††institutetext: Berkeley Center for Theoretical Physics and Department of
Physics
University of California, Berkeley, CA 94720, U.S.A.

# Chern-Simons Theory, Ehrhart Polynomials, and Representation Theory

Chao Ju<EMAIL_ADDRESS>
###### Abstract

The Hilbert space of level $q$ Chern-Simons theory of gauge group $G$ of the
ADE type quantized on $T^{2}$ can be represented by points that lie on the
weight lattice of the Lie algebra $\mathfrak{g}$ up to some discrete
identifications. Of special significance are the points that also lie on the
root lattice. The generating functions that count the number of such points
are quasi-periodic Ehrhart polynomials which coincide with the generating
functions of $SU(q)$ representation of the ADE subgroups of $SU(2)$ given by
the McKay correspondence. This coincidence has roots in a string/M theory
construction where D3(M5)-branes are put along an ADE singularity. Finally, a
new perspective on the McKay correspondence that involves the inverse of the
Cartan matrices is proposed.

††arxiv: 2304.11830

##  1 Introduction

The 3-dimensional Chern-Simons theory is a remarkably rich tool that has
applications in both mathematics and physics. For example, Witten showed that
Chern-Simons theory is connected to the Jones polynomial and knot invariants
wittenjones . It is also applied in the study of S-duality ganorandhong ;
gaiottoandwitten , a subject related to the geometric Langlands program
wittenkapustin . In terms of the more tangible physics applications, Chern-
Simons theory has been used extensively to study physics on 2-dimensional
surfaces. For example, it is used to endow particles in 2+1 dimensions with
fractional statistics wilczek ; wilczekandzee , so that upon exchanging two
identical particles the wave function of the two particles can end up with a
phase different from $\pm 1$. In addition, Laughlin explained fractional
quantum hall effect by applying Chern-Simons theory111For a review of Chern-
Simons theory and fractional quantum hall effect, see zee  and the references
therein. laughlin .

The fact that Chern-Simons theory is important for 2+1 dimensional physics is
no accident. The Chern-Simons action

$S=\frac{q}{4\pi}\int\text{tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge
A\right)$ (1)

is the unique relativistically invariant action of the gauge field $A$ that
has only one derivative acting on $A$. In this paper, we take the convention
that $A$ is $\mathfrak{g}$ valued 1-form and the trace is taken in the
representation such that $q$ is a quantized positive integer. We take
$\mathfrak{g}$ to be simply laced, namely $\mathfrak{su}(N)$,
$\mathfrak{so}(2N)$, $\mathfrak{e}_{6}$, $\mathfrak{e}_{7}$, and
$\mathfrak{e}_{8}$. The global property of the gauge group will not concern us
in this paper. The geometry we are interested in is $T^{2}\times\mathbb{R}$
where $\mathbb{R}$ is the time direction. Upon quantizing the Chern-Simons
theory in this geometry, one gets a discrete set of states that can be
identified with points in the set cs

$\frac{\Lambda_{w}}{W\ltimes q\Lambda_{r}}$ (2)

where $\Lambda_{w}$, $\Lambda_{r}$ are the weight and the root lattice of
$\mathfrak{g}$, and $W$ is the Weyl group.

In this paper, we study the counting of a special class of states that lie in
the set

$\frac{\Lambda_{w}}{W\ltimes q\Lambda_{r}}\cap\Lambda_{r}$ (3)

where the intersection with $\Lambda_{r}$ picks out the states in the set (2)
that are also roots.

We will show that the counting of such states leads to a curious connection to
Ehrhart polynomials, McKay correspondence, and representation theory. Ehrhart
polynomials were first constructed to count lattice points in rational
polytopes ehrhart , a problem that is in general NP-hard deloera  to solve by
computer. Ehrhart polynomials have been found to connect different areas of
mathematics such as number theory, geometry, and topology222For an extensive
introduction, see the book beckandrobins . Since closed-form Ehrhart
polynomials are rare and are only for very special geometries, our computation
in this paper will add more examples to the known collection of Ehrhart
polynomials.

We will also show that the Ehrhart polynomials we obtain from the geometric
point of view can also be obtained from a representation theory point of view.
In fact, the latter comes from a dual formulation of the problem. The two
approaches are connected by the McKay correspondence mckay , which gives a ADE
Dynkin diagram classification of discrete subgroups of $SU(2)$ and their
irreducible representations. The physics behind this dual formulation of the
problem comes from a holographic system of D-branes on ADE singularity. More
precisely, there is a duality between the ground state Hilbert space $SU(q)$
$\mathcal{N}=4$ Supersymmetric Yang-Mills theory on $S^{3}/\Gamma$ and a
certain subspace (3) of the Hilbert space of level $q$ Chern-Simons theory on
$T^{2}$ with gauge algebra $\mathfrak{g}(\Gamma)$ given by the McKay
correspondence.

This paper is organized as follows. Because our construction has to do with
D-branes on ADE singularity, in section 2 we will review this subject and
point out the difference between our construction and the literature. In
section 3, we review the quantization of Chern-Simons theory on $T^{2}$ and
formulate the Hilbert space geometrically in terms of certain points on the
Lie algebra lattice. To illustrate the rather abstract notation, we give an
example of states of $\mathfrak{g}=\mathfrak{su}(3)$. In section 4, we pose
the problem of counting the number of the special class of states in the set
defined in (2). We shall find that the problem is the same as counting lattice
points in rational polytopes, and that the generating function for counting
the states is exactly the corresponding Ehrhart polynomial. In section 5, we
compute the explicit form of the Ehrhart polynomial for a specific case by
using the $\Omega$ operator introduced by MacMahon macmahon . The general case
is solved by reverse-engineering some representation theory formulas in
section 6. We will see that the Ehrhart polynomial that counts the special
states at level $q$ with gauge algebra $\mathfrak{g}$ is the same as the
generating function for the $SU(q)$ representation of the ADE subgroup given
by the McKay correspondence. In section 7, we extend our result to the
D-series where the gauge algebra of Chern-Simons theory is
$\mathfrak{so}(2(N+2))$, $N\geq 1$. In section 8, we discuss some curious
representation theory properties implied by the inverse of Cartan matrices
modulo 1 for ADE Lie algebras by focusing on the exceptional series
$\mathfrak{e}_{6}$, $\mathfrak{e}_{7}$, and $\mathfrak{e}_{8}$. In section 9,
we derive the string/M theory origin of the proposed duality.

The reader who is only interested in the physical aspect of this work can skip
the second half of section 4 and all of section 5.

##  2 String Theory and D-branes on Orbifold

There is a long history of the study of string theory and D-branes on orbifold
singularity. Since the particular orbifold singularity that will concern us in
this paper is the ADE orbifold singularity, we will in this section do a quick
review of some past research on ADE singularity and string theory. We will
discuss how this work is different from past research and motivate the Chern-
Simons/SYM duality mentioned in the introductory section 1.

There are roughly speaking three uses of putting string theory or D-branes on
ADE singularity. The first is to obtain a more realistic compactification of
string theory down to lower dimensions. The second is to use string theory to
probe the topology and geometry of the ADE singularity. The third is to use
string theory construction to obtain new classes of quantum field theory.

One way to obtain ADE singulairty is the “blow-down” of K3 surface stringorb ,
the only nontrivial Calabi-Yau manifold in four (real) dimensions aspinwall .
String theory on K3 surface leads to more realistic models of string
compactification because the $SU(2)$ holonomy of K3 surface helps break half
of the supersymmetry. It also leads to more examples of string dualities
aspinwall . In fact, as we will see in section 9, the duality considered in
this paper is derived using the duality between heterotic string on $T^{4}$
and IIA string on K3.

The second use partially overlaps with the first use because the low energy
spectra of string theory on K3 sheds light on the topological invariants of
the K3 surface stringorb ; aspinwall . To probe the geometry and not just the
topology of K3 surface, D-brane technologies must be used johnsonandmyers . In
johnsonandmyers , the metric of some ALE (asymptotically locally Euclidean)
space was computed by probing the space using D1 branes transverse to the
space. The fact that the moduli space of vacua for the low energy theory
coincides with the ALE space suggests that the metric of the underlying space
coincides with the metric that enters the kinetic term of the low energy
theory.

The third use dates back to stringcomments , where it was mentioned that
M-theory on ADE singulairty $\Gamma$ leads to a seven dimensional super Yang
Mills theory transverse to the 4-dimensional ADE singularity with the gauge
algebra $\mathfrak{g}(\Gamma)$ McKay-dual to the ADE singularity. In douglas
gomis , quiver gauge theories are constructed by putting D3 branes transverse
to the orbifold singularity. In the language of AdS-CFT correspondence, in
this construction, the singularity mods out the $S^{5}$ part of the bulk
geometry and keeps the $AdS_{5}$ part untouched.

The difference between this work and the previous constructions is that here,
the ADE singularity is longitudinal to the D3 brane worldvolume. In other
words, the bulk geometry is $AdS_{5}/\Gamma\times S^{5}$. As will be derived
in section 9, the long distance limit of this holographic system will involve
a Chern-Simons theory in the AdS bulk direction. Constructions similar to this
work can be found in dijkgraaf vafawitten . Differences between our
construction and those in dijkgraaf vafawitten  will be explained in section
9.

##  3 Quantization of Chern-Simons Theory on $T^{2}$ and the States

The Chern-Simons theory with gauge algebra $\mathfrak{g}$ on $T^{2}$ can be
quantized using the standard quantization procedure by choosing a gauge
$A_{0}=0$ and imposing the constraint $\delta S/\delta A_{0}=0$, where $S$ is
defined in equation (1). The constraint gives the condition that the
connections are flat $F=dA+A\wedge A=0$. Imposing the constraint, using the
remaining freedom to gauge transform the connection $A$ into the maximal torus
of $\mathfrak{g}$, and imposing the canonical commutation relation, one
obtains that the states are in one-to-one correspondence with points in the
set cs  described in equation (2).

In words, states are points on the weight lattice of $\mathfrak{g}$ with two
states being identified if they differ by some combinations of the Weyl
transformation and $q$ times the root lattice translation. We shall henceforth
call this set the state set. Notice that because of the identification by
$q\Lambda_{r}$, the number of states is finite, a fact that can also be seen
from the compactness of the Chern-Simons phase space on $T^{2}$. In the large
$q$ limit, the number of states is given by

$\frac{q^{r}\det C}{\absolutevalue{W}}$ (4)

where $r$ is the rank of $\mathfrak{g}$, $C$ is the Cartan matrix, and
$\absolutevalue{W}$ is the size of the Weyl group. This formula can either be
derived from equation (2) by noting that $\det
C=\absolutevalue{\Lambda_{w}/\Lambda_{r}}$ or going back to canonical
quantization and demanding that the phase space contains one state per $2\pi$
cell (we set $\hbar=1$).

To give an example of the state set of $\mathfrak{su}(3)$ Chern-Simons theory
at level $q=1$ and $q=2$, we present the figure (see Fig. 1) from wenandzee
333In wenandzee , a more pedestrian way of quantizing the Chern-Simons theory
on $T^{2}$ for is used..

Figure 1: In both pictures, $e_{1}$ and $e_{2}$ are the simple roots, and
$d_{1}$ and $d_{2}$ are the fundamental weights. In the top picture, three
black dots represent the three unique states of the level $q=1$ theory. In the
bottom picture, the six black dots are the unique states of the level $q=2$
theory. It is easy to convince oneself that one can reach the white dots or
other weight lattice points through a combination of Weyl reflections and $q$
times the root lattice translation on the black dots. Figure retrieved from
wenandzee .

Fig. 1 shows the unique states in the Hilbert space for the level 1 and the
level 2 theory, respectively. Notice that the pattern continues to all levels:
all states lie in the fundamental Weyl chamber, and that as the level
increases, the number of state increases quadratically (roughly as the area of
the shaded region). In fact, the number of states for the $\mathfrak{su}(3)$
theory at level $q$ is

$\frac{q(q+1)}{2}$ (5)

This formula can be derived as follows. For simply laced algebra (which is our
only focus in this paper), the root lattice $\Lambda_{r}$ is the same as the
coroot lattice $\Lambda_{cr}$. A theory in affine Lie algebra shows that the
state set (2) is simply the highest weight representation of the corresponding
affine Lie algebra $\tilde{\mathfrak{g}}$ at level $q$ kac . Let a state be
labeled by the Dynkin label $(a_{1},a_{2},...,a_{r})$, where each $a_{i}$ is
some nonnegative integer. The highest weight states satisfy the inequality

$\sum_{i=1}^{r}c_{i}a_{i}\leq q$ (6)

where $c_{i}\in\mathbb{Z}_{>0}$ is the coefficient multiplying the $i$th
simple root $\alpha_{i}$ in the expression for the highest root $\theta$:

$\theta\equiv\sum_{i=1}^{r}c_{i}\alpha_{i}$ (7)

For the case of $\mathfrak{su}(3)$, the numbers are $c_{1}=c_{2}=1$, so the
number of Chern-Simons states at level $q$ is the same as the number of
solutions to the inequality

$a_{1}+a_{2}\leq q$

and solving for this gives exactly the quadratic formula (5).

Although the example is given for $\mathfrak{su}(3)$, the reader should keep
in mind the generalization of the picture to other ADE gauge algebras. In the
next section, we look at a special class of states on the state lattice.

##  4 A Special Class of States

###  4.1 The Geometry and the Counting Problem

The special states we want to focus on are those that belong to the set

$Q_{q}=\frac{\Lambda_{w}}{W\ltimes q\Lambda_{r}}\cap\Lambda_{r}$ (8)

namely states that also roots. The physical motivation for considering such
states is mentioned in ganorandju  and is explained in the section 9.
Following the $\mathfrak{su}(3)$ example in Fig. 1, at level 1 there is one
state that belongs to $Q_{q}$, the state at the origin. At level 2 there are
two states, the additional one being at the position $d_{1}+d_{2}$.

The question we pose is: given $\mathfrak{g}$ and level $q$, how many states
are in $Q_{q}$? There are two equivalent formulations of this counting
problem. The first formulation uses the Cartan matrix and it naturally leads
to the concept of Ehrhart polynomials. The second (dual) formulation uses the
inverse of the Cartan matrix, and it leads to representation theory and
connections to string theory. The rest of the section focuses on the first
formulation of the problem.

Let the Cartan matrix of $\mathfrak{g}$ be $C$. In the basis of Dynkin labels,
the rows of $C$ give the Dynkin coefficients of the simple roots. Since we are
dealing with simply laced Lie algebras, $C^{T}=C$, so that the columns of $C$
also give the representation of the simple roots. If $y\in Q_{q}$, $y$ being
the Dynkin label (as a column vector) of some state in the theory, then
$y\in\Lambda_{r}$ by definition, so that it can be represented as a linear
combination of the simple roots:

$y=Cx$

where $x\in\mathbb{Z}_{\geq 0}^{r}$ is a column of nonnegative integers. The
nonnegativity of $x$ is because states in $Q_{q}$ are all dominant weights.
What constraints does $y$ have to satisfy? The first constraint on $y$ is that
the entries are nonnegative (note that $x\geq 0$ does not imply $y\geq 0$).
The second constraint is the level $q$ constraint in equation (6). In terms of
$x$, the constraints read

$\displaystyle\sum_{j=1}^{r}C_{ij}x_{j}$ $\displaystyle\geq 0$ (9)
$\displaystyle\sum_{i,j=1}^{r}c_{i}C_{ij}x_{j}$ $\displaystyle\leq q$ (10)

Geometrically, the above constraints define a rational polytope, a polygon
whose vertices have rational coordinates. The solutions to the above
constraints are simply integer points contained inside the rational polytope.
By adding slack variables, any rational polytope can be represented as a
system of linear equations beckandrobins

$Ax^{\prime}=b$ (11)

for some matrix $A$, vector $b$, and unknowns $x^{\prime}$.

For our problem, we need $r+1$ slack variables
$k_{1},k_{2},...,k_{r+1}\in\mathbb{Z}_{\geq 0}$, so that the system of
inequalities reduces to the system of equalities:

$\displaystyle\sum_{j=1}^{r}C_{ij}x_{j}-k_{i}$ $\displaystyle=0$ (12)
$\displaystyle\sum_{i,j=1}^{r}c_{i}C_{ij}x_{j}+k_{r+1}$ $\displaystyle=q$ (13)

Stacking the vector $x$ and $k$ into $x^{\prime}=(x,k)$, we have the system of
linear equation $Ax^{\prime}=b$ where $A$ is $r+1$ by $2r+1$ and $b$ is $r+1$
by 1:

$\displaystyle A$
$\displaystyle=\begin{pmatrix}C_{11}&C_{12}&\ldots&C_{1r}&-1&0&0&\ldots&0\\\
C_{21}&C_{22}&\ldots&C_{2r}&0&-1&0&\ldots&0\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
C_{r1}&C_{r2}&\ldots&C_{rr}&0&0&0&-1&0\\\
\sum_{i}c_{i}C_{i1}&\sum_{i}c_{i}C_{i2}&\ldots&\sum_{i}c_{i}C_{ir}&0&\ldots&0&\ldots&1\end{pmatrix}$
(14) $\displaystyle b$ $\displaystyle=\begin{pmatrix}0\\\ 0\\\ \vdots\\\ 0\\\
q\end{pmatrix}$ (15)

The problem of counting states in $Q_{q}$ is then transformed into counting
the solutions to this system of linear equations.

###  4.2 Ehrhart Polynomials

What we did in the previous section is under the guise of Ehrhart polynomials
ehrhart . They are generating functions for counting lattice points contained
in polytopes. Let us formulate the theory following the notation of
beckandrobins . Let a rational polytope $\mathcal{P}$ be specified by a system
of linear equations with slack variables added:

$\mathcal{P}=\\{x\in\mathbb{R}^{d}_{\geq 0}:Ax=b\\}$ (16)

for some integer valued matrix $A$ and integer valued vector $b$. One
considers the $t$th dilate of $\mathcal{P}$, defined as

$t\mathcal{P}=\\{x\in\mathbb{R}^{d}_{\geq 0}:Ax=tb\\}$ (17)

where $t\in\mathbb{Z}_{>0}$. Let $L_{\mathcal{P}}(t)$ denote the number of
lattice points contained in the $t\mathcal{P}$:

$L_{\mathcal{P}}(t)=\\#\\{x\in\mathbb{Z}^{d}_{\geq 0}:Ax=tb\\}$ (18)

The Ehrhart polynomial associated to the polytope $\mathcal{P}$ is defined as

$\text{Ehr}_{\mathcal{P}}(z)=1+\sum_{t\geq 1}L_{\mathcal{P}}(t)z^{t}$ (19)

In our problem, the polytope is given as the $q$th dilate of

$Ax^{\prime}=\begin{pmatrix}0\\\ 0\\\ \vdots\\\ 0\\\ 1\end{pmatrix}$

with $A$ and $x^{\prime}$ defined in the previous section (see equation (14)
and equation (15)). We call this base polytope $\mathcal{Q}_{\mathfrak{g}}$,
where the dependence on the Lie algebra $\mathfrak{g}$ is made explicit as
each simply laced Lie algebra has its unique polytope. Since we are interested
in computing the number of special states for each level $q>0$, we want to
find the number of lattice points contained in the $q$th dilate of
$\mathcal{Q}_{\mathfrak{g}}$ for each positive q. Therefore, the question
posed in the previous section can be now phrased as finding the Ehrhart
polynomial of $\mathcal{Q}_{\mathfrak{g}}$:

$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)=?$

This concludes the first formulation of our problem. The dual formulation of
the problem in terms of the inverse of Cartan matrices will be introduced in
section 6. In the next section, we develop a formal method to compute
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ in the most general way possible.

##  5 Computation of $\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$

In this section and the next, we compute
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ for all simply laced
$\mathfrak{g}$. We give two approaches to this computation. The first approach
uses MacMahon’s $\Omega$ operator method macmahon  but quickly becomes tedious
when the level $q$ becomes large. However, the merit of this approach is that
it is general: it can be applied to solving for all Ehrhart polynomials given
the constraints, and it will be amiss if we do not discuss the general
solution. Due to the computational difficulty, we shall only use this approach
to give the explicit formula for $\mathcal{Q}_{\mathfrak{g}}$ for the case
$\mathfrak{g}=\mathfrak{su}(2)$. MacMahon’s method is reviewed in section 5.1
and the computation for $\mathfrak{su}(2)$ is done in section 5.2.

The second formulation of the problem uses a hint from representation theory,
and leads to the expression of $\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ in
one full sweep. The second approach is inspired from a duality relation
constructed from string theory (see section 9 and ref dijkgraaf vafawitten ),
without which it is not obvious how one can make a connection of
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ to being solved by representation
theory.

###  5.1 The $\Omega$ Operator

In computing the number of ways of partitioning some integer $u$ into a sum of
$n$ nonnegative integers $a_{1}+a_{2}+...+a_{n}=u$, the order of the integers
in the sum does not matter. Therefore, the problem of counting n-partitions of
$u$ is quite different from the problem of counting solutions to the the
equation $a_{1}+a_{2}+...+a_{n}=u$, where the number of solutions is the
coefficient of the $x^{u}$ term in the generating function

$\frac{1}{(1-x)^{n}}$ (20)

To introduce the $\Omega$ operator, we focus on the number partition problem,
and therefore we can assume an ordering of $a_{i}$ to be $a_{1}\geq
a_{2}\geq...\geq a_{n}$ without loss of generality. One way to impose this
ordering constraint is to consider the expression macmahon

$\underset{\geq}{\Omega}\frac{1}{(1-\lambda_{1}x)(1-\frac{\lambda_{2}}{\lambda_{1}}x)(1-\frac{\lambda_{3}}{\lambda_{2}}x)...(1-\frac{\lambda_{n}}{\lambda_{n-1}}x)}$
(21)

where the notation $\underset{\geq}{\Omega}$ means restricting terms that have
only nonnegative powers of each $\lambda$ and setting each $\lambda$ to be one
in the end. This is easily verified by expanding each fraction in power
series.

The $\underset{\geq}{\Omega}$ leads to many identities. For example, one can
again expand in power series and verify that

$\underset{\geq}{\Omega}\frac{1}{(1-\lambda
x^{p_{1}})(1-\frac{x^{p_{2}}}{\lambda})}=\frac{1}{(1-x^{p_{1}})(1-x^{p_{1}+p_{2}})}$
(22)

Repeated use of this identity shows that equation (21) is equal to

$\frac{1}{(1-x)(1-x^{2})(1-x^{3})...(1-x^{n})}$ (23)

which is exactly the generating function to count the n-partition of some
integer.

One can also compose the $\underset{\geq}{\Omega}$ operation. We modify the
notation accordingly if there is any ambiguity with extra variables:

$\displaystyle\underset{\lambda\geq}{\Omega}\underset{\mu\geq}{\Omega}\frac{1}{(1-\lambda\mu
x)(1-\frac{y}{\lambda^{2}\mu})}$
$\displaystyle=\underset{\lambda\geq}{\Omega}\frac{1}{(1-\lambda
x)(1-\frac{xy}{\lambda})}$ $\displaystyle=\frac{1}{(1-x)(1-x^{2}y)}$

This expression counts the number of partition into two nonnegative integers
$a_{1}$ and $a_{2}$ such that the constraint $a_{1}\geq 2a_{2}$ is
satisfied444Reason: $\underset{\mu\geq}{\Omega}$ implements the condition that
ordering does not matter, and $\underset{\lambda\geq}{\Omega}$ implements the
constraint $a_{1}\geq 2a_{2}$..

In fact, one can also have two other operators $\underset{\leq}{\Omega}$ and
$\underset{=}{\Omega}$, defined in a self-explanatory way. As we shall see, we
will be interested in identities involving $\underset{=}{\Omega}$. The three
operators satisfy some useful algebraic relations listed in macmahon , and can
be used to compute $\underset{=}{\Omega}$. For example, let $F(\lambda)$ be
some polynomial depending on $\lambda$ as the variable used in the $\Omega$
operators, the following expression

$\underset{=}{\Omega}F(\lambda)=\underset{\geq}{\Omega}F(\lambda)+\underset{\geq}{\Omega}F(\lambda^{-1})-F(1)$
(24)

can be used to derive identities involving $\underset{=}{\Omega}$ in terms of
the identities involving $\underset{\geq}{\Omega}$. In the next subsection we
will use an identity $\underset{=}{\Omega}$ to compute
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ for
$\mathfrak{g}=\mathfrak{su}(2)$. We will also sketch the idea of computing
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ for a general simply laced
$\mathfrak{g}$.

###  5.2 Applying the $\Omega$ Operator

To illustrate the use of the $\Omega$ operator, we use it to write down a
formal expression for the Ehrhart polynomial
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$. Recall that the base polytope is
given by

$Ax^{\prime}=\begin{pmatrix}0\\\ 0\\\ \vdots\\\ 0\\\ 1\end{pmatrix}$ (25)

where $A$ ($r+1$ by $2r+1$) was given in equation (14). Let
$\textbf{z}\equiv(z_{1},z_{2},...,z_{r},z)$. The Ehrhart polynomial is
computed by the formal expression

$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)=\underset{z_{1}=}{\Omega}\underset{z_{2}=}{\Omega}...\underset{z_{r}=}{\Omega}\left(\prod_{i=1}^{2r+1}\frac{1}{1-z_{1}^{A_{1i}}z_{2}^{A_{2i}}...z_{r}^{A_{ri}}z^{A_{r+1,i}}}\right)$
(26)

where a composition of $r$ $\underset{=}{\Omega}$ is applied, each time
restricting the polynomial to the $0$th order term of some $z_{i}$. This
formula looks intimidating, but can be easily derived as follows. The term
involving the product is simply the generating function for counting the
combinations of $Ax^{\prime}$. Imposing the constraint that the right hand
side of the equation has $r$ vanishing entries means that one must restrict to
the $0$th order term of $z_{1},...z_{r}$. Since the last entry of the column
on the right hand side is 1, the number of solutions to the $q$th dilate of
the polytope is then the the coefficient of the $z^{q}$ term. This completes
the argument that the formal expression in (26) computes the Ehrhart
polynomial.

It is nice to have a formal expression like eqn. (26) for the Ehrhart
polynomial. If one wants to compute the first few terms, a computer can easily
do the job. However, we want to take a step further and obtain a closed form
solution, which as we shall see exists for all simply laced $\mathfrak{g}$.

To illustrate how one might obtain a closed form solution using the current
formalism, we first focus on the case of $\mathfrak{su}(2)$. The Cartan matrix
for $\mathfrak{su}(2)$ is simply 2, the coefficient for the highest root is
$c_{1}=1$. Therefore the $A$ matrix is

$A=\begin{pmatrix}2&-1&0\\\ 2&0&1\end{pmatrix}$ (27)

Using expression (26), the Ehrhart polynomial is given by

$\text{Ehr}_{\mathcal{Q}_{\mathfrak{su}}(2)}(z)=\underset{z_{1}=}{\Omega}\left(\frac{1}{1-z_{1}^{2}z^{2}}\frac{1}{1-z_{1}^{-1}}\frac{1}{1-z}\right)$
(28)

Using the $\Omega$ operator identity macmahon

$=\underset{\lambda=}{\Omega}\frac{1}{(1-\lambda^{2}x)(1-y\lambda^{-1})}=\frac{1}{1-xy^{2}}$
(29)

we find that the Ehrhart polynomial for $\mathfrak{g}=\mathfrak{su}(2)$ is

$\text{Ehr}_{\mathcal{Q}_{\mathfrak{su}}(2)}(z)=\frac{1}{(1-z)(1-z^{2})}$ (30)

Now let us generalize to arbitrary $\mathfrak{g}$. The first thing to note is
that one needs more general $\Omega$ identities in addition to equation (29),
since a variable $Z_{i}$ can appear more than two times in the product in
equation (26). However, the $\Omega$ identities for a general polynomial are
not documented and do not have clean solutions. This problem has been solved
by andrews , where the authors developed the Omega Package in Mathematica to
compute $\Omega$ identities for a general polynomial

$\frac{P(x_{1},...,x_{n};\lambda_{1},...\lambda_{r})}{\prod_{i=1}^{n}(1-x_{i}\lambda_{1}^{v_{1}(i)}...\lambda_{r}^{v_{r}(i)})}$
(31)

For our problem, we can apply the program $r$ times to eliminate
$z_{1},...z_{r}$ to find $\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ of rank
$r$. In the next section, we will use a trick from representation theory to
compute all $\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ by hand, bypassing
the need for the computer program computation.

##  6 $\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ From Representation Theory

To connect $\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ to representation
theory, we briefly summarize the McKay correspondence mckay , which associates
any simply laced $\mathfrak{g}$ to some discrete subgroup
$\Gamma(\mathfrak{g})$ of $SU(2)$. More specifically,

$\displaystyle\Gamma(\mathfrak{su}(N))$ $\displaystyle=\mathbb{Z}_{N}$ (32)
$\displaystyle\Gamma(\mathfrak{so}(2(N+2)))$ $\displaystyle=\text{Dic}_{N}$
(33) $\displaystyle\Gamma(\mathfrak{e}_{i})$
$\displaystyle=E_{i}\qquad\text{i=6,7,8}$ (34)

where $\mathbb{Z}_{N}$ is the cyclic group of order $N$, $\text{Dic}_{N}$ is
the dicyclic (binary dihedral) group of order $4N$, and $E_{6},E_{7},E_{8}$
are the binary tetrahedral group, binary octahedral group, and the binary
icosahedral group, respectively (also called $2T$, $2O$, and $2I$).

For some simply laced $\mathfrak{g}$, consider the homomorphism
$\Gamma(\mathfrak{g})\to SU(q)$, or, in other words, $SU(q)$ representations
of the group $\Gamma(\mathfrak{g})$. Let the number of Weyl-inequivalent
representations555This means that two diagonal $SU(q)$ matrices are identified
if the diagonal elements differ by some permutation. $\Gamma(\mathfrak{g})\to
SU(q)$ be $b_{q}$, so that one forms the generating function to count the
number of $SU(q)$ representations

$\Phi_{\mathfrak{g}}(z)\equiv 1+\sum_{i=1}b_{i}z^{i}$ (35)

We claim that this generating function is exactly the same as the Ehrhart
polynomial corresponding to the same Lie algebra:

$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)=\Phi_{\mathfrak{g}}(z)$ (36)

This beautiful equation connects geometry (left hand side) with representation
theory (right hand side) via the McKay correspondence and string theory (the
equality sign). The rest of this subsection is devoted to proving this
equality and using this equality to compute the closed form
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ for
$\mathfrak{g}=\mathfrak{su}(N)$ and $\mathfrak{g}=\mathfrak{so}(2(N+2))$. The
formulae for the exceptional Lie algebras are computed in the appendix of
ganorandju . In section 9, we will “prove” this equation using a construction
in string/M theory.

The strategy we use here is to look at the dual formulation by starting with
weights, expressing them in terms of the simple roots, and imposing the
constraints. This reverse process is carried out using the inverse the Cartan
matrix. Since the inverse of Cartan matrices in general has fractional
entries, this dual formulation is less suited for geometric arguments we had
in the previous sections. Instead, we will use purely algebraic arguments to
prove equation (36).

Let $\mathfrak{g}=\mathfrak{su}(r+1)$ and the corresponding Cartan matrix be
$C$. Start with some weight $x$ in the state set (2) which we reproduce here
for convenience

$\frac{\Lambda_{w}}{W\ltimes q\Lambda_{r}}$ (37)

We can represent $x$ as a vector in $\mathbb{Z}^{r}_{\geq 0}$ by using its
Dynkin label $(x_{1},...,x_{r})$. It satisfies the constraint

$\sum_{i=1}^{r}x_{i}\leq q$ (38)

as argued in section 3.

Since the simple roots span $\mathbb{R}^{r}$, the weight $x$ has a unique
expansion in terms of the simple roots $\alpha_{1},...,\alpha_{r}$:

$x=\sum_{i=1}^{r}l_{i}\alpha_{i}$ (39)

The expansion coefficients $l_{j}$ are simply given by multiplying $x$ (as a
vector) by the inverse of the Cartan matrixkac

$l_{i}=\sum_{j=1}^{r}C^{-1}_{ij}x_{j}$ (40)

Since we want $x$ to lie on the root lattice $\Lambda_{r}$ as well, the only
other constraint we need to impose is that each $l_{j}$ is an integer

$l_{i}=\sum_{j=1}^{r}C^{-1}_{ij}x_{j}\in\mathbb{Z}$ (41)

The inverse of the Cartan matrix for $\mathfrak{su}(N)$ algebra has an
interesting modular structure. Here, we display $C^{-1}$ for
$\mathfrak{su}(4)$ and $\mathfrak{su}(7)$:

$\displaystyle C^{-1}_{\mathfrak{su}(4)}$
$\displaystyle=\frac{1}{4}\begin{pmatrix}3&2&1\\\ 2&4&2\\\ 1&2&3\end{pmatrix}$
(42) $\displaystyle C^{-1}_{\mathfrak{su}(7)}$
$\displaystyle=\frac{1}{7}\begin{pmatrix}6&5&4&3&2&1\\\ 5&10&8&6&4&2\\\
4&8&12&9&6&3\\\ 3&6&9&12&8&4\\\ 2&4&6&8&10&5\\\ 1&2&3&4&5&6\\\ \end{pmatrix}$
(43)

In general, the formula for $C^{-1}_{\mathfrak{su}(N)}$ is dynkin

$C^{-1}_{\mathfrak{su}(N),ij}=\frac{1}{N}\left[\min(i,j)\times N-ij\right]$
(44)

Even though we have $r$ number of constraints from equation (41), we shall see
that because of the peculiar property of $C^{-1}_{\mathfrak{su}(N)}$, there is
effectively only one constraint, the one imposed by the last row of the
matrix:

$\frac{1}{N}\sum_{i=1}^{N-1}ix_{i}\in\mathbb{Z}$ (45)

A quick proof that constraint (45) implies the rest of the constraints is as
follows. The matrix elements $C^{-1}_{\mathfrak{su}(N),ki}$ of the $k$th row
are

$\displaystyle
C^{-1}_{\mathfrak{su}(N),ki}=\begin{cases}\frac{(N-k)i}{N}\qquad k>i\\\
\frac{(N-i)k}{N}\qquad k\leq i\end{cases}$

Using this, we obtain the constraint imposed by the $k$th row:

$\displaystyle l_{k}$
$\displaystyle=\sum_{j=1}^{N-1}C^{-1}_{\mathfrak{su}(N),kj}x_{j}$
$\displaystyle=\sum_{i=1}^{k-1}\frac{(N-k)i}{N}x_{i}+\sum_{i=k}^{N-1}\frac{(N-i)k}{N}x_{i}$
$\displaystyle=-\frac{k}{N}\sum_{i=1}^{N-1}ix_{i}\mod 1$

Up to integers, the constraint imposed by the $k$th row is simply $k$ times
that of the constraint imposed by the last row. Therefore, the only unique
constraint in (41) is the one given by the last row. In summary, there are two
constraints on $x$:

$\displaystyle\sum_{i=1}^{N-1}x_{i}\leq q$ (46)
$\displaystyle\frac{1}{N}\sum_{i=1}^{N-1}ix_{i}\in\mathbb{Z}$ (47)

We would now like to argue that this is exactly the same constraints satisfied
by $SU(q)$ representations of $\Gamma(\mathfrak{su}(N))=\mathbb{Z}_{N}$.

###  6.1 $SU(q)$ representations of $\Gamma(\mathfrak{su}(N))=\mathbb{Z}_{N}$

Since finite dimensional representations are built up from irreducible
representations, we first look at the irreducible representations of
$\mathbb{Z}_{N}$. There are in total $N$ 1-dimensional irreducible
representations. The $k$th irreducible representation is given by the $k$th
power of the $N$th root of unity

$\omega_{k}=\exp\left(\frac{2\pi ik}{N}\right)$ (48)

Here, $k$ runs from 0 to $N-1$, with $k=0$ being the trivial representation.
The $SU(q)$ representations are constructed by inserting $x_{k}$ copies of the
$k$th irreducible representation. By definition, the determinant of any
$SU(q)$ matrix must be 1, so we have

$\prod_{k=0}^{N-1}\omega_{k}^{x_{k}}=\exp\left(\frac{2\pi
i}{N}\sum_{k=0}^{N-1}kx_{k}\right)=1$ (49)

But this is equivalent to the constraint 47:

$\frac{1}{N}\sum_{i=1}^{N-1}ix_{i}\in\mathbb{Z}$

The constraint that the dimensions of the irreducible representations add up
to $q$ is

$\sum_{i=0}^{N-1}x_{i}=q$ (50)

However, treating $x_{0}$ as a slack variable, this constraint is equivalent
to (46)

$\sum_{i=1}^{N-1}x_{i}\leq q$

Therefore, we have shown that the two counting problems satisfy the same
constraints and are secretly one and the same. This concludes the proof of
eqn. (36) for the case $\mathfrak{g}=\mathfrak{su}(N)$:

$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)=\Phi_{\mathfrak{g}}(z)$

In the next subsection, we give explicit formulae for
$\text{Ehr}_{\mathcal{Q}_{\mathfrak{g}}}(z)$ by computing
$\Phi_{\mathfrak{g}}(z)$.

###  6.2 Computation of $\Phi_{\mathfrak{g}}(z)$

We want to find the generating function (35)

$\Phi_{\mathfrak{g}}(z)=1+\sum_{i}b_{i}z^{i}$

in which $b_{q}$ counts the number of inequivalent $SU(q)$ representations of
$\mathbb{Z}_{N}$. We saw in the previous subsection that the $N$ irreducible
representations are given by the $N$th roots of unity. Consider the
function666We thank O. Ganor for pointing out this trick.

$\frac{1}{(1-z)(1-wz)(1-w^{2}z)...(1-w^{N-1}z)}$ (51)

The coefficient of $z^{q}$ is in general a sum of $n$ terms ($n$ is some
integer), representing $n$ ways of building a $q$-dimensional representation.
Each of the $n$ terms has some coefficient $w^{m}$ for some integer
$m\in\mathbb{Z}$, which represents the determinant of that representation. We
want to retain terms of determinant 1 only. To project out terms of non-unit
determinant, we simply have to sum over $w$ in $w^{m}$ and divide by $N$,
since we know from Fourier analysis that

$\sum_{w}w^{m}=\begin{cases}N\qquad m=0\mod N\\\ 0\qquad m\neq 0\mod
N\end{cases}$

where the sum is taken over the $N$th root of unity. Therefore, the generating
function is

$\Phi_{\mathfrak{su}(N)}(z)=1+\sum_{i}b_{i}z^{i}=\frac{1}{N}\sum_{i=0}^{N-1}\frac{1}{(1-z)(1-w^{i}z)(1-w^{2i}z)...(1-w^{(N-1)i}z)}$
(52)

One can check that $b_{i}$ is a quasi-polynomial in $i$, a property shared by
Ehrhart polynomial. Finally, we use the theorem proved in the last subsection
to give the explicit expression for the Ehrhart polynomial

$\text{Ehr}_{\mathcal{Q}_{\mathfrak{su}}(N)}(z)=\frac{1}{N}\sum_{i=0}^{N-1}\frac{1}{(1-z)(1-w^{i}z)(1-w^{2i}z)...(1-w^{(N-1)i}z)}$
(53)

For example, let $\mathfrak{g}=\mathfrak{su}(2)$. Then

$\displaystyle\text{Ehr}_{\mathcal{Q}_{\mathfrak{su}}(2)}(z)$
$\displaystyle=\frac{1}{2}\left(\frac{1}{(1-z)(1-z)}+\frac{1}{(1-z)(1+z)}\right)$
$\displaystyle=\frac{1}{(1-z)(1-z^{2})}$

which agrees with what we obtained in equation (30) using the $\Omega$
operator calculus.

##  7 Ehrhart Polynomial for $\mathfrak{so}(2(N+2))$

In the previous subsections, we obtained the Ehrhart polynomial for
$\mathfrak{su}(N)$ by counting the $SU(q)$ representations of
$\mathbb{Z}_{N}$. We now use the same method to compute the Ehrhart polynomial
for $\mathfrak{so}(2(N+2))$, $N\geq 1$. Here, $N$ is shifted by 2 because of
convenience. The Dynkin diagram associated to $\mathfrak{so}(2(N+2))$ is
$D_{N+2}$. According to the McKay correspondence, we should be looking for the
representation of the discrete group $\text{Dic}_{N}$, the dicyclic (or binary
dihedral) group of order $N$. Since $\text{Dic}_{N}$ is less well-known than
$\mathbb{Z}_{N}$, we analyze the group structure in detail and derive the
irreducible representations in the next subsection. After that, we will use
the irreducible representations to construct the $SU(q)$ representation of
$\text{Dic}_{N}$ and obtain the generating function $\Phi(z)$. We prove that,
in a spirit similar to what we did for the $\mathfrak{su}(N)$ case, the
generating function $\Phi(z)$ coincides with the Ehrhart polynomial for
$\mathfrak{so}(2(N+2))$.

###  7.1 $SU(q)$ representation of $\text{Dic}_{N}$

The group $\text{Dic}_{N}$ is defined by the following multiplication rules:

$\displaystyle r^{2N}$ $\displaystyle=e$ $\displaystyle s^{2}$
$\displaystyle=r^{N}$ $\displaystyle s^{-1}rs$ $\displaystyle=r^{-1}$

where $e$ is the identity element. The reader may have noticed a similarity to
the dihedral group of order $2N$, identifying $r$ with the fundamental
rotation and $s$ with the rotation. The difference here is that the reflection
$s$ does not square to the identity. Instead, it squares to a central element
of the group.

To analyze the irreducible representations of this group, we need to
understand the conjugacy classes. For $\text{Dic}_{N}$, there are $N+3$
conjugacy classes:

$\\{e\\},\\{r,r^{-1}\\},\\{r^{2},r^{-2}\\},...,\\{r^{N-1},r^{-N+1}\\},\\{r^{N}\\},\\{sr^{2k}\\},\\{sr^{2k+1}\\}$
(54)

where for the last two conjugacy classes, $k$ is an integer running from 0 to
$2N-1$. Therefore, there are $N+3$ irreducible representations. The affine
Dynkin diagram of $D_{N+2}$ is shown in Fig. 2 kac .

$(1,1)$ $(2,2)$ $(3,2)$ …$(N,2)$ $(N+1,1)$ $(0,1)$ $(N+2,1)$ Figure 2: The
affine Dynkin diagram for $D_{N+2}$. The tuple $(x,y)$ indicates the $x$th
simple root with (co)mark $y$. The mark is the same as the comark here because
the algebra is simply-laced. By the McKay correspondence, the mark also
indicates the dimension of the corresponding irreducible representation.
Therefore, there are four 1-dimensional irreducible representations and $N-1$
2-dimensional irreducible representations.

By the McKay correspondence mckay  or by abelianizing the group, one sees that
there are four 1-dimensional irreducible representations and $N-1$
2-dimensional irreducible representations. The 1-dimensional irreducible
representations for $\text{Dic}_{N}$ behave differently for $N$ even and $N$
odd. In the following, we restrict $N$ to be an even integer, since the case
for $N$ odd can be treated analogously. We present the character table:

$\\{e\\}$ | $\\{r,r^{-1}\\}$ | $\\{r^{2},r^{-2}\\}$ | … | $\\{r^{N-1},r^{-N+1}\\}$ | $\\{r^{N}\\}$ | $\\{sr^{2k}\\}$ | $\\{sr^{2k+1}\\}$
---|---|---|---|---|---|---|---
1 | 1 | 1 | … | 1 | 1 | 1 | 1
1 | 1 | 1 | … | 1 | 1 | -1 | -1
1 | -1 | 1 | … | -1 | 1 | 1 | -1
1 | -1 | 1 | … | -1 | 1 | -1 | 1
2 | $w+w^{-1}$ | $w^{2}+w^{-2}$ | … | $w^{N-1}+w^{-(N-1)}$ | -2 | 0 | 0
2 | $w^{2}+w^{-2}$ | $w^{4}+w^{-4}$ | … | $w^{2(N-1)}+w^{-2(N-1)}$ | 2 | 0 | 0
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
2 | $w^{N-1}+w^{-N+1}$ | $w^{2(N-1)}+w^{-2(N-1)}$ | … | $w^{(N-1)^{2}}+w^{-(N-1)^{2}}$ | -2 | 0 | 0
Table 1: Character table for the $\text{Dic}_{N}$ group. The first four lines
are the characters for the four 1-dimensional irreducible representations.
Note that the first line is the trivial 1-dimensional representation. The rest
of the irreducible representations are 2-dimensional. $w$ represents the
$2N$th root of unity $\exp(\pi i/N)$. One can check that the character
orthogonality relation holds.

For the $N-1$ 2-dimensional irreducible representations, we note that $r$ and
$s$ take on the following form:t

$\displaystyle r$ $\displaystyle=\begin{pmatrix}\exp(m\pi i/N)&0\\\
0&\exp(-m\pi i/N)\end{pmatrix}\quad m=1,...,N-1$ $\displaystyle s$
$\displaystyle=\begin{cases}\begin{pmatrix}0&1\\\ -1&0\end{pmatrix},m\text{
odd}\\\ \begin{pmatrix}0&1\\\ +1&0\end{pmatrix},m\text{ even}\end{cases}$

In particular, for the 2-dimensional irreducible representations, $\det r=1$,
and a non-unit determinant can only come from $s$. The formula for $s$ shows
that the determinant of the 2-dimensional representations777In our case, we
define the determinant of a representation as follows. If all conjugacy
classes in the representation have determinant 1, we say that the
representation has determinant 1. If some conjugacy classes have non-unit
determinant, we pick the determinant $D$ that, in the polar decomposition, has
the smallest angle $\theta\in[0,2\pi)$ and call $D$ the determinant of the
representation. For example, if $\omega$ is the 3rd root of unity,
$\omega=\exp(2\pi i/3)$. Suppose there are two conjugacy classes with non-unit
determinant, one with determinant $\omega$ and the other $\omega^{2}$. Because
$\omega$ has a smaller polar angle than $\omega^{2}$, we say that the
representation has determinant $\omega$.  alternates between 1 and -1. For
example, $\text{Dic}_{2}$ has one 2-dimensional irreducible representation
with determinant 1, whereas $\text{Dic}_{4}$ has three 2-dimensional
irreducible representations, with determinant 1, -1, 1, respectively.

We now use the above information to compute the generating function for the
$SU(q)$ representation of $\text{Dic}_{N}$. The $SU(q)$ representation is
constructed out of the $N+3$ irreducible representations so that the
dimensions add up to $q$. Let the number of the four 1-dimensional
representations be $x_{0},x_{1},x_{N+1},x_{N+2}$, respectively888The numbering
here is to make connection with the numbering of the nodes in the affine
Dynkin diagram.. Let the number of 2-dimensional representations be labeled by
$x_{2},x_{3},...,x_{N}$. The constraint on the size of the representation is

$x_{0}+x_{1}+2x_{2}+2x_{3}+...+2x_{N}+x_{N+1}+x_{N+2}=q$ (55)

In addition to the size constraint, we also have the unit-determinant
constraint. From the character table, we see that a $-1$ determinant can only
come from three columns999Note that the entries in the character table contain
the trace of the conjugacy classes. The determinant coincides with the trace
for 1-dimensional representations. For 2-dimensional representations the
determinant in this case was computed earlier by looking at the explicit
matrix representations. There are other columns that have $-1$ determinant,
but one can show that there are only these three independent columns to
consider.: $\\{r,r^{-1}\\}$, $\\{sr^{2k}\\}$, and $\\{sr^{2k+1}\\}$. The unit-
determinant constraints coming from the three columns are

$\displaystyle\frac{x_{N+1}+x_{N+2}}{2}$ $\displaystyle\in\mathbb{Z}$ (56)
$\displaystyle\frac{x_{1}+x_{3}+x_{5}+...+x_{N-1}+x_{N+2}}{2}$
$\displaystyle\in\mathbb{Z}$ (57)
$\displaystyle\frac{x_{1}+x_{3}+x_{5}+...+x_{N-1}+x_{N+1}}{2}$
$\displaystyle\in\mathbb{Z}$ (58)

Note that the constraints are not independent, since (57) and (58) imply (56).
Therefore, we effectively only have two determinant constraints.

To write down the generating function $\Phi_{\mathfrak{so}(2(N+2))}$, we can
use the independent constraints (57) and (58). The idea is similar to the
$\mathfrak{su}(N)$ case. Consider the function

$\frac{1}{(1-z)(1-w_{1}z)(1-w_{2}z)(1-w_{1}w_{2}z)(1-z^{2})^{N/2}(1-w_{1}w_{2}z^{2})^{N/2-1}}$
(59)

Here, $w_{1},w_{2}\in\\{1,-1\\}$. The first four terms represent the
contributions of inserting the four 1-dimensional representations, and the
last two terms represent the contributions of inserting the $N-1$ copies of
the 2-dimensional representations. If we expand the fractions into a power
series, a generic term would look like

$cw_{1}^{k_{1}}w_{2}^{k_{2}}z^{n}$ (60)

where $c,k_{1},k_{2}\in\mathbb{Z}$. The $w_{2}^{k_{2}}$ term represents the
determinant contribution from the two 1-dimensional representations labeled by
$N+1$ and $N+2$. We want to retain the term that satisfies $w_{2}^{k_{2}}=1$.
This can be done by a projection similar to what we did for the
$\mathfrak{su}(N)$ case, except now we need to sum over
$\omega_{2}\in\\{1,-1\\}$ and divide by 2. We also need to do a similar
projection on $w_{1}$. In fact, the two projections help us retain the terms
that satisfy the constraints (57) and (58). Therefore, we need to compute

$\Phi_{\mathfrak{so}(2(N+2))}(z)=\frac{1}{4}\sum_{w_{1},w_{2}}\frac{1}{(1-z)(1-w_{1}z)(1-w_{2}z)(1-w_{1}w_{2}z)(1-z^{2})^{N/2}(1-w_{1}w_{2}z^{2})^{N/2-1}}$

e which gives us the answer for $\Phi_{\mathfrak{so}(2(N+2))}(z)$:

$\frac{1}{4}\left(\frac{1}{(1-z)^{4}(1-z^{2})^{N-1}}+\frac{2}{(1-z^{2})^{2}(1-z^{2})^{N/2}(1+z^{2})^{N/2-1}}+\frac{1}{(1-z^{2})^{2}(1-z^{2})^{N-1}}\right)$
(61)

As an example, we set $N=2$, so that we are looking at the generating function
for $SU(q)$ representations of Dic2. The first few terms of (61) are

$\Phi_{\mathfrak{so}(8)}(z)=1+z+5z^{2}+...$

which suggests that there are one $SU(1)$ representation and five $SU(2)$
representations. The one $SU(1)$ representation is just the trivial
representation. Even though there are four unitary 1-dimensional
representations for Dic2, only the trivial representation has unit determinant
as can be seen from the character table. The reader can easily work out the
five $SU(2)$ representations, one of which comes from the unique irreducible
2-dimensional representation of Dic2. The rest come from the reducible
2-dimensional representations by combining the 1-dimensional irreducible
representations such that the determinant constraint is satisfied.

We will show in the next subsection that this is exactly the Ehrhart
polynomial $\text{Ehr}_{\mathcal{Q}_{\mathfrak{so}}(2(N+2))}(z)$ for the
${D}_{N+2}$ polytope defined in a similar fashion as in section 4.

###  7.2 Equivalence to the Ehrhart polynomial

We are looking at the level $q$ highest weight representations of
$\mathfrak{so}(2(N+2))$ which also lie on the root lattice. To show that the
Ehrhart polynomial coincides with the generating function obtained in the
previous subsection, we simply show that the two systems have the same
constraints. Let the Dynkin label of some highest weight representation be
$\lambda=(x_{1},x_{2},...,x_{N+2})$, each term some nonnegative integer. The
level $q$ constraint yields

$(\lambda,\theta)=x_{1}+2x_{2}+2x_{3}+...+2x_{N}+x_{N+1}+x_{N+2}\leq q$ (62)

where $\theta$ is the highest root whose expansion coefficients in terms of
the simple roots can be read off from the affine Dynkin diagram. By adding a
slack variable to turn the inequality into an equality, we reproduce the first
constraint (56). Next, we need to demand that the weights are expressed as
integer combinations of simple roots:

$\sum_{j=1}^{N+2}C^{-1}_{ij}x_{j}\in\mathbb{Z}$ (63)

The inverse of the Cartan matrix for the D-series has an interesting form
weiandzou . Since the matrix is symmetric, we only give the values for the
upper half of the matrix. Let $C^{-1}$ be the inverse of the Cartan matrix for
$D_{N+2}$. We have

$C^{-1}_{ij}=\begin{cases}i,&1\leq i\leq j\leq N\\\ i/2,&i\leq N,j=N+1\
\text{or}\ N\\\ \frac{N}{4},&i=N+1,j=N+2\\\ \frac{N+2}{4},&i=j=N+1\ \text{or}\
N+2\end{cases}$

We give an example of $C^{-1}_{ij}$ for $D_{6}$ where $N=4$:

$C^{-1}_{\mathfrak{so}(12)}=\begin{pmatrix}1&1&1&1&1/2&1/2\\\ 1&2&2&2&1&1\\\
1&2&3&3&3/2&3/2\\\ 1&2&3&4&2&2\\\ 1/2&1&3/2&2&3/2&1\\\ 1/2&1&3/2&2&1&3/2\\\
\end{pmatrix}$

Note that the $N$ by $N$ block is integer-valued, so it does not enter into
the constraint (63). We only have to worry about the last two columns and the
last two rows (which are the same as the last two columns by symmetry). For
our case, $N$ is an even number, so the only fraction that enter into the
constraints modulo 1 is $1/2$. Restricting our attention to the last two
columns, we see that as the row number increases from $1$ to $N$, the values
of the last two columns alternate between being half-integer valued and
integer valued. Therefore, from the first $N$ rows of $C^{-1}$, we effectively
get only one constraint:

$\frac{x_{N+1}+x_{N+2}}{2}\in\mathbb{Z}$ (64)

The constraints coming from the last two rows can be deduced similarly.
Restricting ourselves to the last two rows, as the column number $j$ increases
from 1 to $N$, the values of $C^{-1}_{N+1,j}$ and $C^{-1}_{N+2,j}$ alternate
between being half-integer valued and integer valued. Taking into account of
the last two columns, the constraints are

$\displaystyle\frac{x_{1}+x_{3}+x_{5}+...+x_{N-1}+x_{N+2}}{2}$
$\displaystyle\in\mathbb{Z}$ (65)
$\displaystyle\frac{x_{1}+x_{3}+x_{5}+...+x_{N-1}+x_{N+1}}{2}$
$\displaystyle\in\mathbb{Z}$ (66)

We see that (62), (65), and (65) are exactly the same constraints we obtained
in the last subsection. This establishes the equivalence of the generating
function for $SU(q)$ representation of $\text{Dic}_{N}$ group and the Ehrhart
polynomial for the $\mathfrak{so}(2(N+2))$ polytope. In fact, one can repeat
the same argument to show that the equivalence holds for the exceptional Lie
algebras as well. This concludes the proof of remarkable formula (36).

##  8 A New Perspective on the McKay Correspondence

Let $C$ be some Cartan matrix of an ADE type Lie algebra of rank $r$. Let
$G_{C}$ be the discrete $SU(2)$ subgroup corresponding to $C$ according to the
McKay correspondence. Define $[C^{-1}]$ as $C^{-1}$ modulo 1, where modulo 1
is done element-wise. Our computation in the previous two sections shows that
a great deal of information is hidden in this object. In particular,
$[C^{-1}]$ can tell us about the determinant101010The determinant of a
representation is defined in footnote 7. of the irreducible representations of
the group $G_{C}$. To make it more precise, we define the following $\vee$
operator acting on the rational numbers in the congruence class of 1 as

$a\vee b=\begin{cases}a+b,&\text{if}\ a=0\ \text{or}\ b=0\\\ a,&\text{$b$ is a
nonzero integer multiple of $a$}\\\ b,&\text{$a$ is a nonzero integer multiple
of $b$}\end{cases}$

Note that this definition comes with a priority structure: there could be
cases where condition 2 and 3 are both satisfied. In that case, we stick with
condition 2 and demand that $a\vee b=a$. For example, bearing in mind that we
are working within the congruence class of 1, the above rules imply

$\displaystyle 2/3\vee 1/3$ $\displaystyle=2/3\quad\text{because $1/3$ is 2
times $2/3$}$ $\displaystyle 5/7\vee 6/7$ $\displaystyle=5/7\quad\text{because
$6/7$ is 4 times $5/7$ }$ $\displaystyle 1/4\vee 0$ $\displaystyle=1/4$
$\displaystyle 1/2\vee 1/2$ $\displaystyle=1/2$

As we shall see, we never have to worry about the case when $a$ and $b$ do not
satisfy the three cases. To use the $\vee$ operator, let us define $X_{i}$ to
be the $i$th row of $[C^{-1}]$. Let the operator $\vee$ act element-wise on
$X_{i}$. Define the row vector $X$ as

$X\equiv X_{1}\vee X_{2}\vee...\vee X_{r}$ (67)

The dual group $G_{C}$ has $r+1$ irreducible representations, among which 1 of
them is the trivial representation. We claim that the determinant of the rest
of the $r$ non-trivial representations, encoded in the row vector
$D=(d_{1},...d_{r})$, can be found by

$D=\exp 2\pi iX$ (68)

where the exponential is taken element-wise on $X$, producing another row
vector.

This claim can be proved by checking for all ADE Lie algebras. The information
presented in the previous sections is enough for the readers to check the
claim for the $A$ and the $D$ series. Here we will focus on the exceptional
series $\mathfrak{e}_{6}$, $\mathfrak{e}_{7}$, and $\mathfrak{e}_{8}$. The
character tables for the dual group are worked out in ganorandju . We give a
brief summary.

  * •

$\mathfrak{e}_{6}$. The dual group has six nontrivial irreducible
representations, of which two have determinant $\exp(2\pi i/3)$ and two have
determinant $\exp(4\pi i/3)$. The other two have determinant 1.

  * •

$\mathfrak{e}_{7}$. The dual group has seven nontrivial irreducible
representations, of which three have determinant $-1$ while the rest has
determinant 1.

  * •

$\mathfrak{e}_{8}$. All representations of the dual group have unit
determinant.

Let us now compare the prediction made by equation (68) to the facts cited
above.

The most trivial case to check is $\mathfrak{e}_{8}$, whose inverse Cartan
matrix modulo 1 vanishes. In this case, $X_{1}=X_{2}=...=X_{8}=0$, so
according to the formula above, all irreducible representations of the binary
icosahedral group have unit determinant.

For $\mathfrak{e}_{7}$, $[C^{-1}]$ is

$[C^{-1}_{\mathfrak{e}_{7}}]=\begin{pmatrix}0&0&0&0&0&0&0\\\ 0&0&0&0&0&0&0\\\
0&0&0&0&0&0&0\\\ 0&0&0&1/2&0&1/2&1/2\\\ 0&0&0&0&0&0&0\\\
0&0&0&1/2&0&1/2&1/2\\\ 0&0&0&1/2&0&1/2&1/2\end{pmatrix}$

The $\vee$ sum of all the rows gives $(0,0,0,1/2,0,1/2,1/2)$. Equation (68)
shows that the determinants are $(1,1,1,-1,1,-1,-1)$, agreeing with the facts
above.

For $\mathfrak{e}_{6}$, $[C^{-1}]$ is

$[C^{-1}_{\mathfrak{e}_{6}}]=\begin{pmatrix}1/3&2/3&0&1/3&2/3&0\\\
2/3&1/3&0&2/3&1/3&0\\\ 0&0&0&0&0&0\\\ 1/3&2/3&0&1/3&2/3&0\\\
2/3&1/3&0&2/3&1/3&0\\\ 0&0&0&0&0&0\\\ \end{pmatrix}$

The $\vee$ sum of all the rows gives $(1/3,2/3,0,1/3,2/3,0)$. Equation (68)
shows that the determinants are $(w,w^{2},1,w,w^{2},1)$, where $w$ is the
third root of unity. This also agrees with the facts cited above, and
completes the proof of equation (68).

##  9 The Physical Origin of the Duality

In this section, we give an overview of the physics underlying equation (36).
As we shall see in this section, the right hand side of equation (36) will be
interpreted as the generating function for computing the dimension of the
ground state Hilbert space of $\mathcal{N}=4$ $SU(q)$ Yang-Mills theory on
$S^{3}/\Gamma$. The left hand side of equation (36) is the generating function
for computing the dimension of a certain subspace (3) of the Hilbert space of
level $q$ Chern-Simons theory with gauge algebra $\mathfrak{g}(\Gamma)$, where
$\mathfrak{g}(\Gamma)$ is given by the McKay correspondence. The SYM theory
ground states are simply the flat $SU(q)$ Wilson lines wrapping around the
$\Gamma$ singularity, which in the language of representation theory
translates to $SU(q)$ representation of $\Gamma$ up to conjugation and Weyl
group identification. To prove (36), we would like to show that these two
Hilbert spaces indeed have the same dimension for a given $SU(q)$ gauge group
on the SYM side and for any ADE singularity $\Gamma$.

To show this, consider in IIB string theory a stack of $N$ D3 branes that in
the Euclidean signature span the 0123 directions. The ADE singularity $\Gamma$
acts on the world volume111111Note that this setup is different from the
D-brane on orbifold setup considered in douglas , where the orbifold action is
along the transverse direction of the D-brane worldvolume rather than the
longitudinal direction. of the D3 branes $\mathbb{C}^{2}$ to make it
$\mathbb{C}^{2}/\Gamma$. Since M-theory on $T^{2}$ is dual to IIB string
theory, we lift the D3 branes to M5 branes by first taking the T-duality in
the 4 direction to turn them into D4 branes and then blowing up the M-theory
circle in the # direction. The situation can be summarized in Table 2.

| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | #
---|---|---|---|---|---|---|---|---|---|---|---
D3 | N | N | N | N | D | D | D | D | D | D | x
M5 | N | N | N | N | N | D | D | D | D | D | N
Table 2: The string/M theory setup. Here, $D$ denotes “Dirichlet” and $N$
“Neumann”. To avoid confusion, we use # to deonte the 10th direction. Since
the D3 brane exists only in the 9+1 dimensional universe, an “x” is put under
the 10th direction to indicate that the D3 brane does not exist in that
direction.

A stack of $N$ M5 branes whose worldvolume is placed on an ADE singularity
$\mathbb{C}^{2}/\Gamma$ has a near horizon geometry $AdS_{7}/\Gamma\times
S^{4}$. Let the Lie algebra corresponding to the ADE singularity be
$\mathfrak{g}(\Gamma)$. The boundary theory lives on directions 012345. Let
the bulk direction be 6. We claim that there exists a coupling

$-\frac{1}{4\pi}\int\frac{C_{3}}{2\pi}\wedge\tr(F\wedge F)$ (69)

where $C_{3}$ is the 3-form in M-theory and $F$ is the 2-form field strength
taking values in $\mathfrak{g}$. The integral is taken in the 456789#
directions. The 5-plane transverse to the M5 branes can be decomposed into
$\mathbb{R}^{+}\times S^{4}$, where $\mathbb{R}^{+}$ is the direction into the
bulk, and $S^{4}$ is the 4-sphere that surrounds the M5 branes. Integrating by
parts, we obtain

$\frac{1}{4\pi}\int_{T^{2}\times\mathbb{R}^{+}\times
S^{4}}\frac{dC_{3}}{2\pi}\wedge\tr(A\wedge dA+\frac{2}{3}A\wedge A\wedge
A)=\frac{N}{4\pi}\int_{T^{2}\times\mathbb{R}^{+}}\tr(A\wedge
dA+\frac{2}{3}A\wedge A\wedge A)$ (70)

reproducing the Chern-Simons theory on $T^{2}$ at level $N$.

Before we derive this, we mention two similar constructions in the literature.
Ref dijkgraaf  reproduces a part of this duality by considering intersecting
D4-D6 branes in type IIA string theory, so only the A-singularity is probed
since D6 branes cannot create the D- or the E-singularity. Ref vafawitten , on
the other hand, considers Nakajima’s instanton construction nakajima  on
$\mathbb{R}^{4}/\Gamma$ using a certain twisted super Yang-Mills theory
without explicitly mentioning the Chern-Simons theory. In vafawitten , it is
shown that a certain combination of the middle-dimensional cohomology of
$U(q)$ n-instanton moduli space on $\mathbb{R}^{4}/\Gamma$ has the same
structure as the highest weight representation of a level $q$ state of algebra
$\mathfrak{g}(\Gamma)$ (see equation 4.42 in vafawitten ). The link to the
Chern-Simons theory is implicit in their work because of the equivalence
between WZW conformal blocks and Chern-Simons theory as shown in wittenjones .
We derive the duality using a perspective different from these two
constructions. Compared to dijkgraaf , our construction has the virtue of
incorporating the D- and the E-singularity. Compared to vafawitten , our
construction does not require twisting the super Yang-Mills theory and is
easier to understand from a physicist’s perspective.

The strategy we use to derive equation (69) is through the string-string
duality. In the first step, we compactify the 4 direction and take the T-dual
to end up in the Type IIA string theory. We now have $N$ D4 branes along the
01234 directions. The ADE singularity acting on the 0123 directions can be
locally thought of as a certain degeneration limit of a K3 surface. Although
the latter is compact and the former is noncompact, the distinction will not
be important in the following derivation. It is known that heterotic string on
$T^{4}$ is dual to IIA string on K3, with the coupling constants being related
by the equation

$e^{\phi}=e^{-\phi^{\prime}}$ (71)

where unprimed quantities are for the heterotic theory and the primed
quantities are for the IIA theory wittendynamics . In addition, the NS-NS
2-forms are related in 6D as

$dB=e^{-2\phi^{\prime}}\star dB^{\prime}$ (72)

Now, both the $T^{4}$ and the K3 are along the Euclideanized 0123 directions.
Recall that the NS-NS 2-form in IIA theory descends from the 3-form in
M-theory as $dB^{\prime}\wedge dx^{\\#}=dC_{3}$. So if one blows up the
M-theory circle and lift the above relation to 7D (i.e. adding the $\\#$
direction), one gets

$dB=e^{-2\phi^{\prime}}\star dC_{3}$ (73)

We now have $N$ M5 branes along the 01234# directions, where the 4 and the #
directions are circles.

The heterotic theory has a modified action for its 3-form field strength
(setting the curvature contribution to zero121212It is amusing to note that if
one were to include curvature, i.e. the gravitational Chern-Simons term, one
would be able to give a string-theory derivation of the curvature counter term
in the one-loop correction to the Chern-Simons action. In wittenjones , this
counter term arises out of the consideration for canceling the framing
anomaly.):

$\frac{1}{2}\int_{\mathbb{R}^{6}}e^{-2\phi}(dB-CS,dB-CS)$ (74)

where $CS$ denotes the Chern-Simons three form and where we considered the
$T^{4}$ compactified action so the integration is along the directions 456789
in the IIA theory. Notice that $e^{-2\phi}=e^{2\phi^{\prime}}$ is the
heterotic dilaton coupling. If we consider blowing up the M-theory circle to
lift the above action to 7D, and plug in the duality relation, the above term
becomes

$\frac{1}{2}\int_{\mathbb{R}^{7}=T^{2}\times\mathbb{R}^{5}}e^{2\phi^{\prime}}(e^{-2\phi^{\prime}}\star
dC_{3}-CS,e^{-2\phi^{\prime}}\star dC_{3}-CS)$ (75)

where the integral is taken along the directions 456789# and we split the
directions along the torus (4#) and the $\mathbb{R}^{5}$ (56789) transverse to
the $N$ M5 branes (see Table 2). The $\mathbb{R}^{5}$ can be further
decomposed into a radial direction times an $S^{4}$ that surrounds the $N$ M5
branes. From this expression, we get a cross term

$-\int(CS,\star dC_{3})=\int CS\wedge\star\star dC_{3}=\int CS\wedge
dC_{3}=N\int CS$ (76)

where in the last step we integrated the 4-form fluxes along the $S^{4}$ to
pick up the overall prefactor $N$, reproducing (69) and the Chern-Simons
action we sought for.

In the IR limit, the Chern-Simons action is the unique action that has the
lowest number of derivatives, and will characterize the ground state structure
of the theory. Therefore, the ground state Hilbert space of $\mathcal{N}=4$
$U(q)$ SYM on $S^{3}/\Gamma$ is equivalent to that of the holographic dual
which is the Hilbert space of level $q$ Chern-Simons theory with gauge group
$\mathfrak{g}(\Gamma)$. This Chern-Simons Hilbert space is given by the set
(2). In this work, we considered $\mathcal{N}=4$ $SU(q)$ SYM on
$S^{3}/\Gamma$. Changing the gauge group from $U(q)$ to $SU(q)$ reduces the
number of ground states in the SYM ground state Hilbert space. What is the
corresponding reduction on the Chern-Simons side? To see this, we use the
example $\Gamma=\mathbb{Z}_{k}$. The $SU(q)$ SYM ground states Hilbert space
admits a level-rank duality: the dimension of $SU(q)$ ground state Hilbert on
$S^{3}/\mathbb{Z}_{k}$ is the same as that of the $SU(k)$ ground state Hilbert
space on $S^{3}/\mathbb{Z}_{q}$131313A quick way to show this is to use the
$SU(N)$ Ehrhart polynomials computed in (53) and show that the $q$th term of
$\text{Ehr}_{\mathfrak{su}(k)}$ is the same as the $k$th term of
$\text{Ehr}_{\mathfrak{su}(q)}$.. The only nontrivial subspace of the Chern-
Simons Hilbert space that is invariant under the level-rank duality is the one
given by the set (3) kac . This concludes the physical derivation of (36).

##  10 Conclusion

We have shown that the counting of root lattice states of level $q$ Chern-
Simons theory Hilbert space on $T^{2}$ can be solved by computing the exact
generating function using either the $\Omega$ operator calculus or the duality
proposed in ganorandju  that has a similar origin to!dijkgraaf vafawitten .
The reader might wonder why we framed the problem in terms of the Ehrhart
polynomials. First, the number of root lattice states grows as some quasi-
periodic polynomial, a property shared by Ehrhart polynomials. As we saw in
section 4, the geometric formulation of the counting problem leads very
naturally to the idea of Ehrhart polynomials. Second, and most importantly,
recent development of mathematics and physics shows that it is always fruitful
to find connections between different subfields of mathematics and physics.
Since Ehrhart polynomials connect various branches of mathematics such as
number theory and topology, formulating the problem using Ehrhart polynomials
is an attempt at achieving more unification.

Despite being a decades-old subject, Chern-Simons theory is at the heart of
the inter-connectedness explored in this paper. In fact, as shown in the last
section and in ganorandju dijkgraaf vafawitten , the Chern-Simons theory arise
naturally at the long distance limit of some holographic system in string
theory. Another key ingredient is the McKay correspondence mckay . In fact, we
saw in section 8 that the simple Lie algebras know a lot more than the content
of the McKay correspondence through the inverse of the Cartan matrices. The
simple Lie algebras secretly know the determinant of the irreducible
representations of the corresponding ADE subgroup.

Finally, we pose some open questions. In this work, we focused on a special
subspace of the Chern-Simons Hilbert space on $T^{2}$ given by the set (3) and
found that there is a description using a certain Supersymmetric Yang-Mills
theory. Are there other Chern-Simons states that admit a similar dual
description? Chern-Simons theory can be quantized on higher-genus Riemann
surfaces. On the SYM side, compactifying the M5 branes on higher-genus Riemann
surfaces will break another half of the supersymmetry. The building blocks of
the Hilbert space on closed Riemann surfaces are the $N_{ijk}$ fusion
coefficients. What does the SYM dual of the fusion coefficients look like? We
will tentatively propose an answer to this question in a forthcoming work. In
this work, we focused on the $SU(q)$ gauge group for the SYM theory, but one
can also ponder over other compact Lie groups such as $SO(q)$ or $Sp(2q)$.
What special states in the Chern-Simons theory would these representations
correspond to? A standard way to construct $SO(q)$ and $Sp(2q)$ gauge groups
in string theory is by using orientifolds, so we would expect that a similar
brane construction using orientifolds can help shed light on this question.

###### Acknowledgements.

We would like to thank Emil Albrychiewicz, Ori Ganor, Yasunori Nomura, and
Tong Zhou for useful discussions. In particular, we thank Ori Ganor for
pointing out the potential significance of the states considered in this
paper.

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[]Alexander A. Panferov

# On turbulent viscosity in relativistic jets and accretion disks

###### Abstract

The mechanism of turbulent viscosity is the central question in investigations
of turbulence. This is also the case in the accretion disk theory, where
turbulence is considered to be responsible for the outward transport of
angular momentum in the accretion disk. In turbulent flows, vortices transport
momentum over their length scales providing the mechanism of viscosity that is
controlled by mass entrainment. We have earlier proposed an entrainment model
for the particular case of the relativistic jets in the radio galaxy 3C 31. In
this paper, we further constrain the model parameters. The model (in the non-
relativistic part) is successfully tested versus experimental and simulation
data on the Reynolds stresses of free mixing layers and predicts the
Smagorinsky constant $C_{\mathrm{S}}\approx 0.11$, which is consistent with
the experimental range for shear flows $C_{\mathrm{S}}\approx 0.1-0.12$. For
accretion disks, the entrainment model allows us to derive the same accretion
mass rate as in the Shakura & Sunyaev’s $\alpha$-model without appealing to
the turbulent kinematic viscosity $\nu_{\mathrm{t}}$, and the viscosity
parameter $\alpha$ derived in the form $\displaystyle\alpha=-\frac{8}{3}\beta
s_{\mathrm{T}}\frac{\mathrm{v_{t}}^{2}}{c_{\mathrm{s}}^{2}}$ depends on the
power $s_{\mathrm{T}}$ of the temperature slope along the disk radius,
$T\propto r^{s_{\mathrm{T}}}$, and quadratically on the turbulent velocity
$\mathrm{v_{t}}$.

##  1 Introduction

In free shear flows, which jets and accretion disks look like, turbulent
mixing layers develop and entrain matter from the surrounding irrotational
flow by means of large-scale vortices [1, Brown & Roshko 2012]. These vortices
determine transverse transport of momentum, i.e. turbulent stress.

The turbulent stress $\tau$ is usually dubbed as the Reynolds stress ($R$),
which is determined by turbulent pulsations of fluid velocity, e.g.
$\tau_{\mathrm{xy}}=-\rho\overline{\mathrm{v_{x}v_{y}}}$ for the
(xy)-component of the turbulent stress tensor, which is supplementary to the
stress tensor of the Reynolds equations system, where $\rho$ is the fluid
density, $\mathrm{v_{x}}$ and $\mathrm{v_{y}}$ are the x and y-components of
the fluid fluctuating velocity, respectively, thus
$\mathrm{V_{x}}=\overline{\mathrm{V_{x}}}+\mathrm{v_{x}}$ is the whole
x-component of the velocity. The Reynolds stresses cannot be derived from the
system of Reynolds equations because the latter is unclosed. The usual
approach to the turbulent stress is local, after Boussinesq, through the
kinematic turbulent (eddy) viscosity coefficient $\nu_{\mathrm{t}}$ and local
mean strain:
$\displaystyle\tau_{\mathrm{xy}}=-\rho\nu_{\mathrm{t}}\frac{\partial\overline{\mathrm{V_{x}}}}{\partial
y}$ in the simple shear, where the gradient
$\displaystyle\frac{\partial\overline{\mathrm{V_{x}}}}{\partial y}$ is only
non-zero. The turbulent viscosity may be modelled as
$\displaystyle\nu_{\mathrm{t}}=\mathrm{v_{t}}l_{\mathrm{m}}=l_{\mathrm{m}}^{2}\left|\frac{\partial\overline{\mathrm{V_{x}}}}{\partial
y}\right|$ using the concept of Prandtl’s mixing length $l_{\mathrm{m}}$, an
empirical parameter. In this way Shakura & Sunyaev [2, thereafter SS73]
prescribed the turbulent fluid viscosity as
$\nu_{\mathrm{t}}=\mathrm{v_{t}}l_{\mathrm{m}}=\alpha c_{\mathrm{s}}H$ in the
$\alpha$-model of accretion disk, where
$\displaystyle\alpha=\frac{\mathrm{v_{t}}l_{\mathrm{m}}}{c_{\mathrm{s}}H}$ is
the viscosity parameter, $\mathrm{v_{t}}$ the turbulent velocity,
$c_{\mathrm{s}}$ the speed of sound, $H$ the half thickness of the disk.
However, the concept of turbulent viscosity does not represent the mechanism
of turbulent stress, when large-scale vortices dominate the momentum transport
which determines the stress [1, Brown & Roshko 2012].

In our paper [3, Panferov 2017] we have derived a formula for the entrainment
in the jets of the radio galaxy 3C 31 (repeated here in Section 2). Here, we
express the turbulent shear stress in dependency on turbulence intensity,
elaborating the model of Panferov [3], and test this model vs. laboratory and
simulation data (Section 3). The model principles are successfully applied to
explain the experimental value of the Smagorinsky constant, widely used in
large eddy simulations of turbulent flows. After this verification we search
for entrainment in accretion disks (Section 4). Section 5 is concluding.

##  2 Entrainment in jets of 3C 31

Jets of FR I radio galaxies strongly decelerate at kiloparsec scales [4, Laing
and Bridle 2014]. The origin of this is seemingly the turbulent friction of
the jets in the ambient medium, which is associated with entrainment and is
indicated by a shear at the edges in the transverse velocity profiles of the
jets. The profile of the external entrainment along the jet 3C 31 from Wang et
al. [5], which is shown in Fig. 1, we have described using the function of
mass flux density at the jet edge

$q_{\mathrm{t}}=\beta\rho_{\mathrm{a}}\mathrm{v_{t}}\sigma_{\mathrm{t}},$ (1)

in dependency on the ambient medium density $\rho_{\mathrm{a}}$, the
statistical coefficient $\beta$, the mean turbulent velocity $\mathrm{v_{t}}$
and the so-called cross-section of turbulence $\sigma_{\mathrm{t}}$ [3,
Panferov 2017]. In this formula, $\beta\rho_{\mathrm{am}}\mathrm{v_{t}}$ terms
the mean flux density of mass in any direction driven by the turbulence at the
jet edge. And the cross-section

$\sigma_{\mathrm{t}}=\sin
i=\frac{\mathrm{v}_{\perp}}{\sqrt{\mathrm{v}_{\parallel}^{2}+\mathrm{v\mathrm{{}_{\perp}^{2}}}}}=\frac{\eta}{\sqrt{\gamma^{2}+\eta^{2}}}$
(2)

is the efficiency of the capture of matter from vortices, which are imagined
as fluid pulsations, by the vortical turbulent flow of the jet of the velocity
$\mathrm{v_{j}}$. Here $\mathrm{v}_{\parallel}=\mathrm{v_{j}}$,
$\displaystyle\mathrm{v}_{\perp}=\frac{\mathrm{v_{t\perp}}}{\gamma}$ are the
parallel and perpendicular components of the pulsation velocity (e.g.
$\mathrm{v_{t\perp}}^{2}=\mathrm{v_{y}}^{2}+\mathrm{v_{z}}^{2}$), $i$ the
inclination to the jet of the turbulence pulses, in the jet rest frame,
$\displaystyle\eta=\frac{\mathrm{v_{t\perp}}}{\mathrm{v_{j}}}$,
$\displaystyle\gamma=\left(1-\left(\frac{\mathrm{v_{j}}}{c}\right)^{2}\right)^{-1/2}$
the Lorentz factor, $c$ the speed of light.

Figure 1: Profiles of the entrainment along the jets of 3C 31: for the flux
densities of the external (blue solid circles) and of the internal (blue
hollow circles) entrainment after Wang et al. [5], and the theoretical
external entrainment $q_{\mathrm{t}}$ (red solid line). The internal
entrainment is from Fig. 5a of Wang et al. [5], not their approximation
$g_{\mathrm{s}}(x)$. The jet velocity (dashed line) and the turbulence cross-
section (dotted line) are also plotted.

Two parameters of the entrainment model, $\beta$ and $\mathrm{v_{t}}$, were
defined intuitively as $\displaystyle\frac{1}{6}$ and the speed of sound
$c_{\mathrm{s}}$, respectively, and the entrained matter was supposed to just
acquire the jet velocity $\mathrm{v_{j}}$. The latter supposition
straightforwardly gives the turbulent shear stress at the jet edge
$\tau_{\mathrm{t}}=q_{\mathrm{t}}\mathrm{v_{j}}$, or $\tau_{\mathrm{t}}=\alpha
P$, where $\alpha$ now is the parameter dependent on turbulence intensity, not
on turbulent viscosity, and $P$ the pressure. Afterwards, $\beta$ was
constrained to $\displaystyle\frac{1}{5}$ by fitting the function
$q_{\mathrm{t}}$, which is plotted in Fig. 1, to external entrainment of Wang
et al. [5]. The entrainment model is simple and was compared with very
speculative data, therefore its success on this plot seems as miraculous, and
further investigations of the model are required.

##  3 Parameters of entrainment model

For isotropic turbulence the mean velocity of stochastic flow in a direction
is, as follows:

$\mathrm{v_{f}}=\beta\mathrm{v_{t}}=\mathrm{v_{t}}\int_{0}^{\pi/2}\cos{\theta}\frac{2\pi\sin{\theta}d\theta}{4\pi}=\frac{1}{4}\mathrm{v_{t}}=\frac{1}{4}\sqrt{\frac{3}{2}}\mathrm{v}_{\perp},$
(3)

where $\theta$ is the inclination of the velocity of a pulsation to the given
direction. Then Eq. (1) is rewritten as

$q_{\mathrm{t}}=\beta\rho_{\mathrm{a}}\mathrm{v}_{\perp}\sigma_{\mathrm{t}},$
(4)

where $\displaystyle\beta=\frac{1}{3.266}\approx\frac{1}{\pi}$ is accepted,
and $\mathrm{v}_{\perp}$ substitutes $\mathrm{v_{t}}$.

To test this model vs. experimental data on free shear flows, we have to
invent how the entrainment $q_{\mathrm{t}}$ depends on turbulent stresses. For
this we define the drag velocity $\mathrm{v}_{\tau}$ so that

$\tau_{\mathrm{xy}}=q_{\mathrm{t}}\mathrm{v}_{\tau},$ (5)

and the mean flux of the fluctuating streamwise momentum $\rho\mathrm{v_{x}}$,
which is transported by the transverse fluctuations $\mathrm{v_{y}}$, or the
turbulent shear stress, is derived for a self-similar mixing layer as follows:

$\tau_{\mathrm{xy}}=q_{\mathrm{t}}\int_{-\infty}^{y_{0}}\rho\mathrm{V_{x}}(y)^{2}dy\bigg{/}\int_{-\infty}^{y_{0}}\rho\mathrm{V_{x}}(y)dy,$
(6)

where $y_{0}$ is the depth where $\tau_{\mathrm{xy}}(y_{0})$ is maximal, from
which we can express $\mathrm{v}_{\tau}$. Hereafter the term $\mathrm{V_{x}}$
means the streamwise velocity averaged over time. Hence, the drag velocity is
the weighted stream velocity acquired by entrained matter, it determines the
stress, therefore it substitutes the variable $\mathrm{v_{j}}$ in the
turbulent cross-section (2), as we guess.

Figure 2: A graph of the experimental (red full circles) and simulated (blue
hollow symbols) data on the Reynolds stresses of self-similar free mixing
layers in the coordinates of the squared intensity
$\displaystyle\frac{R_{\mathrm{yy}}+R_{\mathrm{zz}}}{\rho\mathrm{V_{x}}^{2}}$
and the normalized shear stress
$\displaystyle\frac{R_{\mathrm{xy}}}{\rho\mathrm{V_{x}}^{2}}$. The
experimentals are from [7, Bell & Mehta 1990] and [8, Gruber et al. 1993],the
data of spatial direct numerical simulations (DNS) are from [6, Zhou et al.
2012] and [9, Barre & Bonnet 2015] (circles), the data of variable-density
temporal DNS are from [10, Almagro et al. 2017] and [11, Baltzer & Livescu
2020] (squares and diamonds, respectively), crosses stand for the density
ratio of the two free streams $s\neq 1$. Line is the function
$\tau_{\mathrm{xy}}=q_{\mathrm{t}}\mathrm{v}_{\tau}$.

We have collected the experimental and simulated data on self-similar free
mixing layers from the literature. To the best of our knowledge, there are
only few experimental data on the needed set of Reynolds stresses
($R_{\mathrm{xy}}$, $R_{\mathrm{yy}}$ and $R_{\mathrm{zz}}$): the data on
$R_{\mathrm{zz}}$ are very scarce, besides they are contradictory to each
other [6, Zhou et al. 2012]. The simulated data on spatial free shear flows
also are scarce, and the temporal ones are not quite adequate for the
entrainment investigations. The collected data are plotted in Fig. 2 in
coordinates of the normalized Reynolds stresses:
$\displaystyle(I_{\mathrm{y}}^{2}+I_{\mathrm{z}}^{2},\
I_{\mathrm{xy}}^{2})\equiv\frac{(R_{\mathrm{yy}}+R_{\mathrm{zz}},\
R_{\mathrm{xy}})}{\rho\mathrm{V_{x}}^{2}}$, where $\mathrm{V_{x}}$ is the
velocity difference over the shear layer width.

The plot of Eq. (5) is well aligned with the plotted data points and exactly
predicts the point of Bell & Mehta. This is the only point among ocean of
experimental data on turbulent mixing layers which is characterized by
uncertainty, and this uncertainty is only $\pm 5\%$, while the observed
experimental scatter is 25-50% for the layer growth rates (and the same is
true for the simulations, [12, Smits & Dussauge 2006, pp. 140, 156]).
Moreover, this point is the most referenced in comparative studies. Thus,
other points have much smaller weights and we don’t discuss them.

To take a more hard probe of the entrainment law (4), we checked out its
capability to predict the Smagorinsky constant $C_{\mathrm{S}}$, an empirical
constant used in turbulent flow models in large eddy simulations (e.g. [13,
Pope 2000, p. 587]). For the simple shear layer we match the kinematic
turbulent viscosity after Smagorinsky to the viscosity trivially derived from
Eq. (5) omitting the square root in $\sigma_{\mathrm{t}}$:

$\nu_{\mathrm{t}}=(C_{\mathrm{S}}\Delta_{\mathrm{S}})^{2}\left|\frac{\mathrm{dV_{x}}}{\mathrm{d}y}\right|\equiv\beta\mathrm{v}_{\perp}^{2}\left(\frac{\mathrm{d}y}{\mathrm{dV_{x}}}\right)^{2}\left|\frac{\mathrm{dV_{x}}}{\mathrm{d}y}\right|\approx\beta\left(\frac{\mathrm{v}_{\perp}}{\Delta\mathrm{V_{x}}}\Delta
y\right)^{2}\left|\frac{\mathrm{dV_{x}}}{\mathrm{d}y}\right|,$ (7)

where $\Delta_{\mathrm{S}}$ is the specific length scale of large eddy
simulations, and the mean gradient over the transverse scale $\Delta y$
substitutes the local gradient. By setting $\Delta y\equiv\Delta_{\mathrm{S}}$
and substituting the experimental turbulent intensities of self-similar
incompressible plane mixing layers from [14, Yoder et al. 2015]

$0.016\leq\frac{\overline{\mathrm{v_{y}}^{2}}}{\Delta\mathrm{V_{x}}^{2}}\leq
0.020,\
0.020\leq\frac{\overline{\mathrm{v_{z}}^{2}}}{\Delta\mathrm{V_{x}}^{2}}\leq
0.022,$ (8)

for the ratio
$\displaystyle\bigg{(}\frac{\mathrm{v}_{\perp}}{\Delta\mathrm{V_{x}}}\bigg{)}^{2}$
in Eq. (7) we get the constant value range:

$C_{\mathrm{S}}=\sqrt{\beta}\frac{\mathrm{v}_{\perp}}{\Delta\mathrm{V_{x}}}=0.107-0.116,$
(9)

which is in the bounds of the experimental range $C_{\mathrm{S}}\simeq
0.1-0.12$ for shear flows [15, Sagaut 2006, p. 124]. Omitting the square root
in $\sigma_{\mathrm{t}}$ we exaggerated the above $C_{\mathrm{S}}$ by $\approx
7\%$, if to take roughly
$\displaystyle\mathrm{v}_{\tau}=\frac{\mathrm{V_{x}}}{2}$, that does not
change positivity of our test of the entrainment law vs. the Smagorinsky
constant. This justifies the setting $\Delta y\equiv\Delta_{\mathrm{S}}$ that,
on other hand, suggested proportionality of the scales of turbulent velocity
$\mathrm{v_{t}}$ to the driving scales $\Delta\mathrm{V_{x}}$, i.e.
$\displaystyle\frac{\mathrm{v}_{\perp}}{\Delta\mathrm{V_{x}}}$ is independent
on $\Delta y$.

##  4 Entrainment in accretion disk

An accretion disk is a rotating shear layer. Nonlaminar flows are inevitable
in accretion disks in view of their huge Reynolds numbers. Pure hydrodynamic
turbulence and the formation of even large long-living vortices in accretion
disk models has been observed [16, Manger & Klahr 2018]. Vortices, we know,
provide turbulent shear stress over macroscopic distances of mixing. We
conjecture that long-living shear layers of intermittent vorticity,
irrespectively of mechanisms of turbulence, form in the accretion disks (see
vorticity maps in [16, Manger & Klahr 2018] and [17, Richard, Nelson & Umurhan
2016]) and entrain the disk stuff from the irrotational layers between them,
i.e. pull the stuff radially inward and outward of the disk, with the net
effect of stuff accretion and angular momentum excretion in the entrainment
processes.

In a steady mode of accretion with the accretion rate of mass inflow
$\dot{M}_{\mathrm{a}}$, the accretion disk between radii $r$ and $r_{0}$ has
to shed the angular momentum

$\dot{M}_{\mathrm{a}}(\mathrm{v}_{\phi}r-\mathrm{v}_{\phi
0}r_{0})=T_{\mathrm{r\phi}}r-T_{\mathrm{r\phi 0}}r_{0},$ (10)

due to the momentum fluxes $T_{\mathrm{r\phi}}$. Here $\dot{M}_{\mathrm{a}}$
and $T_{\mathrm{r\phi}}$ (and the entrainment flux $Q_{\mathrm{t}}$ below) are
the integrated rates over a cylindrical cross-section of the accretion disk of
a radius $r$, e.g. $\displaystyle
T_{\mathrm{r\phi}}=\int_{-\infty}^{\infty}\int_{0}^{2\pi}\tau_{\mathrm{r\phi}}dzd\phi$
in cylindrical coordinates ($r,\,\phi,\,z$), and $\mathrm{v}_{\phi}$ is the
azimuthal velocity. Under the condition $T_{\mathrm{r\phi 0}}=0$, the momentum
flux

$T_{\mathrm{r\phi}}=\dot{M}_{\mathrm{a}}\mathrm{v}_{\phi}\left(1-\left(\frac{r}{r_{0}}\right)^{1/2}\right),$
(11)

which is resolved from Eq. (10) (cf. Eq. (2.4) in [2, SS73], here
$\mathrm{v}_{\phi}$ is allowed to be the Keplerian), can be induced:
a) either by "the friction between the adjacent layers" [2, SS73], i.e.
locally; in this way $\dot{M}_{\mathrm{a}}$ is derived by Shakura & Sunyaev
[2, SS73] as $\dot{M}_{\mathrm{a}}=f(T_{\mathrm{r\phi}})$;
b) or by large-scale vortices, i.e. by entrainment; then
$\dot{M}_{\mathrm{a}}=\Delta
Q_{\mathrm{t}}=Q_{\mathrm{t\,in}}-Q_{\mathrm{t\,out}}$, in this way the
accretion rate is not a function of turbulent viscosity.

The entrainment in the disk is locally seen as that in a flat mixing layer,
for which the formula (4) was designed, if the disk half thickness $H\ll r$.
For convenience, we again use it in the stripped down version:

$q_{\mathrm{t}}=\beta\rho\frac{\mathrm{v}_{\perp}^{2}}{\mathrm{v}_{\tau}}.$
(12)

As in the case discussed above, the fluctuations $\mathrm{v}_{\perp}$ are
proportional to the driving scale $\mathrm{v}_{\tau}$, which, again, roughly
equals a half of the (Keplerian) velocity difference in the ring turbulent
layer of a width $\Delta r$. The disk half thickness $H$ is defined from the
disk hydrostatic balance as
$\displaystyle\frac{H}{r}=\frac{c_{\mathrm{s}}}{\mathrm{v}_{\phi}}$, therefore
the drag velocity, and the turbulent velocity $\mathrm{v}_{\perp}$ too, are
scaled with the temperature of the disk as

$\mathrm{v}_{\tau}=\frac{1}{2}\left|\Delta\mathrm{v}_{\phi}\right|\approx\frac{1}{4}\frac{\mathrm{v}_{\phi}}{r}\Delta
r=\frac{1}{4}\frac{\Delta r}{H}c_{\mathrm{s}}\propto T^{1/2}.$ (13)

In accretion disks, temperature slopes as $\displaystyle T\propto
r^{s_{\mathrm{T}}}$, where the power $s_{\mathrm{T}}<0$. The entrainment
fluxes $Q_{\mathrm{t\,in}}$ and $Q_{\mathrm{t\,out}}$ are the time-averages
over the same surface but of opposing vortical flows. Here we conjecture that
the mean density $\rho$ in the opposing vortices does not change essentially
while they propagate, and their stochastic velocities adjust to the
temperature of the ambient medium (really, to the driving velocity, which is
the property of the background flow). Thus we get the entrainment scaling only
with $r$, $\displaystyle q_{\mathrm{t}}\propto r^{s_{\mathrm{T}}/2}$.
Therefore, the accretion rate is derived by a simple difference of the
entrainment fluxes $Q_{\mathrm{t}}$ of the neighboring mixing layers, each of
the same width $\Delta r\ll r$:

$\dot{M}_{\mathrm{a}}=2\pi
r\Sigma\beta\left(\frac{\mathrm{v}_{\perp}^{2}}{\mathrm{v}_{\tau}}\bigg{|}_{r-\Delta
r}-\frac{\mathrm{v}_{\perp}^{2}}{\mathrm{v}_{\tau}}\bigg{|}_{r+\Delta
r}\right)\approx-2\pi\Sigma\beta\frac{s_{\mathrm{T}}}{2}\frac{\mathrm{v}_{\perp}^{2}}{\mathrm{v}_{\tau}}2\Delta
r,$ (14)

where $\displaystyle\Sigma=\int_{-\infty}^{\infty}\rho dz$ is the surface
density in the disk. Substituting expression (13) for $\mathrm{v}_{\tau}$ in
this equation, we arrive to:

$\dot{M}_{\mathrm{a}}\omega=2\pi\alpha\Sigma c_{\mathrm{s}}^{2},$ (15)

where $\omega$ is the angular velocity, and

$\alpha=-\frac{8}{3}\beta
s_{\mathrm{T}}\frac{\mathrm{v_{t}}^{2}}{c_{\mathrm{s}}^{2}}.$ (16)

The functional form of the accretion rate in Eq. (15) is just the same as in
the $\alpha$-model of accretion disks of Shakura & Sunyaev [2, SS73, Eq.
(2.4)] for $r\gg r_{0}$. Moreover, the viscosity parameter $\alpha$ has the
right functional form
$\displaystyle\alpha\propto\frac{\mathrm{v_{t}}^{2}}{c_{\mathrm{s}}^{2}}$,
which can be envisaged from the physical interpretation of the stress
$\displaystyle\tau_{\mathrm{r\phi}}\propto\rho\mathrm{v_{t}}^{2}$ and the
definition $\displaystyle\tau_{\mathrm{r\phi}}\propto\alpha\rho
c_{\mathrm{s}}^{2}$ of the parameter $\alpha$ [2, SS73, Eq. (1.2)].

##  5 Conclusions

We have elucidated the entrainment model of [3, Panferov 2017] and applied it
to the problems on turbulent shear stress, in other words, on turbulent
viscosity, in a plane mixing layer, in the jets of 3C 31 and in accretion
disks. The model reduces the problem of turbulent shear stress to the question
about turbulence intensity.

The entrainment model allows us to predict the deceleration of the 3C 31 jets,
which is caused by turbulent friction at the jets surface. The model readily
expresses the surface turbulent stress in the form $\tau_{\mathrm{t}}=\alpha
P$, where $\alpha$ can be set.

Interdependence of the Reynolds stresses observed in plane mixing layers under
varying physical conditions is predicted (Fig. 2).

Application of the principles of the entrainment model to the Smagorinsky
prescription of turbulent viscosity gave us the Smagorinsky constant
$C_{\mathrm{S}}\simeq 0.11$, observed in experimental shear flows.

The entrainment model has been shown to be capable to construct the widely
used $\alpha$-model of accretion disks in another way, bypassing the Prandtl’s
assignment of turbulent viscosity.

## Acknowledgements

I wish to thank the anonymous referee.

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