# AGT conjecture and AFLT states: a complete construction

###### Abstract

A complete construction of the AFLT states is proposed. With this construction
and for all the cases we have checked, the AGT conjecture on the equivalence
of Nekrasov Instanton Counting (NIC) to the $Vir\oplus u(1)$ conformal block
has been verified to be true.

Bao Shou<EMAIL_ADDRESS>Jian-Feng Wu<EMAIL_ADDRESS>, and Ming
<EMAIL_ADDRESS>

_Institute of Theoretical Physics,
Chinese Academy of Sciences, Beijing, 100190, China_

PACS: 11.25.Hf, 12.60.Jv, 12.40.Nn, 02.30.Ik

Keywords: Liouville Theory, Conformal Blocks, AGT Conjecture, Supersymmetric
Gauge Theory,
S-duality

##  1 Introduction

Conformal blocks, which are defined on the (punctured) Riemann surfaces,
holomorphic in each $z_{i}$ coordinate except when they meet each other, play
an essential role in building correlation functions in two dimensional
(Euclidean) conformal field theories[1]. They can be best understood as sewing
together chiral vertex operators[2, 3, 4], which by definition, are not local
objects, but the correlation functions are. The later combine both holomorphic
and anti-holomorphic conformal blocks in a consistent way to make modular
covariant objects. On the sphere, the $n$-point conformal block is represented
graphically as in fig.1, where $h_{i}$ is the conformal dimension of the
primary field inserted at coordinate $z_{i}$, and $\tilde{h_{i}}$ labels the
contribution arising from the conformal family descending from a primary field
with the conformal dimension $\tilde{h_{i}}$. The global conformal invariance
is $SL(2)\times SL(2)$, which may be used to fix three coordinates $z_{1}=0$,
$z_{n-1}=1$ and $z_{n}=\infty$. So the independent variables are $z_{i}$,
$i=2,...,n-2$, with the degrees of freedom $n-3$ for the $n$-point conformal
blocks on the sphere.

The calculation of conformal blocks is based on the conformal Ward-identities,

$[L_{n},V_{h}(z)]=(z^{n+1}\partial_{z}+(n+1)hz^{n})V_{h}(z).$

and carried out perturbatively level by level [1, 5, 6]. In some special
cases, the decoupling of the Virasoro null vectors can be implemented as
differential equations for the conformal blocks. For the general case,
recursion relations have been proposed by Zamolodchikov[5, 6] on the
meromorphic structures of the conformal blocks either in complex $c$-plane or
$h$-plane. However, in general, the global perspective of the sewing procedure
for the conformal blocks was still not fully understood until recently when
the AGT duality [13] had been proposed.

AGT conjecture relates 2d Liouville conformal field theories to 4d $N=2$
supersymmetric gauge theories of the $A_{1}$ type. The main idea is coupling
to the Liouville field a $u(1)$ field444In fact, the zero mode of the $u(1)$
field is a gauge symmetry and can be fixed to any desired value., then this
system is dual to a $U(2)=SU(2)\times U(1)$ superconformal 4d theory. In this
case, the partition function by Nekrasov instanton counting(NIC)[15, 16] of
the 4d $U(2)$ theory is to be identified with the conformal blocks of the
$u(1)$ coupled Liouville type. The Liouville CFT is characterized by a 2d one
boson theory with center charge $c\geq 25$. Finally, one can decouple the
$U(1)$ factor and obtain the instanton partition function of the $SU(2)$
theory which duals to Liouville conformal blocks. Liouville interaction breaks
down the charge conservation explicitly and leads to the introduction of the
screening charges. Because of the existence of the screening charges, the
conformal blocks of the Liouville type is much more complicated than its
counterpart of the $u(1)$ free boson theory. However, the AGT conjecture, if
proven true, means that there exists an orthogonal basis upon which the
$Liouville\times u(1)$ conformal blocks are built. From the above reasoning,
there exists a tree-like structure which describes the duality in coupling
space of the $N=2$ 4d superconformal linear quiver gauge theory. The primary
objects for this tree-like structure is the “bifundmental” matter coupling,
which, if translated correctly, should be represented by the inner products of
the bra and ket descendant fields in 2d conformal families sandwiched by a
“primary” vertex operator at position, say, $z$. Such kind of pants-like
diagram can be sewed together to form a linear quiver diagram, which, on the
2d CFT side, is just the $n$-point functions on the sphere for our
consideration. Of course, in the present context, we mean the $Vir\oplus u(1)$
2d CFT.

At first sight, it seems that such duality does not bring in any conveniences.
However, the Nekrasov instanton counting on the 4d field theory shows a rather
compact form for the summands which are completely factorized in “momentum”
$P$. And the summation is well organized into the combinatorial enumeration of
the Young tableuax. This simple structure implies Liouville theory, in
particular, the evaluation of the Liouville conformal blocks, could be
resolved by embedding it into a bigger system. So one may expect a new
construction for the Liouville conformal blocks from the corresponding NIC.

Figure 1: $n$-point conformal block on $S^{2}$

As pointed out by Nakajima[23, 24, 25], the instanton counting for $N=2$ gauge
theory is equivalent to the Hilbert scheme of points on the corresponding
Seiberg-Witten curve (blow-up Riemann surface)[11, 12, 21]. This can be
translated into a topological string description from physicists’ point of
view. By invoking the D4-D0 brane setup[20, 19] for ADHM construction[18] of
the instanton moduli space and the resolving process for ALE
singularities[22], these indicate that the instanton counting is a counting
for D0 branes in a toric Calabi-Yau 3-fold. Actually, there are two kinds of
D0 branes in the Calabi-Yau 3-fold, one is the regular D0 brane, which is in
regular representation of $\Gamma$, the center of the corresponding ADE group.
It carries no flux and can move freely on the Riemann surface. The other is
the fractional D0 brane, which is a D2 brane wrapping on a zero-sized two
sphere. It is always attached to the ALE singularity since it has a nontrivial
monodromy while moving around the singularity. It is these fractional D0
branes that resolve the ALE singularity, and leave fluxes on the blow-up
Riemann surface. This property ensures that one can identify these fractional
instantons as “anyons” on the Riemann surface. On the other hand, the regular
ones are “electric charged” particles on the Riemann surface. So the total
counting is equvalent to solving the problem of “electron gas” system with
insertions of anyons at the blow-up singularity on the Riemann surface. This
point of view is partialy included in Dijkgraaf and Vafa’s article[17]. For
each pants of the pants decomposition for the (punctured) Riemann surface, one
can guess that the instanton partition function can be rewritten as summation
over all the intermediate states passing through the sewn holes[4]. For the
interests of the present paper, we concern ourselves only with the special
pants diagram that one of the tubes is replaced by the blow-up singularity.
Then the summand in the instanton partition function represents itself as an
inner product of the bra and ket states, sandwiched by the anyonic vertex
operator. These bra and ket states should come from the interacting
“electronic”555For each simple root of an ADE group, one should introduce a
kind of “electronic” field. particles. A candidate description of the
“electronic gas” system is the integrable system of multiple Calogero-
Sutherland model, each living on a cycle. The whole (punctured) Riemann
surface, can be obtained by sewing together these pants on nonintersecting
cycles.

There are many efforts on relating the conformal blocks to the NIC[14, 26, 27,
28, 29, 30, 31, 32, 33, 34, 35] from various points of views, and these works
confirm the validity of the AGT duality. However, the explicit construction
for the Liouville conformal blocks has remained largely unclear until the
recent work [7] by Alba, Fateev, Litvinov and Tarnopolsky. In [7], they have
put forward the AGT duality in a more explicit form

$\displaystyle\dfrac{{}_{\vec{Y}^{\prime}}\langle{P^{\prime}}|V_{\alpha}|{P}\rangle_{\vec{Y}}}{\langle{P^{\prime}}|V_{\alpha}|{P}\rangle}$
$\displaystyle=$ $\displaystyle
Z_{bif}(\alpha|P^{\prime},\vec{Y}^{\prime};P,\vec{Y})\,\,,$ (1)

here specifically for a free field realization,

$V_{\alpha}(z)=e^{2i(Q-\alpha)\tilde{\varphi}_{-}(z)}e^{-2i\alpha\tilde{\varphi}_{+}(z)}S^{n}:e^{2i\alpha\varphi(z)}:\,,$

with $P+P^{\prime}+\alpha+nb=0$ , and $S=\oint e^{2ib\varphi(z)}\rm dz$ is the
screening charge in the Virasoro sector. The l.h.s. of eq(1) is the pants-like
(with one of the tubes labeled by $\alpha$ shrinks to a line) conformal block.
The r.h.s. of (1) reproduces $Z_{bif}$ for the instanton counting, which is
given by

$\displaystyle
Z_{bif}(\alpha|P^{\prime},\vec{Y}^{\prime};P,\vec{Y})=\prod_{i,j=1}^{2}\prod_{s\in
Y_{i}}\left(Q-E_{Y_{i},Y_{j}^{\prime}}(P_{i}-P_{j}^{\prime}|s)-\alpha\right)\prod_{t\in
Y_{j}^{\prime}}\left(E_{Y_{j}^{\prime},Y_{i}}(P^{\prime}_{j}-P_{i}|t)-\alpha\right)\,,$
(2)

where $\vec{P}=(P,-P)$, $\vec{P^{\prime}}=(P^{\prime},-P)$ and

$\displaystyle E_{Y,Y^{\prime}}(P|s)\equiv
P+b^{-1}(a_{Y}(s)+1)-bl_{Y^{\prime}}(s)\,.$ (3)

Here $a_{Y}(s)$ and $l_{Y}(s)$ resp. are the arm length and the leg length
resp. of the box $s$ in the Young tableau $Y$, defined as

$a_{Y}(s)|_{s=(i,j)}:=\lambda_{i}-j,\ \ \ \
l_{Y}(s)|_{s=(i,j)}:=\lambda_{j}^{t}-i\,,$

$\lambda_{i}$ and $\lambda_{j}^{t}$ resp. are the $i$-th part of the partition
$\lambda=(\lambda_{1},\lambda_{2},\cdots),\,\,\lambda_{i}\geq\lambda_{i+1}$
and the $j$-th part of the transpose partition $\lambda^{t}$ respectively .

(1) means that the matrix elements of a special “chiral vertex operator”
$V_{\alpha}$ in a suitably chosen basis, can be translated into a 4d theory as
an instanton contribution for a special bifundamental contribution of the NIC.
By sewing together pants-like diagrams one gets any desired duality diagrams
in the coupling space of the linear quiver gauge theory. So, the checking of
the AGT duality reduces to the construction of the states
$|{P}\rangle_{\vec{Y}}$, which we shall call the AFLT states[7], with
$\vec{Y}\equiv(Y_{1},Y_{2})$ the Young tableaux. Here the $Y$’s, the
partitions of natural numbers, or equivalently represented by Young tableaux,
are labels for the orthogonal basis for the descendant fields (Verma modules)
in a $Vir\oplus u(1)$ conformal family from the 2d CFT point of view. By
definition, the AFLT states form a complete set of states for the family
members in a given $Vir\oplus u(1)$ conformal family and the inner products
between them, sandwiched by a vertex operator of the particular form,
$V_{\alpha}(z)$, at position,say, $z=1$, is factorized exactly as the NIC
$Z_{bif}$ presented on the r.h.s. of (1). The explicit formula, (1), puts
strong constraints on the possible forms of the AFLT states and make a
systematic construction of them unaccessible at first glance. In [7], only the
explicit form of the state $|{P}\rangle_{Y,\varnothing}$ has been found,

$|{P}\rangle_{Y,\varnothing}=J^{+}_{-Y}|{P}\rangle\Omega_{Y}(P),$

with $J^{+}_{-Y}$ the creator $(-ib)^{-1}a^{+}_{-n}$’s valued Jack symmetric
function, and $\Omega_{Y}(P)$ the normalization constant.

In our opinion, the AGT conjecture, written in the form of (1), strongly
suggests that the $Vir\oplus u(1)$ conformal family is a Hamiltonian system
with $|{P}\rangle_{\vec{Y}}$ the Hamiltonian eigenstates. So the construction
of the AFLT states becomes a quantum mechanical problem of solving the
Schrodinger equation. Put things in this way, we propose a possible form of
the Hamiltonian $H$ and construct its eigenstates explicitly. We shall
identify those eigenstates as the AFLT states desired. For in all the cases we
have checked, (1) is verified to be true, using the AFLT states we have
constructed. We shall present now as the main results of our present paper the
explicit form of the Hamiltonian $H$ along with the complete construction of
the AFLT states, $|{P}\rangle_{\vec{Y}}$. More elaborated exposition will come
in the subsequent sections.

$\displaystyle H$ $\displaystyle=$ $\displaystyle H_{0}+H_{I}$ (4)
$\displaystyle|{P}\rangle_{\vec{Y}}$ $\displaystyle=$
$\displaystyle\frac{1}{1-\frac{1}{E_{\vec{Y}}(P)-H_{0}}H_{I}}J_{-\vec{Y}}|P\rangle\Omega_{\vec{Y}}(P).$

Here, $J^{\pm}_{\pm Y}$ are the Jack states constructed in terms of the
oscillators $a^{\pm}_{n}$’s or $a^{\pm}_{-n}$’s ($n>0$) solely, $H^{\pm}$ the
corresponding Hamiltonian for the Jack symmetric functions, $H_{0}\equiv
H^{+}+H^{-}$. Thus the eigenstate of $H_{0}$ is just
$J_{-\vec{Y}}|{P}\rangle\equiv J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle$ with
the eigenvalue $E_{\vec{Y}}(P)$. $H^{\pm}$ in our formalism is defined to
include zero modes $a^{\pm}_{0}$ also, $-ia^{\pm}_{0}|{P}\rangle=\pm
P|{P}\rangle$. It is important that $H_{I}$ is strictly triangular with
respect to the basis vectors of the $H_{0}$ eigenstates. By “strictly
triangular” we mean the (upper or lower) triangular matrix with zero diagonal
entries. It is easy to see that if the interaction term $H_{I}$ is strictly
triangular, then the eigenvalue spectrum of $H_{0}$ remains unperturbed and
$|{P}\rangle_{\vec{Y}}$ in (4) well defined for non-degenerate $H_{0}$
spectrum descending from a mother state $J_{-\vec{Y}}|{P}\rangle$ for generic
values of $P$’s. Putting things all together, we have

$\displaystyle H=H_{0}+H_{I},\ \ \ H_{0}=H^{+}+H^{-},\ \ \
H_{I}=\sum_{n=1}^{\infty}2Qna_{-n}^{+}a_{n}^{-},$ (5) $\displaystyle
H^{\pm}=\dfrac{-i}{3}\oint\left(z\partial_{z}\varphi^{\pm}\right)^{3}\frac{dz}{2\pi
iz}+\sum_{n=1}^{\infty}Qna_{-n}^{\pm}a_{n}^{\pm},$ $\displaystyle
E_{\vec{Y}}(P)=E_{Y_{1}}+E_{Y_{2}}+2P(|Y_{1}|-|Y_{2}|),\ \ \
E_{Y}=\sum_{i}(y^{2}_{i}b^{-1}+(2i-1)y_{i}b),$
$\displaystyle\Omega_{\vec{Y}}(P)=(-)^{|Y_{1}|}b^{|Y_{1}|+|Y_{2}|}\prod_{Y_{1}}\left(2P+(a_{Y_{1}}+1)b^{-1}-l_{Y_{2}}b\right)\prod_{Y_{2}}\left(2P-a_{Y_{2}}b^{-1}+(l_{Y_{1}}+1)b\right),$
$\displaystyle|{P}\rangle_{\vec{Y}}=\frac{1}{1-\frac{1}{E_{\vec{Y}}(P)-H_{0}}H_{I}}J_{-\vec{Y}}|P\rangle\Omega_{\vec{Y}}(P),$
$\displaystyle
H_{0}J_{-\vec{Y}}|{P}\rangle=E_{\vec{Y}}(P)J_{-\vec{Y}}|{P}\rangle,\ \ \
H|{P}\rangle_{\vec{Y}}=E_{\vec{Y}}(P)|{P}\rangle_{\vec{Y}},\ \ \
-ia^{\pm}_{0}|{P}\rangle=\pm P|{P}\rangle$

Notice that
1) $|{P}\rangle_{Y,\varnothing}$ constructed in [7] are included in our
construction as subcases.
2) The Hamiltonian $H$ constructed by us, albeit in a disguised form, turns
out to coincide up to some trivial factor with $I_{3}$, one of the integrals
of motion found in a different context in appendix C of [7]. $I_{3}$ in [7],
written in the form of $Vir\oplus u(1)$, makes the Virasoro symmetry manifest,
but is not suitable for solving a perturbation theory with perturbation
parameter $Q=b+b^{-1}$. The Hamiltonian $H$ written in terms of the
interacting bi-Jack polynomial system as in (5), shows Virasoro symmetry only
implicitly, but makes the perturbation theory exactly solvable as we shall see
soon after.

The procedure is outlined as follows. On the 2d CFT side, the $Vir\oplus u(1)$
theory can be represented as a theory of two independent scalars
$\tilde{\varphi}(z)$ and $\varphi(z)$. $\tilde{\varphi}(z)$ part is
essentially a free theory of timelike oscillators, while the scalar field
$-i\varphi(z)$ is spacelike but engaged in a Liouville type interaction. The
two scalars can be linearly combined to form the “light-cone” scalars.
$\varphi^{+}(z)$ and $\varphi^{-}(z)$. The labeling $\vec{Y}$ of the basis
vectors strongly suggests that there exist a bi-Jack polynomial structure,
plus possibly some interactions between these two sectors. That is, the “free”
$H^{\pm}$ spectrums should be described by $J_{Y_{1}}^{+}$ and $J_{Y_{2}}^{-}$
respectively, here $J_{Y}$ denotes Jack states related to Young tableau $Y$.
First we construct the “unperturbed” energy operator $H_{0}$ which just sums
up the “energies” in $J^{\pm}_{Y}$ sectors, $H_{0}=H^{+}+H^{-}$. The next
thing is to specify the interaction between these two sectors. Strictly
speaking, $H_{0}$ does not describe a free theory, since it also contains the
interaction terms proportional to $Q$. But the new interaction term $H_{I}$
further mixes the $J^{\pm}_{Y}$’s and the coupling is also a first order in
$Q$. It is good to see that $H_{I}$ is strictly triangular with respect to the
basis vectors of $H_{0}$ eigenstates. Our method can be easily generalized to
wider classes of integrable models, in which the interacting Hamiltonian
splits into two parts, $H^{//}$ and $H^{\perp}$, representing respectively the
shift of energies and the rotations (mixings) of states. The later keeps the
eigenvalue spectrum untouched[37].

Besides being triangular, the form of the interaction term is however much
restricted, also by the Virasoro symmetry. Since the total Hamiltonian is of
the form $Vir\oplus u(1)$, an “interaction energy operator” $H_{I}$ is needed
to make the “full Energy operator” $H=H_{0}+H_{I}$ the combination of
$a_{n}$’s and $L_{n}$’s only. Once the Hamiltonian structure is determined,
then the construction of the Hamiltonian eigenstate $|{P}\rangle_{\vec{Y}}$ is
just a quantum mechanical problem. $H_{0}$ and $H$ share the same eigenvalue
spectrum, but only the eigenstates of $H$, represented by
$|{P}\rangle_{\vec{Y}}$’s, form a complete set of basis vectors for the
$Vir\oplus u(1)$ conformal family.

We have checked by examples the corresponding AGT duality formula, (1) up to
level 4, and have found that indeed Nekrasov instanton counting can be
reproduced with this construction, (5). In fact, we have also checked more
general cases and all get positive answers. But those more general results
will appear elsewhere due to lacking of space to include them in this paper.

The insertions of the screening charges play an important role in checking the
AGT duality. However, in the present work we concern ourselves only with the
cases in which the screening charges can be detached away from the vertex
operator $V_{\alpha}$ and moved on to act on the AFLT states (similar to the
Felder’s BRST operators)[8, 9]. The more general cases in which screening
charges can not be moved away from $V_{\alpha}$ will be under our future
studies.

It is well known that it is possible to map the Liouville theory to the
analytic continuation of the Calogero-Sutherland(CS) model, which was
originally and in most cases considered to be a theory with the parameter
$\beta>0$, while in the Liouville case $\beta<0$ is required. Some explanation
is given in [36]. The physical space of the CS model are created by Jack
polynomials, which are symmetric functions studied in great detail in
mathematics and physics literatures[38, 39]. The integrability of the CS model
may be derived in different ways, e.g., from the knowledge of the hidden
$W_{1+\infty}$ symmetry of the model. A recursion relation related to the
Virasoro singular vectors and an integral representation based on it has
appeared recently in [36], in which more references can be found on the
subjects of the CS model and the Jack symmetric functions. It should be
stressed again that for $\beta>0$, there is no null vectors in the CS model.
So the “null” vectors are not the true null vectors of the CS model, since the
Virasoro algebra based on which the null vectors are constructed is not the
true conformal algebra of the CS model in that case. But for $\beta<0$, yes,
there are null vectors in the CS model. It is possible to describe the
$Liouville\times u(1)$ theory in terms of the Jack polynomials considered as
analytic continuation from $\beta>0$ to $\beta<0$.

There is another hint that the $Liouville\times u(1)$ theory has something to
do with $\beta<0$ CS model. It can be found from the Nekrasov partition
function, in which each term in the summation can be written in the form of
the Carlsson-Okounkov formula[10], for the special cases when no screening
charges are inserted. Carlsson-Okounkov formula is a formula for the inner
products between the bra Jack states and the ket Jack states sandwiched with a
modified vertex operator. This extraordinary formula is of great help in
checking the AGT duality with our construction for the orthogonal basis
vectors $|{P}\rangle_{\vec{Y}}$’s defined in (1).

We notice that the construction we found shares many similarities with the
construction of the Jack functions themselves. Namely, we take the state
$J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle$ as the mother state and its
descendants are constructed in such a manner that two partitions are
“squeezed” into other pairs. The squeezing does not change the total level of
the two partitions, but does make the inner products of the descendants a
triangular form.

Although the 4d to 2d duality has just begun to be understood, it has been
known for sometime that 2d conformal blocks can be equivalently described as
insertions of Wilson lines in 3d pure Chern-Simons topological gauge theory.
In fact, we can interpret the $n$-point conformal block represented by fig.1
as a Wilson line insertion inside a three-ball. The path integral in Chern-
Simons-Witten gauge theory thus creates a state living on the boundary of the
three ball, which is punctured $S^{2}$. So it should not be a too big surprise
that 2d conformal field theory has something to do with higher dimensional
quantum field theories. Taking into account that Jack symmetric polynomials
can be taken as some special limit of the two parameter Macdonald symmetric
polynomials, one natural guess is that our construction can be generalized to
the case of Macdonald symmetric polynomials. In that case there should be a 5d
to 3d duality.

This paper is organized in the following way. Our general formalism on the
construction of the AFLT states is presented in the introduction. In section
2, we explore the general structure of the $Vir\oplus u(1)$ conformal family.
We found in some cases it is more convenient to work with the bi-Jack function
basis. Section 3 contains the major derivation of our construction. Section 4
is the conclusion. And in appendix A the explicit construction of the AFLT
states up to level 3 is presented.

##  2 Exploring the $Vir\oplus u(1)$ Structure

We are dealing with a 4d N=2 $U(2)$ linear quiver gauge theory coupled to
special bi-fundmental matter in a superconformal way. According to the
standard AGT duality dictionary, the corresponding 2d conformal block is of
the $Vir\oplus u(1)$ type, which reproduces the instanton part of the Nekrasov
partition function for the $U(2)$ theory. There are two sets of Young diagrams
which measure the partitions of the instantons. If one wants to extract the
Virasoro basis of the conformal blocks, one need to factor out the $u(1)$
factor.

In this section we shall mainly explore the Hilbert space for the $Vir\oplus
u(1)$ theory and find the requirements that the energy operator $H$ should
meet. Our procedure depends heavily on the Nekrasov instanton counting formula
written more suitably for the construction of the conformal blocks, (1).
First, the 2d $u(1)$ conformal block, realized in terms of the oscillators of
the scaler field $\tilde{\varphi}$, is essentially of free theory with center
charge $c=1$. The zero modes can be integrated out trivially and does not play
any significant role here. The vertex operators for $\tilde{\varphi}$, take a
peculiar form

$e^{2i(Q-\alpha)\tilde{\varphi}_{(-)}(z)}e^{-2i\alpha\tilde{\varphi}_{(+)}(z)}$

Here, $\tilde{\varphi}_{(\pm)}$ means the positive (negative) mode part of the
scalar field $\tilde{\varphi}$ . Although the above vertex operator is not the
standard one in 2d CFT, its contribution to the conformal block can be easily
read off and factored out. Second, the $Vir$ part is a Liouville conformal
field theory of the $\varphi(z)$ scalar field and is more complicated because
of the existence of the screening charges.

We have the following mode expansion for the scalar fields $\varphi(z)$ and
$\tilde{\varphi}(z)$,

$\displaystyle\varphi(z)$ $\displaystyle=$ $\displaystyle
q+c_{0}\log(z)+\sum_{n\in Z,n\neq 0}\frac{c_{-n}}{n}z^{n}\,,$ (6)
$\displaystyle\tilde{\varphi}(z)$ $\displaystyle=$
$\displaystyle\tilde{q}+a_{0}\log(z)+\sum_{n\in Z,n\neq
0}\frac{a_{-n}}{n}z^{n}\,,$ $\displaystyle\left[c_{n},c_{m}\right]$
$\displaystyle=$ $\displaystyle\frac{n}{2}\delta_{n+m,0}\,,\ \ \
\,(c_{-n})^{\dagger}=c_{n}\,,\ \ \ \left[c_{0},q\right]=\frac{1}{2}$
$\displaystyle\left[a_{n},a_{m}\right]$ $\displaystyle=$
$\displaystyle\frac{n}{2}\delta_{n+m,0}\,,\ \ \
\,(a_{-n})^{\dagger}=-a_{n}\,,\ \ \
\left[a_{0},\tilde{q}\right]=\frac{1}{2}\,.$

Virasoro generators in the $Vir$ part, $L_{n}$, thus reads

$\displaystyle L_{n}$ $\displaystyle=$
$\displaystyle\sum_{k\in\mathbb{Z}}c_{k}c_{n-k}-inQc_{n}=\sum_{k\neq
0,n}c_{k}c_{n-k}+i(2\hat{P}-nQ)c_{n},$ (7) $\displaystyle L_{0}$
$\displaystyle=$
$\displaystyle\frac{Q^{2}}{4}-\hat{P}^{2}+2\sum_{k>0}c_{-k}c_{k}\,,$ (8)

here, $c_{0}=i\hat{P}$,

$-ic_{0}|{P}\rangle=\hat{P}|{P}\rangle=P|{P}\rangle,\,\,\,\langle{P}|(-ic_{0})=\langle{P}|\hat{P}=-P\langle{P}|.$

By this construction, $L_{n}$ defined in (7-8) is obviously unitary,

$L_{-n}=L_{n}^{\dagger}.$

In 2d CFT, one frequently meets another (more conventional) definition of the
Virasoro generators,

$L^{0}_{n}=\sum_{k\in\mathbb{Z}}c_{k}c_{n-k}-iQ(n+1)c_{n}\,.$ (9)

If (7) and (8) are combined in this way,

$T(z)=\partial\varphi\partial\varphi+iQ\partial^{2}\varphi+iQz^{-1}\partial\varphi-\frac{Q^{2}}{4}\frac{1}{z^{2}}=\sum_{n}L_{n}z^{-n-2}\,,$
(10)

then $T(z)$ differs from the more conventional one
$T^{0}(z)=\partial\varphi\partial\varphi+iQ\partial^{2}\varphi$ by a
similarity transformation

$\displaystyle T(z)$ $\displaystyle=$ $\displaystyle
e^{-iQq}(\partial\varphi\partial\varphi+iQ\partial^{2}\varphi)e^{iQq}=\sum_{n}\tilde{L}_{n}z^{-n-2}$
(11) $\displaystyle\tilde{L}_{n}$ $\displaystyle=$
$\displaystyle\sum_{k\in\mathbb{Z}}\tilde{c}_{k}\tilde{c}_{n-k}-iQ(n+1)\tilde{c}_{n}\,\,.$

Comparing (10) and (11), we have

$\displaystyle\tilde{c_{n}}=\left\\{\begin{array}[]{l}c_{n},\,\,\,n\neq 0\\\
c_{0}+\frac{i}{2}Q,\,\,\,n=0\end{array}\right.\,.$ (14)

Viewing the $Vir\oplus u(1)$ model as a 2d sigma model, since under
conjugation $c_{n}$ and $a_{n}$ transform differently we recognize that
$-i\varphi$ is spacelike and $\tilde{\varphi}$ timelike when they are
considered as coordinates in target space. So the target space of the sigma
model under consideration is curved in space and flat in time direction. The
two scalars can also be linearly combined to form the “light-cone” scalars
$\varphi^{\pm}(z)$,

$\displaystyle\varphi^{\pm}(z)=\tilde{\varphi}(z)\pm\varphi(z)\,,$ (15)
$\displaystyle\varphi^{\pm}(z)\varphi^{\pm}(z^{\prime})=\log(z-z^{\prime})\,,$
$\displaystyle\varphi^{\pm}(z)\varphi^{\mp}(z^{\prime})=0\,,$
$\displaystyle\varphi^{\pm\dagger}(z)=\varphi^{\mp}(z)\,.$

The descendant states in the conformal family split into sub-spaces of
different levels, which are measured by
$I_{2}=L_{0}+\sum_{n=1}^{\infty}a_{-n}a_{n}$. Within the sub-space of given
level $N\equiv\sum_{n>0}(a_{-n}a_{n}+c_{-n}c_{n})$, states can be labeled
either by linear conbinations of $a_{-X}L_{-Y}$’s or $J^{+}_{-X}J^{-}_{-Y}$’s,
with $X$, $Y$ the Young tableau, $|X|+|Y|=N$, $J^{\pm}_{\pm Y}$ the
annihilator $(-ib)^{-1}a_{Y}^{\pm}$’s (or creator $(-ib)^{-1}a_{-Y}^{\pm}$’s)
valued Jack symmetric functions. In either case, one can infer from AGT
duality that there exist a Hermitian operator $H$, which commutes with $I_{2}$
and diagonizes this subspace. Hence the eigenstates of $H$ form an orthogonal
basis. We know $I_{2}$ acts on this subspace trivially like an identity
operator. So in order to eliminate the degeneracy, the next candidate $H$ we
are looking for should be at least cubic in the oscillators $a_{n}$’s and
$c_{n}$’s. Once $H$ is introduced, the descendant states will organize
themselves into an orthogonal basis labeled by two sets of Young tableaux
$\\{Y_{1},Y_{2}\\}$. In our opinion, it is better to start with the
$J^{\pm}_{Y}$ system, since there is already a Hamiltonian structure $H^{\pm}$
acting separately on them. But $H_{0}=H^{+}+H^{-}$ does not commute with the
screening charges $S^{\pm}$ pertaining to $L_{n}$,

$S^{\pm}=\oint e^{2ib^{\pm}\varphi(z)}\rm dz\,,$

here $b^{+}\equiv b,\,b^{-}\equiv b^{-1}$ . We then add a new term $H_{I}$ to
$H_{0}$, $H=H_{0}+H_{I}$ and require that $\left[S^{\pm},H\right]=0$. If
$H_{I}$ are chosen correctly, the eigenstates of $H$ will coincide with the
unique orthogonal basis $|{P}\rangle_{\vec{Y}}$, which we call AFLT states and
are defined to satisfy (1), in which the matrix elements
$\frac{{}_{\vec{Y^{\prime}}}\langle{P^{\prime}}|V_{\alpha}|{P}\rangle_{\vec{Y}}}{\langle{P^{\prime}}|V_{\alpha}|{P}\rangle}$
is factorized in a consistent way.

On the 4d theory side, one can decouple a single massless bifundamental
matter666The massless condition implies
$|Y^{\prime}_{1}|+|Y^{\prime}_{2}|=|Y_{1}|+|Y_{2}|$. $(\vec{a}=(P,P),m=0)$. We
shall show that under this condition the contributions can be written as the
orthogonality condition for the $|{P}\rangle_{\vec{Y}}$’s, provided (1) is
satisfied.

######  Proposition 1.

777This is actually Proposition 2.4 in [7], but here we give more details.

If AGT conjecture is true, then the AFLT states, $|{P}\rangle_{\vec{Y}}$’s
defined in (1), form an orthogonal basis.

$\\!\\!{}_{Y^{\prime}_{1},Y^{\prime}_{2}}\langle P|{P}\rangle_{Y_{1},Y_{2}}$
$\displaystyle=$ $\displaystyle
Z_{bifund}^{U(2)inst}(\vec{a},\vec{Y},\vec{a},\vec{Y^{\prime}};0)\propto\delta_{\vec{Y},\vec{Y^{\prime}}}.$
(16)

Proof: We proceed, from (1),

${}_{Y^{\prime}_{1},Y^{\prime}_{2}}\langle
P|{P}\rangle_{Y_{1},Y_{2}}=\,\,\\!\\!_{Y^{\prime}_{1},Y^{\prime}_{2}}\langle{P}|V_{\alpha=0}|{P}\rangle_{Y_{1},Y_{2}}$
$\displaystyle=$
$\displaystyle\prod_{Y_{1}}\\{Q-[(a_{Y_{1}}+1)b^{-1}-l_{Y^{\prime}_{1}}b]\\}\prod_{Y^{\prime}_{1}}\\{(a_{Y^{\prime}_{1}}+1)b^{-1}-l_{Y_{1}}b\\}$
$\displaystyle\times$
$\displaystyle\prod_{Y_{1}}\\{Q-[2P+(a_{Y_{1}}+1)b^{-1}-l_{Y^{\prime}_{2}}b]\\}\prod_{Y^{\prime}_{2}}\\{-2P+(a_{Y^{\prime}_{2}}+1)b^{-1}-l_{Y_{1}}b\\}$
$\displaystyle\times$
$\displaystyle\prod_{Y_{2}}\\{Q-[-2P+(a_{Y_{2}}+1)b^{-1}-l_{Y^{\prime}_{1}}b]\\}\prod_{Y^{\prime}_{1}}\\{2P+(a_{Y^{\prime}_{1}}+1)b^{-1}-l_{Y_{2}}b\\}$
$\displaystyle\times$
$\displaystyle\prod_{Y_{2}}\\{Q-[(a_{Y_{2}}+1)b^{-1}-l_{Y^{\prime}_{2}}b]\\}\prod_{Y^{\prime}_{2}}\\{(a_{Y^{\prime}_{2}}+1)b^{-1}-l_{Y_{2}}b\\}\,.$

We shall prove now that under this situation, if the result is non-zero, one
can conclude

$\vec{Y}=\vec{Y^{\prime}}.$

If

${}_{Y^{\prime}_{1},Y^{\prime}_{2}}\langle P|{P}\rangle_{Y_{1},Y_{2}}\neq 0,$

one gets

$y_{1,1}\leq y^{\prime}_{1,1}.$ 888Here we use the notation $y_{1,r}$ to label
the $r$-th part of the partition $Y_{1}$.

Since otherwise there must exist a box in the tableau $Y_{1}$ satisfying

$a_{Y_{1}}=0,\,\,l_{Y_{1}^{\prime}}=-1\,\,.$

This will lead to

$Q-[(a_{Y_{1}}+1)b^{-1}-l_{Y^{\prime}_{1}}b]=0,$ (18)

This argument cycles and one finally conclude:

$y_{1,i}\leq y^{\prime}_{1,i}\,\,,i=1,2,\dots\,.$

For $Y_{2}$, similarly, the argument follows, and gives:

$y_{2,i}\leq y^{\prime}_{2,i}\,\,,i=1,2,\dots\,.$

However, the original condition

$|Y^{\prime}_{1}|+|Y^{\prime}_{2}|=|Y_{1}|+|Y_{2}|$

then forces $Y_{1}=Y^{\prime}_{1},Y_{2}=Y^{\prime}_{2}$. Q.E.D.

The orthogonality condition, (16), strongly suggests the existence of mutually
commuting Hermitian operators, whose common eigenstates form a complete
orthogonal basis of the Hilbert space. One of the operators, called the energy
operator, probably cubic in $a_{n}$’s and $c_{n}$’s (since this is most likely
the case beyond $I_{2}$), is the first object we are going to construct.
However, hermiticity alone is not enough to constrain the possible forms of
the construction. For example, $H_{0}=H^{+}+H^{-}$ is Hermitian, but does not
belong to $Vir\oplus u(1)$. In fact, we shall show that altogether there
should be at least 3 conditions $H$ are to meet in order the orthogonality of
the $H_{0}$ eigenstates play an important role here.

$\displaystyle i)$ $\displaystyle\hskip 42.67912ptHermiticity$ $\displaystyle
ii)$ $\displaystyle\hskip 42.67912ptTriangularity$ $\displaystyle iii)$
$\displaystyle\hskip 42.67912ptReflection-invariance\,.$

Now we explain what the other two conditions means. Condition ii),
triangularity, means that $H$, in its matrix form,
$H_{\vec{Y^{\prime}},\vec{Y}}(P)=\langle{\vec{Y^{\prime}}}|H(P)|{\vec{Y}}\rangle$,
where $|{\vec{Y}}\rangle$’s are the eigenstates of $H_{0}=H^{+}+H^{-}$(the
“unperturbed” energy operator)999Here we have fixed $\hat{P}$ eigenvalue
equals $P$ ., is lower(or upper)-triangular with $H_{I}\equiv H-H_{0}$
strictly triangular (with zero diagonal entries). Under such circumstances,
the spectrum of $H$ coincides with that of $H_{0}$, and the eigenstates of $H$
can be expressed as
$|{P}\rangle_{Y_{1},Y_{2}}=\Omega_{Y_{1},Y_{2}}(P)R(E)|{\vec{Y},P}\rangle$,
here the normalization constant $\Omega_{Y_{1},Y_{2}}(P)$ will be specified
later on. $R(E)=1+\tilde{R}(E)$, a unitriangular matrix, is again triangular
with identity diagonal entries following the triangularity of $H$. $H^{\pm}$
is the collective mode Hamiltonian for the Calogero-Sutherland model in terms
of the oscillators ${a^{\pm}_{n}}$’s. Thus the eigenstate of $H_{0}$ is just
$J^{+}_{-Y_{1}}J^{-}_{-Y_{2}}|{P}\rangle$. $H^{\pm}$ in our formalism
(including the zero modes $a^{\pm}_{0}$) is defined as

$H^{\pm}=-i\frac{1}{3}\oint(z\partial_{z}\varphi^{\pm}(z))^{3}dz/z+\sum_{n=1}^{\infty}Qna^{\pm}_{-n}a^{\pm}_{n}$

Since $\varphi^{\pm}\dagger=\varphi^{\mp}$, we have
$J^{+}_{Y}\dagger=J^{-}_{-Y}$. There is a natural question on how to define
the inner products between $J^{\pm}_{Y}$’s. The answer is that we need the
condition iii) Reflection-invariance. Notice that
$\langle{P^{\prime}}|P\rangle$ not zero means $P+P^{\prime}=0$. In order to
get a non-vanishing result, we need to shift $\langle{P}|$ to $\langle{-P}|$.
We thus expect that there exists an operation which changes
${}_{Y_{1},Y_{2}}\langle{P}|$ to ${}_{Y_{2},Y_{1}}\langle{-P}|$. We call this
operation reflection following the terminology in a similar situation in [29].
Actually, by looking closer to the NIC formula i.e. the r.h.s of (1), one can
find that there exist an apparent symmetry

${}_{Y_{1},Y_{2}}\langle{P}|\leftrightarrow_{Y_{2},Y_{1}}\langle{-P}|.$ (19)

If we change either bra state $\langle{P}|_{Y_{1},Y_{2}}$ to
${}_{Y_{2},Y_{1}}\langle{-P}|$, or ket state $|{P}\rangle_{Y_{1},Y_{2}}$ to
$|{-P}\rangle_{Y_{2},Y_{1}}$, on the l.h.s. of (1), the factors on the r.h.s.
of eq(1) get reshuffling but the final result keep invariant. We may name this
symmetry “reflection” or “flipping” symmetry. On the 2d CFT side, from general
reasoning that such an operation should be conformally invariant, it is
natural to identify the insertions of the screening charges as this
“reflection” operation. For Liouville theory (or Coulomb gas model), we can
attach to $V_{\alpha=0}$ some screening charges101010We suppose originally
there is no screening charge attached to $V_{\alpha=0}$ for simplicity., such
that

${}_{Y_{1}^{\prime},Y_{2}^{\prime}}\langle{P}|V_{0}S^{n}|{P}\rangle_{Y_{1},Y_{2}}\neq
0$ (20) $\displaystyle S=\oint e^{2ib\varphi(z)}\rm dz\,.$ (21)

Now the neutrality condition forces $2P+nb=0\,\,\,.$ If this is satisfied,
then Felder’s contour for the integration of the screening charges actually
closes and $S^{n}$ becomes a floating charge[8, 9]. Now $S^{n}$ can move away
from $V_{\alpha}$ and communing through $L_{n}$’s and finally acts on the
vacuum state $\langle{P}|$. Since $S^{n}$ acts by not changing the conformal
weight, we deduce

$\langle{P}|S^{n}=\langle{-P}|$

for a suitable normalization of $S^{n}$. Similar arguments apply to the case
of $V_{\alpha},\ \alpha\neq 0$, and one can always move a subset of screening
charges, $S^{\frac{-2P}{b}}$ away from $V_{\alpha}(z)$. Since AGT duality
formula is valid for any $P$, we may assume that $n$ can take arbitrary real
value, as analytical continuation away from integer $n$. This flipping is due
to the fact that $S^{n}$ can be detached from $V_{\alpha}$, and act on the
vacuum directly. Similar operation exists in Felder’s BRST cohomology [8].

We are going to identify the reflection symmetry in NIC as the Hamiltonian
symmetry in 2d CFT for the insertions of the screening charges $S^{n}$ with
$2P=-nb$. Since Hamiltonian $H\in Vir\oplus u(1)$, satisfies
$\left[H,S^{n}\right]=0$ it has the property of double degeneracy. So $S^{n}$
with $2P=-nb$ should map one AFLT state to its partner state. If we require

${}_{Y_{1},Y_{2}}\langle{P}|H$ $\displaystyle=$
$\\!\\!{}_{Y_{1},Y_{2}}\langle{P}|E_{Y_{1},Y_{2}}$ (22)
${}_{Y_{1},Y_{2}}\langle{P}|S^{n}H$ $\displaystyle=$
${}_{Y_{1},Y_{2}}\langle{P}|S^{n}E_{Y_{1},Y_{2}}\,,$ (23)

then we can identify

${}_{Y_{1},Y_{2}}\langle{P}|S^{n}=_{Y_{2},Y_{1}}\langle{-P}|\,,$

since reflection symmetry means $E_{Y_{1},Y_{2}}(P)=E_{Y_{2},Y_{1}}(-P)$.
Notice that nothing has changed for the $u(1)$ part. Define
$P^{\pm}=-ia_{0}\mp ic_{0},$ then we have111111We have set
$a_{0}|{P}\rangle=0$ throughout this paper.:

$P^{\pm}|{P}\rangle=\pm
P|{P}\rangle,\,\,\,\langle{-P}|P^{\pm}=\langle{-P}|(\pm P)\,,$

which obviously shows that $\langle-P|{P}\rangle\neq 0$. So reflection
invariance means that we can identify the inner product
${}_{Y_{1}^{\prime},Y_{2}^{\prime}}\langle{P}|P\rangle_{Y_{1},Y_{2}}$ with
either ${}_{Y_{2}^{\prime},Y_{1}^{\prime}}\langle{-P}|P\rangle_{Y_{1},Y_{2}}$
or ${}_{Y_{1}^{\prime},Y_{2}^{\prime}}\langle{P}|-P\rangle_{Y_{2},Y_{1}}$ by
the incertions of screening charges satisfying $n=-2Pb^{-1}$.

Having determined that $|{P}\rangle_{\vec{Y}}$ form a normalizable orthogonal
basis, the next step is the determination of their normalization. Before doing
this, let’s review the so-called Carlsson-Okounkov formula[10] which is useful
for our formulation. First, define

$\displaystyle
E=1+e_{1}+e_{2}+\dots=e^{-\sum_{n}\frac{(-)^{n}}{n}p_{n}}=e^{-\frac{1}{k}\varphi_{(-)}(-1)}$
(24) 121212For infinitely many arguments $z_{i}$’s, $i=1,2,\cdots,\infty$, one
may identify $p_{n}\equiv\sum_{i}z_{i}^{n}$ with $\frac{a_{-n}}{k}$,
$k^{2}=\beta$ and $J_{Y}^{1/\beta}(\\{p_{n}\\})$ with
$J_{Y}^{1/\beta}(\\{\frac{a_{-n}}{k}\\})$. Here our convention is that
$\frac{a_{-n}}{k}|{0}\rangle$ creates a state $|{p_{n}}\rangle$. As a
consequence, $e_{n}$ is to be identified with
$P_{-1^{n}}\equiv\dfrac{J_{1^{n}}^{1/\beta}(\\{\frac{a_{-n}}{k}\\})}{n!}\equiv
J_{-1^{n}}/n!$ . Such kind of identification is justified because they share
the same values of their inner products. For more details see [36].

which is a vertex operator, and also a generating function for $J_{-1^{n}}$

$e^{-\frac{1}{k}\varphi_{(-)}(z)}|{0}\rangle=\sum_{n}(-)^{n}\dfrac{J_{-1^{n}}}{n!}z^{n}|{0}\rangle\,,$

here $e_{i}$ are elementary symmetric functions, $p_{n}$ is the power sum
symmetric function. Then

$e^{-\frac{1}{k}\varphi_{(-)}(-1)}|{0}\rangle=\sum_{n}\dfrac{J_{-1^{n}}}{n!}|{0}\rangle\equiv\sum_{n}P_{-1^{n}}|{0}\rangle.$
(25)

The conjugation of $E$ reads

$E^{\ast}=e^{\frac{1}{k}\varphi_{(+)}(-1)}\,,$ (26)

and we have

$\langle{0}|E^{\ast}=\langle{0}|\sum_{n}P_{1^{n}}\,.$

Now the Carlsson-Okounkov formula reads

$\displaystyle\langle E^{m}(E^{\ast})^{\beta-m-1}J_{-Y_{1}},J_{-Y_{2}}\rangle$
$\displaystyle=$
$\displaystyle(-)^{|Y_{1}|}\beta^{-|Y_{1}|-|Y_{2}|}\prod_{Y_{1}}(m+(a_{Y_{1}}+1)+\beta
l_{Y_{2}})\prod_{Y_{2}}(m-a_{Y_{2}}-\beta(l_{Y_{1}}+1))$ $\displaystyle=$
$\displaystyle\langle J_{Y_{1}}E^{\beta-m-1}(E^{\ast})^{m}J_{-Y_{2}}\rangle$
$\displaystyle=$ $\displaystyle\langle
J_{Y_{1}}e^{(-k+k^{-1}+\frac{m}{k})\varphi_{(-)}(-1)}e^{\frac{m}{k}\varphi_{(+)}(-1)}J_{-Y_{2}}\rangle\,.$

For Liouville theory, $k=-ib$. If we set $\frac{m}{-ib}=-2i\alpha$, then the
Carlsson-Okounkov formula reads

$\displaystyle\langle
J_{Y_{1}}e^{i(Q-2\alpha)\varphi_{(-)}(-1)}e^{-2i\alpha\varphi_{(+)}(-1)}J_{-Y_{2}}\rangle$
$\displaystyle=$
$\displaystyle(-)^{|Y_{2}|}b^{-|Y_{1}|-|Y_{2}|}\prod_{Y_{1}}(-2\alpha+(a_{Y_{1}}+1)b^{-1}-l_{-Y_{2}}b)\prod_{Y_{2}}(-2\alpha-
a_{Y_{2}}b^{-1}+(l_{Y_{1}}+1)b)\,.$

The normalization of the AFLT states is inherited from AFLT’s version of the
AGT duality formula, (1) and the orthogonality condition, (2),

${}_{Y_{1},Y_{2}}\langle
P|{P}\rangle_{Y_{1},Y_{2}}=_{Y_{2},Y_{1}}\langle-P|{P}\rangle_{Y_{1},Y_{2}}$
$\displaystyle=$
$\displaystyle\prod_{Y_{1}}\\{-a_{Y_{1}}b^{-1}+(l_{Y_{1}}+1)b\\}\\{(a_{Y_{1}}+1)b^{-1}-l_{Y_{1}}b\\}$
$\displaystyle\times$
$\displaystyle\prod_{Y_{2}}\\{-a_{Y_{2}}b^{-1}+(l_{Y_{2}}+1)b\\}\\{(a_{Y_{2}}+1)b^{-1}-l_{Y_{2}}b\\}$
$\displaystyle\times$
$\displaystyle\prod_{Y_{1}}\\{-2P-a_{Y_{1}}b^{-1}+(l_{Y_{2}}+1)b)\\}\prod_{Y_{2}}\\{-2P+(a_{Y_{2}}+1)b^{-1}-l_{Y_{1}}b\\}$
$\displaystyle\times$
$\displaystyle\prod_{Y_{2}}\\{2P-a_{Y_{2}}b^{-1}+(l_{Y_{1}}+1)b)\\}\prod_{Y_{1}}\\{2P+(a_{Y_{1}}+1)b^{-1}-l_{Y_{2}}b\\}$
$\displaystyle=$ $\displaystyle(-)^{|Y_{1}|+|Y_{2}|}j_{Y_{1}}j_{Y_{2}}$
$\displaystyle\times$ $\displaystyle\prod_{Y_{1}}\\{-2Pb-
a_{Y_{1}}+(l_{Y_{2}}+1)b^{2}\\}\prod_{Y_{2}}\\{-2Pb+(a_{Y_{2}}+1)-l_{Y_{1}}b^{2}\\}$
$\displaystyle\times$ $\displaystyle\prod_{Y_{2}}\\{2Pb-
a_{Y_{2}}+(l_{Y_{1}}+1)b^{2}\\}\prod_{Y_{1}}\\{2Pb+(a_{Y_{1}}+1)-l_{Y_{2}}b^{2}\\}$
$\displaystyle\equiv$ $\displaystyle
j_{Y_{1}}j_{Y_{2}}\Omega_{Y_{2},Y_{1}}(-P)\Omega_{Y_{1},Y_{2}}(P)$
$\displaystyle=$ $\displaystyle j_{Y_{1}}j_{Y_{2}}\langle
J_{Y_{2}}e^{i(Q-2P)\varphi_{(-)}(-1)}e^{-i2P\varphi_{(+)}(-1)}J_{-Y_{1}}\rangle$
$\displaystyle\times$ $\displaystyle(b^{4})^{|Y_{1}|+|Y_{2}|}\langle
J_{Y_{1}}e^{i(Q+2P)\varphi_{(-)}(-1)}e^{i2P\varphi_{(+)}(-1)}J_{-Y_{2}}\rangle\,.$

In reaching the last line in the above eqation, Carlsson-Okounkov formula has
been applied, and we have defined

$\displaystyle\Omega_{Y_{1},Y_{2}}(P)$ $\displaystyle=$
$\displaystyle(-)^{|Y_{1}|}b^{|Y_{1}|+|Y_{2}|}\prod_{Y_{1}}\left(2P+(a_{Y_{1}}+1)b^{-1}-l_{Y_{2}}b\right)\prod_{Y_{2}}\left(2P-a_{Y_{2}}b^{-1}+(l_{Y_{1}}+1)b\right)$
$\displaystyle=$ $\displaystyle(-b^{2})^{(|Y_{1}|+|Y_{2}|)}\langle
J_{Y_{1}}e^{i(Q+2P)\varphi_{(-)}(-1)}e^{i2P\varphi_{(+)}(-1)}J_{-Y_{2}}\rangle$
$\displaystyle=$ $\displaystyle b^{2(|Y_{1}|+|Y_{2}|)}\langle
J_{Y_{1}}e^{i(Q+2P)\varphi_{(-)}(1)}e^{i2P\varphi_{(+)}(1)}J_{-Y_{2}}\rangle$
$\displaystyle\equiv$ $\displaystyle\Omega_{\vec{Y}}(P)\,.$

Notice that $\Omega_{\vec{Y}}(P)$ is just a generalization of $\Omega_{Y}(P)$
defined in [7].

##  3 The Construction of the AFLT States

Now we come to our main problem of the construction of the Hamiltonian $H$
with the requirement that its eigenstates be identified with ALFT states
satisfying (1). We prefer to work first on the basis of Jack symmetric
functions $J_{\vec{Y}}$, which already form an orthogonal basis. We found that
if $H_{I}$ matrix elements are strictly triangular on this basis, then the
orthogonality of the $H=H_{0}+H_{I}$ eigenstates follows immediately from the
orthogonality of the $H_{0}$ eigenstates. This is just the simplest way to go
from one orthogonal basis to another one. To see this, let’s introduce an
operator $R(E)$ which map $H_{0}$ eigenstates to $H$ eigenstates, with
$\Omega_{Y_{1},Y_{2}}(P)$ the normalization constant

$\displaystyle|{P}\rangle_{Y_{1},Y_{2}}$ $\displaystyle=$ $\displaystyle
R(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle\Omega_{Y_{1},Y_{2}}(P)\,$ (31)
${}_{Y_{2}^{\prime},Y_{1}^{\prime}}\langle{-P}|$ $\displaystyle=$
$\displaystyle\langle{-P}|J_{Y_{2}^{\prime}}^{-}J_{Y_{1}^{\prime}}^{+}R(E^{\prime})^{\dagger}\Omega_{Y_{2}^{\prime}Y_{1}^{\prime}}(-P)$
$\displaystyle R(E)$ $\displaystyle=$ $\displaystyle
1+\cdots=1+\tilde{R}(E)\,,$

where the reflection symmetry has been applied to the AFLT states

${}_{Y_{1}^{\prime},Y_{2}^{\prime}}\langle{P}|S^{n}=\langle{P}|S^{n}J_{Y_{1}^{\prime}}^{-}J_{Y_{2}^{\prime}}^{+}R(E^{\prime})^{\dagger}\Omega_{Y_{1}^{\prime}Y_{2}^{\prime}}(P)=\\\
_{Y_{2}^{\prime},Y_{1}^{\prime}}\langle{-P}|=\langle{-P}|J_{Y_{2}^{\prime}}^{-}J_{Y_{1}^{\prime}}^{+}R(E^{\prime})^{\dagger}\Omega_{Y_{2}^{\prime}Y_{1}^{\prime}}(-P),$

and $\tilde{R}(E)$ is strictly lower(or upper)-triangular $\Rightarrow$
$\tilde{R}_{\vec{Y},\vec{Y}}(E)=0$. The way $R(E)$ is expanded in (31) follows
from the normalization condition, (2-2) as we shall see in (3).

The Hermitian operator $H$ should satisfy:

$\displaystyle H|{P}\rangle_{Y_{1},Y_{2}}$ $\displaystyle=$ $\displaystyle
E_{Y_{1},Y_{2}}(P)|{P}\rangle_{Y_{1},Y_{2}}$ (32)
${}_{Y_{2}^{\prime},Y_{1}^{\prime}}\langle{-P}|H$ $\displaystyle=$
${}_{Y_{2}^{\prime},Y_{1}^{\prime}}\langle{-P}|E_{Y_{2}^{\prime},Y_{1}^{\prime}}(-P),$

where the energy eigenvalue has the double degeneracy:

$E_{Y_{2},Y_{1}}(-P)=E_{Y_{1},Y_{2}}(P)\,,$ (33)

due to the orthogonality condition,

${}_{Y_{1}^{\prime},Y_{2}^{\prime}}\langle
P|{P}\rangle_{Y_{1},Y_{2}}=_{Y_{2}^{\prime},Y_{1}^{\prime}}\langle{-P}|P\rangle_{Y_{1},Y_{2}}\propto\delta_{\vec{Y},\vec{Y^{\prime}}}\,.$
(34)

Then the next step is to determine if we get the right normalization for
$|{P}\rangle_{Y_{1},Y_{2}}$

${}_{Y_{2},Y_{1}}\langle{-P}|{P}\rangle_{Y_{1},Y_{2}}$ $\displaystyle=$
$\displaystyle\langle{-P}|J_{Y_{2}}^{-}J_{Y_{2}}^{+}R(E)^{\dagger}R(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle\Omega_{Y_{1},Y_{2}}(P)\Omega_{Y_{2},Y_{1}}(-P)$
$\displaystyle=$
$\displaystyle\langle{-P}|J_{Y_{2}}^{-}J_{Y_{1}}^{+}(1+\tilde{R}(E)^{\dagger})(1+\tilde{R}(E))J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle\Omega_{Y_{1},Y_{2}}(P)\Omega_{Y_{2},Y_{1}}(-P)$
$\displaystyle=$
$\displaystyle\Omega_{Y_{1},Y_{2}}(P)\Omega_{Y_{2},Y_{1}}(-P)$
$\displaystyle\times$
$\displaystyle\left[j_{Y_{1}}j_{Y_{2}}+\langle{-P}|J_{Y_{2}}^{-}J_{Y_{1}}^{+}(\tilde{R}(E)^{\dagger}+\tilde{R}(E)+\tilde{R}(E)^{\dagger}\tilde{R}(E))J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle\right]$
$\displaystyle=$
$\displaystyle\Omega_{Y_{1},Y_{2}}(P)\Omega_{Y_{2},Y_{1}}(-P)j_{Y_{1}}j_{Y_{2}}\,.$

It is in agreement with (2). In deriving this we have used the fact that if
$R(E)$ is a unitriangular matrix131313A unitriangular matrix is a triangular
matrix with the diagonal entries equal to 1 .,

$R(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle=J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle+\sum_{\begin{subarray}{c}|Y_{1}^{\prime}|>|Y_{1}|\\\
|Y_{2}^{\prime}|<|Y_{2}|\\\
|Y_{1}^{\prime}|+|Y_{2}^{\prime}|=|Y_{1}|+|Y_{2}|\end{subarray}}R_{Y_{1},Y_{2}}^{Y_{1}^{\prime}Y_{2}^{\prime}}(E)J_{-Y_{1}^{\prime}}^{+}J_{-Y_{2}^{\prime}}^{-}|{P}\rangle\,,$
(36)

then it is easy to check that

$\displaystyle\langle{-P}|J_{Y_{2}}^{-}J_{Y_{1}}^{+}\tilde{R}(E)^{\dagger}J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle$
$\displaystyle=\langle{-P}|J_{Y_{2}}^{-}J_{Y_{1}}^{+}\tilde{R}(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle$
$\displaystyle=\langle{-P}|J_{Y_{2}}^{-}J_{Y_{1}}^{+}\tilde{R}(E)^{\dagger}\tilde{R}(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle$
$\displaystyle=0\,.$

Now we summarize the requirements for $R(E)$

$\displaystyle i)\hskip 28.45274pt\text{$R(E)$ is unitriangular}$
$\displaystyle ii)\hskip 28.45274pt\text{$R(E)$ creates the eigenstate for
$H$}$ $\displaystyle\hskip
28.45274ptHR(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle=E_{Y_{1},Y_{2}}(P)R(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle$
$\displaystyle iii)\hskip 28.45274pt\text{Reflection invariant}$
$\displaystyle\hskip
28.45274ptS^{n}R(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle\Omega_{Y_{1},Y_{2}}(P)=R(E)J_{-Y_{2}}^{+}J_{-Y_{1}}^{-}|{-P}\rangle\Omega_{Y_{2},Y_{1}}(-P)$
$\displaystyle\hskip 28.45274pt\left[S^{n},H\right]=0,\ \ \ \
E_{Y_{1},Y_{2}}(P)=E_{Y_{2},Y_{1}}(-P)$

This means that the Hamiltonian $H$ should also be triangular, and $H_{I}$
strictly triangular,

$\displaystyle H$ $\displaystyle=$ $\displaystyle H^{+}+H^{-}+H_{I}$ (37)
$\displaystyle H^{\pm}$ $\displaystyle=$
$\displaystyle\dfrac{-i}{3}\oint\left(z\partial_{z}\varphi^{\pm}\right)^{3}+\sum_{n>0}Qna_{-n}^{\pm}a_{n}^{\pm}\,,$
(38)

here $H_{I}$ is to be determined later on.

$\displaystyle H^{+}+H^{-}$ $\displaystyle=$
$\displaystyle\dfrac{-i}{3}\oint\left[\left(z\partial_{z}(\varphi+\tilde{\varphi})\right)^{3}+\left(z\partial_{z}(\varphi-\tilde{\varphi})\right)^{3}\right]\dfrac{\rm
dz}{z}+$
$\displaystyle\sum_{n>0}Qna_{-n}^{+}a_{n}^{+}+\sum_{n>0}Qna_{-n}^{-}a_{n}^{-}$
$\displaystyle=$
$\displaystyle\dfrac{-i}{3}\oint\left[2\left(z\partial_{z}\tilde{\varphi}\right)^{3}+6(z\partial_{z}\tilde{\varphi})(z\partial_{z}\varphi)^{2}\right]\dfrac{\rm
dz}{z}$ $\displaystyle+$ $\displaystyle\sum_{n>0}2Qn(a_{-n}a_{n}+c_{-n}c_{n})$
$\displaystyle=$
$\displaystyle\dfrac{-2i}{3}\oint\left(z\partial_{z}\tilde{\varphi}\right)^{3}\frac{\rm
dz}{z}+\sum_{n>0}2Qna_{-n}a_{n}-2i\oint\left(z\partial_{z}\tilde{\varphi}\right)\left(z\partial_{z}\varphi\right)^{2}\frac{\rm
dz}{z}+2\sum_{n>0}Qnc_{-n}c_{n}\,.$

Now the requirement that $H$ commute with $S^{n}$ is equivalent to say that
$H$ can be written in terms of $L_{n}$’s and $a_{n}$’s. To make $H$
triangular, we may try

$\displaystyle H_{I}$ $\displaystyle\propto$
$\displaystyle\sum_{n}Qna_{-n}^{+}a_{n}^{-}$ $\displaystyle=$
$\displaystyle\sum_{n}Qn(a_{-n}a_{n}-c_{-n}c_{n}-a_{-n}c_{n}+c_{-n}a_{n})$
$\displaystyle=$ $\displaystyle\sum_{n}Qn(a_{-n}a_{n}-c_{-n}c_{n})+Q\oint
z\partial_{z}\tilde{\varphi}(z\partial_{z})^{2}\varphi\frac{\rm dz}{z}\,.$

If we now make use of (10) and choose

$H_{I}=\sum_{n}2Qna_{-n}^{+}a_{n}^{-},\,$

then we get

$\displaystyle H$ $\displaystyle=$
$\displaystyle-\dfrac{2i}{3}\oint(z\partial_{z}\tilde{\varphi})^{3}\dfrac{\rm
dz}{z}+4Q\sum_{n\in\mathbb{N}^{+}}na_{-n}a_{n}-2i\oint(z\partial_{z}\tilde{\varphi})z^{2}T(z)\dfrac{\rm
dz}{z}+2ia_{0}\dfrac{Q^{2}}{4}$ $\displaystyle=$
$\displaystyle-\dfrac{2i}{3}\oint(z\partial_{z}\tilde{\varphi})^{3}\dfrac{\rm
dz}{z}+4Q\sum_{n\in\mathbb{N}^{+}}na_{-n}a_{n}-2i\sum_{n\in\mathbb{Z}}a_{-n}L_{n}+2ia_{0}\dfrac{Q^{2}}{4}$
$\displaystyle=$
$\displaystyle-i\left\\{\sum_{n,m\in\mathbb{N}^{+}}\left(a_{-n-m}^{+}a_{n}^{+}a_{m}^{+}+a_{-n-m}^{-}a_{n}^{-}a_{m}^{-}\right)\right.$
$\displaystyle+$
$\displaystyle\left.\sum_{n,m\in\mathbb{N}^{+}}\left(a_{-n}^{+}a_{-m}^{+}a_{n+m}^{+}+a_{-n}^{-}a_{-n}^{-}a_{n+m}^{-}\right)\right\\}$
$\displaystyle+$
$\displaystyle\sum_{n\in\mathbb{N}^{+}}Qn(a_{-n}^{+}a_{n}^{+}+a_{-n}^{-}a_{n}^{-}+2a_{-n}^{+}a_{n}^{-})$
(40) $\displaystyle+$
$\displaystyle\sum_{n\in\mathbb{N}^{+}}-2ia_{0}^{+}a_{-n}^{+}a_{n}^{+}-2ia_{0}^{-}a_{-n}^{-}a_{n}^{-}-\dfrac{i}{3}((a_{0}^{+})^{3}+(a_{0}^{-})^{3})\,.$
(41)

Clearly, $H$ indeed satisfies the three requirements proposed in the previous
section.

$\displaystyle i)$ $\displaystyle\hskip 42.67912ptHermitian$ $\displaystyle
ii)$ $\displaystyle\hskip 42.67912ptTriangular$ $\displaystyle iii)$
$\displaystyle\hskip 42.67912ptReflection-invarint\,.$

Besides, we found that $H\propto I_{3}$, where $I_{3}$ is defined in Appendix
C of [7] as one of the infinitely many commuting operators which may makes the
system integrable. The authors of [7] have checked for the first a few levels
that the AFLT states, $|{P}\rangle_{Y_{1},Y_{2}}$, which satisfies AGT duality
formula, (1), are also the eigenstates of $I_{3}$. But a general formula for
the $I_{3}$ eigenstates is missing in [7].

Now the next question is : how to find all the eigenstates of $H$? First,
let’s consider $H^{+}$

$\displaystyle H^{+}$ $\displaystyle=$
$\displaystyle-i\sum_{n,m\in\mathbb{N}^{+}}\left\\{a_{-n-m}^{+}a_{n}^{+}a_{m}^{+}+a_{-n}^{+}a_{-m}^{+}a_{n+m}^{+}\right\\}$
$\displaystyle+$
$\displaystyle\sum_{n\in\mathbb{N}^{+}}\left\\{nQa_{-n}^{+}a_{n}^{+}+2a^{+}_{0}(-i)a_{-n}^{+}a_{n}^{+}\right\\}-\frac{i(a_{0}^{+})^{3}}{3}\,,$

Its eigenvalue

$H^{+}J^{+}_{-Y}|P^{+}\rangle=E^{+}_{Y}(P^{+})J^{+}_{-Y}|P^{+}\rangle$
$E^{+}_{Y}(P^{+})=\sum_{i}\left\\{y^{2}_{i}b^{-1}+(2i-1)y_{i}b\right\\}+2P^{+}|Y|-\frac{(P^{+})^{3}}{3}.$

Here we have assumed the zero modes take the following eigenvalues,

$a_{0}=iP^{a}\,,\,c_{0}=iP^{c}\,,\,a^{\pm}_{0}=iP^{\pm}=i(P^{a}\pm P^{c})$
(43)

For the bi-Jack system, we have the following eigenequation,

$HR(E)J^{+}_{-Y_{1}}J^{-}_{-Y_{2}}|P^{+},P^{-}\rangle=E_{Y_{1},Y_{2}}(P^{+},P^{-})R(E)J^{+}_{-Y_{1}}J^{-}_{-Y_{2}}|P^{+},P^{-}\rangle\,.$

Triangularity means

$\displaystyle E_{Y_{1},Y_{2}}(P^{+},P^{-})=$ $\displaystyle
E^{+}_{Y_{1}}(P^{+})+E^{-}_{Y_{2}}(P^{-})$ $\displaystyle=$
$\displaystyle\sum_{i}\left\\{y_{1,i}^{2}b^{-1}+(2i-1)y_{1,i}^{2}b\right\\}+\sum_{i}\left\\{y_{2,i}^{2}b^{-1}+(2i-1)y_{2,i}^{2}b\right\\}$
$\displaystyle+2P^{+}|Y_{1}|+2P^{-}|Y_{2}|-\frac{(P^{+})^{3}+(P^{-})^{3}}{3}$

Since $H$ can be constructed in terms of $L_{n}$’s and $a_{n}$’s, so
$S^{n}|P^{+},P^{-}\rangle_{Y_{1},Y_{2}}$ dose not change the eigenvalue. But
$S^{n}$ changes $P^{c}\rightarrow-P^{c}$ and $P^{+}\leftrightarrow P^{-}$ and
from $E_{Y_{1},Y_{2}}(P^{+},P^{-})=E_{Y_{2},Y_{1}}(P^{-},P^{+})$. We conclude

$S^{n}|P^{+},P^{-}\rangle_{Y_{1},Y_{2}}\propto|P^{-},P^{+}\rangle_{Y_{2},Y_{1}}.$
(44)

Next, since $P^{a}$ does not play any important role, we may consider it as a
gauge symmetry and can be fixed to any desired value. For convenience, we fix
$P^{a}=0$, hence, $P^{+}=P^{c}\equiv P$, $P^{-}=-P^{c}\equiv-P$ and

$\displaystyle E_{Y_{1},Y_{2}}(P,-P)$ $\displaystyle\equiv E_{Y_{1},Y_{2}}(P)$
$\displaystyle=E_{Y_{1}}+E_{Y_{2}}+2P(|Y_{1}|-|Y_{2}|)$

Here

$E_{Y}=\sum_{i}\left\\{y^{2}_{i}b^{-1}+(2i-1)y_{i}b\right\\}$

If we define

$\displaystyle|P\rangle$ $\displaystyle\equiv$ $\displaystyle|P,-P\rangle$
$\displaystyle|P\rangle_{Y_{1},Y_{2}}$ $\displaystyle\equiv$ $\displaystyle
R(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle\Omega_{Y_{1},Y_{2}}(P)$

Then we infer from from (44)

$S^{n}|P\rangle_{Y_{1},Y_{2}}=|-P\rangle_{Y_{2},Y_{1}}$

with the proper normalization for $S^{n}$. Now we are going to determine
$R(E)$ which satisfies

$HR(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle=E_{Y_{1},Y_{2}}(P)R(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle.$

######  Proposition 2.

$R(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle=\frac{1}{1-\frac{1}{E-H_{0}}H_{I}}J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle,$

here $H_{0}=H^{+}+H^{-},E=E_{Y_{1},Y_{2}}(P)$. $R(E)$ defined in such a way
should be understood as

$\displaystyle R(E)$ $\displaystyle=$
$\displaystyle\frac{1}{1-\frac{1}{E-H_{0}}H_{I}}$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}(\frac{1}{E_{Y_{1},Y_{2}}(P)-H_{0}}H_{I})^{n}$

Proof: First, we rewrite $H$ as

$\displaystyle H$ $\displaystyle=$ $\displaystyle H_{0}+H_{I}$
$\displaystyle=$ $\displaystyle E+H_{0}+H_{I}-E$ $\displaystyle=$
$\displaystyle E+(H_{0}-E)(1+\frac{1}{H_{0}-E}H_{I})\,.$

Then from

$\displaystyle HR(E)$ $\displaystyle=$
$\displaystyle(E+(H_{0}-E)(1+\frac{1}{H_{0}-E}H_{I}))\frac{1}{1-\frac{1}{E-H_{0}}H_{I}}$
$\displaystyle=$ $\displaystyle
E\frac{1}{1-\frac{1}{E-H_{0}}H_{I}}+H_{0}-E\,,$ $\displaystyle=$
$\displaystyle ER(E)+H_{0}-E\,,$

one gets

$HR(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle=ER(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle+(H_{0}-E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle\,.$

Since $J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle$ is an eigenstate of $H_{0}$
with eigenvalue $E=E_{Y_{1},Y_{2}}(P)$, we have

$(H_{0}-E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle=0.$

Hence, we conclude that $R(E)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}|{P}\rangle$ is an
eigenstate of $H$ with eigenvalue $E$,

$E\equiv
E_{Y_{1},Y_{2}}(P)=\left(\sum_{i=1,l=1}^{i=2,l=y_{i,1}^{t}}(y_{i,l})^{2}+(2i-1)y_{i,l}\right)+2P(|Y_{1}|-|Y_{2}|)\,,$

Q.E.D.
Now we shall address the question raised in [7] on the possible degeneracy of
$H$. The authors of [7], argued that $I_{3}$ has some degeneracy at level 4
and higher. We have analyzed what causes such kind of degeneracy. After
analyzing the spectrum of $H$, we believe that such degeneracy happens when
$|Y_{1}|=|Y_{2}|$, and we have $2P(|Y_{1}|-|Y_{2}|)=0$,

$E_{Y_{1},Y_{2}}(P)=E_{Y_{1}}+E_{Y_{2}}=E_{Y_{2}}+E_{Y_{1}}=E_{Y_{2},Y_{1}}(P)$

This can happen, for $Y_{1}\neq Y_{2}$, first at level 4,
$|Y_{1}|+|Y_{2}|\equiv|\vec{Y}|=4$, and $Y_{1}=2$, $Y_{2}=1^{2}$. Such
degeneracy can happen at any even level higher or equal to 4. For example at
$\textrm{level}=6$:

$Y_{1}=3,Y_{2}=1^{3},\textrm{or}~{}Y_{1}=3,Y_{2}=\\{2,1\\},\textrm{or}~{}Y_{1}=1^{3},Y_{2}=\\{2,1\\},$

or simply, we have $(Y_{1},Y_{2})$ pair

$(3,1^{3}),(3,\\{2,1\\}),(1^{3},\\{2,1\\})$

Such degeneracy does not cause any problem in constructing the eigenstate of
$H$ for the following reasons.

i) The mother state $J^{+}_{-Y_{1}}J^{-}_{-Y_{2}}|P\rangle$ is uniquely
determined by the Young diagram, even for the degenerate $E$.

ii) Consider power expansion

$R(E)=\sum_{n=0}^{\infty}(\frac{1}{E_{Y_{1},Y_{2}}(P)-H_{0}}H_{I})^{n}.$

For an intermediate state.

$\displaystyle E_{Y_{1},Y_{2}}(P)-H_{0}$ $\displaystyle\sim
E_{Y_{1},Y_{2}}(P)-E_{Y^{{}^{\prime}}_{1},Y^{{}^{\prime}}_{2}}(P)$
$\displaystyle=E_{Y_{1}}+E_{Y_{2}}-E_{Y^{{}^{\prime}}_{1}}-E_{Y^{{}^{\prime}}_{2}}+2P(|Y_{1}|-|Y_{2}|-|Y^{{}^{\prime}}_{1}|+|Y^{{}^{\prime}}_{2}|)$

Since $|Y^{{}^{\prime}}_{1}|>|Y_{1}|$, $|Y^{{}^{\prime}}_{2}|<|Y_{2}|$ and
$|Y_{1}|-|Y^{{}^{\prime}}_{1}|+|Y^{{}^{\prime}}_{2}|-|Y_{2}|<0$ because of
strictly triangularity of $H_{I}$, so for a general value of $P$,
$\frac{1}{E_{Y_{1},Y_{2}}(P)-H_{0}}$ is not singular, and
$R(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle$ is well defined.

iii) The construction given above leads to the orthogonality of the state
$|P\rangle_{Y_{1},Y_{2}}$ for distinct $Y_{1},Y_{2}$ even for the degenerate
values of $E$, cf. eqs.(2,2,3).

iv) It can be proven that the eigenstate of $H$, constructed as in proposition
2, is actually the common eigenstate for all the conseved charges which
commute with $H$, with the mild assumption that all the conserved charges are
triangular in a similar way as $H$ is. Due to lack of space for the present
paper, we shall give a proof on this statement elsewhere.

Finally, we shall make a comment on the possible poles of $R(E)$ in the
complex $p-$plane. The $R(E)$ matrix elements is calculated based on the
following formula,

$\displaystyle|P\rangle_{Y_{1},Y_{2}}$
$\displaystyle=R(E)J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle\Omega_{Y_{1},Y_{2}}(P)$
$\displaystyle=\sum_{n=0}^{\infty}(\frac{1}{E_{Y_{1},Y_{2}}(P)-H_{0}}H_{I})^{n}J^{+}_{-Y_{1}}{J^{-}_{-Y_{2}}}|P\rangle\Omega_{Y_{1},Y_{2}}(P)\,.$

which always ends up with finite order perturbation because $H_{I}$ is
strictly triangular. We found, by the explicit calculations carried out so
far, that there is no pole in the finite $P$ complex plane. The poles in
$R(E)$ either cancels the zeros in $\Omega_{Y_{1},Y_{2}}(P)$ or simply cancels
by summing over all the relevant terms. Of course, this property is also the
necessary condition if $|{P}\rangle_{Y_{1},Y_{2}}$’s satisfy (1). Now the
general AFLT state can be written as

$\displaystyle|{P}\rangle_{Y_{1},Y_{2}}$ $\displaystyle=$
$\displaystyle\left\\{\Omega_{Y_{1},Y_{2}}(P)J_{-Y_{1}}^{+}J_{-Y_{2}}^{-}+\sum_{\begin{subarray}{|}|Y_{1}^{\prime}|=|Y_{1}|+1\\\
|Y_{2}^{\prime}|=|Y_{2}|-1\end{subarray}}C_{Y_{1},Y_{2}}^{Y_{1}^{\prime},Y_{2}^{\prime}}J_{-Y_{1}^{\prime}}^{+}J_{-Y_{2}^{\prime}}^{-}\right.$
$\displaystyle+$
$\displaystyle\sum_{\begin{subarray}{|}|Y_{1}^{\prime\prime}|=|Y_{1}|+2\\\
|Y_{2}^{\prime\prime}|=|Y_{2}|-2\end{subarray}}C_{Y_{1},Y_{2}}^{Y_{1}^{\prime\prime},Y_{2}^{\prime\prime}}J_{-Y_{1}^{\prime\prime}}^{+}J_{-Y_{2}^{\prime\prime}}^{-}$
$\displaystyle+$
$\displaystyle\left.\cdots+\sum_{|Y|=|Y_{1}|+|Y_{2}|}C_{Y_{1},Y_{2}}^{Y,\varnothing}J_{-Y}^{+}\right\\}|{P}\rangle,$

here $C_{Y_{1},Y_{2}}^{Y_{3},Y_{4}}$ is the transition coefficient which
measures the changing from the Young tableau vector $(Y_{1},Y_{2})$ to
$(Y_{3},Y_{4})$.

We have calculated those coefficients up to level 4, the explicit results(up
to level 3) are included in Appdix A. With the coefficients we calculated, one
can check that:

$\displaystyle Z_{bif}(\alpha|P^{\prime},\vec{X};P,\vec{Y})=$ (46)
$\displaystyle\sum_{(X_{1}^{\prime},X_{2}^{\prime}),(Y_{1}^{\prime},Y_{2}^{\prime})}\langle{P^{\prime}}|J^{-}_{X_{1}^{\prime}}J^{+}_{X_{2}^{\prime}}C_{X_{1},X_{2}}^{X_{1}^{\prime},X_{2}^{\prime}}V_{\alpha}C_{Y_{1},Y_{2}}^{Y_{1}^{\prime},Y_{2}^{\prime}}J_{-Y_{1}^{\prime}}J_{-Y_{2}^{\prime}}|{P}\rangle\,,$

holds true, thus (1) is verified. Here for simplicity, we have only verified
the cases without the incertions of the screening charges, i.e.
$P+P^{\prime}+\alpha=0$.

##  4 Conclusion and Perspective

The present work can be generalized in different ways. First, since the one
parameter Jack symmetric function is a special limit of the two parameter
Macdonald symmetric function, we expect that much of our work can be
generalized to the cases where Macdonald symmetric function plays a role. In
that case, we expect a similar relation to the NIC for 5d theory. Second, the
Calogero-Sutherland model is an integrable system. And consequently, Jack
symmetric function is the common eigenstate of the infinitely many commuting
charges which are deformed $W^{\infty}$ charges. And for the construction of
the AFLT states, the conserved charges are further deformed from those for the
Jack symmetric functions. The final construction should give the same results
as $I_{n}$ proposed in [7], which are constructed from integrable KdV
equations. We find in this case, the AFLT states remain to be the eigenstates
for all the conserved charges. However, it is desirable to have infinitely
many conserved charges constructed explicitly. Third, the reflection symmetry
studied in this paper is actually powerful enough to give a closed form for
the construction of the AFLT states. We shall present this result in our
future work. Another interesting idea related to our work is to consider the
Jack function as a perturbation away from the Schur function, we have found
that similar formalism applies [37]. Finally, it is very interesting to see
how we present the full pants diagram for the conformal blocks, comparing to
the one we have considered with one external leg.

##  5 Acknowledgments

We are grateful to Yi-hong Gao for detailed explanations on the the AGT
conjecture and the related topics.
This work is part of the CAS program “Frontier Topics in Mathematical Physics”
(KJCX3-SYW-S03) and is supported partially by a national grant NSFC(11035008).

##  Appendix A Coefficients for AFLT States(up to level 3)

Now we give the explicit construction of the AFLT states up to level 3. The
transition coefficients $C_{Y_{1},Y_{2}}^{Y_{1}^{\prime},Y_{2}^{\prime}}$ are
defined as,

$C_{Y_{1},Y_{2}}^{Y_{1}^{\prime},Y_{2}^{\prime}}\equiv
R_{Y_{1},Y_{2}}^{Y_{1}^{\prime},Y_{2}^{\prime}}(E)\Omega_{Y_{1},Y_{2}}(P),\ \
\ \ C_{Y_{1},Y_{2}}^{Y_{1},Y_{2}}\equiv\Omega_{Y_{1},Y_{2}}(P)\,.$

Level 2 coefficients:

$C_{0,1^{2}}^{1,1}=C_{0,2}^{1,1}=-4b\left(1+b^{2}\right)P,\\\
C_{1,1}^{1^{2},0}=\frac{\left(1+b^{2}\right)(1+2bP)}{-1+b^{2}},\\\
C_{1,1}^{2,0}=-\frac{b^{2}\left(1+b^{2}\right)(1+2bP)}{-1+b^{2}},\\\
C_{0,1^{2}}^{12,0}=1+b^{2}\left(3+2b\left(b-\frac{2\left(1+b^{2}\right)P}{-1+b^{2}}\right)\right),\\\
C_{0,1^{2}}^{2,0}=\frac{4b^{3}\left(1+b^{2}\right)P}{-1+b^{2}},\\\
C_{0,2}^{2,0}=\frac{\left(1+b^{2}\right)P^{2}\left(-2+b^{2}+b^{4}+4bP\right)}{-1+b^{2}},\\\
C_{0,2}^{1^{2},0}=\frac{4b\left(1+b^{2}\right)P}{-1+b^{2}}$

Level 3 coefficients:

$C_{2,1}^{3,0}=\left\\{-\frac{b^{3}\left(1+b^{2}\right)(b+2P)\left(1+b^{2}+2bP\right)}{-2+b^{2}}\right\\},\\\
C_{2,1}^{\\{2,1\\},0}=\left\\{\frac{4\left(1+b^{2}\right)(1+bP)\left(1+b^{2}+2bP\right)}{-2+b^{2}}\right\\},\\\
C_{1^{2},1}^{1^{3},0}=\left\\{\frac{\left(1+b^{2}\right)(1+2bP)\left(1+b^{2}+2bP\right)}{-1+2b^{2}}\right\\},\\\
C_{1^{2},1}^{\\{2,1\\},0}=\left\\{-\frac{4b^{3}\left(1+b^{2}\right)(b+P)\left(1+b^{2}+2bP\right)}{-1+2b^{2}}\right\\},\\\
C_{1,2}^{2,1}=\left\\{-\frac{2b^{2}\left(1+b^{2}\right)(1+2bP)\left(-1+b^{2}+2bP\right)}{-1+b^{2}}\right\\},\\\
C_{1,2}^{1^{2},1}=\left\\{\frac{4b\left(1+b^{2}\right)P(1+2bP)}{-1+b^{2}}\right\\},\\\
C_{1,2}^{3,0}=\left\\{\frac{b^{3}\left(1+b^{2}\right)\left(b+b^{3}+4P\right)\left(-1+b^{2}+2bP\right)}{2-3b^{2}+b^{4}}\right\\},\\\
C_{1,2}^{\\{2,1\\},0}=\left\\{-\frac{4\left(1+b^{2}\right)^{2}(1+2bP)(-1+2b(b+P))}{2-5b^{2}+2b^{4}}\right\\},\\\
C_{1,2}^{1^{3},0}=\left\\{\frac{4b\left(1+b^{2}\right)P(1+2bP)}{1-3b^{2}+2b^{4}}\right\\},\\\
C_{1,1^{2}}^{2,1}=\left\\{-\frac{4b^{4}\left(1+b^{2}\right)P(b+2P)}{-1+b^{2}}\right\\},\\\
C_{1,1^{2}}^{1^{2},1}=\left\\{-\frac{2b\left(1+b^{2}\right)(b+2P)\left(-1+b^{2}-2bP\right)}{-1+b^{2}}\right\\},\\\
C_{1,1^{2}}^{3,0}=\left\\{\frac{4b^{6}\left(1+b^{2}\right)P(b+2P)}{2-3b^{2}+b^{4}}\right\\},\\\
C_{1,1^{2}}^{\\{2,1\\},0}=\left\\{\frac{4b^{3}\left(1+b^{2}\right)^{2}(b+2P)\left(-2+b^{2}-2bP\right)}{2-5b^{2}+2b^{4}}\right\\},\\\
C_{1,1^{2}}^{1^{3},0}=\left\\{-\frac{\left(1+b^{2}\right)\left(-1+b^{2}-2bP\right)\left(1+b^{2}+4b^{3}P\right)}{1-3b^{2}+2b^{4}}\right\\},\\\
C_{0,3}^{1,2}=\left\\{-6b\left(1+b^{2}\right)P(-1+2bP)\right\\},\\\
C_{0,3}^{2,1}=\left\\{\frac{6b\left(1+b^{2}\right)P\left(-4+b^{2}+b^{4}+4bP\right)}{-1+b^{2}}\right\\},\\\
C_{0,3}^{1^{2},1}=\left\\{-\frac{12b\left(1+b^{2}\right)P(-1+2bP)}{-1+b^{2}}\right\\},\\\
C_{0,3}^{3,0}=\left\\{-\frac{\left(1+b^{2}\right)\left(12+b\left(b\left(1+b^{2}\right)\left(-8+b^{2}+b^{4}\right)+12\left(-3+b^{2}+b^{4}\right)P+24bP^{2}\right)\right)}{2-3b^{2}+b^{4}}\right\\},\\\
C_{0,3}^{1^{3},0}=\left\\{-\frac{12b\left(1+b^{2}\right)P(-1+2bP)}{1-3b^{2}+2b^{4}}\right\\},\\\
C_{0,3}^{\\{2,1\\},0}=\left\\{\frac{12b\left(1+b^{2}\right)P\left(-5+3b^{2}+2b^{4}+6bP\right)}{2-5b^{2}+2b^{4}}\right\\},\\\
C_{0,1^{3}}^{1,1^{2}}=\left\\{6b^{2}\left(1+b^{2}\right)(b-2P)P\right\\},\\\
C_{0,1^{3}}^{2,1}=\left\\{-\frac{12b^{4}\left(1+b^{2}\right)(b-2P)P}{-1+b^{2}}\right\\},\\\
C_{0,1^{3}}^{1^{2},1}=\left\\{\frac{6b\left(1+b^{2}\right)\left(-1+b^{2}(-1+4b(b-P))\right)P}{-1+b^{2}}\right\\},\\\
C_{0,1^{3}}^{3,0}=\left\\{\frac{12b^{6}\left(1+b^{2}\right)(b-2P)P}{2-3b^{2}+b^{4}}\right\\},\\\
C_{0,1^{3}}^{1^{3},0}=\left\\{-\frac{1}{1-3b^{2}+2b^{4}}\left(1+b^{2}\right)\left(1+b^{2}(2+b(12P+b(-7+4b(b(-2+3(b-2P)(b-P))+3P))))\right)\right\\},\\\
C_{0,1^{3}}^{\\{2,1\\},0}=\left\\{-\frac{12b^{3}\left(1+b^{2}\right)P\left(-2+b^{2}\left(-3+5b^{2}-6bP\right)\right)}{2-5b^{2}+2b^{4}}\right\\},\\\
C_{0,\\{2,1\\}}^{1,1^{2}}=\left\\{\frac{2b^{2}\left(-2+b^{2}\right)\left(1+b^{2}\right)(b-2P)P}{-1+b^{2}}\right\\},\\\
C_{0,\\{2,1\\}}^{1,2}=\left\\{-\frac{2b\left(-1+b^{2}+2b^{4}\right)P(-1+2bP)}{-1+b^{2}}\right\\},\\\
C_{0,\\{2,1\\}}^{2,1}=\left\\{\frac{4b\left(1+b^{2}\right)^{2}P\left(-1+b^{2}+2bP\right)}{-1+b^{2}}\right\\},\\\
C_{0,\\{2,1\\}}^{1^{2},1}=\left\\{\frac{4b\left(1+b^{2}\right)^{2}P\left(-1+b^{2}-2bP\right)}{-1+b^{2}}\right\\},\\\
C_{0,\\{2,1\\}}^{3,0}=\left\\{-\frac{2b^{3}\left(1+b^{2}\right)P\left(-3+2b\left(b+b^{3}+3P\right)\right)}{2-3b^{2}+b^{4}}\right\\},\\\
C_{0,\\{2,1\\}}^{\\{2,1\\},0}=\left\\{-\frac{\left(1+b^{2}\right)\left(4-17b^{4}+4b^{8}-2b\left(4+3b^{2}+3b^{4}+4b^{6}\right)P-36b^{4}P^{2}\right)}{2-5b^{2}+2b^{4}}\right\\},\\\
C_{0,\\{2,1\\}}^{1^{3},0}=\left\\{\frac{2b\left(1+b^{2}\right)\left(-2+b^{2}(-2+3b(b-2P))\right)P}{1-3b^{2}+2b^{4}}\right\\}$

In the above expressions, 0 labels $\\{\varnothing\\}$.

## References

  * [1] A. A. Belavin, A. M. Polyakov and A. B. Zamolodchikov, “Infinite Conformal Symmetry in Two-Dimensional Quantum Field Theory,” Nucl. Phys. B 241, 333 (1984).
  * [2] H. Sonoda, “Sewing Conformal Field Theories,” Nucl. Phys. B311, 401 (1988).
  * [3] H. Sonoda, “Sewing Conformal Field Theories. 2.,” Nucl. Phys. B311, 417 (1988).
  * [4] G. W. Moore, N. Seiberg, “Classical and Quantum Conformal Field Theory,” Commun. Math. Phys. 123, 177 (1989).
  * [5] Al. B. Zamolodchikov, “Conformal symmetry in two-dimensional space: Recursion representation of conformal block , Teoret. Mat. Fiz., 73:1 (1987), 103 C110
  * [6] Al. B. Zamolodchikov, “Conformal symmetry in two dimensions: An explicit recurrence formula for the conformal partial wave amplitude”, Commun. Math. Phys. 96, 3, 419-422 (1984)
  * [7] V. A. Alba, V. A. Fateev, A. V. Litvinov and G. M. Tarnopolsky, “On combinatorial expansion of the conformal blocks arising from AGT conjecture,” arXiv:1012.1312 [hep-th].
  * [8] G. Felder, “BRST Approach to Minimal Methods,” Nucl. Phys. B317, 215 (1989).
  * [9] D. Bernard, 2, G. Felder, “Fock Representations and BRST Cohomology in SL(2) Current Algebra,” Commun. Math. Phys. 127, 145 (1990).
  * [10] E. Carlsson and A. Okounkov, “Exts and Vertex Operators,” arXiv: 0801.2565v2.
  * [11] N. Seiberg, E. Witten, “Electric - magnetic duality, monopole condensation, and confinement in N=2 supersymmetric Yang-Mills theory,” Nucl. Phys. B426, 19-52 (1994). [hep-th/9407087].
  * [12] N. Seiberg, E. Witten, “Monopoles, duality and chiral symmetry breaking in N=2 supersymmetric QCD,” Nucl. Phys. B431, 484-550 (1994). [hep-th/9408099].
  * [13] L. F. Alday, D. Gaiotto and Y. Tachikawa, “Liouville Correlation Functions from Four-dimensional Gauge Theories,” Lett. Math. Phys. 91, 167 (2010) [arXiv:0906.3219 [hep-th]].
  * [14] N. A. Nekrasov and S. L. Shatashvili, “Quantization of Integrable Systems and Four Dimensional Gauge Theories,” arXiv:0908.4052 [hep-th].
  * [15] N. A. Nekrasov, “Seiberg-Witten prepotential from instanton counting,” Adv. Theor. Math. Phys. 7, 831 (2004) [arXiv:hep-th/0206161].
  * [16] N. Nekrasov and A. Okounkov, “Seiberg-Witten theory and random partitions,” arXiv:hep-th/0306238.
  * [17] R. Dijkgraaf and C. Vafa, “Toda Theories, Matrix Models, Topological Strings, and N=2 Gauge Systems,” arXiv:0909.2453 [hep-th].
  * [18] M. F. Atiyah, N. J. Hitchin, V. G. Drinfeld, Y. .I. Manin, “Construction of Instantons,” Phys. Lett. A65, 185-187 (1978).
  * [19] N. Seiberg, E. Witten, “String theory and noncommutative geometry,” JHEP 9909, 032 (1999). [hep-th/9908142].
  * [20] M. R. Douglas, “Branes within branes,” arXiv:hep-th/9512077.
  * [21] E. Witten, “Solutions of four-dimensional field theories via M theory,” Nucl. Phys. B500, 3-42 (1997). [hep-th/9703166].
  * [22] S. H. Katz, A. Klemm, C. Vafa, “Geometric engineering of quantum field theories,” Nucl. Phys. B497, 173-195 (1997). [hep-th/9609239].
  * [23] H. Nakajima, K. Yoshioka, “Instanton counting on blowup. 1.,” [math/0306198 [math-ag]].
  * [24] H. Nakajima, K. Yoshioka, “Instanton counting on blowup. II. K-theoretic partition function,” [math/0505553 [math-ag]].
  * [25] H. Nakajima, K. Yoshioka, “Lectures on instanton counting,” [math/0311058 [math-ag]].
  * [26] L. F. Alday and Y. Tachikawa, “Affine SL(2) conformal blocks from 4d gauge theories,” Lett. Math. Phys. 94, 87 (2010) [arXiv:1005.4469 [hep-th]].
  * [27] M. C. N. Cheng, R. Dijkgraaf and C. Vafa, “Non-Perturbative Topological Strings And Conformal Blocks,” arXiv:1010.4573 [hep-th].
  * [28] H. Itoyama and T. Oota, “Method of Generating q-Expansion Coefficients for Conformal Block and N=2 Nekrasov Function by beta-Deformed Matrix Model,” Nucl. Phys. B 838, 298 (2010) [arXiv:1003.2929 [hep-th]].
  * [29] A. Belavin and V. Belavin, “AGT conjecture and Integrable structure of Conformal field theory for c=1,” Nucl. Phys. B 850, 199 (2011) [arXiv:1102.0343 [hep-th]].
  * [30] V. Belavin and B. Feigin, “Super Liouville conformal blocks from N=2 SU(2) quiver gauge theories,” arXiv:1105.5800 [hep-th].
  * [31] N. Dorey, S. Lee and T. J. Hollowood, “Quantization of Integrable Systems and a 2d/4d Duality,” arXiv:1103.5726 [hep-th].
  * [32] H. Awata, H. Fuji, H. Kanno, M. Manabe and Y. Yamada, “Localization with a Surface Operator, Irregular Conformal Blocks and Open Topological String,” arXiv:1008.0574 [hep-th].
  * [33] A. Mironov, A. Morozov and S. Shakirov, “On ’Dotsenko-Fateev’ representation of the toric conformal blocks,” J. Phys. A 44, 085401 (2011) [arXiv:1010.1734 [hep-th]].
  * [34] A. Mironov, A. Morozov and S. Shakirov, “Conformal blocks as Dotsenko-Fateev Integral Discriminants,” Int. J. Mod. Phys. A 25, 3173 (2010) [arXiv:1001.0563 [hep-th]].
  * [35] J. f. Wu, “Note on refined topological vertex, Jack polynomials and instanton counting,” arXiv:1012.2147 [hep-th].
  * [36] J. F. Wu, Y. Y. Xu and M. Yu, “Recursions in Calogero-Sutherland Model Based on Virasoro Singular Vectors,” arXiv:1107.4234 [hep-th].
  * [37] J. F. Wu and M. Yu, work in progress.
  * [38] R. P. Stanley, “Some Combinatorial Properties of Jack Symmetric Functions”, Advances in Mathematics 77, 76-115, (1989)
  * [39] I. G. Macdonald, Symmetric Functions and Hall Polynomials, 1995, 2nd Edition, Clarendon Press Oxford