On Goulden-Jackson's determinantal expression for the immanant

0.12 MB·English

by  Matjaz Konvalinka

7 ON GOULDEN-JACKSON’S DETERMINANTAL EXPRESSION FOR
0
THE IMMANANT
0
2

n MATJAZˇ KONVALINKA
a
J
2 Abstract. In1992,GouldenandJacksonfoundabeautifuldeterminantalexpres-
1 sion for the immanantof a matrix. This paper provesthe same result combinatori-
ally.

]
O
C
.
h 1. Introduction
t
a
Take a matrix A = (a )m , its characteristic polynomial χ (t) = det(A−tI) and
m ij i,j=1 A
its eigenvalues ω ,...,ω . Vieta’s formulas tell us that the elementary symmetric
[ 1 m
functions
1
e (ω ,...,ω ) = ω ···ω
v i 1 m X j1 ji
6
j1<...<ji
6
are easily expressible in terms of a :
3 ij
1 e tm −e tm−1 +...+(−1)me = (t−ω )···(t−ω ) =
0 0 1 m 1 m
7
m
0 = det(tI −A) = (−1)itm−i detA ,
/ X X J
h
t i=0 J∈([m])
a i
m with A = (a ) , i.e.
J ij i,j∈J
:
v e (ω ,...,ω ) = detA .
i i 1 m X J
X J∈([m])
i
r
a The complete homogeneous symmetric functions
h (ω ,...,ω ) = ω ···ω
i 1 m X j1 ji
j1≤...≤ji
are also easy. We know that
1 1 1 1
h ti = = = =
X i (−1)ie ti tm (−1)ie ti−m tmdet(t−1I −A) det(I −tA)
i Pi i Pi i
and by MacMahon Master Theorem [MM, page 98]
(1) hi(ω1,...,ωm) = Xaλ′1λ1aλ′2λ2 ···aλ′iλi,
λ
Date: February 2, 2008.
1
2 MATJAZˇ KONVALINKA
where:
• λ = λ ···λ runs over all sequences of i letters from [m], and
1 i
• λ′ = λ′ ···λ′ is the non-decreasing rearrangement of λ.
1 i
Goulden and Jackson [GJ1] proved the following.
Theorem 1 Denote h (ω ,...,ω ) by ∆ , choose a partition λ = (λ ,...,λ ) of n,
i 1 m i 1 p
and write a = a a ···a for a permutation π ∈ S . Then
π 1π(1) 2π(2) nπ(n) n
(2) [a ]det(∆ ) = χλ(π),
π λi−i+j p×p
where χλ is the irreducible character of the symmetric group S corresponding to λ.
n
Goulden and Jackson’s result is stated in the (clearly equivalent) language of im-
manants. Theirs is one of the many papers in the early 1990’s that brought about fas-
cinating conjectures and results on immanants; see for example [GJ2], [Gr], [StaSte],
[Ste] for details and further references.
We will give two more proofs of this result. The first gives a recursion that specializes
to Murnaghan-Nakayama’s rule, and the second is a simple combinatorial proof of a
statement equivalent to (2).
2. A recursive proof of Theorem 1
Since ∆ is equal to h (ω ,...,ω ), det(∆ ) can be expressed as
λi−i+j i 1 m λi−i+j
s (ω ,...,ω )
λ 1 m
by the Jacobi-Trudi identity (see [Sta, Theorem 7.16.1]). Here s is the Schur sym-
λ
metric function corresponding to the partition λ.
Note first that all the terms a = a ···a a ··· of
1∗ 1∗ 2∗
sλ(ω1,...,ωm) = det(eλ′−i+j)
i
are balanced in the sense that each i appears as many times among a as among a
i∗ ∗i
(this also follows from (1) and the formula s = det(h )); here we are using the
λ λi−i+j
dual Jacobi-Trudi identity [Sta, Corollary 7.16.2].
Suppose we want to find the coefficient of a = a ···a a ···. Assume that C =
1∗ 1∗ 2∗
{1,...,k}, D = {k + 1,...,m} and that a does not contain a for i ∈ C,j ∈ D or
ij
i ∈ D,j ∈ C. The coefficient of a does not change if we put all a that do not appear
ij
in a to 0; the matrix A then has a block diagonal form A ⊕A , and if ξ ,...,ξ are
1 2 1 k
the eigenvalues of A and ζ ,...,ζ are the eigenvalues of A , the eigenvalues of A
1 k+1 m 2
are (ω ,...,ω ) = (ξ ,...,ξ ,ζ ,...,ζ ). By definition,
1 m 1 k k+1 m
s (ω ,...,ω ),
λ 1 m
is the sum of ωT = ωα1(T)ωα2(T)··· over all semistandard Young tableau (SSYT) T
1 2
of shape λ; here α (T) is the number of i’s in T. See [Sta, §7.10] for definitions and
i
ON GOULDEN-JACKSON’S DETERMINANTAL EXPRESSION FOR THE IMMANANT 3
details. In every such T, the numbers 1,...,k form a SSYT of some shape ν ⊆ λ,
and the numbers k +1,...,m form a SSYT of shape λ/ν. Therefore
(3) [a]s (ω ,...,ω ) = [a ]s (ξ ,...,ξ )[a ]s (ζ ,...,ζ ).
λ 1 m X 1 ν 1 k 2 λ/ν k+1 n
ν⊆λ
Since s is homogeneous of degree |λ| − |ν|, we can restrict the sum to partitions
λ/ν
ν ⊢ k.
The second term in the product can be calculated explicitly by using the reverse
Jacobi-Trudi identity (for example fora “small”) andthefirst termcanbecalculated
2
recursively.
Example 2 Let us calculate the coefficient of a a a a2 a a . By (3), we have to
11 12 21 22 34 43
find the coefficient of a a in s for ν = 32,311,221.
34 43 λ/ν
We have
s (ζ ,ζ ) = s (ζ ,ζ ) = h (ζ ,ζ ) = a2 +a a +a a +a2 ,
322/32 3 4 2 3 4 2 3 4 33 33 44 34 43 44
s (ζ ,ζ ) = s (ζ ,ζ ) = e (ζ ,ζ ) = a a −a a ,
322/311 3 4 11 3 4 2 3 4 33 44 34 43
s (ζ ,ζ ) = s (ζ ,ζ ) = e2(ζ ,ζ ) = a2 +2a a +a2
322/221 3 4 21/1 3 4 1 3 4 33 33 44 44
and therefore
[a a a a2 a a ]s (ω ,...,ω ) = [a a a a2 ](s (ζ ,ζ )−s (ζ ,ζ )).
11 12 21 22 34 43 322 1 4 11 12 21 22 32 1 2 311 1 2
Furthermore,
e 0 0
2
(cid:12) (cid:12)
s (ζ ,ζ ) = (cid:12)e e 0(cid:12) = e2e = (a a −a a )2(a +a )
32 1 2 (cid:12) 1 2 (cid:12) 2 1 11 22 12 21 11 22
(cid:12)0 e e (cid:12)
0 1
(cid:12) (cid:12)
(cid:12) (cid:12)
and
0 0 0
(cid:12) (cid:12)
s (ζ ,ζ ) = (cid:12)1 e 0(cid:12) = 0.
311 1 2 1
(cid:12) (cid:12)
(cid:12)0 1 e (cid:12)
1
(cid:12) (cid:12)
Therefore (cid:12) (cid:12)
[a a a a2 a a ]s (ω ,...,ω ) = −2.
11 12 21 22 34 43 322 1 4
Assume that a is of the form a a ···a =: b ···b . The correspond-
2 k+1,k+2 k+2,k+3 m,k+1 1 l
ing matrix A is
2
0 b 0 ... 0
1
 
0 0 b ... 0
2
(4) ... ... ... ... ... 
 
0 0 0 ... b 
 l−1
b 0 0 ... 0 
l
4 MATJAZˇ KONVALINKA
and its characteristic polynomial is (−1)l(tl − b b ···b ). If the zeros are denoted
1 2 l
η ,...,η, then e (η ,...,η ) = 1, e (η ,...,η ) = (−1)l−1b ···b , e (η ,...,η) = 0
1 l 0 1 l l 1 l 1 l i 1 l
for 1 ≤ i ≤ l−1.
Recall that a border strip tableau is a connected skew shape with no 2 × 2 square,
and that the height ht of a border strip tableau is defined to be one less than the
number of rows. The following result is well known; we include the proof for the sake
of completeness.
Lemma 3 If λ = (λ ,...,λ ) is a partition and ν ⊆ λ with |λ|−|ν| = l, then:
1 p
(1) λ/ν is a border-strip tableau if and only if λ′/ν′ is a border-strip tableau;
(2) λ/ν is a border-strip tableau if and only if λ +p = l+1;
1
(3) λ/ν is a border-strip tableau if and only if λ = ν +1 for 2 ≤ i ≤ p.
i i−1
Proof: (1) is obvious. (2) The squares of a border-strip tableau λ/µ form a NE-path
from (1,p) to (λ ,1). Each such path has p−1+λ −1+1 squares. Conversely, a
1 1
partition from (1,p) to (λ ,1) with λ +p−1 squares must be a NE-path and hence
1 1
its squares form a border-strip tableau. (3) A 2 × 2 square in lines i − 1,i implies
that λ ≥ ν + 2. It is clear that if the squares of a tableau λ/µ form a NE-path,
i i−1
we have λ = ν +1. (cid:3)
i i−1
Proposition 4 If the zeros of (−1)l(tl − b b ···b ) are denoted by η ,...,η, we
1 2 l 1 l
have s (η ,...,η ) = 0 unless λ/ν is a border-strip tableau, and s (η ,...,η ) =
λ/ν 1 l λ/ν 1 l
(−1)ht(λ/ν)b b ···b for a border-strip tableau λ/ν.
1 2 l
Proof: We can assume that λ 6= ν and that ν = 0. The indices of the entries of
i i p
sλ/ν = det(eλ′i−νj′−i+j)λ1×λ1
are strictly increasing in rows; the largest possible index is λ + p − 1 = l; and the
1
indices of the diagonal elements are eλ′i−νi′ with 1 ≤ λ′i − νi′ ≤ l (and λ′i − νi′ = l
only in the trivial case λ = l, ν = ∅). By the lemma and the fact that e = 0 for
1 i
1 ≤ i ≤ l−1, the matrix is 0 on and above the diagonal unless λ/ν is a border-strip
tableau. If it is a border-strip tableau, the elements at (i,j), i ≤ j < λ , are 0 (in
1
particular, all the entries in the first row are 0 except the last, which is e ), and by
l
the lemma, the subdiagonal elements are 1 and the element (1,λ ) is (−1)l−1b ···b .
1 1 l
Therefore s (η ,...,η ) = (−1)l−1+λ1−1b b ···b = (−1)p−1b b ···b , with p−1 =
λ/ν 1 l 1 2 l 1 2 l
ht(λ/ν). (cid:3)
In other words, by (3), [a ]s (ω ,...,ω ) satisfy the Murnaghan-Nakayama’s rule,
π λ 1 m
see [Sag, Theorem 4.10.2]. Together with the fact that Murnaghan-Nakayama’s rule
completely determines the irreducible characters χλ, this shows that
[a ]s (ω ,...,ω ) = χλ(π).
π λ 1 m
ON GOULDEN-JACKSON’S DETERMINANTAL EXPRESSION FOR THE IMMANANT 5
Note that this also gives us the coefficient of a = a ···a in p (ω ,...,ω ):
π 1π(1) nπ(n) λ 1 m
we know that
p = χµ(λ)s
λ X µ
µ
and hence
[a ]p (ω ,...,ω ) = χµ(λ)χµ(π)
π λ 1 m X
µ
is (by the orthogonality of the table of characters) equal to
z = 1j1j !2j2j !···
λ 1 2
if the partition corresponding to π is λ = h1j12j2···i, and 0 otherwise; see e.g. [Sta,
Proposition 7.17.6].
3. A proof of Theorem 1 via scalar product
Let us find the coefficient of a in e (ω ,...,ω ) = e e ···e . If we pick a
π λ 1 m λ1 λ2 λp iπ(i)
from e , we must also pick a from e because every term in e is balanced.
λ1 π(i)j λ1 λ1
But since a is the only term of a with π(i) as the first index, we must have
π(i)π2(i) π
j = π2(i). In other words, each of the cycles of π must be chosen from one of the
e ’s. We know that for J = {j < j < ... < j } and τ a permutation of j ,...,j ,
λi 1 2 i 1 i
a ···a appears in e with coefficient equal to the sign of τ.
j1τ(j1) jiτ(ji) i
This reasoning implies that the coefficient of a = a ···a with π of cycle type
π 1π(1) nπ(n)
µ = (µ ,...,µ ) in e is equal to ε R , where
1 q λ µ µλ
• ε is equal to (−1)j2+j4+... for µ = h1j12j2···i, and
µ
• R is the number of ordered partitions (B ,...,B ) of the set {1,...,q} such
µλ 1 p
that
λ = µ
j X i
i∈Bj
for 1 ≤ j ≤ p.
But we know that if h·,·i is the standard scalar product in the space of symmetric
functions defined by hh ,m i = δ and ω is the standard (scalar product preserving)
λ µ λµ
involution given by ω(h ) = e , then p = R m and ω(p ) = ε p (see [Sta,
λ λ µ Pν µν ν µ µ µ
§7.4 – §7.9]) imply
he ,p i = hω(e ),ω(p )i = ε hh ,p i = ε R .
λ µ λ µ µ λ µ µ µλ
Since e form a vector-space basis of the space of symmetric functions and since both
λ
the scalar product and the operator [a ] are linear, we have proved the following.
π
Proposition 5 For any symmetric function f and for π a permutation of cycle type
µ, we have
[a ]f(ω ,...,ω ) = hf,p i
π 1 m µ
6 MATJAZˇ KONVALINKA
In particular,
[a ]s (ω ,...,ω ) = hs ,p i = χλ(µ). (cid:3)
π λ 1 m λ µ
The proposition of course also implies that
[a ]p (ω ,...,ω ) = hp ,p i = z δ ,
π λ 1 m λ µ λ λµ
which is what we proved at the end of the last section.
4. An application to irreducible characters of S
n
It is worthwhile to note the following determinantal description of irreducible char-
acters of the symmetric group.
Corollary 6 Let λ,µ be partitions of m and define f ,...,f via the formula
0 m
(tµ1 −u )(tµ2 −u )···(tµq −u ) = f tm −f tm−1 +...±f .
1 2 q 0 1 m
Then
χλ(µ) = [u1···uq]det(fλ′−i+j).
i
Example 7 Let λ = (2,2,2,1) and µ = (3,2,2). Then
(t3−u )(t2−u )(t2−u ) = t7−(u +u )t5−u t4+u u t3+(u u +u u )t2−u u u
1 2 3 2 3 1 2 3 1 2 1 3 1 2 3
and so f = 1, f = 0, f = −u −u , f = u , f = u u , f = −u u −u u , f = 0,
0 1 2 2 3 3 1 4 2 3 5 1 2 1 3 6
f = u u u . Hence
7 1 2 3
f f
χ2221(322) = [u u u ](cid:12) 4 5(cid:12) = [u u u ](f f −f f ) = 1−2 = −1. (cid:3)
1 2 3 (cid:12)f f (cid:12) 1 2 3 4 3 2 5
2 3
(cid:12) (cid:12)
(cid:12) (cid:12)
Proof: Take the permutation π = (1,2,...,µ )(µ + 1,µ + 2,...,µ + µ )··· and
1 1 1 1 2
form the block diagonal A = A ⊕...⊕A with blocks of the form (4) corresponding
1 q
to cycles of π. For µ = (3,2,2), the permutation is (123)(45)(67) and the matrix is
0 a 0 0 0 0 0
12
 0 0 a 0 0 0 0 
23
a 0 0 0 0 0 0
 31 
 0 0 0 0 a 0 0 .
 45 
 0 0 0 a 0 0 0 
 54 
 0 0 0 0 0 0 a 
 67
 0 0 0 0 0 a 0 
76
The characteristic polynomial of A is
(tµ1 −u )(tµ2 −u )···(tµq −u )
1 2 q
for u = a a ···a , u = a a ··· ,a , etc. In other words,
1 12 23 µ11 2 µ1+1,µ1+2 µ1+2,µ1+3 µ1+µ2,µ1+1
if ω1,...,ωm are the eigenvalues of A, ei(ω1,...,ωm) = fi. But then det(fλ′−i+j) =
i
s (ω ,...,ω ) and equation (2) implies
λ 1 m
[u1···uq]det(fλ′−i+j) = [aπ]sλ(ω1,...,ωm) = χλ(µ). (cid:3)
i
ON GOULDEN-JACKSON’S DETERMINANTAL EXPRESSION FOR THE IMMANANT 7
Acknowledgements. TheauthorwouldliketothankIgorPak, AlexanderPostnikov
and Richard Stanley for helpful discussions and suggestions.
References
[GJ1] I. P. Goulden, D. M. Jackson,Immanants, Schur functions, and the MacMahonmaster theo-
rem, Proc. Amer. Math. Soc. 115 (1992), no. 3, 605–612
[GJ2] I. P. Goulden, D. M. Jackson, Immanants of combinatorial matrices, J. Algebra 148 (1992),
no. 2, 305–324
[Gr] C. Greene, Proof of a conjecture on immanants of the Jacobi-Trudi matrix, Linear Algebra
Appl. 171 (1992), 65–79
[MM] P. A. MacMahon, Combinatory Analysis, 2 vols., Cambridge Univ. Press, 1915 and 1916;
reprinted in one volume by Chelsea, New York, 1960
[Sag] B.E.Sagan,The symmetric group: representations, combinatorial algorithms, and symmetric
functions, Springer-Verlag New York, 2001
[Sta] R. P. Stanley, Enumerative combinatorics, Vol. 2, Cambridge University Press, Cambridge,
1999
[StaSte] R.P.Stanley, J.R.Stembridge,OnimmanantsofJacobi-Trudimatricesandpermutations
with restricted position, J. Combin. Theory Ser. A 62 (1993), no. 2, 261–279
[Ste] J.R.Stembridge,Immanantsoftotallypositivematricesarenonnegative,Bull. London Math.
Soc. 23 (1991), no. 5, 422–428
Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139
[email protected]
http://www-math.mit.edu/~konvalinka/