Symmetry, Integrability and Geometry: Methods and Applications SIGMA 7 (2011), 026, 48 pages
Vector-Valued Jack Polynomials from Scratch
Charles F. DUNKL † and Jean-Gabriel LUQUE ‡
† Dept. of Mathematics, University of Virginia, Charlottesville VA 22904-4137, USA
E-mail: cfd5z@virginia.edu
URL: http://people.virginia.edu/~cfd5z/
‡ Universite´ de Rouen, LITIS Saint-Etienne du Rouvray, France
E-mail: jean-gabriel.luque@univ-rouen.fr
URL: http://www-igm.univ-mlv.fr/~luque/
Received September 21, 2010, in final form March 11, 2011; Published online March 16, 2011
doi:10.3842/SIGMA.2011.026
Abstract. Vector-valued Jack polynomials associated to the symmetric group SN are
polynomials with multiplicities in an irreducible module of SN and which are simultaneous
eigenfunctions of the Cherednik–Dunkl operators with some additional properties concerning
the leading monomial. These polynomials were introduced by Griffeth in the general setting
of the complex reflections groups G(r, p,N) and studied by one of the authors (C. Dunkl) in
the specialization r = p = 1 (i.e. for the symmetric group). By adapting a construction due
to Lascoux, we describe an algorithm allowing us to compute explicitly the Jack polynomials
following a Yang–Baxter graph. We recover some properties already studied by C. Dunkl
and restate them in terms of graphs together with additional new results. In particular,
we investigate normalization, symmetrization and antisymmetrization, polynomials with
minimal degree, restriction etc. We give also a shifted version of the construction and we
discuss vanishing properties of the associated polynomials.
Key words: Jack polynomials; Yang–Baxter graph; Hecke algebra
2010 Mathematics Subject Classification: 05E05; 16T25; 05C25; 33C80
1 Introduction
The Yang–Baxter graphs are used to study Jack polynomials. In particular, these objects
have been investigated in this context by Lascoux [14] (see also [15, 16] for general properties
about Jack and Macdonald polynomials). Vector-valued Jack polynomials are associated with
irreducible representations of the symmetric group SN , that is, to partitions of N . A Yang–
Baxter graph is a directed graph with no loops and a unique root, whose edges are labeled by
generators of a certain subsemigroup of the extended affine symmetric group. In this paper the
vertices are labeled by a pair consisting of a weight in NN and the content vector of a standard
tableau. The weights are the labels of monomials which are the leading terms of polynomials, and
the tableaux all have the same shape. There is a vector-valued Jack polynomial associated with
each vertex. These polynomials are special cases of the polynomials introduced by Griffeth [7, 8]
for the family of complex reflection groups denoted by G (r, p,N) (where p|r). This is the group
of unitary N × N matrices such that their nonzero entries are rth roots of unity, the product
of the nonzero entries is a (r/p)th root of unity, and there is exactly one nonzero entry in each
row and each column. The symmetric group is the special case G (1, 1, N). The vector space in
which the Jack polynomials take their values is equipped with the nonnormalized basis described
by Young, namely, the simultaneous eigenvectors of the Jucys–Murphy elements.
The labels on edges denote transformations to be applied to the objects at a vertex. Vector-
valued Jack polynomials are uniquely determined by their spectral vector, the vector of eigen-
values under the (pairwise commuting) Cherednik–Dunkl operators. This serves to demonstrate
2 C.F. Dunkl and J.-G. Luque
the claim that different paths from one vertex to another produce the same result, a situation
which is linked to the braid or Yang–Baxter relations. These refer to the transformations.
Following Lascoux [14] we define the monoid ŜN , a subsemigroup of the affine symmetric
group, with generators {s1, s2, . . . , sN−1,Ψ} and relations:
sisj = sjsi, |i− j| > 1,
sisi+1si = si+1sisi+1, 1 ≤ i < N − 1,
s1Ψ
2 = Ψ2sN−1,
siΨ = Ψsi−1, 2 ≤ i ≤ N − 1.
The relations s2i = 1 do not appear in this list because the graph has no loops.
The main objects of our study are polynomials in x = (x1, . . . , xN ) ∈ RN with coefficients
in Q (α), where α is a transcendental (indeterminate), and with values in the SN -module cor-
responding to a partition λ of N . The Yang–Baxter graph Gλ is a pictorial representation of
the algorithms which produce the Jack polynomials starting with constants. The generators
of ŜN correspond to transformations taking a Jack polynomial to an adjacent one. At each
vertex there is such a polynomial, and a 4-tuple which identifies it. The 4-tuple consists of
a standard tableau denoting a basis element of the SN -module, a weight (multi-index) descri-
bing the leading monomial of the polynomial, a spectral vector, and a permutation, essentially
the rank function of the weight. The spectral vector and permutation are determined by the
first two elements. For technical reasons the standard tableaux are actually reversed, that is,
the entries decrease in each row and each column. This convention avoids the use of a reversing
permutation, in contrast to Griffeth’s paper [8] where the standard tableaux have the usual
ordering.
The symmetric and antisymmetric Jack polynomials are constructed in terms of certain
subgraphs of Gλ. Furthermore the graph technique leads to the definition and construction of
shifted inhomogeneous vector-valued Jack polynomials.
Here is an outline of the contents of each section.
Section 2 contains the basic definitions and construction of the graph Gλ. The presentation
is in terms of the 4-tuples mentioned above. It is important to note that not every possible label
need appear on edges pointing away from a given vertex: if the e weight at the vertex is v ∈ NN
then the transposition (i, i+ 1) (labeled by si) can be applied only when v [i] ≤ v [i+ 1], that
is, when the resulting weight is greater than or equal to v in the dominance order. The action
of the affine element Ψ is given by vΨ = (v [2] , v [3] , . . . , v [N ] , v [1] + 1).
The Murphy basis for the irreducible representation of SN along with the definition of the
action of the simple reflections (i, i+ 1) on the basis is presented in Section 3 Also the vector-
valued polynomials, their partial ordering, and the Cherednik–Dunkl operators are introduced
here.
Section 4 is the detailed development of Jack polynomials. Each edge of the graph Gλ
determines a transformation that takes the Jack polynomial associated with the beginning vertex
to the one at the ending vertex of the edge. There is a canonical pairing defined for the vector-
valued polynomials; the pairing is nonsingular for generic α and the Cherednik–Dunkl operators
are self-adjoint. The Jack polynomials are pairwise orthogonal for this pairing and the squared
norm of each polynomial can be found by use of the graph.
In Section 5 we investigate the symmetric and antisymmetric vector-valued Jack polynomials
in relation with connectivity of the Yang–Baxter graph whose affine edges have been removed.
Also, in this section one finds the method of producing coefficients so that the corresponding
sum of Jack polynomials is symmetric or antisymmetric. The idea is explained in terms of
certain subgraphs of Gλ.
Vector-Valued Jack Polynomials from Scratch 3
Vertices of Gλ satisfying certain conditions may be mapped to vertices of a graph related
to SM , M < N , by a restriction map. This topic is the subject of Section 6. This section also
describes the restriction map on the Jack polynomials.
In Section 7 the shifted vector-valued Jack polynomials are presented. These are inhomoge-
neous and the parts of highest degree coincide with the homogeneous Jack polynomials of the
previous section. The construction again uses the Yang–Baxter graph Gλ; it is only necessary
to change the operations associated with the edges.
Throughout the paper there are numerous figures to concretely illustrate the structure of the
graphs.
2 Yang–Baxter type graph associated to a partition
2.1 Sorting a vector
Consider a vector v ∈ NN , we want to compute the unique decreasing partition v+, which is
in the orbit of v for the action of the symmetric group SN acting on the right on the position,
using the minimal number of elementary transpositions si = (i i+ 1).
If v is a vector we will denote by v[i] its ith component. Each σ ∈ SN will be associated to
the vector of its images [σ(1), . . . , σ(N)]. Let σ be a permutation, we will denote `(σ) = min{k :
σ = si1 · · · sik} the length of the permutation. By a straightforward induction one finds:
Lemma 2.1. Let v ∈ NN be a vector, there exists a unique permutation σv such that v = v+σv
with `(σv) minimal.
The permutation σv is obtained by a standardization process: we label with integer from 1
to N the positions in v from the largest entries to the smallest one and from left to right.
Example 2.2. Let v = [2, 3, 3, 1, 5, 4, 6, 6, 1], the construction gives:
σv = [ 7 5 6 8 3 4 1 2 9 ]
v = [ 2 3 3 1 5 4 6 6 1 ]
We verify that vσ−1v = [6, 6, 5, 4, 3, 3, 2, 1, 1] = v
+.
The definition of σv is compatible with the action of SN in the following sense:
Proposition 2.3.
1. σvsi =
{
σv if v = vsi,
σvsi otherwise.
2. If v[i] = v[i+ 1] then σvsi = sσv [i]σv.
This can be easily obtained from the construction.
Define the affine operation Ψ acting on a vector by
[v1, . . . , vN ]Ψ = [v2, . . . , vN , v1 + 1],
and more generally let Ψα by
[v1, . . . , vN ]Ψ
α = [v2, . . . , vN , v1 + α].
Denote also by θ := Ψ0 the circular permutation [2, . . . , N, 1].
Again, one can prove easily that the computation of σv is compatible (in a certain sense)
with the action of Ψ:
4 C.F. Dunkl and J.-G. Luque
Proposition 2.4.
σvΨ = σvθ.
Example 2.5. Consider v = [2, 3, 3, 2, 5, 4, 6, 6, 1], one has
σv = [7, 5, 6, 8, 3, 4, 1, 2, 9]
and
σvθ = [5, 6, 8, 3, 4, 1, 2, 9, 7].
But
v′ := v[2, 3, 4, 5, 6, 7, 8, 9, 1] = [3, 3, 2, 5, 4, 6, 6, 1, 2]
and σv′ = [5, 6, 7, 3, 4, 1, 2, 9, 8]; here an underlined integer means that there is a difference with
the same position in σvθ and in σv′ . This is due to the fact that v[1] is the first occurrence of 2
in v while v′[9] is the last occurrence of 2 in v′. Adding 1 to v′[9] one obtains
vΨ = [3, 3, 2, 5, 4, 6, 6, 1, 3].
The last occurrence of 2 becomes the last occurrence of 3 (that is the number of the first
occurrence of 2 minus 1). Hence,
σvΨ = [5, 6, 8, 3, 4, 1, 2, 9, 7] = σvθ.
2.2 Construction and basic properties of the graph
Definition 2.6. A tableau of shape λ is a filling with integers weakly increasing in each row
and in each column. In the sequel row-strict means increasing in each row and column-strict
means increasing in each column.
A reverse standard tableau (RST) is obtained by filling the shape λ with integers 1, . . . , N
and with the conditions of strictly decreasing in the row and the column. We will denote by
Tabλ, the set of the RST with shape λ.
Let τ be a RST, we define the vector of contents of τ as the vector CTτ such that CTτ [i] is
the content of i in τ (that is the number of the diagonal in which i appears; the number of the
main diagonal is 0, and the numbers decrease from down to up). In other words, if i appears in
the box [col, row] then CTτ [i] = col− row.
Example 2.7. Consider the tableau τ =
2
5 4
6 3 1
, we obtain the vector of contents by labeling
the numbers of the diagonals
−2
−1 0
0 1 2
. So one obtains,
CT 2
5 4
6 3 1
= [2,−2, 1, 0,−1, 0].
We construct a Yang–Baxter-type graph with vertices labeled by 4-tuples (τ, ζ, v, σ), where
τ is a RST, ζ is a vector of length N with entries in Z[α] (ζ will be called the spectral vector),
v ∈ NN and σ ∈ SN , as follows: First, consider a RST of shape λ and write a vertex labeled by
Vector-Valued Jack Polynomials from Scratch 5
the 4-tuple (τ,CTτ , 0N , [1, . . . , N ]). Now, we consider the action of the elementary transposition
of SN on the 4-tuple given by
(τ, ζ, v, σ)si =



(τ, ζsi, vsi, σsi) if v[i+ 1] 6= v[i],
(
τ (σ[i],σ[i+1]), ζsi, v, σ
)
if v[i] = v[i+ 1] and τ (σ[i],σ[i+1]) ∈ Tabλ,
(τ, ζ, v, σ) otherwise,
where τ (i,j) denotes the filling obtained by permuting the values i and j in τ .
Consider also the affine action given by
(τ, ζ, v, σ)Ψ = (τ, ζΨα, vΨ, σ[2, . . . , N, 1]) = (τ, [ζ2, . . . , ], [v2, . . . ], [σ2, . . . ]).
Example 2.8.
1. ( 31
542 , [1, 0, 2α, α+ 2, α− 1], [0, 0, 2, 1, 1], [45123]
)
s2 =( 31
542 , [1, 2α, 0, α+ 2, α− 1], [0, 2, 0, 1, 1], [41523]
)
2. ( 31
542 , [1, 0, 2α, α+ 2, α− 1], [0, 0, 2, 1, 1], [45123]
)
s4 =( 21
543 , [1, 0, 2α, α− 1, α+ 2], [0, 0, 2, 1, 1], [45123]
)
3. ( 31
542 , [1, 0, 2α, α+ 2, α− 1], [0, 0, 2, 1, 1], [45123]
)
s1 =( 31
542 , [1, 0, 2α, α+ 2, α− 1], [0, 0, 2, 1, 1], [45123]
)
4. ( 31
542 , [1, 0, 2α, α+ 2, α− 1], [0, 0, 2, 1, 1], [45123]
)
Ψ =
( 31
542 , [0, 2α, α+ 2, α− 1, α+ 1], [0, 2, 1, 1, 1], [51234]
)
Definition 2.9. If λ is a partition, denote by τλ the tableau obtained by filling the shape λ
from bottom to top and left to right by the integers {1, . . . , N} in the decreasing order.
The graph Gλ is an infinite directed graph constructed from the 4-tuple
(τλ,CTτλ , [0
N ], [1, 2, . . . , N ]),
called the root and adding vertices and edges following the rules
1. We add an arrow labeled by si from the vertex (τ, ζ, v, σ) to (τ ′, ζ ′, v′, σ′) if (τ, ζ, v, σ)si =
(τ ′, ζ ′, v′, σ′) and v[i] < v[i+1] or v[i] = v[i+1] and τ is obtained from τ ′ by interchanging
the position of two integers k < ` such that k is at the south-east of ` (i.e. CTτ (k) ≥
CTτ (`) + 2).
2. We add an arrow labeled by Ψ from the vertex (τ, ζ, v, σ) to (τ ′, ζ ′, v′, σ′) if (τ, ζ, v, σ)Ψ =
(τ ′, ζ ′, v′, σ′).
3. We add an arrow si from the vertex (τ, ζ, v, σ) to ∅ if (τ, ζ, v, σ)si = (τ, ζ, v, σ).
An arrow of the form
(τ, ζ, v, σ) (τ, ζ′, v′, σ′)si or Ψ
will be called a step. The other arrows will be called jumps, and in particular an arrow
(τ, ζ, v, σ) ∅si
will be called a fall; the other jumps will be called correct jumps.
As usual a path is a sequence of consecutive arrows in Gλ starting from the root and is denoted
by the sequence if the labels of its arrows. Two paths P1 = (a1, . . . , ak) and P2 = (b1, . . . , b`)
are said to be equivalent (denoted by P1 ≡ P2) if they lead to the same vertex.
6 C.F. Dunkl and J.-G. Luque
We remark that from Proposition 2.3, in the case v[i] = v[i+1], the part 1 of Definition 2.9 is
equivalent to the following statement: τ ′ is obtained from τ by interchanging σv[i] and σv[i+1] =
σv[i] + 1 where σv[i] is to the south-east of σv[i] + 1, that is, CTv[σv[i]]− CTv[σv[i] + 1] ≥ 2.
Example 2.10. The following arrow is a correct jump
31
542 ,[1,0,2α,α+2,α−1]
[0,0,2,1,1],[45123]
21
543 ,[1,0,2α,α−1,α+2]
[0,0,2,1,1],[45123]
s4
whilst
31
542 ,[1,0,2α,α+2,α−1]
[0,0,2,1,1],[45123]
31
542 ,[1,2α,0,α−1,α+2]
[0,2,0,1,1],[41523]
s2
is a step.
The arrows
31
542 ,[1,0,2α,α+2,α−1]
[0,0,2,1,1],[45123]
21
543 ,[1,0,2α,α−1,α+2]
[0,0,2,1,1],[45123]
s4
and
31
542 ,[1,0,2α,α+2,α−1]
[0,0,2,1,1],[45123]
31
542 ,[1,2α,0,α−1,α+2]
[0,2,0,1,1],[41523]
s2
are not allowed.
Example 2.11. Consider the partition λ = 21, the graph G21 in Fig. 1 is obtained from the
4-tuple
(
2
3 1
, (1,−1, 0), (0, 0, 0), 1
)
by applying the rules of Definition 2.9. In Fig. 1, the
steps are drawn in orange, the jumps in blue and the falls have been omitted.
For a reverse standard tableau τ of shape λ, a partition of N , let
inv(τ) = #{(i, j) : 1 ≤ i < j ≤ N, rw(i, τ) > rw(j, τ)},
where rw(i, τ) is the row of τ containing i (also we denote the column containing i by cl(i, τ)).
Then a correct jump from τ to τ ′ implies inv(τ ′) = inv(τ)+1 (the entries σ[i] and σ[i+1] = σ[i]+1
are interchanged in τ to produce τ ′). Thus the number of correct jumps in a path from the root
to (τ, ζ, v, σ) equals inv(τ)− inv(τλ).
So we consider the number of steps in a path from 0N to v;1 recall that one step links v to v′
where either v[i] < v[i+ 1] and v′ = vsi or v′ = vΨ.
For x ∈ Z (or R) let (x) := 12(|x| + |x + 1| − 1), then (x) = x for x ≥ 0, (x) = 0 for
−1 ≤ x ≤ 0, and (x) = −x− 1 for x ≤ −1.
There is a symmetry relation: (x) = (−x− 1).
Definition 2.12. For v ∈ NN let |v| :=
N∑
i=1
v[i] and set
S(v) :=
∑
1≤i<j≤N
(v[i]− v[j]).
The above formula can be written as
S(v) =
1
2
∑
1≤i<j≤N
(|v[i]− v[j]|+ |v[i]− v[j] + 1|)−
N(N − 1)
4
.
1The other components of the label of the vertices are omitted here.
Vector-Valued Jack Polynomials from Scratch 7
2
31 ,[000],
[1,−1,0],123
1
32 ,[000],
[−1,1,0],123
s1
1
32 ,[001],
[1,0,α−1],231
1
32 ,[010],
[1,α−1,0],213
1
32 ,[100],
[α−1,1,0],123
2
31 ,[001],
[−1,0,α+1],231
2
31 ,[010],
[−1,α+1,0],213
2
31 ,[100],
[α+1,−1,0],123
s
2
s
1
s 2
s 1
Ψ Ψ
1
32 ,[011],
[0,α−1,α+1],312
2
31 ,[011],
[0,α+1,α−1],312
1
32 ,[101],
[α−1,0α+1],132
2
31 ,[101],
[α+1,0α−1],132
1
32 ,[002],
[1,0,2α−1],231
2
31 ,[002],
[−1,0,2α+1],231
Ψ
Ψ
Ψ
Ψ
Ψ
Ψ
s2
s
1 s 1Ψ
1
32 ,[110],
[α−1,α+1,0],123
2
31 ,[110],
[α+1,α−1,0],123
s 2
s
2
s1
1
32 ,[020],
[1,2α−1,0],213
1
32 ,[200],
[2α−1,1,0],123
2
31 ,[020],
[−1,2α+1,0],213
2
31 ,[200],
[2α+1,−1,0],123
s
2
s
1
s 2
s 1
Figure 1. The first vertices of the graph G21 where we omit to write the vertex ∅ and the associated
arrows.
Proposition 2.13. The number of steps in any path joining 0N to v equals |v|+ S(v).
Proof. The base point satisfies |0N | = 0 and |S(0N )| = 0. Since jumps do not modify v,
consider a step of the form v′ = vsm, then |v′| = |v|, v[m + 1] − v[m] ≥ 1 and S(v′) − S(v)
involves only the pair (m,m+ 1) in the sum over all pairs (i, j), 1 ≤ i < j ≤ N . Indeed
S(v′)− S(v) = (v[m+ 1]− v[m])− (v[m]− v[m+ 1])
= (v[m+ 1]− v[m])− (−v[m] + v[m+ 1]− 1) = 1.
8 C.F. Dunkl and J.-G. Luque
It remains to show that S(vΨ) = S(v) (because |vΨ| = |v|+ 1). Note (vΨ)[N ] = v[1] + 1. Then
S(v)− S(vΨ) =
N∑
j=2
(v[1]− v[j])−
N∑
i=2
(v[i]− v[1]− 1)
=
N∑
j=2
(v[1]− v[j])− (v[j]− v[1]− 1) = 0.
This completes the proof. 
As a straightforward consequences, Proposition 2.13 implies
Corollary 2.14. All the paths joining two given vertices in Gλ have the same length.
This suggests that some properties could be shown by induction on the common length of all
the all the paths joining two given vertices.
For a given 4-tuple (τ, ζ, v, σ) the values of ζ and σ are determined by those of τ and v, as
shown by the following proposition.
Proposition 2.15. If (τ, ζ, v, σ) is a vertex in Gλ, then σ = σv and ζ[i] = viα+CTτ [σ[i]]. We
will set ζv,τ := ζ.
Proof. We prove the result by induction on the length k of a path (a1, . . . , ak) (from Corol-
lary 2.14 all the paths have the same length) from the root to (τ, ζ, v, σ) and set
(τ ′, ζ ′, v′, σ′) =
(
τλ,CTτλ , 0
N , [1, . . . , N ]
)
a1 · · · ak−1.
Suppose that ak = Ψ is the affine operation. More precisely, τ = τ ′, ζ = ζ ′Ψα, v = v′Ψ and σ =
σ′[2, . . . , N, 1]. Using the induction hypothesis one has σ′ = σv′ and ζ ′[i] = v′[i]α + CTτ [σv′ [i]].
Hence, Proposition 2.4 gives σ = σv′Ψ = σv. Suppose that i < N then
ζ[i] = ζ ′[i+ 1] = v′[i+ 1]α+ CTτ [σv′ [i+ 1]] = v[i]α+ CTτ [σv[i]].
If i = N then again
ζ[i] = ζ ′[1] + a = (v′[1] + 1)α+ CTτ [σv′ [1]] = v[N ]α+ CTτ [σv[N ]].
Suppose now that ak is not an affine operation. Using the induction hypothesis one has σ′ = σv′
and ζ ′[j] = v′[j]α+ CTτ [σv′ [j]] for each j. If ak = si is a step then τ = τ ′, ζ = ζ ′si, v = v′si and
σ = σ′si. Hence, Proposition 2.3 gives σ = σv′si = σv. If j 6= i, i + 1 then one has ζ[j] = ζ
′[j],
v[j] = v′[j] and σ[j] = σ′[j], hence ζ[j] = v[j]α+CTτ [σv[j]]. If j = i then one has ζ[i] = ζ ′[i+1],
v[i] = v′[i+ 1] and σ[i] = σ′[i+ 1], and again the result is straightforward. And similarly when
j = i+ 1 one finds the correct value for ζ[i+ 1].
Suppose now that ak = si is a jump. That is τ = τ ′(σ[i],σ[i+1]), ζ = ζ ′si, v = v′ and σ = σ′.
Straightforwardly, σ = σv′ = σv and if j 6= i, i+ 1 then
ζ[j] = ζ ′[j] = v′[j]α+ CTτ ′ [σv′ [j]] = v[j]α+ CTτ [σv[j]].
Suppose that j = i, since ak is a jump v′[i] = v′[i+ 1] and
ζ[i] = ζ ′[i+ 1] = v′[i+ 1]α+ CTτ ′ [σv′ [i+ 1]] = v[i]α+ CTτ [σv[i]].
Similarly, when j = i+ 1, one obtains the correct value for ζ[j]. This ends the proof. 
Vector-Valued Jack Polynomials from Scratch 9
Example 2.16. Consider the RST τ =
3
7 4 1
8 6 5 2
and the vector v = [6, 2, 4, 2, 2, 3, 1, 4].
One has σv = [1, 5, 2, 6, 7, 4, 8, 3] and CTτ = [1, 3,−2, 0, 2, 1,−1, 0] and then
ζv,τ = [6α+ 1, 2α+ 2, 4α+ 3, 2α+ 1, 2α− 1, 3α, α, 4α− 2].
Hence, the 4-tuple


3
7 4 1
8 6 5 2
, [6α+ 1, 2α+ 2, 4α+ 3, 2α+ 1, 2α− 1, 3α, α, 4α− 2], [6, 2, 4, 2, 2, 3, 1, 4], [1, 5, 2, 6, 7, 4, 8, 3]


labels a vertex of G431.
As a consequence,
Corollary 2.17. Let (τ, v) be a pair constituted with a RST τ of shape λ (a partition of N)
and a vector v ∈ NN . Then there exists a unique vertex in Gλ labeled by a 4-tuple of the form
(τ, ζ, v, σ). We will denote Vτ,ζ,v,σ := (τ, v).
We point out that all the information can be retrieved from the spectral vector ζ – the
coefficients of α give v, the rank function of v gives σ, and the constants in the spectral vector
give the content vector which does uniquely determine the RST τ .
Definition 2.18. We define the subgraph Gτ as the graph obtained from Gλ by erasing all the
vertices labeled by RST other than τ and the associated arrows. Such a graph is connected.
Note that the graph Gλ is the union of the graphs Gτ connected by jumps. Furthermore,
if Gτ and Gτ ′ are connected by a succession of jumps then there is no step from Gτ ′ to Gτ .
Example 2.19. In Fig. 1, the graph G21 is constituted with the two graphs G 1
32
and G 2
31
connected by jumps (in blue).
3 Vector-valued polynomials
3.1 About the Young seminormal representation of the symmetric group
We consider the space Vλ spanned by reverse tableaux of shape λ and the (Young) action of the
symmetric group as defined by Murphy2 in [17] by
τsi =



bτ [i]τ if bτ [i]2 = 1,
bτ [i]τ + τ (i,i+1) if 0 < bτ [i] ≤ 12 ,
bτ [i]τ + (1− bτ [i]2)τ (i,i+1) otherwise,
(3.1)
where bτ [i] := 1CTτ [i]−CTτ [i+1] . Note that when |bτ [i]| < 1, τ
(i,i+1) is always a reverse standard
tableau when τ is a reverse standard tableau.
Murphy showed [17] that the RST are the simultaneous eigenfunctions of the Jucys–Murphy
elements:
ωi =
N∑
j=i+1
sij ,
where sij denotes the transposition exchanging i and j. More precisely:
2The Young seminormal representation was defined in Young’s last papers but himself apparently underesti-
mated the importance of the construction. G. Murphy rediscovered it when reading the Jucys’ paper [11]. See [18]
for more details about the seminormal representation and its relation with the notions of Gelfand–Tsetlin basis.
10 C.F. Dunkl and J.-G. Luque
Proposition 3.1.
τωi = CTτ [i]τ.
As usual, a polynomial representation for the Murphy action on the RST can be computed
through the Yang–Baxter graph. We start from τλ and we construct the associated polynomial
in the variables t1, . . . , tN :
Pτλ =
∏
i
∏
k>l
(i,k),(i,l)∈λ
(tτλ[i,k] − tτλ[i,l]),
where τ [i, j] denotes the integer belonging at the column i and the row j in τ . Such a polynomial
is a simultaneous eigenfunction of the Jucys–Murphy idempotents:
Pτλωi = CTτλ [i]Pτλ .
Suppose that Pτ is the polynomial associated to τ . Suppose also that 0 < bτ [i] < 1. Hence,
the polynomial Pτ (i,i+1) is obtained from the polynomial Pτ by acting with si − bτ [i] (with the
standard action of the transposition si on the variables tj).
Example 3.2.
P 21
43
= P 31
42
(
s2 − 12
)
= (t3 − t4)(t1 − t2)
(
s2 − 12
)
= t1t2 − 12 t4t1 −
1
2 t3t2 + t4t3 −
1
2 t4t2 −
1
2 t3t1
Let us remark that in [13], Lascoux simplified the Young construction by having recourse
to the covariant algebra (of SN ) C[x1, . . . , xN ]/Sym+ where Sym+ is the ideal generated by
symmetric functions without constant terms. Note that the covariant algebra is isomorphic to
the regular representation C[SN ]. In the aim to adapt his construction to our notations, we
replace each polynomial with its dominant monomial represented by the vectors of its exponents.
The vector associated to the root of the graph is the vector exponent of the leading monomial
in the product of the Vandermonde determinants associated to each column and is obtained by
putting the number of the row minus 1 in the corresponding entry.
Example 3.3. The vector associated to 452
631
is [010210].
In fact, the covariant algebra being isomorphic to the regular representation of SN , the
computation of the polynomials is completely encoded by the action of the symmetric group on
the leading monomials, as shown in the following example. Observe that we do not replace the
representation by the orbit of the leading monomial (since the space generated by the orbit is
in general bigger), but we consider the projection which completely determines the elements.
Example 3.4. Consider the RST of shape 221, one has
3
41
52
2
41
53
1
42
53
2
31
54
1
32
54
[10210]
[12010]
[21010] [12100]
[21100]
s
2
−
1 3
s
1 −
1
2 s3
−
1
2
s3
−
1
2
s
1 −
1
2
s
2
s
1 s3
s3
s
1
Vector-Valued Jack Polynomials from Scratch 11
For instance, one has
P 1
42
53
= t21t4t2 +
1
2 t4t
2
2t3 +
1
2 t
2
2t5t1 −
1
2 t
2
2t5t3 +
1
2 t4t
2
3t1 +
1
2 t
2
4t1t3 + t
2
1t5t3 − t
2
1t5t2 +
1
2 t
2
5t4t3
+ 12 t5t
2
4t2 +
1
2 t
2
5t1t2 −
1
2 t4t
2
2t1 −
1
2 t
2
5t4t2 +
1
2 t
2
3t5t2 −
1
2 t
2
4t1t2 − t
2
1t4t3 −
1
2 t
2
3t5t1
− 12 t5t
2
4t3 −
1
2 t4t
2
3t2 −
1
2 t
2
5t1t3,
whose leading monomial is t21t2t4.
From the construction, the leading monomial of Pτ is the product of all the t
rw(i,τ)−1
i . For
example, the leading monomial in P 51
732
9864
is t21t2t3t
2
5t7.
3.2 Definition and dominance properties of vector-valued polynomials
Consider the space
MN = spanC
{
xv[1]1 · · ·x
v[N ]
N ⊗ τ : v ∈ N
N , τ ∈ Tabλ, λ ` N
}
,
where Tab(N) denotes the set of the reverse standard tableaux on {1, . . . , N}.
This space splits into a direct sum MN =
⊕
λ`N Mλ, where
Mλ = spanC
{
xv[1]1 · · ·x
v[N ]
N ⊗ τ | v ∈ N
N , τ ∈ Tabλ
}
.
The algebra C[SN ] ⊗ C[SN ] acts on these spaces by commuting the vector of the powers on
the variables on the left component and the action on the tableaux defined by Murphy (equa-
tion (3.1)) on the right component.
Example 3.5.
x31x
1
2 ⊗
2
3 1
(s2 ⊗ s1) =
1
2
x31x
1
3 ⊗
2
3 1
+ x31x
1
3 ⊗
1
3 2
.
For simplicity we will denote xv = xv[1]1 · · ·x
v[N ]
N and x
v,τ := xv ⊗ τσv. By abuse of lan-
guage xv,τ will be referred to as a polynomial. Note that the space Mλ is spanned by the set of
polynomials
Mλ :=
{
xv,τ : v ∈ NN , τ ∈ Tabλ
}
,
which can be naturally endowed with the order 
 defined by
xv,ττ 
 xv
′,τ ′ iff v 
 v′,
with v
v′ means that v+ ≺ v′+ or v+ = v′+ and v ≺ v′, where ≺ denotes the classical dominance
order on vectors:
v  v′ iff ∀ i, v[1] + · · ·+ v[i] ≤ v′[1] + · · ·+ v′[i].
Example 3.6.
1. x
031, 2
3 1 
 x
310, 1
3 2 since 031 ≺ 310.
2. x
220, 2
3 1 
 x
301, 1
3 2 since 220 ≺ 310.
3. The polynomials x
031, 2
3 1 and x
031, 1
3 2 are not comparable.
12 C.F. Dunkl and J.-G. Luque
The partial order  will provide us a relevant dominance notion.
Definition 3.7. The monomial xv,τ is the leading monomial of a polynomial P if and only if P
can be written as
P = αvx
v,τ +
∑
xv′,τ ′
xv,τ
αv′,τ ′x
v′,τ ′
with αv 6= 0.
As in [9], we define Ψ := (θ ⊗ θ)xN , with θ = s1s2 · · · sN−1. The following proposition
describes the transformation properties of leading monomials with respect to the si and Ψ.
Proposition 3.8. Suppose that xv,τ is the leading monomial in P then
1. If v[i] < v[i+ 1] then xvsi,τ is the leading monomial in P (si⊗ si). Its leading monomial is
xv˜ ⊗ (τ.sij), where xv˜ is the dominant term in ∂ijxv˜.
2. xvΨ,τ is the leading monomial in PΨ.
3.3 Dunkl and Cherednik–Dunkl operators for vector-valued polynomials
We define the Dunkl operators
Di :=
∂
∂xi
⊗ 1 +
1
α
∑
i 6=j
∂ij ⊗ sij ,
where sij denotes the transposition which exchanges i and j and
∂ij = (1− sij)
1
xi − xj
is the divided difference.
This definition is the same as in [4], but our operators act on their left. One has
Lemma 3.9. If Di denotes the Dunkl operator, one has
(si ⊗ si)Di = Di+1(si ⊗ si).
Proof. Straightforward from the definition of Di and the equalities
sisij = si+1,jsi, si∂ij = ∂i+1jsi and si
∂
∂xi
=
∂
∂xi+1
si. 
The Cherednik–Dunkl operators are pairwise commuting operators defined by [4]
Ui := xiDi −
1
α
i−1∑
j=1
si,j ⊗ si,j .
We do not repeat the proof of the commutation [Ui,Uj ] = 0 which can be found in [4]. But, as we
will see in the next section, this property is not used to prove the existence of the vector-valued
Jack polynomials.
One has
Lemma 3.10.
1. (si ⊗ si)Ui = Ui+1(si ⊗ si) + 1α .
Vector-Valued Jack Polynomials from Scratch 13
2. (si ⊗ si)Uj = Uj(si ⊗ si), j 6= i, i+ 1.
3. (si ⊗ si)Ui+1 = Ui(si ⊗ si)− 1α .
Proof. The three identities are of the same type. We prove only the first one which follows
from the equalities
(si ⊗ si)Ui = (si ⊗ si)xiDi −
1
α
i−1∑
j=1
si,j ⊗ si,j
=

xi+1Di+1 −
1
α
i−1∑
j=1
si+1,j ⊗ si+1,j

 (si ⊗ si) = Ui+1(si ⊗ si) +
1
α
. 
The affine operator Ψ has the following commutation properties with the Dunkl operators:
Lemma 3.11.
1. Di+1Ψ = ΨDi + (θ ⊗ θ)(si,N ⊗ si,N ), i < 1.
2. D1Ψ = ΨDN + (θ ⊗ θ)
(
N−1∑
j=1
(sN,j ⊗ sN,j)− 1
)
.
As a consequence, one finds.
Lemma 3.12.
ΨUi = Ui+1Ψ, i 6= N and ΨUN = (U1 + 1) Ψ.
The action on the RST is given by
Lemma 3.13.
(1⊗ τ)Ui =
(
1 +
1
α
CTτ [i]
)
(1⊗ τ).
Proof. One has
(1⊗ τ)Ui = (1⊗ τ)xiDi −
1
α
i−1∑
j=1
(1⊗ τ)(si,j ⊗ si,j) = (1⊗ τ)
(
1 +
1
α
1⊗ ωi
)
,
where ωi :=
N∑
j=i+1
(i j) denotes a Jucys–Murphy element. Since the RST are eigenfunctions of
the Jucys–Murphy elements and the associated eigenvalues are given by the contents, the lemma
follows. 
For convenience, define ξ˜i := αUi − α. From the preceding lemmas, one obtains
Proposition 3.14.
(si ⊗ si)ξ˜i = ξ˜i+1(si ⊗ si) + 1,
(si ⊗ si)ξ˜i+1 = ξ˜i(si ⊗ si)− 1,
(si ⊗ si)ξ˜j = ξ˜j(si ⊗ si), j 6= i, i+ 1,
Ψξ˜i = ξ˜i+1Ψ, i 6= N,
Ψξ˜N =
(
ξ˜1 + α
)
Ψ.
14 C.F. Dunkl and J.-G. Luque
4 Nonsymmetric vector-valued Jack polynomials
In this section we recover the construction, due to one of the authors [4], of a basis of vector-
valued polynomials Jv,τ . This construction belongs to a large family of vector-valued Jack
polynomials associated to the complex reflection groups G(r, 1, n) defined by Griffeth [8]. We
will denote by ζv,τ their associated spectral vectors. We will see also that many properties of
this basis can be deduced from the Yang–Baxter structure.
4.1 Yang–Baxter construction associated to Gλ
Let λ be a partition and Gλ be the associated graph. We construct the set of polynomials
(JP)P path in Gλ using the following recursive rules:
1. J[] := (1⊗ τλ).
2. If P = [a1, . . . , ak−1, si] then
JP := J[a1,...,ak−1]
(
si ⊗ si +
1
ζ[i+ 1]− ζ[i]
)
,
where the vector ζ is defined by
(τλ,CTτλ , 0
N , [1, 2, . . . , N ])a1 . . . ak−1 = (τ, ζ, v, σ).
3. If P = [a1, . . . , ak−1,Ψ] then
JP = J[a1,...,ak−1]Ψ.
One has the following theorem.
Theorem 4.1. Let P = [a0, . . . , ak] be a path in Gλ from the root to (τ, ζ, v, σ). The polyno-
mial JP is a simultaneous eigenfunctions of the operators ξ˜i whose leading monomial is xv,τ .
Furthermore, the eigenvalues of ξ˜i associated to JP are equal to ζ[i].
Consequently JP does not depend on the path, but only on the end point (τ, ζ, v, σ), and will
be denoted by Jv,τ . The family (Jv,τ )v,τ forms a basis of Mλ of simultaneous eigenfunctions of
the Cherednik operators.
Furthermore, if P leads to ∅ then JP = 0.
Proof. We will prove the result by induction on the length k. If k = 0 then the result follows
from Proposition 3.13. Suppose now that k > 0 and let
(τ ′, ζ ′, v′, σv′) =
(
τλ,CTτλ , 0
N , [1, . . . , N ]
)
a1 · · · ak−1.
By induction, J[a1,...,ak−1] is a simultaneous eigenfunctions of the operators ξ˜i such that the
associated vector of eigenvalues is given by
J[a1,...,ak−1]ξ˜i = ζ
′[i]J[a1,...,ak−1]
and the leading monomial is xv
′,τ ′ .
If ak = Ψ is an affine arrow, then τ = τ ′, ζ = ζ ′.Ψα, v = v′Ψ, σv = σv′ [2, . . . , N, 1] and
JP = J[a1,...,ak−1]Ψ. If i 6= N
JPξ˜i = J[a1,...,ak−1]Ψξ˜i = J[a1,...,ak−1]ξ˜i+1Ψ = ζ
′[i+ 1]JP = ζ[i]JP.
Vector-Valued Jack Polynomials from Scratch 15
If i = N then,
JPξ˜N = J[a1,...,ak−1]Ψξ˜N = J[a1,...,ak−1](˜ξ1 + α)Ψ = (ζ
′[1] + α)JP = ζ[N ]JP.
The leading monomial is a consequence of Proposition 3.8.
Suppose now that ak = si is a non affine arrow, then ζ = ζ ′si, v = v′si and
JP = J[a1,...,ak−1]
(
si ⊗ si +
1
ζ ′[i+ 1]− ζ ′[i]
)
.
If j 6= i, i+ 1 then
JPξ˜j = J[a1,...,ak−1]
(
si ⊗ si +
1
ζ ′[i+ 1]− ζ ′[i]
)
ξ˜j
= J[a1,...,ak−1]ξ˜j
(
si ⊗ si +
1
ζ ′[i+ 1]− ζ ′[i]
)
= ζ ′[j]JP = ζ[j]JP.
If j = i then
JPξ˜i = J[a1,...,ak−1]
(
si ⊗ si +
1
ζ ′[i+ 1]− ζ ′[i]
)
ξ˜i
= J[a1,...,ak−1]
(
ξ˜i+1(si ⊗ si) + 1 + ξ˜i
1
ζ ′[i+ 1]− ζ ′[i]
)
= J[a1,...,ak−1]
(
ζ ′[i+ 1](si ⊗ si) + 1 +
ζ ′[i]
ζ ′[i+ 1]− ζ ′[i]
)
= ζ ′[i+ 1]J[a1,...,ak−1]
(
si ⊗ si +
1
ζ ′[i+ 1]− ζ ′[i]
)
= ζ[i]JP.
If j = i+ 1 then
JPξ˜i+1 = J[a1,...,ak−1]
(
si ⊗ si +
1
ζ ′[i+ 1]− ζ ′[i]
)
ξ˜i
= J[a1,...,ak−1]
(
ξ˜i(si ⊗ si)− 1 + ξ˜i+1
1
ζ ′[i+ 1]− ζ ′[i]
)
= J[a1,...,ak−1]
(
ζ ′[i](si ⊗ si)− 1 +
ζ ′[i+ 1]
ζ ′[i+ 1]− ζ ′[i]
)
= ζ ′[i]J[a1,...,ak−1]
(
si ⊗ si +
1
ζ ′[i+ 1]− ζ ′[i]
)
= ζ[i+ 1]JP.
Let us examine the leading monomials. First, suppose that ak = si is a step then τ = τ ′
and σv = σv′si. From Proposition 3.8, the leading monomial in JP equals the leading term in
xv
′τ ′
(
si ⊗ si + 1ζ′[i+1]−ζ′[i]
)
that is xv
′si ⊗ (τ ′σv′si) = xv,τ .
Suppose that ak = si is not a step and set Q := xv
′,τ ′
(
si ⊗ si + 1ζ′[i+1]−ζ′|i]
)
. One has
JP = Q+
∑
xv′′,τ ′′
xv,τ
αv′′,τ ′′x
v′′,τ ′′ .
If ak = si is a jump then τ = τ ′
(σv′ [i],σv′ [i]+1) and σv = σ′v. But
Q = xv
′si ⊗ (τ ′σv′si) +
1
ζ ′[i+ 1]− ζ ′[i]
xv
′
⊗ (τ ′σv′)
16 C.F. Dunkl and J.-G. Luque
= xv ⊗ (τ ′sσv′ [i]σv′) +
1
ζ ′[i+ 1]− ζ ′[i]
xv
′
⊗ (τ ′σv′)
= xv ⊗ (τσv) +
(
b′τ [σv′ [i]] +
1
ζ ′[i+ 1]− ζ ′[i]
)
xv
′
⊗ (τ ′σv′)
= xv,τ +
(
b′τ [σv′ [i]] +
1
ζ ′[i+ 1]− ζ ′[i]
)
xv
′,τ ′ .
But ζ ′[i] = CTτ ′ [σv′ [i]] and ζ ′[i + 1] = CTτ ′ [σv′ [i + 1]] = CTτ ′ [σv′ [i] + 1], hence bτ ′ [σv′ [i]] =
− 1ζ′[i+1]−ζ′[i] . And the leading monomial is Q = x
v,τ as expected.
This proves the first part of the theorem and that the family (Jv,τ )v,τ forms a basis of Mλ of
simultaneous eigenfunctions of the Cherednik operators.
Finally, if ak = si is a fall, Q is proportional to xv
′
⊗ (τ ′σv′) and then JP is proportional to
J[a1,...,ak−1]. But clearly, the two polynomials are eigenfunction of the Cherednik operators with
different eigenvalues from the cases j = i and j = i+ 1. This proves that JP = 0. 
As a consequence, we will consider the family of polynomials (Jv,τ )v,τ indexed by pairs (v, τ)
where v ∈ NN is a weight and τ is a tableau.
Example 4.2. For λ = 21, the first polynomials Jv,τ are displayed in Fig. 2. The spectral
vectors can be read on Fig. 1.
Note that if [a1, . . . , ak−1] leads to a vertex other than ∅ and [a1, . . . , ak−1, si] leads to ∅,
the last part of Theorem 4.1 implies that J[a1,...,ak−1] is symmetric or antisymmetric under the
action of si ⊗ si.
The recursive rules of this section first appeared in [8]. The Lemma 5.3 and the Yang–Baxter
graph constitute essentially what Griffeth called calibration graph in that paper.
4.2 Partial Yang–Baxter-type construction associated to Gτ
To compute an expression for a polynomial Jv,τ it suffices to find the good path in the sub-
graph Gτ as shown by the following examples.
Example 4.3. Consider τ =
1
3 2
, Fig. 3 explains how to obtains the values of Jv, 132
from the
graph G 1
32
.
Example 4.4. For the trivial representation (i.e., λ has a single part), note that the Cherednik
operators (in [14]) have the same eigenspaces as the Cherednik–Dunkl operators Ui (in [4]). In
the notations of [14], ξi reads
ξi = αxi
∂
∂xi
+
N∑
j=1
j 6=i
piij + (1− i),
where
piij =
{
xi∂ij if j < i,
xj∂ij if i < j,
where ∂ij denotes the divided difference on the variables xi and xj . Noting that xi∂ij = ∂ijxi−1,
xj∂ij = ∂ijxi − (ij) and xi ∂∂xi =
∂
∂xi
xi − 1, one finds
ξi = αUi − (α+N − 1) = ξ˜i − (N − 1) .
Vector-Valued Jack Polynomials from Scratch 17
J000, 231
J
001, 231
J
010, 231
J
100, 231
s 2
+
1
α
+
1
s 1
+
1
α
+
2
Ψ
J
011, 231
J
101, 231
J
002, 231
Ψ
Ψ
Ψ
s 1
+
1
α
+
1
J
110, 231
s
2 +
1α
−
1
J
020, 231
J
200, 231
s
2 +
12α
+
1
s
1 +
12α
+
2
J
000, 132
J
001, 132
J
010, 132
J
100, 132
J
011, 132
J
101, 132
J
002, 132
J
110, 132
J
020, 132
J
200, 132
s1 +
1
2
s
2 +
1α
−
1
s
1 +
1α
−
2
Ψ
Ψ
Ψ
Ψ
s2 +
1
2
s
1 +
1α
−
1
Ψ
s 2
+
1
α
+
1
s1 +
1
2
s 2
+
1
2α
−
1
s 1
+
1
2α
−
2
Figure 2. First values of the polynomials Jv,τ for λ = 21 (si means si ⊗ si).
Example 4.5. Consider sign representation associated to the partition [1N ]. The set Tab[1N ]
contains a unique element τ =
1
...
N
. Hence, we can omit τ when we write the polynomials
of M[1N ]. One can see that the corresponding Jack polynomials are equal to the standard ones
for the coefficient −α. Indeed, since τsij = −τ one has
PDi ' (P ⊗ τ)Di = (P ⊗ τ)


∂
∂xi
+
1
α
∑
i 6=j
∂ij ⊗ si,j

= (P ⊗ τ)


∂
∂xi
−
1
α
∑
i 6=j
∂ij ⊗ 1

 .
18 C.F. Dunkl and J.-G. Luque
J000, 132
J
001, 132
J
010, 132
J
100, 132
s 2
+
1
α
−
1
s 1
+
1
α
−
2
Ψ
J
011, 132
J
101 132
J
002, 132
Ψ
Ψ
Ψ
s 1
+
1
α
−
1
J
110, 132
s
2 +
1α
+
1
J
020, 132
J
200, 132
s
2 +
12α
+
1
s
1 +
12α
+
2
Figure 3. First values of the polynomials Jv, 132 .
Hence, the Cherednik–Dunkl operator Ui = xiDi− 1α
i−1∑
j=1
sij⊗sij acts on M[1N ] as the operator Ui
acts on M[N ] but for the parameter −α.
Example 4.6. Let us explain the method on a bigger example: J[0,0,2,1,1,0],τ , for τ :=
4 3
6 5 2 1
.
First, we obtain the vector [0, 0, 2, 1, 1, 0] from [0, 0, 0, 0, 0, 0] by the following sequence of opera-
tions:
[0, 0, 0, 0, 0, 0]
Ψ
→ [0, 0, 0, 0, 0, 1]
s5→ [0, 0, 0, 0, 1, 0]
s4→ [0, 0, 0, 1, 0, 0]
s3→ [0, 0, 1, 0, 0, 0]
s2→ [0, 1, 0, 0, 0, 0]
s1→ [1, 0, 0, 0, 0, 0]
Ψ
→ [0, 0, 0, 0, 0, 2]
s5→ [0, 0, 0, 0, 2, 0]
Ψ
→ [0, 0, 0, 2, 0, 1]
Vector-Valued Jack Polynomials from Scratch 19
s5→ [0, 0, 0, 2, 1, 0]
Ψ
→ [0, 0, 2, 1, 0, 1]
s5→ [0, 0, 2, 1, 1, 0].
Replace Ψ by Ψα in the list of the operations, the associated sequence is
ζ[0,0,0,0,0,0] = [3, 2, 0,−1, 1, 0]
Ψα
→ ζ[0,0,0,0,0,1] = [2, 0,−1, 1, 0, α+ 3]
s5→ ζ[0,0,0,0,1,0] = [2, 0,−1, 1, α+ 3, 0]
s4→ ζ[0,0,0,1,0,0] = [2, 0,−1, α+ 3, 1, 0]
s3→ ζ[0,0,1,0,0,0] = [2, 0, α+ 3,−1, 1, 0]
s2→ ζ[0,1,0,0,0,0] = [2, α+ 3, 0,−1, 1, 0]
s1→ ζ[1,0,0,0,0,0] = [α+ 3, 2, 0,−1, 1, 0]
Ψα
→ ζ[0,0,0,0,0,2] = [2, 0,−1, 1, 0, 2α+ 3]
s5→ ζ[0,0,0,0,2,0] = [2, 0,−1, 1, 2α+ 3, 0]
Ψα
→ ζ[0,0,0,2,0,1] = [0,−1, 1, 2α+ 3, 0, α+ 2]
s5→ ζ[0,0,0,2,1,0] = [0,−1, 1, 2α+ 3, α+ 2, 0]
Ψα
→ ζ[0,0,2,1,0,1] = [−1, 1, 2α+ 3, α+ 2, 0, α]
s5→ ζ[0,0,2,1,1,0] = [−1, 1, 2α+ 3, α+ 2, α, 0].
Now, to obtain the vector-valued Jack polynomial, it suffices to start from 1⊗
4 3
6 5 2 1
and act successively with the affine operator Ψ (when reading Ψα) and with si ⊗ si + 1ζ[i+1]−ζ[i]
(when reading si).
In conclusion, the computation of vector-valued Jack for a given RST is completely indepen-
dent of the computations of the vector-valued Jack indexed by the other RST with the same
shape.
4.3 Normalization
The space Vλ spanned by the RST τ of the same shape λ is naturally endowed (up to a mul-
tiplicative constant) by SN -invariant scalar product 〈 , 〉0 with respect to which the RST are
pairwise orthogonal. As in [4], we set
||τ ||2 =
∏
1≤i<j≤N
CTτ [i]<CTτ [j]−1
(CTτ [i]− CTτ [j]− 1)(CTτ [i]− CTτ [j] + 1)
(CTτ [i]− CTτ [j])2
.
As in [4], we consider the contravariant form 〈 , 〉 on the space Mλ which is the symmetric
SN -invariant form extending 〈 , 〉0 and such that the Dunkl operator Di is the adjoint to the
multiplication by xi (see appendix in [4] for more details).
The operator xiDi is self adjoint and the adjoint of σ ∈ SN is σ−1. Since sij = s−1ij is self
adjoint, Ui is self-adjoint for the form 〈 , 〉 and the polynomials Jv,τ are pairwise orthogonal.
Let us compute their squared norms ||Jv,τ ||2 (the bilinear form is nonsingular for generic α
and positive definite for α in some subset of R [6]). The method is essentially the same as in [4]
and we show that the result can be read in the Yang–Baxter graph. More precisely, one has
Proposition 4.7.
1. ||J(v,τ)si ||
2 =
(ζv,τ [i+ 1]− ζv,τ [i]− 1)(ζv,τ [i+ 1]− ζv,τ [i] + 1)
(ζv,τ [i+ 1]− ζv,τ [i])2
||Jv,τ ||
2.
2. ||J(v,τ)Ψ||
2 =
(
1
α
ζv,τ [1] + 1
)
||Jv,τ ||
2.
Proof. 1. Since
J(v,τ)si = Jv,τ
(
si ⊗ si +
1
ζv,τ [i+ 1]− ζv,τ [i]
)
,
20 C.F. Dunkl and J.-G. Luque
J000, 132
J
001, 132
J
010, 132
J
100, 132
×α(α+2)
(α+1)2
× (α+1)(α+3)
(α+2)2
×
(
1
α + 1
)
J
011, 132
J
101 132
J
002, 132
×
(
− 1α + 1
)
×
(
− 1α + 1
)
×
(
1
α + 2
)
×α(α+2)
(α+1)2
J
110, 132
× (α−2)α
(α−1)2
J
020, 132
J
200, 132
× 2α(2α+2)
(2α+1)2
× (2α+1)(2α+3)
(2α+2)2
Figure 4. Computation of ||J020, 132 ||
2 using the graph G 1
32
.
one obtains
||Jv,τ (si ⊗ si)||
2 = ||J(v,τ)si ||
2 +
1
(ζv,τ [i+ 1]− ζv,τ [i])2
||Jv,τ ||
2.
But ||Jv,τ (si ⊗ si)||2 = ||Jv,τ ||2, which gives the result.
2. One has
||J(v,τ)Ψ||
2 = ||Jv,τ (θ ⊗ θ)xN ||
2 = 〈Jv,τ (θ ⊗ θ), Jv,τ (θ ⊗ θ)xNDN 〉
= 〈Jv,τ (θ ⊗ θ), Jv,τx1D1(θ ⊗ θ)〉 = 〈Jv,τ , Jv,τx1D1〉;
recall that θ = s1s2 · · · sN−1. Since U1 = x1D1, one obtains the results. 
Example 4.8. Let again τ =
2
3 1
, we compute the normalization following the Yang–Baxter
graph (see Fig. 4).
For instance:
||J[020],τ ||
2 = ||τ ||2
(
1 +
1
α
)(
α(α+ 2)
(α+ 1)2
)(
(α+ 1)(α+ 3)
(α+ 2)2
)(
2 +
1
α
)(
2α(2α+ 2)
(2α+ 1)2
)
=
4(α+ 3)(α+ 1)
(2α+ 1)(α+ 2)
.
Vector-Valued Jack Polynomials from Scratch 21
5 Symmetrization and antisymmetrization
In [2], Baker and Forrester investigated the coefficients and the norm of the symmetric Jack
polynomials by symmetrizing the nonsymmetric Jack polynomials. The symmetrization method
was used in [5] for the polynomials associated with the complex groups G(r, p,N). In this
section, we generalize their results and obtain symmetric and antisymmetric vector-valued Jack
polynomials.
5.1 Non-affine connectivity
Let us denote by Hλ the graph obtained from Gλ by removing the affine edges, all the falls and
the vertex ∅. The purpose of this section is to investigate the connected components of Hλ.
Recall that v+ is the unique decreasing partition obtained by permuting the entries of v.
Definition 5.1. Let v ∈ NN and τ ∈ Tabλ (λ partition). We define the filling T (τ, v) obtained
by replacing i by v+[i] in τ for each i.
Proposition 5.2. Two 4-tuples (τ, ζ, v, σ) and (τ ′, ζ ′, v′, σ′) are in the same connected compo-
nent of Hλ if and only if T (τ, v) = T (τ ′, v′).
Proof. Remark first that the steps and correct jumps preserve T (τ, v). Indeed steps leave
invariant the pairs (τ, v+) whilst the correct jumps act on the RST by τsσv [i] where v[i] = v[i+1]
(or equivalently by τsj where v+[j] = v+[j + 1]. Hence, we show that if (τ, ζ, v, σ) is connected
to (τ ′, ζ ′, v′, σ′) then T (τ, v) = T (τ ′, v′).
Let us prove the converse. Suppose that T (τ, v) = T (τ ′, v′). Since (τ, ζ, v, σ) (resp. (τ ′, ζ ′,
v′, σ′)) is connected to (τ, ζ, v+, Id) (resp. (τ ′, ζ ′, v′+, Id)) by steps, it suffices to prove the result
for T (τ, µ) = T (τ ′, µ) when µ is a (decreasing) partition. Let ρ ∈ SN such that τρ (the tableau τ
where the entries have been permuted by ρ) equals τλ. By construction T (τ, µ) = T (τλ, µρ−1)
and then it suffices to show the result for T (τλ, µρ−1) = T (τ ′
ρ, µρ−1). Again the connectivity by
steps implies that it suffices to prove the result for T (τλ, µ) = T (τ ′, µ) when µ is a partition. We
will show that if T (τλ, µ) = T (τ, µ) then there exists a series of correct jumps from (τ, ζ ′, µ, Id)
to (τλ, ζ, µ, Id) when µ is a partition. We prove the result by induction on the length of the
shortest permutation ω such that τω = τλ for the weak order. The base point of the induction
is straightforward. Now, choose i such that ω[i] > ω[i + 1] and i and i + 1 are neither in the
same row nor in the same column in τ then ω = siω′ where `(ω′) < `(ω). Since T (τ, µ) =
T (τ ′, µ), this means that µ[i] = µ[i+ 1] and hence, there is a correct jump from (τ, ζ ′, µ, Id) to
(τ (i,i+1), ζ ′si, µ, Id). By the induction hypothesis, this shows the result. 
This shows that the connected components of Hλ are indexed by the T (τ, µ) where µ is
a partition.
Definition 5.3. We will denote by HT the connected component associated to T in Hλ. The
component HT will be said to be 1-compatible if T is a column-strict tableau. The component HT
will be said to be (−1)-compatible if T is a row-strict tableau.
Example 5.4. Let µ = [2, 1, 1, 0, 0] and λ = [3, 2]. There are four connected components with
vertices labeled by permutations of µ in Hλ (see Fig. 5). The possible values of T (τ, µ) are
12
001
,
02
011
,
01
012
and
11
002
,
squared in red in Fig. 5. The 1-compatible components are H 12
001
and H 11
002
while there is only
one (−1)-compatible component H 01
012
. The component H 02
011
is neither 1-compatible nor (−1)-
compatible.
The component H 12
001
contains vertices of G 31
542
and G 21
543
connected by jumps.
22 C.F. Dunkl and J.-G. Luque
41
532
[00112]
42
531
[00112]
32
541
[00112]
41
532
[00121]
41
532
[01012]
02
011
41
532
[00211]
41
532
[01021]
41
532
[10012]
s
4 s 2
s
3 s 2
s
4 s 1
42
531
[00121]
42
531
[01012]
01
012
42
531
[00211]
42
531
[01021]
42
531
[10012]
s
4 s 2
s
3 s 2
s
4 s 1
11
002
32
541
[00121]
32
541
[01012]
32
541
[00211]
32
541
[01021]
32
541
[10012]
s
4 s 2
s
3 s 2
s
4 s 1
31
542
[00112]
21
543
[00112]
31
542
[00121]
31
542
[01012]
31
542
[00211]
31
542
[01021]
31
542
[10012]
12
001
21
543
[00121]
21
543
[01012]
21
543
[00211]
21
543
[01021]
21
543
[10012]
s
4 s 2
s
3 s 2
s
4 s 1
s
4 s 2
s
3 s 2
s
4 s 1
s3
s4
. . . . . .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . .
. . . . . .. . . . . . . . . . . . . . . . . .. . . . . . . . . . . .
Figure 5. Some connected components of H32.
We we use the following result in the sequel, its proof is easy and left to the reader.
Proposition 5.5. Let (τ, ζ, v, σ) be a vertex of HT such that (τ, ζ, v, σ)si = ∅. One has
1. If HT is 1-compatible then σ[i] and σ[i+ 1](= σ[i] + 1) are in the same row in τ .
2. If HT is (−1)-compatible then σ[i] and σ[i+ 1](= σ[i] + 1) are in the same column in τ .
The following definition is used to find a RST corresponding to a filling of a shape.
Definition 5.6. Let T be a filling of shape λ, the standardization std(T ) of T is the reverse
standard tableau with shape λ obtained by the following process:
1. Denote by |T |i the number of occurrences of i in T
2. Read the tableau T from the left to the right and the bottom to the top and replace
successively each occurrence of i by the numbers N − |T |0 − · · · − |T |i−1, N − |T |0 − · · · −
|T |i−1 − 1, . . . , N − |T |0 − · · · − |T |i.
Alternatively, one has
std(T ) [i, j] := # {(k, l) : T [k, l] > T [i, j]}+ # {(k, l) : l > j, T [k, l] = T [i, j]}
+ # {(k, j) : k ≥ i, T [k, j] = T [i, j]} .
We will denote by λT the unique partition obtained by sorting in the decreasing order all the
entries of T .
Vector-Valued Jack Polynomials from Scratch 23
Example 5.7. Pictorially, reading 01002 one obtains
0 0 2 0 1
0 0 . 0 .
. . . . 1
. . 2 . .
Renumbering in increasing order from the bottom to the top and the right to the left, one reads
0 0 2 0 1
5 4 . 3 .
. . . . 2
. . 1 . .
Hence, we have std
( 01
002
)
= 32541 and λ 01002
= [21000].
Note that each HT has a unique sink (that is a vertex with no outward edge) and this vertex
is labeled by (std(T ), ζT , λT , Id) for a certain vector ζT and a unique root.
Example 5.8. Consider the tableau T = 0100 . Its standardization is std(T ) =
21
43 and the
graph HT is:
31
42
[0001]
31
42
[0010]
31
42
[0100]
31
42
[1000]
21
43
[0001]
21
43
[0010]
21
43
[0100]
21
43
[1000]
s 1 s 1 s 2
s3 s2 s1
s3 s2 s1
The sink is denoted by a red disk and the root by a green disk.
5.2 Symmetric and antisymmetric Jack polynomials
For convenience, let us define:
(v, τ)si = (v
′, τ ′) if (τ, ζ, v, σ)si = (τ
′, ζ ′, v′, σ′)
and
(v, τ)si = ∅ if (τ, ζ, v, σ)si = ∅.
Denote also, J∅ := 0.
Let (τ, ζ, v, σ) be a vertex of HT , set bv,τ [i] = 1ζv,τ [i+1]−ζv,τ [i] and cv,τ [i] =
ζv,τ [i]−ζv,τ [i+1]
ζv,τ [i]−ζv,τ [i+1]+1
.
Note that
1 + cv,τ [i]bv,τ [i] = cv,τ [i] (5.1)
and
cv,τ [i](1− bv,τ [i]
2)− bv,τ [i] = 1. (5.2)
Let HT be a 1-compatible component of Gλ. For each vertex (τ, ζ, v, σ) of HT , we define the
coefficient Ev,τ by the following induction:
24 C.F. Dunkl and J.-G. Luque
1. Ev,τ = 1 if there is no arrows of the form
(τ ′, ζ′, v′, σ′) (τ, ζ, v, σ)si
in HT .
2. Ev,τ =
ζ′[i+1]−ζ′[i]
ζ′[i+1]−ζ′[i]−1Ev′,τ ′ =
ζ[i+1]−ζ[i]
ζ[i+1]−ζ[i]+1Ev′,τ ′ = cv′,τ ′Ev′,τ ′ if there is an arrow
(τ ′, ζ′, v′, σ′) (τ, ζ, v, σ)si
in HT .
The symmetric group acts on the spectral vectors ζ by permuting their components. Hence the
value of Ev,τ does not depend on the path used for its computation and the Ev,τ are well defined.
Indeed, it suffices to check that the definition is compatible with the commutations sisj = sjsi
with |i− j| > 1 and the braid relations sisi+1si = si+1sisi+1.
Let us first prove the compatibility with the commutation relations. Suppose
(τ0, ζ0, v0, σ0)sisj = (τ1, ζ1, v1, σ1)sj = (τ2, ζ2, v2, σ2),
with |i− j| > 1 and
(τ0, ζ0, v0, σ0)sjsi = (τ
′
1, ζ
′
1, v
′
1, σ
′
1)si = (τ
′
2, ζ
′
2, v
′
2, σ
′
2).
Note that τ ′2 = τ2, ζ
′
2 = ζ2, v
′
2 = v2 and σ
′
2 = σ2. But, since the symmetric group acts on ζ by
permuting its components, one has
ζ2[j + 1] = ζ
′
1[j + 1], ζ2[j] = ζ
′
1[j], ζ1[j + 1] = ζ
′
2[j + 1] and ζ1[j] = ζ
′
2[j].
Hence,
ζ2[j + 1]− ζ2[j]
ζ2[j + 1]− ζ2[j] + 1
.
ζ1[i+ 1]− ζ1[i]
ζ1[i+ 1]− ζ1[i] + 1
=
ζ1[i+ 1]− ζ1[i]
ζ1[i+ 1]− ζ1[i] + 1
.
ζ2[j + 1]− ζ2[j]
ζ2[j + 1]− ζ2[j] + 1
=
ζ ′1[i+ 1]− ζ
′
1[i]
ζ ′1[i+ 1]− ζ
′
1[i] + 1
.
ζ ′2[j + 1]− ζ
′
2[j]
ζ ′2[j + 1]− ζ
′
2[j] + 1
,
and the definition of Ev,τ is compatible with the commutations.
Now, let us show that the definition is compatible with the braid relations and set
(τ0, ζ0, v0, σ0)sisi+1si = (τ1, ζ1, v1, σ1)si+1si = (τ2, ζ2, v2, σ2)si = (τ3, ζ3, v3, σ3),
and
(τ0, ζ0, v0, σ0)si+1sisi+1 = (τ
′
1, ζ
′
1, v
′
1, σ
′
1)sisi+1 = (τ
′
2, ζ
′
2, v
′
2, σ
′
2)si+1 = (τ
′
3, ζ
′
3, v
′
3, σ
′
3).
Note that τ ′3 = τ3, ζ
′
3 = ζ3, v
′
3 = v3 and σ
′
3 = σ3. Since the symmetric group acts on ζ by
permuting its components, one has
ζ3[i+ 1] = ζ
′
1[i+ 1], ζ3[i] = ζ
′
1[i+ 1],
ζ2[i+ 2] = ζ
′
2[i+ 1], ζ2[i+ 1] = ζ
′
2[i],
ζ1[i+ 1] = ζ
′
2[i+ 2] and ζ1[i] = ζ
′
3[i+ 1].
Vector-Valued Jack Polynomials from Scratch 25
Hence,
ζ3[i+ 1]− ζ3[i]
ζ3[i+ 1]− ζ3[i] + 1
.
ζ2[i+ 2]− ζ2[i+ 1]
ζ2[i+ 2]− ζ2[i+ 1] + 1
.
ζ1[i+ 1]− ζ1[i]
ζ1[i+ 1]− ζ1[i] + 1
=
ζ1[i+ 1]− ζ1[i]
ζ1[i+ 1]− ζ1[i] + 1
.
ζ2[i+ 2]− ζ2[i+ 1]
ζ2[i+ 2]− ζ2[i+ 1] + 1
.
ζ3[i+ 1]− ζ3[i]
ζ3[i+ 1]− ζ3[i] + 1
=
ζ ′3[i+ 2]− ζ
′
3[i+ 1]
ζ ′3[i+ 2]− ζ
′
3[i+ 1] + 1
.
ζ ′2[i+ 1]− ζ
′
2[i]
ζ ′2[i+ 1]− ζ
′
2[i] + 1
.
ζ ′1[i+ 2]− ζ
′
1[i+ 1]
ζ ′1[i+ 2]− ζ
′
1[i+ 1] + 1
,
and the definition is compatible with the braid relations.
Define the symmetrization operator
S :=
∑
ω∈SN
ω ⊗ ω.
We will say that a polynomial is symmetric if it is invariant under the action of si ⊗ si for
each i < N .
Theorem 5.9.
1. Let HT be a connected component of Gλ. For each vertex (τ, ζ, v, σ) of HT , the polyno-
mial Jv,τS equals JλT ,std(T )S up to a multiplicative constant.
2. One has JλT ,std(T )S 6= 0 if and only if HT is 1-compatible.
3. More precisely, when HT is 1-compatible, the polynomial
JT =
∑
(τ,ζ,v,σ) vertex of HT
Ev,τJv,τ
is symmetric.
Proof. 1. Let us prove the first assertion by induction on the length of a path from (τ, ζ, v, σ)
to (std(T ), ζT , λT , σ) in HT . Let (τ ′, ζ ′, v′, σ′) such that
(τ, ζ, v, σ) (τ ′, ζ′, v′, σ′)si
is not a jump in HT (hence, −1 < bv,τ [i] < 1). It follows that
Jv,τS =
1
1− bv,τ [i]2
(
Jv′,τ ′(si ⊗ si) + bv,τ [i]Jv′,τ ′
)
S =
1
1− bv,τ [i]2
(1 + bv,τ [i]) Jv′,τ ′S.
By induction Jv′,τ ′S is proportional to JλT ,std(T ), which ends the proof.
2. If HT is not 1-compatible, then there exists si such that JλT ,std(T )(si ⊗ si) = −JλT ,std(T ).
Hence, since S = (si ⊗ si)S, one obtains JλT ,std(T )S = 0.
3. Let us prove that, when HT is 1-compatible, JT (si ⊗ si) = JT for any i. Fix i and
decompose JT := J+ + J0 + J− where
J+ =
∑
τ,v
+
Eτ,vJv,τ ,
where
∑+ means that the sum is over the pairs (τ, v) such that there exists an arrow
(τ ′, ζ′, v′, σ′) (τ, ζ, v, σ)si
26 C.F. Dunkl and J.-G. Luque
in HT ,
J− =
∑
τ,v
−
Eτ,vJv,τ
where
∑− means that the sum is over the pairs (τ, v) such that there exists an arrow
(τ, ζ, v, σ) (τ ′, ζ′, v′, σ′)si
in HT and
J0 =
∑0
Eτ,vJv,τ ,
where
∑0 means that the sum is over the pairs (τ, v) such that there exists an arrow
(τ, ζ, v, σ) ∅si
in GT (equivalently there is no arrow from (τ, ζ, v, σ) labeled by si in HT ). Suppose that
(τ, ζ, v, σ) ∅si
is a fall in GT , then
Jv,τ (si ⊗ si) = J(v,τ)si − bv,τ [i]Jv,τ = −bv,τ [i]Jv,τ .
Since, HT is 1-compatible Proposition 5.5 implies that i and i+ 1 are in the same row. Hence,
bv,τ [i] = −1 and Jv,τ (si ⊗ si) = Jv,τ . It follows that J0(si ⊗ si) = J0.
Now, let
(τ, ζ, v, σ) (τ ′, ζ′, v′, σ′)si
be an arrow in HT , then
Jv,τ (si ⊗ si) = Jv′,τ ′ − bv,τ [i]Jv,τ ,
Jv′,τ ′(si ⊗ si) = bv,τ [i]Jv′,τ ′ + (1− bv,τ [i]
2)Jv,τ and Ev′,τ ′ = cv,τEv,τ .
Hence, equalities (5.1) and (5.2) imply
(Ev,τJv,τ + Ev′,τ ′Jv′,τ ′)(si ⊗ si) = Ev,τ (Jv,τ + cv,τ [i]Jv′,τ ′)(si ⊗ si)
= Ev,τ
(
((cv,τ [i](1− bv,τ [i]
2)− bv,τ [i])Jv,τ + (1 + cv,τ [i]bv,τ [i])Jv′,τ ′
)
= Ev,τ (Jv,τ + cv,τ [i]Jv′,τ ′) = (Ev,τJv,τ + Ev′,τ ′Jv′,τ ′).
This proves that (J+ + J−)(si ⊗ si) = J+ + J−. Hence, JT (si ⊗ si) = JT for each i and JT is
symmetric. 
Example 5.10. Consider the graph H 11
00
21
43
[0011]
21
43
[0101]
21
43
[0110]
21
43
[1001]
21
43
[1010]
21
43
[1100]
s2
× αα−1
s
1× α
−
1α−
2
s 3
×
α
−
1
α
−
2
s
1× α
−
1α−
2
s 3
×
α
−
1
α
−
2
s2
×α−2α−3
Vector-Valued Jack Polynomials from Scratch 27
The polynomial
J 11
00
= J0011, 2143
+
α
α− 1
J0101, 2143
+
α
α− 2
J0110, 2143
+
α
α− 2
J1001, 2143
+
α(α− 1)
(α− 2)2
J1010, 2143
+
α(α− 1)
(α− 2)(α− 3)
J1100, 2143
is symmetric.
Let HT be a connected component, denote by root(T ) the only vertex of HT without inward
edge and by sink(T ) = (std(T ), ζT , λT , Id) the only vertex of HT without outward edge. De-
note by #HT the number of vertices of HT . The following proposition allows to compare the
polynomial JT to the symmetrization of Jroot(T ).
Proposition 5.11. One has
JT =
#HT
N !
Esink(T )Jroot(T )S.
Proof. It suffices to compare the coefficient of Jsink(T ) in JT and in Jroot(T ).S. The coefficient
of Jsink(T ) in JT equals Esink(T ) while the coefficient of Jsink(T ) in Jroot(T )S equals
N !
#H . Indeed
N !
#H is the order of the stabilizer of λT . The leading monomial of Jsink(T ) does not appear in any
other Jv,τ so its coefficient in the symmetrization of Jroot(T ) equals the order of the stabilizer. 
Let HT be a (−1)-compatible component of Gλ. For each vertex (τ, ζ, v, σ) of HT , we define
the coefficient Fv,τ by the following induction:
1. Fv,τ = 1 if there is no arrow of the form
(τ ′, ζ′, v′, σ′) (τ, ζ, v, σ)si
in HT .
2. Fv,τ = −
ζ[i]−ζ[i+1]
ζ[i]−ζ[i+1]+1Fv′,τ ′ = −
ζ′[i+1]−ζ[i]
ζ′[i+1]−ζ[i]+1Fv′,τ ′ if there is an arrow
(τ ′, ζ′, v′, σ′) (τ, ζ, v, σ)si
in HT .
Again the Fv,τ are well defined since the symmetric group acts on the spectral vectors by
permuting their components. Define also the antisymmetrization operator
A :=
∑
ω∈SN
(−1)`(ω)(ω ⊗ ω).
We will say that a polynomial is antisymmetric if it vanishes under the action of 1− si ⊗ si
for each i < N .
Theorem 5.12.
1. Let HT be a connected component of Gλ. For each vertex (τ, ζ, v, σ) of HT , the polyno-
mial Jv,τA equals JλT ,std(T )A up to a multiplicative constant.
2. One has JλT ,std(T )A 6= 0 if and only if HT is (−1)-compatible.
28 C.F. Dunkl and J.-G. Luque
3. More precisely, when HT is (−1)-compatible, the polynomial
J ′T =
∑
(τ,ζ,v,σ) vertex of HT
Fv,τJv,τ
is antisymmetric.
Example 5.13. Consider the graph H 01
01
31
42
[0011]
31
42
[0101]
31
42
[0110]
31
42
[1001]
31
42
[1010]
31
42
[1100]
s2
×− αα+1
s
1
×
−
α
+
1
α
+
2
s 3
×
−
α
+
1
α
+
2
s
1
×
−
α
+
1
α
+
2
s 3
×
−
α
+
1
α
+
2
s2
×−α+2α+3
The polynomial
J ′01
01
= J0011, 3142
−
α
α+ 1
J0101, 3142
+
α
α+ 2
J0110, 3142
+
α
α+ 2
J1001, 3142
−
α(α+ 1)
(α+ 2)2
J1010, 3142
+
α(α+ 1)
(α+ 2)(α+ 3)
J1100, 3142
is antisymmetric.
And, as in the symmetric case, one has:
Proposition 5.14. One has
JT =
#HT
N !
Fsink(T )Jroot(T ).A.
5.3 Normalization
As a consequence of Proposition 4.7, one deduces the following result using Theorems 5.9
and 5.12.
Corollary 5.15. Let HT be a connected component and (τ, ζ, v, σ) be a vertex of HT . Denote
by `Tτ,v the length of a path from root(T ) to (τ, ζ, v, σ). One has,
||Jv,τ ||
2 = (−1)`
T
τ,vE−1v,τF
−1
v,τ ||Jroot(T )||
2.
From Theorems 5.9 and 5.12, vector-valued symmetric and antisymmetric Jack polynomials
are also pairwise orthogonal.
Proposition 5.16.
1. Let HT1 and HT2 be two 1-compatible connected components. If T1 6= T2 then 〈JT1 , JT2〉 = 0.
Vector-Valued Jack Polynomials from Scratch 29
2. Let HT1 and HT2 be two (−1)-compatible connected components. If T1 6= T2 then
〈J ′T1 , J
′
T2〉 = 0.
Proof. It suffices to remark that from Theorem 5.9 (resp. Theorem 5.12) each JT (resp. J ′T )
is a linear combination of Jv,τ for (τ, ζ, v, σ) vertex in the connected component HT . 
In the special cases when HT is ±1-compatible, the value of ||JT ||2 admits a remarkable equality.
Proposition 5.17. One has:
1. If HT is a 1-compatible connected component then
||JT ||
2 = #HTEsink(T )||Jroot(T )||
2.
2. If HT is a (−1)-compatible connected component then
||J ′T ||
2 = #HTFsink(T )||Jroot(T )||
2.
Proof. The two cases being very similar, let us only prove the symmetric case. From Proposi-
tion 5.11, one has:
||JT ||
2 =
#HT
N !
Esink(T )〈JT , Jroot(T ).S〉 =
#HT
N !
Esink(T )
∑
σ∈SN
〈JT , Jroot(T )(σ ⊗ σ)〉
= #HTEsink(T )||Jroot(T )||
2. 
From Corollary 5.15 and Theorem 5.17, one obtains the surprising equalities:
Corollary 5.18. If HT is 1-compatible, one has:
∑
(τ,ζ,v,σ) vertex of HT
(−1)`
T
v,τ
Ev,τ
Fv,τ
= #HTEsink(T ). (5.3)
If HT is (−1)-compatible, one has:
∑
(τ,ζ,v,σ) vertex of HT
(−1)`
T
v,τ
Fv,τ
Ev,τ
= #HTFsink(T ).
Example 5.19. Consider the graph H 11
00
, the sum (5.3) gives
1 +
α+ 1
α− 1
(
1 +
α
α− 2
(
2 +
α
α− 2
(
1 +
α− 1
α− 3
)))
= 6
α (α− 1)
(α− 2) (α− 3)
as expected.
5.4 Symmetric and antisymmetric polynomials with minimal degree
Since the irreducible characters of SN are real it follows that the tensor product of an irreducible
module with itself contains the trivial representation exactly once. The tensor product of the
module corresponding to a partition λ with the module for tλ (the transpose) contains the sign
representation exactly once. We demonstrate these facts explicitly. Using the concepts from
Section 4.1 let
ζ1 =
∑
τ∈Tabλ
a (τ) (τ ⊗ τ) ∈ Vλ ⊗ Vλ
30 C.F. Dunkl and J.-G. Luque
be symmetric with (rational) coefficients a (τ) to be determined. We impose the conditions
ζ1(si⊗si) = ζ1 for i = 1, . . . , N −1. Fix some i and split the sum as suggested by equation (3.1)
ζ1 =
∑
bτ [i]=±1
a (τ) (τ ⊗ τ) +
∑
0<bτ [i]≤ 12
(
a (τ) (τ ⊗ τ) + a
(
τ (i,i+1)
)(
τ (i,i+1) ⊗ τ (i,i+1)
))
.
In the first sum (τ ⊗ τ) (si ⊗ si) = bτ [i]
2 (τ ⊗ τ) = τ ⊗ τ . For the second sum, note that
τ (i,i+1)si =
(
1− bτ [i]
2 )τ − bτ [i] τ (i,i+1). Simple computations show that
(
a (τ) (τ ⊗ τ) + a
(
τ (i,i+1)
)(
τ (i,i+1) ⊗ τ (i,i+1)
))
(si ⊗ si)
= a (τ) (τ ⊗ τ) + a
(
τ (i,i+1)
)(
τ (i,i+1) ⊗ τ (i,i+1)
)
exactly when a (τ) =
(
1 − bτ [i]
2 )a
(
τ (i,i+1)
)
. The unique (up to a constant multiple) SN -
invariant norm on Vτ satisfies
∥
∥τ (i,i+1)
∥
∥2 =
(
1 − bτ [i]
2 ) ‖τ‖2 (see Section 4.3); thus a (τ) =
c/ ‖τ‖2 for some constant c.
Consider the module Vtλ. The transpose map takes each RST τ with shape λ to the RST
tτ
of shape tλ. Thus btτ [i] = −bτ [i] for 1 ≤ i ≤ N . Suppose 0 < bτ [i] ≤
1
2 for some τ and i, then
−12 ≤ btτ [i] < 0 and the following transformation rules apply:
tτsi = btτ [i]
tτ +
(
1− btτ [i]
2 )(tτ (i,i+1)
)
,
tτ (i,i+1)si =
tτ − btτ [i]
tτ (i,i+1).
Let
ζdet =
∑
τ∈Tabλ
a (τ)
(tτ
)
⊗ τ ∈ Vtλ ⊗ Vλ
be antisymmetric with (rational) coefficients a (τ) to be determined. We impose the conditions
ζdet(si ⊗ si) = −ζdet for i = 1, . . . , N − 1. Fix some i and write
ζdet =
∑
bτ [i]=±1
a (τ)
(tτ
)
⊗ τ +
∑
0<bτ [i]≤ 12
(
a (τ)
(tτ
)
⊗ τ + a
(
τ (i,i+1)
)(t
τ (i,i+1)
)
⊗ τ (i,i+1)
)
.
In the first sum
(
τ ⊗
(
tτ
))
(si ⊗ si) = bτ [i] btτ [i]
(
tτ
)
⊗ τ = −
(
tτ
)
⊗ τ . We find that
(
a (τ)
(tτ
)
⊗ τ + a
(
τ (i,i+1)
)(t
τ (i,i+1)
)
⊗ τ (i,i+1)
)
(si ⊗ si)
= −
(
a (τ)
(tτ
)
⊗ τ + a
(
τ (i,i+1)
)(t
τ (i,i+1)
)
⊗ τ (i,i+1)
)
exactly when a (τ) = −a
(
τ (i,i+1)
)
.
Thus a (τ) = c (−1)inv(τ) (recall inv (τ) = # {(i, j) : 1 ≤ i < j ≤ N, rw (i, τ) > rw (j, τ)}, and
0 < bτ [i] ≤ 12 implies inv
(
τ (i,i+1)
)
= inv (τ) + 1).
We can now write down the symmetric and antisymmetric Jack polynomials of lowest degree,
by replacing the first factors in ζ1 and ζdet by the corresponding polynomials Pτ (x) and Ptτ (x)
(as constructed in Section 3). Let l = ` (λ) = tλ [1].
In the symmetric case let v =
[
(l − 1)λ[l] , (l − 2)λ[l−1] , . . . , 1λ[2], 0λ[1]
]
(using exponents to
indicate the multiplicity of an entry) The corresponding tableau is
T1 :=
l − 1 . . . l − 1 (λ[l]×)
...
. . .
...
1 . . . . . . . . . 1 (λ[2]×)
0 . . . . . . . . . . . . 0 (λ[1]×)
and std (T1) contains the numbers N,N − 1, . . . , 2, 1 entered row-by-row.
Vector-Valued Jack Polynomials from Scratch 31
Example 5.20. If λ = [4, 3, 2] then v = [221110000]. The corresponding tableau is
T1 =
2 2
1 1 1
0 0 0 0
and std(T1) =
2 1
5 4 3
9 8 7 6
.
In
ζ1 (x) =
∑
τ∈Tabλ
c
‖τ‖2
Pτ (x)⊗ τ,
the monomial xv occurs only when τ = std (T1), with coefficient c/ ‖std (T1)‖
2. This polynomial
is a multiple of JT1 (see Theorem 5.9).
For the antisymmetric case let
Tdet :=
0 1 . . . λl − 1
...
...
...
. . .
0 1 . . . λl − 1 . . . λ2 − 1
0 1 . . . λl − 1 . . . λ2 − 1 . . . λ1 − 1
Thus std (Tdet) = τλ and v =
[
(λ [1]− 1)
tλ[λ[1]] , (λ [1]− 2)
tλ[λ[1]−1] , . . . , 0
tλ[1]
]
.
Example 5.21. If λ = [4, 3, 2] then tλ = [3, 3, 2, 1] and v = [322111000]. The corresponding
tableau is
Tdet =
0 1
0 1 2
0 1 2 3
and std(Tdet) =
7 4
8 5 2
9 6 3 1
= τ[4,3,2].
Let
ζdet (x) =
∑
τ∈Tabλ
(−1)inv(τ) Ptτ (x)⊗ τ.
The monomial xv occurs only in the term τ = τλ (see Definition 2.9). This polynomial is
a constant multiple of J ′Tdet (see Theorem 5.12).
We summarize the results of this section in the following theorem.
Theorem 5.22. The subspace of Mλ of the symmetric (resp. antisymmetric) polynomials with
minimal degree is spanned by only one generator: the symmetric (resp. antisymmetric) Jack
polynomial JT1 (resp. JTdet).
As a consequence one observes a remarkable property.
Corollary 5.23. The Jack polynomial JT1 (resp. JTdet) is equal to a polynomial which does
not depend on the parameter α multiplied by the global multiplicative constant Esink(T1) (resp.
Fsink(Tdet)).
Proof. The first part of the sentence is a consequence of Theorem 5.22 since the dimension of
the space is 1. The values of the multiplicative constants follow from Theorems 5.9 and 5.12
together with the fact that the coefficient of the leading terms in a Jack polynomials Jv,τ is 1
(see Theorem 4.1). 
Note also that T1 (resp. Tdet) is not the only tableau for which the corresponding symmetric
(resp. antisymmetric) Jack does not depend on α (up to a global multiplicative constant).
32 C.F. Dunkl and J.-G. Luque
Example 5.24. Consider the partition λ = [221] together with the vector v = [2, 1, 1, 0, 0]. The
corresponding symmetric Jack 1E
21100,
1
32
54
J 2
11
00
does not depend on α.
There are two symmetric Jack polynomials in degree 5: J 2
12
00
and J 3
11
00
. Note that the (non
minimal) polynomial 1E
22100,
1
32
54
J 2
12
00
does not depend on α whilst the parameter α appears in
1
E
31100,
1
32
54
J 3
11
00
even after simplifying the expression.
6 Restrictions
6.1 Restrictions on Yang–Baxter graphs
Consider the operator ↓
M
acting on the Yang–Baxter graphs Gλ by producing a new graph Gλ ↓
M
following the rules below:
1. Add all the possible edges of the form
(τ, ζ, [v[1], . . . , v[M ], 0, . . . , 0], σ) (τ, ζ′, [v[2], . . . , v[M ], v[1] + 1, 0, . . . , 0], σ′)Ψ′
More precisely, the action of Ψ′ on the 4-tuples is given by
Ψ′ = ΨsN−1 · · · sM .
2. Suppress the vertices labeled by (τ, ζ, v, σ) with v[i] 6= 0 for some i > M , with the associ-
ated inward and outward edges.
3. Relabel the remained vertices (τ, ζ, v, σ) ↓
M
:= (τ ↓
M
, ζ ↓
M
, v ↓
M
, σ ↓
M
) with
(a) τ ↓
M
is obtained from τ by removing the nodes labeled by M + 1, . . . , N . Note that
the shape of τ ↓
M
could be a skew partition.
(b) v ↓
M
= [v[1], . . . , v[M ]].
(c) σ ↓
M
= [σ[1], . . . , σ[M ]].
(d) ζ ↓
M
= [ζ[1]− CTτ [M ], . . . , ζ[M ]− CTτ [M ]].
4. Relabel by Ψ the edges labeled by Ψ′.
Example 6.1. Consider the partition λ = 21 and M = 2, the graph G21 in Fig. 1 with edges Ψ′
added. We obtain the graph G21 ↓
M
(Fig. 7) applying the other rules.
Definition 6.2. A RST τ has the property R(M) if the removal of the nodes labeled by
M + 1, . . . , N in τ produces a RST whose Ferrers diagram is a partition.
Example 6.3. The RST
5 2
7 3 1
8 6 4
has the property R(3) while the RST 231 does not have property R(2).
Vector-Valued Jack Polynomials from Scratch 33
2
31 ,[000],
[1,−1,0],123
1
32 ,[000],
[−1,1,0],123
s1
1
32 ,[001],
[1,0,α−1],231
1
32 ,[010],
[1,α−1,0],213
1
32 ,[100],
[α−1,1,0],123
2
31 ,[001],
[−1,0,α+1],231
2
31 ,[010],
[−1,α+1,0],213
2
31 ,[100],
[α+1,−1,0],123
s
2
s
1
s 2
s 1
Ψ Ψ
1
32 ,[011],
[0,α−1,α+1],312
2
31 ,[011],
[0,α+1,α−1],312
1
32 ,[101],
[α−1,0α+1],132
2
31 ,[101],
[α+1,0α−1],132
1
32 ,[002],
[1,0,2α−1],231
2
31 ,[002],
[−1,0,2α+1],231
Ψ
Ψ
Ψ
Ψ
Ψ
Ψ
s2
s
1 s 1Ψ
1
32 ,[110],
[α−1,α+1,0],123
2
31 ,[110],
[α+1,α−1,0],123
s 2
s
2
s1
1
32 ,[020],
[1,2α−1,0],213
1
32 ,[200],
[2α−1,1,0],123
2
31 ,[020],
[−1,2α+1,0],213
2
31 ,[200],
[2α+1,−1,0],123
s
2
s
1
s 2
s 1
Ψ
′
Ψ
′
Ψ
′
Ψ
′
Ψ
′
Ψ
′
Figure 6. The first vertices of the graph G21 with edges Ψ′ for M = 2.
Denote by Gτ the subgraph of Gλ whose root is τ . In particular, one has
Proposition 6.4. Let τ have the property R(M) and satisfy τ ↓
M
=τλ ↓
M
where λ ↓
M
denotes the Fer-
rers diagram of τ ↓
M
. The graph Gλ ↓
M
is identical to the subgraph Gτ ↓
M
of Gλ ↓
M
whose root is τ ↓
M
.
Proof. Obviously, since the Ferrers diagram of τ ↓
M
is a partition, all the spectral vectors ζ
labeling the vertices of Gτ ↓
M
are obtained by subtracting the same integer (that is CTτ [M ])
34 C.F. Dunkl and J.-G. Luque
2
1 ,[0],
[0,2],12
1
2 ,[00],
[−2,0],12
s1
1
2 ,[01],
[0,α−2],21
1
2 ,[10],
[α−2,0],12
2
1 ,[01],
[0,α+2],21
2
1 ,[10],
[α+2,0],12
s
1 s 1
1
2 ,[11],
[α−2,α],12
2
1 ,[11],
[α+2,α],12s1
1
2 ,[02],
[0,2α−2],21
1
2 ,[20],
[2α−2,0],12
2
1 ,[02],
[0,2α+2],21
2
1 ,[20],
[2α+2,0],12
s
1s 1
Ψ
Ψ
Ψ
Ψ
Ψ
Ψ
Figure 7. The first vertices of the graph G21↓
2
.
from the corresponding spectral vector in Gλ. It follows that the action of the si permutes the
components of the spectral vectors in Gτ ↓
M
.
Let v′ = (τ ′, ζ ′, [v′[1], . . . , v′[M ], 0, . . . , 0], σ′) be a vertex of Gτ . Let us prove by induction on
the length of a path from the root to v′ that
1. There is a vertex labeled by v′ ↓
M
:= (τ ′ ↓
M
, ζ ′ ↓
M
, [v′[1], . . . , v′[M ]], σ′ ↓
M
) in Gλ ↓
M
.
Vector-Valued Jack Polynomials from Scratch 35
2. If there is a non affine edge labeled by si with i < M from
v′′ = (τ ′′, ζ ′′, [v′′[1], . . . , v′′[M ], 0, . . . , 0], σ′′)
to v′ in Gτ then there is the same edge from v′′ ↓
M
to v′ ↓
M
in Gλ ↓
M
.
3. If there is an edge from v′′ = (τ ′′, ζ ′′, [v′′[1], . . . , v′′[M ], 0, . . . , 0], σ′′) to v′ in Gτ labeled by
Ψ′ then there is an edge labeled by Ψ from v′′ ↓
M
to v′ ↓
M
in Gλ ↓
M
.
First, observe that if τ ′ = τ and v′[i] = 0 for each i (i.e. v′ is a component of the label of the
root of Gτ ) then the construction gives, straightforwardly the result.
Suppose that there is a non affine edge
v′′ v′si
in Gτ . By induction v′′ ↓
M
labels a vertex of Gλ ↓
M
. We verify that
(v′′si) ↓
M
= (v′′) ↓
M
si = v′ ↓
M
Hence, v′ ↓
M
labels a vertex of Gλ ↓
M
and the assertion (2) is recovered.
Suppose now, that there is a affine edge
v′′ v′Ψ′
in Gτ . By induction v′′ ↓
M
labels a vertex of Gλ ↓
M
. We verify that
(v′′Ψ′) ↓
M
= (v′′ΨsM . . . sN ) ↓
M
= (v′′) ↓
M
Ψ = v′ ↓
M
.
Hence, v′ ↓
M
labels a vertex of Gλ ↓
M
and the assertion (3) is recovered.
Conversely, if v′ labels a vertex of Gλ ↓
M
, there exists a vertex labeled by v′(N) in Gτ veri-
fying v′(N) ↓
M
= v′. Indeed, suppose v′ = (τ ′, ζ ′, v′, σ′) then v′(N) = (τ ′(N), ζ ′(N), v′(N), σ′(N)),
where τ ′(N) is obtained from τ ′ by adding the nodes of τ labeled by M + 1, . . . , N , v′(N) =
[v′[1], . . . , v′[M ], 0, . . . , 0], ζ ′(N) = ζv′(N),τ ′(N)) and σ
′(N) = σv′(N) . Furthermore if v
′si = v′′ then
v′(N)si = v′′
(N) and if v′Ψ = v′′ then v′(N)Ψ′ = v′′(N). This concludes the proof. 
Example 6.5. Consider in Fig. 8 the restriction problem for
τ =
5 2
7 3 1
8 6 4
and M = 3. The subgraph of G 5 2
7 3 1
8 6 4
obtained using only the root and the arrows labeled
by Ψ′, s1 and s2 is isomorphic to the graph G 2
31
(see Fig. 1).
36 C.F. Dunkl and J.-G. Luque
52
731
864
,[000... ],
[1,−1,0,... ],123...
51
732
864
,[000... ],
[−1,1,0,... ],123...
s1
51
732
864
,[001... ],
[1,0,α−1,... ],231...
51
732
864
,[010... ],
[1,α−1,0,... ],213...
51
732
864
,[100... ],
[α−1,1,0,... ],123...
52
731
864
,[001... ],
[−1,0,α+1,... ],231...
52
731
864
,[010... ],
[−1,α+1,0,... ],213...
52
731
864
,[100... ],
[α+1,−1,0,... ],123...
s
2
s
1
s 2
s 1
Ψ
′ Ψ ′
51
732
864
,[011... ],
[0,α−1,α+1,... ],312...
52
731
864
,[011... ],
[0,α+1,α−1,... ],312...
51
732
864
,[101... ],
[α−1,0α+1,... ],132...
52
731
864
,[101... ],
[α+1,0α−1,... ],132...
51
732
864
,[002... ],
[1,0,2α−1,... ],231...
52
731
864
,[002... ],
[−1,0,2α+1,... ],231...
Ψ
′
Ψ
′
Ψ
′
Ψ
′
Ψ
′
Ψ
′
s2
s
1 s 1Ψ
′
51
732
864
,[110... ],
[α−1,α+1,0,... ],123...
52
731
864
,[110... ],
[α+1,α−1,0,... ],123...
s 2
s
2
s1
51
732
864
,[020... ],
[1,2α−1,0,... ],213...
51
732
864
,[200... ],
[2α−1,1,0,... ],123...
52
731
864
,[020... ],
[−1,2α+1,0,... ],213...
52
731
864
,[200... ],
[2α+1,−1,0,... ],123...
s
2
s
1
s 2
s 1
Figure 8. The first vertices of the subgraph of G332 associated to the restriction for M = 3 whose root
is 52731
864
with the edges Ψ′ added.
6.2 Restrictions on tableaux
In the sequel, as in [15], we will denote a skew partition by λ/µ.
Let τ be a RST of shape λ = [λ1, . . . , λk] and M < N . Consider Pτ as a polynomial
in C[t1, . . . , tM ][tM+1, . . . , tN ]. Let µ the sub-partition of λ which is the shape of the RST
obtained by removing the nodes labeled by 1, . . . ,M in τ and denote by τ (M) the associated
RST. Consider also the skew-RST τ ↓
M
of shape λ/µ obtained by removing the nodes labeled
Vector-Valued Jack Polynomials from Scratch 37
M + 1, . . . , N in τ . Let C(τ,M) := {ρ : ρ(M) = τ (M)} and T (τ,M) := {ρ ↓
M
: ρ ∈ C(ρ,M)}.
To each skew-RST ρ ↓
M
in T (τ,M) we associate the polynomial P ′
ρ ↓
M
which is the coefficient of
the monomial
N∏
i=M+1
t
λcl(i,ρ)−rw(i,ρ)
i in Pρ (recall that cl(i, τ) is the column of τ containing i and
rw(i, τ) is the row of τ containing i).
Example 6.6. Consider τ =
2
4 1
5 3
and M = 3. Then τ (3) = 45 , τ↓
3
=
2
1
3
,
C(τ, 3) =



3
4 1
5 2
,
2
4 1
5 3
,
1
4 2
5 3



and
T (τ, 3) =



3
1
2
,
2
1
3
,
1
2
3



.
The polynomial P ′
2
1
3
is the coefficient of t25t4 in P 2
4 1
5 3
.
Suppose that γ := λ/µ is a partition (τ ↓
M
is a RST). When i < M , since si does not act on
the variables tM+1, . . . , tN , the Murphy rules (equation 3.1) give
P ′
ρ ↓
M
si =



bρ[i]P ′ρ ↓
M
if bρ[i]2 = 1,
bρ[i]P ′ρ ↓
M
+ P ′
ρ(i,i+1) ↓
M
if 0 < bρ[i] ≤ 12 ,
bρ[i]P ′ρ ↓
M
+ (1− bρ[i]2)P ′ρ(i,i+1) ↓
M
otherwise.
(6.1)
Since the action si (i < M) on the RST commutes with the restriction ↓
M
, one has P ′
ρ(i,i+1) ↓
M
=
P ′
(ρ ↓
M
)(i,i+1)
and Proposition 3.1 and equation (6.1) imply that the polynomials P ′
ρ ↓
M
are simul-
taneous eigenfunctions of the Jucys–Murphy operators
ω(M)i =
M∑
i+1
sij ,
with eigenvalues CTρ ↓
M
[i]. Since the P ′
ρ ↓
M
for ρ ∈ C(τ,M) span a polynomial representation of
the symmetric group SM with minimal degree, the polynomials P ′ρ ↓
M
are equal up to a global
multiplicative coefficient to the polynomials Pρ ↓
M
.
To summarize:
Proposition 6.7. When τ ↓
M
is a RST, the coefficient of
N∏
i=M+1
t
λcl(i,τ)−rw(i,τ)
i in Pτ is propor-
tional to Pτ ↓
M
.
38 C.F. Dunkl and J.-G. Luque
Example 6.8. The coefficient of t29t
2
8t7t6 in P 51732
9864
equals
1
6
t1 −
1
12
t2 −
1
12
t3 =
1
6
P 1
32
.
6.3 Restrictions on Jack polynomials
Consider the linear map ↓
M
which sends each xi to 0 when i > M and each τ to τ ↓
M
.
Theorem 6.9. Let τ have property R(M). Then
J[v[1],...,v[M ],0,...,0],τ ↓
M
= J[v[1],...,v[M ]],τ ↓
M
.
Proof. From Proposition 6.4 the graphs Gτ ↓
M
and Gshape(τ ↓
M
) are the same. Remark that:
1. The result is correct for the roots:
J([0,...,0],τroot) ↓
M
= J[0,...,0],τroot ↓
M
where τroot denotes the RST obtained from τ by replacing subtableau constituted with the
nodes labeled 1, . . . ,M by τshape(τ ↓
M
).
2. The action of the edges are compatible with the restriction:
(a) The non-affine edges: We use only the difference between the content of two boxes.
Since, the values of the spectral vector ζ in Gτroot ↓
M
are obtained from the values of
the spectral vector in Gτroot by adding to each component the same integer. Hence,
the differences are the same and then the action of the non-affine edges are the same.
(b) The affine edges: One has to verify that
J[v[1],...,v[M ],0,...,0],τΨN (sN−1 ⊗ sN−1 + (∗)) · · · (sM ⊗ sM + (∗)) ↓
M
= J[v[1],...,v[M ],0,...,0],τ ↓
M
ΨM ,
where (∗) denote the correct rational numbers corresponding to the edges of the
Yang–Baxter graph and ΨM means that one applies the operator Ψ for an alphabet
of size M . This identity is easy to obtained from the construction: since Ψ gives
a polynomial whose a factor is xN , the only non vanishing part of
J[v[1],...,v[M ],0,...,0],τΨN (sN−1 ⊗ sN−1 + (∗)) · · · (sM ⊗ sM + (∗)) ↓
M
is
J[v[1],...,v[M ],0,...,0],τΨN (sN−1 · · · sM ⊗ sN−1 · · · sM ) ↓
M
= J[v[1],...,v[M ],0,...,0],τΨM ↓
M
,
the last part of the proof follows from the commutation ΨM ↓
M
= ↓
M
ΨM .
This shows that the polynomials J[v[1],...,v[M ],0,...,0],τ ↓
M
are inductively generated following the
same Yang–Baxter graph as the polynomials J[v[1],...,v[M ]],τ ↓
M
with the same initial conditions.
Hence J[v[1],...,v[M ],0,...,0],τ ↓
M
= J[v[1],...,v[M ]],τ ↓
M
as expected. 
Vector-Valued Jack Polynomials from Scratch 39
7 Shifted vector-valued Jack polynomials
7.1 Knop and Sahi operators for vector-valued polynomials
Let us define the following operators which are the vector-valued versions of the operators
defined in [9]:
1. ςi := ∂i ⊗ 1 + si ⊗ si, where ∂i := ∂i,i+1 = (1− si) 1xi−xi+1 is a divided difference.
2. Denote by Φ the operator sending each xi to xi−1 for i > 1 and x1 to xN − α and
T := Φ⊗ (s1s2 · · · sN−1).
3. Ψˆ := T (xN +N − 1).
Proposition 7.1. The operators ςi satisfy the braid relations
ςiςi+1ςi = ςi+1ςiςi+1, ςiςj = ςjςi, |i− j| > 1,
and the relations between the ςi and the multiplication by the indeterminates are given by the
Leibniz rules:
xiςi = ςi+1xi + 1, xi+1ςi = ςixi − 1, xjςi = ςixj , j 6= i, i+ 1.
Proof. The Leibniz rules are straightforward from the definition while the braid relations are
a direct consequence of
1) the braid relations on the si and the braid relations on the ∂i,
2) ∂i+1∂isi + si+1∂i∂i+1 = ∂isi+1∂i,
3) ∂i∂i+1si + si∂i+1∂i = ∂i+1si∂i+1,
4) si∂i+1si = si+1∂isi+1,
5) ∂isi+1si = si+1si∂i+1,
6) sisi+1∂i = ∂i+1sisi+1. 
Since the ςi verify the braid relations, they realize the braid group: given a permutation
ω ∈ SN and a reduced decomposition ω = si1 · · · sik , the product ςi1 · · · ςik is independent of the
choice of the reduced decomposition. We will denote ςω := ςi1 · · · ςik .
Furthermore, the algebra generated by the ςi and the xj is isomorphic to the degenerate Hecke
affine algebra generated by the operators si + ∂i and the variables.
Our goal is to find a basis of simultaneous eigenvectors of the following operators
ξˆi := xi +N − 1− ςi · · · ςN−1Ψˆς1 · · · ςi−1.
These operators commute and play the role of Cherednik elements for our representation of
the degenerate Hecke affine algebra. As a consequence, one has the following relations:
Proposition 7.2.
1. ςiξˆi+1 = ξˆiςi − 1,
2. ςiξˆi = ξˆi+1ςi + 1,
3. ςiξˆj = ξˆjςi for j 6= i, i+ 1,
4. Ψˆξˆi = ξˆi−1Ψˆ for i 6= 1,
5. Ψˆξˆ1 = (ξˆN − α)Ψˆ.
40 C.F. Dunkl and J.-G. Luque
Furthermore, the RST are simultaneous eigenfunctions of the operators ξˆi. More precisely
Proposition 7.3.
τ ξˆi = CTτ [i]τ.
Proof. The action of ξˆi on polynomials with degree 0 in the xi equals the action of the
operators ξ˜i. Hence, the result follows from the non-shifted version of the equality (Proposi-
tion 3.13). 
Straightforwardly, the operators ςi and Ψˆ are compatible with the leading monomials in the
following sense:
Proposition 7.4. Suppose that P a polynomial such that its highest degree component has the
leading monomial xv,τ then
1. If v[i] < v[i+ 1] then the highest degree component of Pςi has the leading monomial xvsi,τ .
2. The highest degree component of P Ψˆ has the leading monomial xvΨ,τ .
7.2 The Yang–Baxter graph
Let λ be a partition and Gλ the associated graph. We construct the set of the polynomials(
JˆP
)
P path in Gλ
using the following recurrence rules:
1. Jˆ[] := 1⊗ τλ.
2. If P = [a1, . . . , ak−1, si] then
JˆP := Jˆ[a1,...,ak−1]
(
ςi +
1
ζ[i+ 1]− ζ[i]
)
,
where the vector ζ is defined by
(
τλ,CTτλ , 0
N , [1, 2, . . . , N ]
)
a1 · · · ak−1 = (τ, ζ, v, σ).
3. P = [a1, . . . , ak−1,Ψ] then then
JˆP = Jˆ[a1,...,ak−1]Ψˆ.
As expected one obtains
Theorem 7.5. Let P = [a0, . . . , ak] be a path in Gλ from the root to (τ, ζ, v, σ). The polyno-
mial JˆP is a simultaneous eigenfunction of the operators ξˆi whose leading monomial in the
highest degree component is xv,τ . Furthermore, the eigenvalue of ξˆi associated to JˆP equals ζ[i].
Consequently JˆP does not depend on the path, but only on the end point (τ, ζ, v, σ), and will
be denoted by Jˆv,τ . The family (Jˆv,τ )v,τ forms a basis of Mλ of simultaneous eigenfunctions of
the Cherednik operators.
Furthermore, if P leads to ∅ then JˆP = 0.
Proof. The proof goes as in Theorem 4.1 using respectively Propositions 7.2, 7.3 and 7.4 instead
of Propositions 3.14, 3.13 and 3.8. 
In consequence, we will consider the family of polynomials (Jˆv,τ )v,τ indexed by pairs (v, τ) where
v ∈ NN is a weight and τ is a tableau.
Vector-Valued Jack Polynomials from Scratch 41
Jˆ000, 132
Jˆ
001, 132
Jˆ
010, 132
Jˆ
100, 132
ς
2
+
1
1
+
α
ς
1
+
1
2
+
α
Ψˆ Jˆ011, 132
Jˆ
101 132
Jˆ
002, 132
Ψˆ
Ψˆ
Ψˆ
ς
1
+
1
1
+
α
Jˆ
110, 132
ς
2
+
1
α
−
1
Jˆ
020, 132
Jˆ
200, 132
ς
2
+
1
1
+
2
α
ς
1
+
1
2
+
2
α
Figure 9. First values of Jˆv, 231 .
Example 7.6. Let again τ =
2
3 1
and consider the Yang–Baxter Graph G 2
31
(see Fig. 9).
The eigenvalues of τ are [1,−1, 0]. Let τ ′ =
1
3 2
, one has
Jˆ[001],τ = τΨˆ = (x3 + 2)⊗
(
−
1
2
τ + τ ′
)
with associated eigenvalues [−1, 0, α+ 1] and
Jˆ[010],τ = Jˆ[001],τ
(
(s2 + ∂2,3)⊗ s2 +
1
α+ 1
)
=
1
2
(3α+ 1 + x2 + αx2 − x3)
α+ 1
τ +
(x3 + 3 + x2 + αx2 + α)
α+ 1
τ ′
with eigenvalues [−1, α+ 1, 0].
42 C.F. Dunkl and J.-G. Luque
The non-shifted vector-valued Jack polynomials can be recovered easily from the shifted one:
Proposition 7.7. The restriction of Jˆv,τ to its component of top degree equals Jv,τ .
Proof. It suffices to remark that ςi = si ⊗ si + ς
−
i and Ψˆ = Ψ + Ψ
− where ς− and Ψ− are
operators which decrease the degree in the xj . Hence, one computes the component of the top
degree of Jˆv,τ following the Yang–Baxter graph replacing ςi by si ⊗ si and Ψˆ by Ψ, that is the
Yang–Baxter graph which was used to obtain the polynomials Jv,τ . This ends the proof. 
7.3 Symmetrization, antisymmetrization
We will say that a polynomial is symmetric if it is invariant under the action of ςi for each
i < N − 1. Denote also Sˆ =
∑
ω∈SN ςω. As for non shifted Jack, one has:
Theorem 7.8.
1. Let HT be a connected component of Gλ. For each vertex (τ, ζ, v, σ) of HT , the polyno-
mial Jˆv,τ Sˆ equals JˆλT ,std(T )Sˆ up to a multiplicative constant.
2. One has JˆλT ,std(T )Sˆ 6= 0 if and only if HT is 1-compatible.
3. More precisely, when HT is 1-compatible, the polynomial
JˆT =
∑
(τ,ζ,v,σ) vertex of HT
Ev,τ Jˆv,τ
is symmetric.
Proof. The proof is identical to the non-shifted case. 
In the same way, for the antisymmetrization, denoting Aˆ :=
∑
ω∈SN (−1)
`(ω)ςω, we have:
Theorem 7.9.
1. Let HT be a connected component of Gλ. For each vertex (τ, ζ, v, σ) of HT , the polyno-
mial Jˆv,τ Aˆ equals JˆλT ,std(T )Aˆ up to a multiplicative constant.
2. One has JˆλT ,std(T )Aˆ 6= 0 if and only if HT is (−1)-compatible.
3. More precisely, when HT is (−1)-compatible the polynomial
Jˆ ′T =
∑
(τ,ζ,v,σ) vertex of HT
Fv,τ Jˆv,τ
is antisymmetric.
7.4 Propagation of vanishing properties
Some phenomena of propagation of vanishing properties can be deduced from the classical case
(see [14]).
Lemma 7.10 (Lascoux). Let f(x, y) be a function of two variables. Suppose that f(b, a) = 0
with a 6= b then
f
(
sx,y + γ.∂x,y +
γ
b− a
)
(a, b) = 0.
Indeed, similar properties occur for vector-valued polynomials:
Vector-Valued Jack Polynomials from Scratch 43
Lemma 7.11. Let f ∈ C[x, y]⊗ Vλ then
f
(
sx,y ⊗ sx,y + γ∂x,y ⊗ 1 +
γ
b− a
)
(a, b) = 0
when f(b, a) = 0.
Proof. Write f(x, y) =
∑
τ f
τ (x, y) ⊗ τ and remark that f(b, a) = 0 implies f τ (a, b) = 0 for
each τ . But fsx,y =
∑
τ f
τsx,y⊗τsx,y and by Lemma 7.10 f τsx,y(a, b) = 0. Hence fsx,y(a, b) = 0.
In the same way, Lemma 7.10 implies f τ
(
∂x,y+ 1b−a
)
(a, b) = 0 and then f
(
∂x,y⊗1+ 1b−a
)
(a, b) = 0.
This proves the result. 
Example 7.12. Consider the polynomial Jˆ001, 231
. This polynomial vanishes for x3 = −2:
Jˆ001, 231
(x1, x2,−2) = 0. Since
Jˆ010, 231
= Jˆ001, 231
(
ς2 +
1
α+ 1
)
one has Jˆ010, 231
(x1,−2, α− 1) = 0. Finally,
Jˆ100, 231
= Jˆ010, 231
(
ς1 +
1
α+ 2
)
implies Jˆ100, 231
(−2, α, α− 1) = 0.
Denote by Φ+ the operator defined by [x1, . . . , xN ]Φ+ = [x2, x3, . . . , xN , x1 + α]. The action
of the affine operator Ψˆ propagates information about vanishing properties:
Lemma 7.13. One has
fΨˆ
(
[a1, . . . , aN ]Φ
+) = 0
when f(a1, . . . , aN ) = 0.
Hence, for each pair (v, τ) one can compute at least one N -tuple (av,τ1 , . . . , a
v,τ
N ) such that
Jv,τ (a
v,τ
1 , . . . , a
v,τ
N ) = 0.
Proposition 7.14. Denote by Vv,τ the vector whose i-th component is
Vv,τ [i] := α(v+[1]− v[i]) + (CTτ [1]− CTτ [σv[i]])−N + 1.
Let m be the smallest integer such that v[m] = max{v[i] : 1 ≤ i ≤ N}. One has
Jˆv,τ (x1, . . . , xm−1,Vv,τ [m], . . . ,Vv,τ [N ]) = 0
Proof. Denote by Φ˜ the operator defined by
[x1, . . . , xN ]Φ˜ =
{
[x1, . . . , xN ]Φ if x1 = aα+ b with a > 0,
[x2 + α, . . . , xN + α, x1] otherwise.
Observe first that the vectors Vv,τ are obtained by substituting Ψ by Φ˜ in Gτ :
Lemma 7.15. Recall the notation [v1, . . . , vN ]Ψ = [v2, . . . , vN , v1 + 1]. One has
VvΨ,τ = Vv,τ Φ˜.
44 C.F. Dunkl and J.-G. Luque
Proof. Let m be the smallest integer such that v[m] = max{v[i] : 1 ≤ i ≤ N}. If m > 1 then
v+[1] = vΨ+[1]. It follows that VvΨ,τ [i] = Vv,τ [i + 1] if i < N and, since vΨ[N ] = v[1] + 1
one has VvΨ,τ [N ] = Vv,τ [1] − α. That is VvΨ,τ = Vv,τΦ. If m = 1 then v+[1] = (vΨ)
+[1] − 1.
Hence, VvΨ,τ [i] = Vv,τ [i+ 1] + α for i < N and VvΨ,τ [N ] = Vv,τ [1]. This ends the proof. 
Hence, using Lemma 7.15, a straightforward induction shows that
Vv,τ [m] = 1−N. (7.1)
Let us prove the proposition by induction on the length of the path from Jˆ0...01,τ to Jˆv,τ . First
note that
V0...01,τ [N ] = −N + 1,
implies Jˆ0...01,τ (x1, . . . , xN−1,V0...01,τ [N ]) = 0, since
Jˆ0...01,τ = Jˆ0...0,τ Ψˆ = Jˆ0...0,τT (xN −N + 1).
Suppose that there is an arrow
(τ, ζ′, v′, σ′) (τ, ζ, v, σ)si
in Gτ . Let m′ be the smallest integer such that v′[m′] = max{v′[i] : 1 ≤ i ≤ N}. We have by
induction,
Jˆv′,τ
(
x1, . . . , xm′−1,Vv′,τ [m
′], . . . ,Vv′,τ [N ]
)
= 0. (7.2)
By hypothesis, i 6= m′, otherwise v′[i] ≥ v′[i + 1] and the arrow is not in Gτ . Furthermore we
have:
Jˆv,τ = Jˆv′,τ
(
si ⊗ si + ∂i ⊗ 1 +
1
(v′[i+ 1]− v′[i])α+ (CTτ [σv′ [i+ 1]]− CTτ [σv′ [i]])
)
= Jˆv′,τ
(
si ⊗ si + ∂i ⊗ 1 +
1
(v[i]− v[i+ 1])α+ (CTτ [σv[i]]− CTτ [σv[i+ 1]])
)
.
Consider three cases
1. If i < m′ − 1 then Lemma 7.11 implies
Jˆv,τ
(
x1, . . . , xm−1,Vv′,τ [m
′], . . . ,Vv′,τ [N ]
)
= 0.
But Vv′,τ [m′,m′+1, . . . , N ] = Vv,τ [m′,m′+1, . . . , N ] and m′ = m. This proves the result.
2. If i = m′ − 1 then m = m′ − 1 and, as a special case of (7.2), one has
Jˆv′,τ
(
x1, . . . , xm′−2,Vv,τ [m],Vv′,τ [m
′], . . . ,Vv′,τ [N ]
)
= 0.
But
Vv′,τ [m
′]−Vv,τ [m] = Vv,τ [m+ 1]−Vv,τ [m]
= (v[m]− v[m+ 1])α+ (CTτ [σv[m]]− CTτ [σv[m+ 1]]).
The result is, now, a direct consequence of Lemma 7.11.
Vector-Valued Jack Polynomials from Scratch 45
3. If i > m′ then Vv′,τ [m′,m′+1, . . . , N ] = Vv,τ [m′,m′+1, . . . , N ].si and m′ = m. Remarking
that,
Vv′,τ [i]−Vv,τ [i+ 1] = Vv,τ [i+ 1]−Vv,τ [i]
= (v[i]− v[i+ 1])α+ (CTτ [σv[i]]− CTτ [σv[i+ 1]]).
Lemma 7.11 gives the result.
Suppose that there is an arrow
(τ, ζ′, v′, σ′) (τ, ζ, v, σ)Ψ
in Gτ . Let m′ be the smallest integer such that v′[m′] = max{v′[i] : 1 ≤ i ≤ N}. We have by
induction,
Jˆv′,τ
(
x1, . . . , xm′−1,Vv′,τ [m
′], . . . ,Vv′,τ [N ]
)
= 0. (7.3)
We need to consider two cases:
1. If m′ > 1 then, since, Jˆv,τ = Jˆv′,τ .Ψˆ Lemma 7.13 and equation (7.3) imply
Jˆv,τ
(
x2, . . . , xm,Vv,τ [m], . . . ,Vv,τ [N − 1], x1
)
= Jˆv,τ
(
x2, . . . , xm′−1,Vv′,τ [m
′], . . . ,Vv′,τ [N ], x1
)
= 0.
In particular,
Jˆv,τ
(
x1, . . . , xm−1,Vv,τ [m], . . . ,Vv,τ [N ]
)
= 0.
2. If m′ = 1 then, from (7.1), one has Vv′,τ [1] = Vv,τ [N ] = 1 − N . But, since Jˆv,τ =
Jˆv′,τT (xN −N + 1), one has
Jˆv,τ
(
x1, . . . , xN − 1,Vv,τ [N ]
)
= 0,
and the result is just a special case obtained from this equality by specializing the values
of the xi. 
Example 7.16. One has:
V
[0,2,2,1,0,3,5,1],
1
3
75
8642
= [5α− 9, 3α− 8, 3α− 12, 4α− 10, 5α− 10, 2α− 13,−7, 4α− 11].
Indeed,
σ[0,2,2,1,0,3,5,1] = [7, 3, 4, 5, 8, 2, 1, 6] and [0, 2, 2, 1, 0, 3, 5, 1]
+ = [5, 3, 2, 2, 1, 1, 0, 0]
and the values of v+[1] − v[i] and (CTτ [1] − CTτ [σv[i]]) − N + 1 are computed by taking the
corresponding values in the RST
5
3
0 5
4 2 4 3
and
−9
−12
−7 −10
−11 −13 −10 −8
Hence,
Jˆ
[0,2,2,1,0,3,5,1],
1
3
75
8642
(x1, x2, x3, x4, x5, x6,−7, 4α− 11) = 0.
46 C.F. Dunkl and J.-G. Luque
V
001, 231
=[α,α−1,−2]
[x1,x2,−2]
[α,−2,α−1]
[x1,−2,α−1]
[−2,α,α−1]
[−2,α−1,α−1]
s 2
s 1
[α−1,−2,0]
[x1,−2,0]
[−2,α−1,0]
[−2,α−1,0]
[2α,2α−1,−2]
[x1,x2,−2]
Φ˜
Φ˜
Φ˜s 1
[−2,0,α−1]
[−2,0,α−1]
s
2
[2α,−2,2α−1]
[x1,−2,2α−1]
[−2,2α,2α−1]
[−2,2α,2α−1]
s
2
s
1
V
001, 132
=[α−4,α−3,−2]
[x1,x2,−2]
[α−4,−2,α−3]
[x1,−2,α−3]
[−2,α−4,α−3]
[−2,α−4,α−3]
[α−3,−2,−4]
[x1,−2,−4]
[−2,α−3,−4]
[−2,α−3,−4]
[2α−4,2α−3,−2]
[x1,x2,−2]
[−2,−4,α−3]
[−2,−4,α−3]
[2α−4,−2,2α−3]
[x1,−2,2α−3]
[−2,2α−4,2α−3]
[−2,2α−4,2α−3]
s
2
s
1 Φ˜
Φ˜
Φ˜
s
1Φ˜
s 2
s 2
s 1
Figure 10. Propagation of some vanishing properties for λ = 21.
The vanishing properties described in Proposition 7.14 are obtained by combining the actions
of the si and Φ˜ on the initial vectors V0N ,τ .
Example 7.17. Consider the propagation of vanishing properties described in Fig. 10.
Lemma 7.13 and Proposition 7.14 suggest that one can compute other vanishing properties
by combining the actions of the si and Φ+. A general closed formula remained to be found and,
unfortunately, the vanishing properties obtained by propagation from V0N ,τ are not sufficient
to characterize the shifted Jack polynomials.
Example 7.18. One has
Jˆ[002], 231
(
V[001], 231
s2s1Φ
+) = Jˆ[002], 231
(α, α− 1, α− 2) = 0.
Note that the only vanishing property obtained by propagation for Jˆ[100], 231
is
Jˆ[100], 231
(−2, α, α− 1) = 0,
and this is not sufficient to characterize the polynomial Jˆ[100], 231
.
Vector-Valued Jack Polynomials from Scratch 47
8 Conclusion
In this paper we used the Yang–Baxter graph technique to produce a structure describing the
nonsymmetric Jack polynomials whose values lie in an irreducible SN -module. The graph is
directed with no loops and has exactly one root or base point. Any path joining the root to
a vertex is essentially an algorithm for constructing the Jack polynomial at that vertex, and
the edges making up the path are the steps of the algorithm. The edges are labeled by the
generators of the braid group or by an affine operation.
These techniques are used to analyze restriction to a subgroup SM and also to construct
symmetric and antisymmetric Jack polynomials. These are associated with certain subgraphs.
Finally the graph technique is used to construct shifted, or inhomogeneous, vector-valued
Jack polynomials.
The theory is independent of the numerical value of the parameter α provided that the
eigenspaces of ξ˜i all have multiplicity one, that is, that no two vertices of the graph have the
same spectral vector. Future work is needed to analyze situations where this condition is violated,
in particular when α has a singular value,
{
n
m : 2 ≤ n ≤ N,m ∈ Z,
m
n /∈ Z
}
. There may not be
a basis of Jack polynomials for the space of all polynomials. For particular choices of λ there
may exist symmetric Jack polynomials of highest weight, that is, those annihilated by
N∑
i=1
Di.
It seems plausible that any graph describing such a special case would be significantly different
from Gλ. As a final remark, note that in the case of the trivial representation, some families of
highest weight Jack polynomials have been found (see e.g. [3, 10, 1]) and related to the theory
of the fractional quantum Hall effect [12].
Acknowledgments
The authors are grateful to A. Lascoux for fruitful discussions about the Yang–Baxter Graphs.
The authors acknowledge the referees for their meticulous reading and valuable comments. This
paper is partially supported by the ANR project PhysComb, ANR-08-BLAN-0243-04.
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