =

∑

π∈Sm

επψ(ℓπ(1)+1−m, ℓπ(2)+2−m, ... , ℓπ(m))(α)

The idea now is to show thatωλ=χλwheneverλ⊢n.

(4) Lemma(Character formula)

sα 11 ..αnn

∣

∣xm− 1 ,..., 1

∣

∣=

∑

λ∈Λ+(m,n)

ωλ(α)

∣

∣xℓ 1 ,...,xℓm

∣

ProofThe left-hand side is equal to

∑

λ∈Zm

ψλ(α)xλ 11 ..λmm

∑

τ∈Sm

ετxmτ(1)− 1 .. 0 τ(m)

which, by combining terms, is equal to

∑

λ∈Zm

∑

τ∈Sm

ετψλ(α)x λτ(1)+m− 1 τ(1) ..

λτ(m) τ(m)

Letℓi=λτ(i)+m−i—note that this is not our usual convention—then since theψλ(α) are symmetric, our expression is equal to

∑

λ∈Zm

∑

τ∈Sm

ψ(ℓ 1 +1−m, ... , ℓm)(α)xℓτ 1 (1)..ℓτm(m)

=

∑

λ∈Zm

ψ(ℓ 1 +1−m, ... , ℓm)(α)

∣

∣xℓ 1 ,...,xℓm

∣

Whenever theℓiare not distinct, the corresponding term gives zero. So settingλi=ℓi+i−mand using the fact thatℓ 1 &gt;···&gt; ℓmif and only ifλ 1 ≥···≥λm, our expression is equal to

∑

λ 1 ≥···≥λm

∑

π∈Sm

ψ(ℓπ(1)+1−m, ... , ℓπ(m))επ

∣

∣xℓ 1 ,...,xℓm

∣

The terms for whichλis not a partition ofnare zero, for ifλm&lt;0 thenℓm+π− 1 (m)−m &lt;0. Thus we have the desired equality.

The number of permutations inRof this type is ( λ 1! 1 α 11 ..α 1 nα 11 !...α 1 n!

)(

λ 2! 1 α 21 ..α 2 nα 21 !...α 2 n!

)

···

(

λm! 1 αm 1 ..αmnαm 1 !...αmn!

)

and so

χ(α) =

∑[( α 1! 1 α 11 ..α 1 nα 11 !...α 1 n!

)

···

(

αm! 1 αm 1 ..αmnαm 1 !...αmn!

)]

(†)

where the sum runs over allαijsatisfying (∗). By the multinomial theorem,

sα 11 ..αnn=

∑[( α 1 !xα 111 ..αm 1 m 1 α 11 ..α 1 nα 11 !...α 1 n!

)

···

(

αm!xα 1 n 1 ..αmnm 1 αm 1 ..αmnαm 1 !...αmn!

)]

where the sum runs over allαij∈Nsatisfying

α 1 j+α 2 j+···+αmj=αj

for all 1≤j≤n. Soψλ(α), which is the coefficient ofxλ 11 ..λmm, is equal to the right-hand side of (†), and hence it equalsχ(α).

(4) Lemma Letλ,μ ⊢nwithμ ≤λ. The irreducible moduleCSnhμ is isomorphic to a submodule ofCSnrλif and only ifλ=μ.

Proof Withμ &lt; λandσ∈Sn, by(2)there exist two integers in the same row oft◦λand the same column ofσ− 1 t◦μ, so ifτis their transposition thenστσ− 1 ∈Ct◦μ. And

hμσrλ=hμστσ− 1 στrλ=−hμσrλ ⇒ hμσrλ= 0

Thus 0 =hμCSnrλ∼=HomCSn(CSnhμ,CSnrλ) by(1).

Conversely supposeμ=λ. Then hλrλ

∑

σ∈Ct◦λ

εσσ=h 2 λ 6 = 0

so 0 6 =hλCSnrλ∼=HomCSn(CSnhλ,CSnrλ).

Note that, in general,CSnhλis not a submodule ofCSnrλ.

(4) Lemma Ifλ∈Λ+(n,m) thenωλ=χλ.

Proof We’ll take this in steps.

Step 1 ωλis aZ-linear combination ofψνwithν≥λand with coefficient ofψλequal to 1.

Ifπ∈Snandμπis the partition with parts

ℓπ(1)+ 1−m,...,ℓπ(m)

i. with partsλi+π− 1 (i)−1. Sinceψλ(α) is symmetric in theλi, we can write

ωλ=

∑

π∈Sn

επψμπ

Ifπ= 1 thenμπ=λ, so we’re done.

Ifπ 6 = 1, thenλ 1 +π− 1 (1)− 1 ≥λ(with equality if and only ifπ− 1 (1) = 1; and thenλ 2 +π− 1 (2)− 2 ≥λ 2 with equality if and only ifπ− 1 (2) = 2; and so on. Thusμπ&gt; λ. ⊳

Step 2 ωλ=

∑

ν⊢n

kλνχνwithkλν∈Z,kλλ&gt;0 andkλν= 0 ifν &lt; λ.

By(4)and(4),ψλis anN-linear combinationχμs withμ≥λand with nonzero coefficient of χλ. Thusωλis aZ-linear combination ofχνs withν≥λand with positive coefficient ofχλ. ⊳

Step 3 The result follows.

We know by(4)that ∑

cclsα

nαωλ(α)ωμ(α) =

{

n! ifλ=μ 0 ifλ 6 =μ

In the caseλ=μ, orthogonality ofχλimplies

∑

ν⊢n

k 2 λν= 1, and thuskλν= 0 ifλ 6 =μandkλλ= 1. ⊳

So we’re done.

Thekλνs in the above proof are sometimes referred to asKostka numbersin the literature.

At long last, takem∈N,x 1 ,...,xm∈Carbitrary,ℓi=λi+m−iforλ⊢n, andsi=xi 1 +···+xim.

(4) Theorem sα 11 ..αnn

∣

∣xm− 1 ,..., 1

∣

∣=

∑

λ∈Λ+(m,n)

χλ(α)

∣

∣xℓ 1 ,...,xℓn

∣

Proof (4)and(4).

Remarks

(1) Withm≥nwe can ensure that the right-hand side of the equation in(4)involves all partitions ofn.

(2) Ifλ∈Λ+(m,n) thenχλ(α) is the coefficient of the monomialxℓ 11 ..ℓmmin the expansion of sα 11 ..αnn.

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Representation Theory 2013 Lecture Notes & Questions

Module: Representation Theory

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Part III Representation Theory

Lectured by Stuart Martin, Lent Term 2013

Please send any comments and corrections to cn319@cam.ac.uk.

Contents

0 Preliminaries 2

1 Semisimple algebras 4

2 Irreducible modules for the symmetric group 8

3 Standard basis of Specht modules 12

4 Character formula 14

5 The hook length formula 21

6 Multilinear algebra and algebraic geometry 24

Interlude: some reminders about affine varieties 29

7 Schur–Weyl duality 31

8 Tensor decomposition 34

9 Polynomial and rational representations of GL(V)37

Interlude: some reminders about characters of GLn(C)40

10 Weyl character formula 41

11 Introduction to invariant theory and first examples 44

12 First fundamental theorem of invariant theory 50

Example sheet 1 53

Example sheet 2 57

Last updated Thursday 21 March 2013