Notes on Schubert Polynomials
Chapter 3

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 27 June 2013

Multi-Schur functions

For the time being we shall work in an arbitrary λ-ring R, but we shall use the notation of symmetric functions [Mac1979] rather than that of λ-rings. Thus for XR we shall write er(X) in place of λr(X) for the rth exterior power, and hr(X) in place of σr(X) (=(-1)rλr(-X)) for the rth symmetric power of X. We have e0(X)=h0(X)=1; e1(X)=h1(X)=X; and er(X)=hr(X)=0 if r<0.

Recall that if λ,μ are partitions and XR, the skew Schur function sλ/μ(X) is defined by the formula

sλ/μ(X)=det (hλi-μj-i+j(X)) 1i,jn

where nmax((λ),(μ)). It is zero unless λμ.

We generalize this definition as follows: let X1,,XnR and let λ,μ be partitions of length n; then the multi-Schur function sλ/μ(X1,,Xn) is defined by

(3.1) sλ/μ (X1,,Xn) =det (hλi-μj-i+j(X)) 1i,jn .

We also define

(3.1') sα(X1,,Xn) =det (hαi-i+j(X)) 1i,jn

for any sequence α=(α1,,αn) of integers of length n.

Remark. In the definition (3.1) the argument Xi is constant in each row of the determinant. We might therefore also define

sˆλ/μ (X1,,Xn)= det (hλi-μj-i+j(X)) 1i,jn

with arguments constant in each column. However, we get nothing essentially new: if we define partitions λˆ,μˆ by

λˆi=N- λn+1-i, μˆi=N- μn+1-i (1in)

where Nmax(λ1,μ1) (so that λˆ and μˆ are the respective complements of λ and μ in the rectangle (Nn)), then we have

sˆλ/μ (X1,,Xn)= sμˆ/λˆ (Xn,,X1)

as one sees by replacing (i,j) by (n+1-j,n+1-i) in the determinant (3.1).

(3.2) We have

sλ/μ (X1,,Xn)=0

unless λμ.


If λμ then λr<μr for some rn, and hence

λi-μj-i+j λr-μr<0

whenever ir and jr. It follows that the matrix (hλi-μj-i+j(X)) has an (n-r+1)×r block of zeros in the south-west corner, and hence its determinant vanishes.

(3.3) If λμ and (λ)=r<n then

sλ/μ (X1,,Xn)= sλ/μ (X1,,Xr)


We have λs=μs=0 for r+1sn. Hence for each s>r the sth row of the matrix (hλi-μj-i+j(X)) has zeros in the first s-1 places, and 1 in the sth place.

An element xR is said to have finite rank if en(X)=0 for all sufficiently large n. We then define the rank rk(X) of X to be the largest r such that er(X)0. If X,Y both have finite rank, the formula

er(X+Y)= p+q=r ep(X) eq(Y)

shows that X+Y has finite rank, and that

rk(X+Y)rk (X)+rk(Y).

(3.4) Let X1,,Xn, Y1,,YnR with rk(Yj) j-1 (1jn) (so that Y1=0). Then for all αn,

sα (X1,,Xn)= det ( hαi-i+j (Xi-Yj) ) .


We have

hαi-i+j (Xi-Yj)= k=1j hαi-i+k (Xi) hj-k (-Yj),

since hr(-Yj)= (-1)rer(Yj) =0 if rj. Hence the matrix

( hαi-i+j (Xi-Yj) ) 1i,jn

is the product of the matrix

( hαi-i+j (Xi) ) 1i,jn

and the matrix

( hi-j (-Yj) ) 1i,jn

which is unitriangular. Now take determinants.

So far the Xi have been arbitrary elements of the λ-ring R. But it seems that sα(X1,,Xn) is mainly of interest when X1,,Xn is an increasing sequence in R, in the sense that rk(Xi+1-Xi)< for 1in-1.

(3.5) Let xi,yi (i1) be elements of R, each of rank 1, and let

Xi=x1++xi, Yi=y1++yi

for each i0. Then for all αn we have

sα ( X1-Yα1,, Xn-Yαn ) =i=1n j=1αi (xi-yj).

In particular, if λ is a partition of length n,

sλ ( X1-Yλ1,, Xn-Yλn ) =(i,j)λ (xi-yj).


From (3.4) we have

(*) sα ( X1-Yα1,, Xn-Yαn ) =det ( hαi-i+j (Xi-Yαi-Xj-1) ) 1i,jn .

If j>i, then

hαi-i+j (Xi-Yαi-Xj-1) =±eαi-i+j (Yαi+Xj-1-Xi)

which is 0 because

rk(Yαi+Xj-1-Xi) αi+(j-1)-i< αi-i+j.

Hence the determinant at (*) is triangular, with diagonal elements

hαi ( Xi-Yαi- Xi-1 ) = hαi(xi-Yαi) = r0hr (-Yαi) hαi-r (xi) = r0(-1)r er(Yαi) xiαi-r = j=1αi (xi-yj).

The formula (3.5) now follows.

In particular, when all the yi are zero we have

(3.5') sα(X1,,Xn) =i=1n xiαi=xα

for all αn. Also, when all the xi are zero (and α is a partition λ) we have

sλ ( -Yλ1,, -Yλn ) = (-1)|λ| (i,j)λ yj = (-1)|λ| yλ.

If we replace the y's by x's, and λ by λ, this becomes

(3.5'') xλ= (-1)|λ| sλ ( -Xλ1,, -Xλn ) .

(3.6) Let λ=(λ1,λ2,) be a partition of length n, and X1,,Xn elements of a λ-ring R. Suppose that i<j are such that λi=λi+1==λj and

rk(Xj-Xk) j-kforik j.


sλ/μ (X1,,Xn)= sλ/μ ( X1,,Xi-1, Xj,,Xj, Xj+1,,Xn ) ,

that is to say we can replace each Xk (ikj) by Xj without changing the value of the multi-Schur function.


Let Y=Xj-Xi, so that rk(Y)j-1. For all m0 we have

hm(Xi) = hm(Xj-Y) = k=0j-i (-1)kek (Y)hm-k (Xj).

It follows that if we replace the ith row of the determinant sλ/μ ( X1,,Xi-1, Xj,,Xj, Xj+1,,Xn ) by

k=0j-i (-1)kek (Y)rowi+k

we shall obtain

sλ/μ ( X1,,Xi-1, Xi,Xj,,Xj Xj+1,,Xn )

with j-i arguments equal to Xj. The proof is now completed by induction on j-i.


Let (Xn)n be a sequence in the λ-ring R such that

rk(Xn-Xn-1) 1

for all n.

(3.7) Let I be any interval in . Then the inverse of the matrix

H= (hi-j(-Xi)) i,jI

is H-1= (hi-j(Xj+1)) i,jI .


Let K denote the matrix (hi-j(Xj+1)). The (i,k) element of HK is then

jhi-j (-Xi) hj-k (Xj+1)= hi-k (-Xi+Xk+1).

If i<k this is zero; if i=k it is equal to 1; and if i>k it is equal to

(-1)i-k ei-k (Xi-Xk+1)

which is zero because rk(Xi-Xk+1) i-(k+1)<i-k.

(3.8) (Duality Theorem, 1st version) Let λμ be partitions of length n, such that (λ)m. Then

sλ/μ ( -Xλ1-1,, -Xλn-n ) =(-1)|λ-μ| sλ/μ ( X1-λ1,, Xm-λm ) .



ξi=λi-i, ηi=μi-i (1in), ξj= λj-j, ηj= μj-j (1jn),

Then the integers ξi (1in) and -ξj-1 (1jm) fill up the interval [-m,n-1], and so do the ηi and the -ηj-1.

The (ξ1,,ξn;η1,,ηn) minor of the matrix H is

det ( hηi-ηj (-Xξi) ) =sλ/μ ( -Xξ1,, -Xξn ) .

The complementary cofactor of (H-1)= (hj-i(Xi+1)) -mi,jn-1 has row indices -ξi-1 (1im) and column indices -ηj- (1jm). Hence it is

(-1)|λ-μ| sλ/μ ( -Xξ1,, -Xξm ) .

Since each minor of H is equal to the complementary cofactor of (H-1) (because detH=1) the result follows.

Remark. Observe that

rk ( Xλi-i- Xλi+1-i-1 ) (λi-i)- (λi+1-i-1)= λi-λi+1+1.

Hence (3.8) gives a duality theorem for the multi-Schur function sλ/μ (Y1,,Yn) provided that rk(Yi+1-Yi) λi-λi+1+1 for 1in-1.

At first sight the formula (3.8) is disconcerting, because the arguments -Xλi-i on the left are not in general the negatives of the arguments Xi-λi on the right. However, we can use (3.6) to rewrite (3.8), as follows. As in Chapter I, let us write the partition λ in the form

λ= ( p1m1, p2m2,, pkmk )

where p1>p2>>pk>0 and each mi1. Then in

sλ/μ ( -Xλ1-1,, -Xλn-n )

the first m1 arguments are

-Xp1-1,, -Xp1-m1

which by (3.6) may all be replaced by -Xc1, where c1=p1-m1. The next m2 arguments are

-Xp2-m1-1,, -Xp2-m1-m2

which by (3.6) may all be replaced by -Xc2, where c2=p2-m1-m2. In general, for each i=1,2,,k the ith group of mi arguments may all be replaced by -Xci, where ci=pi-(m1++mi). Now if

λ= ( q1n1, q2n2,, qknk )

is the conjugate partition, we have m1++mi=qk+1-i, and ci=π-qk+1-i is the content of the square si=(qk+1-i,pi) in the diagram of λ. The squares s1,,sk are the "salients" of the border of λ, read in sequence from north-east to south-west. Hence the duality theorem (3.8) now takes the form

(3.8') (Duality Theorem, 2nd version). With the above notation, we have

sλ/μ ( (-Xc1)m1,, (-Xck)mk ) =(-1)|λ-μ| sλ/μ ( (Xck)n1,, (Xc1)nk ) .

Finally, if we set Zi=-Xci (1ik) we have

(3.8'') sλ/μ ( Z1m1,, Zkmk ) =(-1)|λ-μ| sλ/μ ( (-Zk)n1,, (-Z1)nk )


rk(Zi+1-Zi) mi+1+ nk+1-i (1ik-1).

Let now x1,x2, be independent indeterminates over . We may regard [x1,x2,] as a λ-ring by requiring that each xi has rank 1. Let Xi=x1++xi for each i1. Then we have

(3.9) ihr(Xi) = hr-1(Xi+1), ier(Xi) = er-1(Xi-1), πihr(Xi) = hr(Xi+1).


Consider the generating functions: ihr(Xi) is the coefficient of tr in

i ( r0 hr(Xi) tr ) = ij=1i (1-xjt)-1 = j=1i-1 (1-xjt)-1 ·i (11-xit),


i(11-xit) = 1xi-xi+1 ( 11-xit- 11-xi+1t ) = t (1-xit) (1-xi+1t)

so that

i ( rhr (Xi)tr ) =tj=1i+1 (1-xjt)-1= shs (Xi+1) ts+1

in which the coefficient of tr is hr-1(Xi+1).

The other two relations are proved similarly.

(3.10) Let αn and let r1,,rn0. If i is such that rirj for all ji then

risα (Xr1,,Xrn) = sα-εi ( Xr1,, Xr+1,, Xrn ) , risα ( -Xr1,, -Xrn ) = -sα-εi ( -Xr1,, -Xri-1,, -Xrn ) , πrisα (Xr1,,Xrn) = sα ( Xr1,, Xr+1,, Xrn ) ,

where εi has ith coordinate equal to 1, and all other coordinates zero.


By definition, we have sα=det(hαi-i+j(Xri)) and ri acts only on the ith row of the determinant, the entries in the other rows being symmetrical in xri and xri+1 (because of the condition rjri if ji). Hence the first of the relations (3.10) follows from the first of the relations (3.9), and the other two are proved similarly.

Remark. We can use the relations (3.10) to give another proof of duality (3.8) in the form

(3.8''') Let λ be a partition such that λ1m and λ1n. Then

(*) sλ ( Xm+1-λ1,, Xm+n-λn ) =(-1)|λ| sλ ( -Xm+λ1-1 ,, -Xλm ) .

Let (i,j) be a corner square of the diagram of λ, so that j=λi and i=λj. Let μ be the partition obtained from λ by removing the square (i,j). By operating on either side of (*) with m+i-j we obtain the same relation with μ replacing λ. Hence it is enough to show that (*) is true when λ=(mn), but in that case both sides are equal to (X1Xm)n, by (3.5'), (3.5'') and (3.6).

(3.11) Let w0 be the longest element of Sn. Then for any αn we have

w0sα ( X1+Z1,, Xn+Zn ) = sα-δ ( Xn+Z1,, Xn+Zn ) , πw0sα ( X1+Z1,, Xn+Zn ) = sα ( Xn+Z1,, Xn+Zn ) ,

where Xi=x1++xi (1in) and the Zi are independent of x1,,xn.


The sequence

( n-1, n-2, n-1,,2,3,, n-1,1,2,3,,n -1 )

is a reduced word for w0, so that

πw0=πn-1 (πn-2πn-1) (π2π3πn-1) (π1π2πn-1)

and likewise for w0. By (3.10), π1π2πn-1 applied to sα(X1+Z1,,Xn+Zn) will produce

sα ( X2+Z1, X3+Z2, , Xn+Zn-1, Xn+Zn ) .

We have next to operate on this with π2π3πn-1, which will produce

sα ( X3+Z1, X4+Z2,, Xn+Zn-2, Xn+Zn-1, Xn+Zn ) .

By repeating this process we shall obtain the formula for πw0sα. That for w0sα is proved similarly.

Remark. Let αn and Z1==Zn=0 in (3.11). Then by (3.5') we have

w0(xα) = w0sα (X1,,Xn) = sα-δ(Xn), πw0(xα) = πw0sα (X1,,Xn) = sα(Xn).

Thus we have independent proofs of (2.11) and (2.16') and hence (by linearity) of (2.10) and (2.16).

Sergeev's formula

Let x1,,xm,y1,,yn be independent variables and let

Xi=x1++xi, Yi=y1++yi

for all i1, with the understanding that xj=0 if j>m and yj=0 if j>n.

(3.12) (Sergeev) For all partitions λ we have

sλ(Xm-Yn)= wSm×Snw ( fλ(x,y)/ D(x)D(y) )


fλ(x,y)= (i,j)λ (xi-yj), D(x)=1i<jm (1-xi-1xj), D(y)=1i<jn (1-yi-1yj).


Let w0(m) (resp. w0(n)) be the longest element of Sm (resp. Sn) and let πx (resp. πy) denote πw0(m) acting on the x's (resp. πw0(n) acting on the y's). From (3.5) we have, if r=(λ),

(1) fλ(x,y)=sλ ( X1-Yλ1,, Xr-Yλr )

and in view of (2.16) Sergeev's formula may be restated in the form

(3.12') sλ(Xm-Yn) =πyπxfλ (x,y).

From (3.11) and (1) above we have

(2) πxfλ(x,y) =sλ ( Xm-Yλ1,, Xm-Yλr ) .

If λ=(p1m1,,pkmk), (2) can be rewritten in the form

(3) πxfλ(x,y)= sλ(Z1m1,,Zkmk)

where Zi=Xm-Ypi. Since

rk(Zi+1-Zi)= rk(Ypi-Ypi+1) =pi-pi+1,

the duality theorem (3.8'') applies, and gives

(4) sλ (Z1m1,,Zkmk) = (-1)|λ| sλ ( (-Zk)n1,, (-Z1)nk ) = (-1)|λ| sλ ( (Ypk-Xm)n1,, (Yp1-Xm)nk ) = (-1)|λ| sλ ( Y1-Xm, Y2-Xm,, Ys-Xm )

where s=n1++nk=(λ). We can now apply (3.11) again and obtain from (3) and (4)

πyπxfλ = (-1)|λ| πysλ ( Y1-Xm,, Ys-Xm ) = (-1)|λ| sλ ( Yn-Xm ) = sλ(Xm-Yn).

Notes and References

This is a typed excerpt of the book Notes on Schubert Polynomials by I. G. Macdonald.

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