Representation theory Lecture Notes: Chapter 5

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 20 November 2013

Symmetric functions

Let P = { (γ1,,γn) |γi0 } , P+ = { (γ1γn) |γi0 } ,and P++ = { (γ1>>γn) |γi0 } . The map P++ 1-1 P+ λ+δ λ whereδ= (n-1,n-2,,2,1,0), is a bijection. There is an Sn action on P given by wγ= ( γw(1),, γw(n) ) ,forwSn, and the elements of P+ are representatives of the Sn orbits on P.

The ring of polynomials in n variables [x1,,xn] has basis { xγ=x1γ1 xnγn |γP } . and there is an action of Sn on [x1,,xn] given by wxγ=xwγ, for allwSn, γP. i.e. Sn acts by permuting the variables. The ring of symmetric polynomials is [x1,,xn]Sn= { f[x1,,xn] |wf=ffor allw Sn } . The monomial symmetric functions mλ=μSnλ xμ,λP+, are a basis of [x1,,xn]Sn.

The vector space of alternating polynomials is A= { f[x1,,xn] |wf=ε(w)f for allwSn } , where ε(w) is the sign of the permutation w. If γP define aγ=wSn ε(w)wxγ=det (xiγj). Then

(a) aγ is an alternating polynomial,
(b) aγ=0 is γi=γj for some pair (i,j),
(c) {aμ|μP++} is a basis of A.

Let f be an alternating polynomial and let γ=(γ1,,γn)n be such that γi=γj. Then f evaluated at γ is 0 since f|γi=γj= ((i,j)f) |γi=γj=- f|γi=γj, where (i,j) is the transposition in Sn which switches i and j. Since f|γi=γj=0 it follows that xi-xj divides the polynomial f in [x1,,xn]. Since the polynomials xi-xj are coprime in the polynomial ring [x1,,xn] it follows that the product i<j(xi-xj) divides f.

In particular, i<j(xi-xj) divides the alternating polynomial aδ. Since aδ and i<j(xi-xj) are both homogeneous polynomials of the same degree the quotient must be 1. Thus we have proved the following.

(a) aδ=i<j(xi-xj).
(b) If f is an alternating polynomial the f is divisible by aδ.
(c) There is a vector space isomorphism A [x1,,xn]Sn f faδ

The quotient f/aδ is a symmetric polynomial since w(f/aδ)= (wf)/(waδ)= (ε(w)f)/(ε(w)aδ)= f/aδ, for all wSn.

Identify λ=(λ1λ2λn)P+ with the partition which has λi boxes in row i. A column strict tableau of shape λ is a filling of the boxes of λ with elements of the set {1,2,,n} such that

(a) The rows are weakly increasing left to right,
(b) The columns are strictly increasing top to bottom.
If T is a column strict tableau then xT= x1(# of 1s inT) x2(# of 2s inT) xn(# ofns inT) . The Schur function is the generating function sλ=TxT, where the sum is over all column strict tableaux of shape λ. In Theorem ??? we will show that, under the isomorphism in (???), A [x1,,xn]Sn aλ+δ sλ, i.e. that the image of basis {aλ+δ|λP+} of A is the basis {sλ|λP+} of [x1,,xn]Sn.

Define er(x), hr(x) and pr(x) by er(x1,,xn) = i1<<ir xi1xir, hr(x1,,xn) = i1<<ir xi1xir, pr(x1,,xn) = ixir. Then hr(x)=s(r)(x), and er(x)=s(1r)(x).

(a) i=1n j=1m 11-xiyj = λ sλ(x) sλ(y) = μ pμ(x)pμ(y) zμ = νhν(x) mν(y)
(b) i=1n j=1m (1+xiyj) = λ sλ(x) sλ(y) = νeν(x) mν(y)

Proof.

(a) The first equality is proved by Schensted insertion. i=1n j=1m 11-xiyj = i,jexp (log(11-xiyj)) = exp(i,j-ln(1-xiyj)) = exp ( i,j r>0 xiryjrr ) ,since- ln(1-x)= r>0 xrr, = exp ( r>0 pr(x)pr(y) r ) ,sincei,j xiryjr= pr(x)pr(y), = r>0exp (pr(x)pr(y)r) = r>0 ( k0 pr(x)k pr(y)k rkk! ) and expanding the product gives i=1n j=1m 11-xiyj= μ pμ(x)pμ(y)zμ. In order to obtain the third equality in (a) write i=1n j=1m 11-xiyj= j=1m ( r0 hr(x) yjr ) Then, expanding the product gives i=1n j=1m 11-xiyj= νhν(x) mν(y).

Define qr(x1,,xn;t,q) by the generating function 1+(q-t)r>0 qrzr=i=1n 1-txiz 1-qxiz . Then

(a) qr= i1<<ikir (-t)#(ij<ij+1) q#(ijij+1) xi1xir.
(b) qr= i1ir (q-t)#(ij>ij+1) q#(ij=ij+1) xi1xir.
(c) qr(x;0,1)= hr(x).
(d) qr(x;1,0)= (-1)r-1er(x).
(e) qr(x;1,1)= pr(x).
(f) qr(x)sμ(x)= λsλ(x) wt(λ/μ)
where the sum is over all λ such that |λ/μ|=r and wt(λ/μ)= Qwt(Q), where the sum is over all standard tableaux Q of shape λ/μ and wt(Q)= i=1r-1 di(Q), where di(Q)= { q, ifi+1is northeast ofi, 0, ifi+1is northwest ofi andi+2is southwest ofi+1, -t, ifi+1is southwest ofi.

(a) hr(x)sλ(x)= λsλ(x),
where the sum is over all λ such that λ/μ is a horizontal strip of length r.
(b) er(x)sλ(x)= λsλ(x),
where the sum is over all λ such that λ/μ is a vertical strip of length r.
(c) pr(x)sλ(x)= λ(-1)# rows(λ/μ)-1sλ(x),
where the sum is over all λ such that λ/μ is a border strip of length r.

The unitary group

The unitary group is the real Lie group Un= { gMn() |ggt =1 } . Note that Un is not a complex Lie group in a natural way. The maximal torus in Un is T = {diagonal matrices inUn} = { ( z1 z2 zn ) ||zi|=1 } = { ( eiθ1 eiθ2 eiθn ) |0θi<2π } The Weyl group is W=N(T)/T=Sn= the symmetric group, since, under conjugation, the eigenvalues of diagonal matrices can only get permuted. Let Xεi: T * ( z1 z2 zn ) zi . Then the irreducible representations of T are Xλ= Xλ1ε1++λnεn =(Xε1)λ1 (Xεn)λn, λi, and so the weight lattice is L= { λ1ε1+ λ2ε2++ λnεn| λ1,,λn } n. Then 𝔲n=Lie(Un)= { xMn()| x+xt=0 } which has basis {Eij-Eji,iEij+iEji|i<j} {iEii|1in} as a vector space over . Then 𝔤=𝔲n i𝔲n=𝔤𝔩n ()=Mn() which has basis {Eij|1i,jn} as a vector space over . Then 𝔱 = Lie(T)=-span {iEii|1in} ,and 𝔥 = 𝔱i𝔱=𝔱= { ( h1 h2 hn ) |hi } . So 𝔤=𝔥 (αR𝔤α) whereR= {εi-εj|1i,jn,ij} and𝔤εi-εj =Eij, since [h,Eij]=hEij -Eijh= (xi-xj)Eij= (εi-εj)(h) Eij. Then Xεk-ε(t) =zkz-1= eiθkeiθ =ei(θk-θ). So L+ = { λ1ε1+ λ2ε2++ λnεn |λ1 λ2λn } = (ε1++εn)+ { λ1ε1++ λn-1 εn-1| λ1λn-1 0 } L++ = { μ1ε1+ μ2ε2++ μnεn |μ1> μ2>>μn } = (ε1++εn)+ { μ1ε1+ μ2ε2++ μn-1 εn-1| μ1>μ2>> μn-1>0 } The bijection L+ L++ k+λ k+(n-12)+λ+δ restricted to those elements of L+ with k=0 is the same as the bijection 𝒫 𝒫s λ λ+δ = ( λ1+n-1, λ2+n-2,, λn-1+1 ) where 𝒫 = {partitions with lengthn-1} = { (λ1,,λn-1) |λ1λn-1 0 } and 𝒫s = {strict partitions with lengthn-1} = { (μ1,,μn-1) |μ1>>μn-1 >0 } The chamber in 𝔱=n can be chosen as C= { (x1,,xn) n|x1 x2xn } the hyperplanes are Hεi-εj { (x1,,xn) n| xi=xj } ,1i<jn. The positive roots are R+= {εi-εj|1i<jn} andR=R+ (-R+). Then R(T)L=-span {eλ|λL} =-span {eλ1ε1++λnεn|λi}. We write eλ1ε1++λnεn= (eε1)λ1 (eεn)λn= z1λ1znλn. Then (P)W=symmetric polynomials in the zi is the set of polynomials such that p(z1,,zn)= p(zw(1),,zw(n)), for all wSn. ρ = 12α>0α =121i<jn (εi-εj) = (n-12)ε1+ (n-32)ε2+ +(-(n-1)2) εn = -(n-12) (ε1++εn) (n-1)ε1+ (n-2)ε2++ εn-1+0εn and aρ = (z1zn)-(n-1)/2 wSnε(w)w ( z1n-1 z2n-2 zn-1 ) = det ( z1n-1 z1n-2 z1 1 z2n-1 z2n-2 z2 1 znn-1 znn-2 zn 1 ) (the Vandermonde determinant!) = (z1zn)-(n-1)/2 i<j (zi-zj) = z1-(n-1)/2 z2-(n-3)/2 zn(n-1)/2 i<j (zizj-1-1) = X-ρ(t) α>0 (Xα(t)-1). Then χλ = aλ+ρaρ= wSn ε(w)w ( z1λ1+n-1 z2λ2+n-2 zn-1λn-1+1 znλn ) wSn ε(w)w ( z1n-1 z2n-2 zn-11 zn0 ) = det(ziλj+n-j) det(zin-j) (the Schur function!).

(Kostka 18??) If (λ1λ2λn)L+ then sλ(z1,,zn) =(z1zn)λ TzT, where the sum is over all column strict tableaux of shape λ.

The Littelmann theorem is the generalization of this theorem to all compact Lie groups.

Think about the derivation of the Weyl dimension formula in the context of Schur functions. sλ(ehδ) = sλ ( eh(n-1), eh(n-2),, eh,1 ) =sλ ( q(n-1), q(n-2),,q,1 ) = det((qn-i)λj+n-j) det((qn-i)n-j) = det((qλj+n-j)n-i) det((qn-i)n-j) = aδ(qλ1+n-1,,qλn+n-n) aδ(qn-1,,qn-n) = i<j qλj+n-j- qλi+n-i qn-j-qn-i =q(i-1)λi ( i<j 1-q(λj+n-j)-(λi+n-i) q(n-j)-(n-i) ) = qn(λ) i1 k=1λi+n-i (1-qk) (bλ1-qh(b)) i<j(1-qi-j) = qn(λ) bλ 1-qn+c(b) 1-qh(b) For example, if λ=5ε1+4ε2+4ε3+3ε4+2ε5, 2 3 5 7 9 8 6 4 1 λ1+n-1=9 1 3 5 7 6 4 2 λ2+n-2=7 2 4 6 5 3 1 λ3+n-3=6 2 4 3 1 λ4+n-4=4 2 1 λ5+n-5=2 Then 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 1 2 So dim(Uλ)= 5·6·7·8·9·4·5·6·7·3·4·5·6·2·3·4·1·2 9·8·6·4·1·7·6·4·2·5·4·3·1·4·3·1·2·1 = 5·7·5·66= 5·7·5=175.

The general linear group GLn()

GLn()= {invertible matrices inMn()}. Note that GLn() is dense in Mn. Writing g=(gk)1k,n for matrices in GLn() let fk be the coordinate functions on GLn(), fk: GLn() g gk and Xk= (fk|g=1), (3.1) the corresponding basis of tangent vectors (derivatives on G) in the tangent space T1(G). Let Eij denote the matrix with a 1 in the (i,j)th entry and 0 everywhere else. Then one parameter subgroups corresponding to the tangent vectors in ??? are γij: GLn() t (Id+tEij) = ( 1 t 1 ) and γii: GLn() t ( 1 1 et 1 1 ) since ( 1 s 1 ) ( 1 t 1 ) = ( 1 s+t 1 ) and ( 1 1 es 1 1 ) ( 1 1 et 1 1 ) = ( 1 1 es+t 1 1 ) . In general, the vector field given by δX=i,j=1n xijfij ,xij, corresponds to the one parameter subgroup γX: GLn() t etX=k0tkXkk!, where X=(xij) is the corresponding matrix in Mn(). In particular, etEij= ( 1 1 ) +t ( 1 1 1 ) +t22! ( 1 1 1 ) 2 = ( 1 t 1 ) and etEii= ( 1 1 et 1 1 ) as they should. If X=(xij) is a matrix in Mn() and dX is the corresponding tangent vector dXT1(G) then dXfk= i,j=1n xijfij fk|g=1 =xk. (3.2) If X is the corresponding left invariant vector field (Xfk)(g) = (Lg-1Xfk) (1)= (XLg-1fk) (1) = dX(Lg-1fk)= dX ( j=1n gkjfj ) =j=1n gkjxj= j=1n xjfkj (g) (3.3) since (Lg-1fk) (h)=fk(gh)= (gh)k= j=1ngkj hj=j=1n gkjfj(h) (3.4) for all hG. We can express formula ??? more concisely as Xf=fX,wheref =(fk)andX =(xij)Mn(). It follows that [X,Y]f= (XY-YX)f= f(XY-YX) and so we may identify 𝔤=Lie(GLn()) with the Lie algebra Mn()with bracket [X,Y]=XY-YX, forX,YMn(). If gG and XMn() then (AdgX)f= RgXRg-1 f=fgXg-1, wheref=(fk). Thus, for G=GLn() with 𝔤=Lie(G) identified with Mn(), Adg: Mn() Mn() X gXg-1, for allgGLn(),and adX: Mn() Mn() Y [X,Y] = XY-YX, for allX,YMn(),

The special unitary group

The special unitary group is the real Lie group SUn= { gMn()| det(g)=1,ggt =1 } . A maximal torus in SUn is T={diagonal matrices inSUn} andW=N(T)/ T=Sn=the symmetric group, is the Weyl group. Then 𝔰𝔲n = { xMn()| xxt=0, tr(x)=0 } and 𝔰𝔩n() = 𝔰𝔲ni𝔰𝔲n= { xMn()| tr(x)=0 } . Then 𝔤=𝔥(αR𝔤α) where R = { εi-εj| 1i,jn,ij } , 𝔤εi-εj = Eij, and 𝔥=𝔱i𝔱= 𝔱= { ( h1 h2 hn ) |hi, i=1nhi=0 } n, The lattice of representations of T is P= { λ=λ1ε1+ +λnεn| λ1++λn=0 ,λi } n-1.

The special orthogonal groups

Let SO(2r+1,) = { gM2r+1 ()| ggt=1,det (g)=1 } . A maximal torus of SO(2r+1,) is T= { ( ( cosθ1 sinθ1 -sinθ1 cosθ1 ) ( cosθr sinθr -sinθr cosθr ) 1 ) |0θi<2π } . The Weyl group W=N(T)/T=WBr= { r×rmatrices with exactly one nonzero entry in each row and each column, and nonzero entries±1 } and is generated by permuting the blocks of a matrix in T and the transformations ( cosθj sinθj -sinθj cosθj ) ( cosθj -sinθj sinθj cosθj ) which come from conjugation by the block (0110). The Lie algebra 𝔲=𝔰𝔬(2r+1,) = { xM2r+1() |x+xt=0 andtr(x)=0 } = { xM2r+1() |x+xt=0 } has basis {Eij-Eji|1i<jr} and 𝔥=-span { E2i-1,2i- E2i,2i-1 |1ir } . Note that e(0θ-θ0)= (cosθsinθ-sinθcosθ). Then G = SO2r+1()= { AMn()| AAt=1,det(A)=1 } , 𝔤 = 𝔰𝔬2r+1()= { xMn()| AAt=1,det(A)=1 } , = 𝔲i𝔲=𝔲= -span {Eij-Eji|1i<jn}, 𝔥 = 𝔱i𝔱=-span { E2i-1,2i- E2i,2i-1 |1ir } . Let Xεi-εj= ( 0(1i-i1) (-1i-i-1)0 ) X-εi+εj= ( 0(1-i-i1) (-1-i-i-1)0 ) Xεi+εj= ( 0(1-i-i-1) (-1ii1)0 ) X-εi-εj= ( 0(1ii-1) (-1-i-i1)0 ) Xεi= ( 0(1-i) (-1i)0 ) X-εi= ( 0(1i) (-1-i)0 ) where, in the first four matrices, the ith and jth blocks are shown and, in the last two, the ith block and the 2r+1 row are shown. This is yucky, typeset this into the matrices as ith block, jth block Then 𝔤=𝔥 (αR𝔤α) where𝔤α=Xα andR= { ±(εi-εj), ±(εi+εj), ±εi } .

Let SO(2r,)= { gM2r+1 ()| ggt=1,det (g)=1 } . A maximal torus of SO(2r+1,) is T= { ( ( cosθ1 sinθ1 -sinθ1 cosθ1 ) ( cosθr sinθr -sinθr cosθr ) ) |0θi<2π } . The Weyl group W=N(T)/T=WDr= { r×rmatrices with exactly one nonzero entry in each row and each column, nonzero entries±1, and an even number of-1s } and is generated by permuting the blocks of a matrix in T and conjugation by the matrices ( 0110 0110 ) . The rest of the story is exactly as in the SO(2r+1,) case except that 𝔤=𝔥 (αR𝔤α) where𝔤α=Xα andR= {±(εi-εj),±(εi+εj)}.

The symplectic group Sp2r

Let U=Sp2r = {gUn()|gtJg=J} = {gUn()|Jg=aJ} = { gUn()|g= (xy-yx) } whereJ= ( 1 0 1 -1 0 -1 ) . A maximal torus in Sp2r is T= { ( z1 0 zr z1 0 zr ) |zi* ,|zi|=1 } and the Weyl group is W=N(T)/T=WCr= { r×rmatrices with exactly one nonzero entry in each row and each column, and nonzero entries±1 } and is generated by permuting the zi and the transformations ( z1 0 zr z1 0 zr ) ( z1 z2 zr 0 0 z1 z2 zr ) which comes from conjugation by the matrix ( 0 1 1 1 0 0 1 0 0 0 1 1 ) . The Lie algebra 𝔲 = { x𝔲n| xtJ+Jx=0 } = { xMn()| xt+x=0, xtJ+Jx=0 } = { xMn()| xt+x=0, -xJ+Jx=0 } = { xMn()| x=(ab-ba) ,at=-a, b=bt } has basis { Eij+Ei+n,J+n, -Eji-Ej+n,i+n |1i<jn } { Eij-Eji | 1in, n+1j2n } . Then 𝔤 = 𝔲i𝔲=𝔲 =𝔰𝔭2r() = { xMn() |xtJ+ Jx=0 } = { xMn() |x= (abc-at), b=bt,c=ct } and 𝔤 has basis Xεi-εj= Eij- Ei+n,j+n, Xεj-εi= Eji- Ej+n,i+n, 1i<jn, Xεi+εj= Ei,j+n+ Ej,i+n, X-εi-εj= Ei+n,j+ Ej+n,i, 1i<jn, X2εk= Ek,k+m, X-2εk= Ek+n,k, 1kn. Then 𝔱 = { ( ih1 ihr 0 0 -ih1 -ihr ) |hi } 𝔥 = 𝔱i𝔱=𝔱 =-span { Eii-Ei+n,i+n |1ir } . Then 𝔤=𝔥 (αR𝔤α) where𝔤α=Xα andR= { ±(εi-εj), ±(εi+εj), ±2εi } .

Examples of Lie groups

GLn()
SLn()
On()
SOn()
SP2n
Un= { ( 1 * 1 ) GLn() } .
Bn= { ( * * * ) GLn() } .
Tn= { ( * * ) GLn() } .
Any closed subgroup of GLn().
Any finite group.
S1={e2πix|x}=/.
S1××S1= torus.
Un()= {gMn()|ggt=1}.

Complex groups

GLn()= {gMn()|det(g)0}.
SLn()= {gMn()|det(g)=1}.
SOn()= {gMn()|ggt=1,det(g)=1}.
SP2n()= {gM2n()|gJgt=J}.
On()= {gMn()|ggt=1}.

Compact subgroups

SOn()= {gMn()|ggt=1,det(g)=1}.
Un= {gMn()|ggt=1}.
SUn= {gMn()|ggt=1,det(g)=1}.
SP2n= {gM2n()|ggt=1,gJgt=J}.

Real noncompact groups

GLn()
SLn()
SLn()
SOn()
SO(m,n)
SU(m,n)
Sp(m,n)
SO*(2n)

Nilpotent groups

Heisenberg groups
Circle groups
Tori

Covers

Spin groups
Pin groups
Metaplectic groups

Real reductive groups

GLn()
SLn()
GLn() (as a subgroup of GL(2n,))
SLn()
O(p,q)
SO(p,q)
U(p,q)
SU(p,q)
Sp(2n,)

Linear algebraic groups

GLn(K)
SLn(K)
On(K)
SOn(K)
Sp2n(K)
𝔾a
𝔾m
Any finite subgroup of GLn(K)
Tn= { ( x1 xn ) GLn(K) } .
Bn the group of upper triangular matrices in GLn(K)
Un the group of upper unitriangular matrices in GLn(K)

K could be an algebraically closed field of characteristic p,
a p-adic field,
a locally compact topological field,
a local field,
a global field.

Finite Chevalley groups

GLn(𝔽q)
SLn(𝔽q)
On(𝔽q)
SOn(𝔽q)
Sp2n(𝔽q)
where 𝔽q is a field with q=pk elements.

Notes and references

This is a typed exert of Representation theory Lecture notes: Chapter 5 by Arun Ram. Research supported in part by National Science Foundation grant DMS-9622985.

page history