# An Ehrhart Theory For Tautological Intersection Numbers

Adam Afandi

###### Abstract

We discover that tautological intersection numbers on
$\overline{\mathcal{M}}_{g,n}$, the moduli space of stable genus $g$ curves
with $n$ marked points, are evaluations of Ehrhart polynomials of partial
polytopal complexes. In order to prove this, we realize the Virasoro
constraints for tautological intersection numbers as a recursion for integer-
valued polynomials. Then we apply a theorem of Breuer that classifies Ehrhart
polynomials of partial polytopal complexes by the nonnegativity of their
$f^{*}$-vector. In dimensions 1 and 2, we show that the polytopal complexes
that arise are _inside-out polytopes_ i.e. polytopes that are dissected by a
hyperplane arrangement.

###### Contents

  1. 1 Introduction
    1. 1.1 Main Result
    2. 1.2 Strategy of Proof
    3. 1.3 Outline of Paper
    4. 1.4 Ackowledgments
  2. 2 Ehrhart Theory
    1. 2.1 Ehrhart Polynomials of Convex Integral $d$-Polytopes
    2. 2.2 Ehrhart Polynomials of Partial Polytopal Complexes
    3. 2.3 Useful Properties of Ehrhart Polynomials
  3. 3 Tautological Intersection Numbers on $\overline{\mathcal{M}}_{g,n}$
    1. 3.1 The Moduli of Stable Curves
    2. 3.2 The Tautological Ring
    3. 3.3 Computing $\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$
  4. 4 Proof of Theorem 1
    1. 4.1 $L_{\vec{d}}(g)$ Is An Integer-Valued Polynomial
    2. 4.2 The $f^{*}$-Vector of $L_{\vec{d}}(g+m(\vec{d}))$ Is Nonnegative
    3. 4.3 Putting the Pieces Together
  5. 5 Examples
    1. 5.1 $|\vec{d}|=1$
    2. 5.2 $|\vec{d}|=2$
  6. 6 What Next?

##  1 Introduction

The purpose of this paper is to present a novel perspective concerning
tautological intersection numbers on $\overline{\mathcal{M}}_{g,n}$, the
moduli space of stable $n$-pointed genus $g$ curves. This perspective uses
ideas from a (seemingly) distant subfield of mathematics, namely, _Ehrhart
theory_.

The main idea behind the present paper can be summarized as follows:
_tautological intersection numbers can be organized into evaluations of
Ehrhart polynomials of partial polytopal complexes_.

Our intent is to present an exposition that is accessible to both algebraic
geometers interested in $\overline{\mathcal{M}}_{g,n}$, and combinatorialists
and discrete geometers coming from Ehrhart theory.

###  1.1 Main Result

An important algebraic object attached to the moduli space of pointed stable
curves is the _tautological ring_ :

$R^{*}(\overline{\mathcal{M}}_{g,n})$

This ring is a subring of $A^{*}(\overline{\mathcal{M}}_{g,n})$, the Chow ring
of $\overline{\mathcal{M}}_{g,n}$. Beginning with the work of Mumford [Mum83],
great strides have been made in our understanding of
$R^{*}(\overline{\mathcal{M}}_{g,n})$. In particular, many important cycles in
$A^{*}(\overline{\mathcal{M}}_{g,n})$ have been shown to be tautological.

Let $\alpha\in R^{3g-3+n}(\overline{\mathcal{M}}_{g,n})$. Since
$\text{dim}(\overline{\mathcal{M}}_{g,n})=3g-3+n$, we can integrate $\alpha$
against the fundamental class of $\overline{\mathcal{M}}_{g,n}$ to obtain a
_tautological intersection number_ :

$\left(\int_{\overline{\mathcal{M}}_{g,n}}\alpha\right)\in\mathbb{Q}$

Let $\mathbb{L}_{i}$ be the $i^{th}$ universal cotangent line bundle on
$\overline{\mathcal{M}}_{g,n}$ i.e. the line bundle whose fiber over a point
$[C,p_{1},\ldots,p_{n}]\in\overline{\mathcal{M}}_{g,n}$ is $T_{p_{i}}^{*}C$,
the cotangent space to the $i^{th}$ marked point. Define $\psi_{i}$ to be the
first Chern class of $\mathbb{L}_{i}$,

$\psi_{i}:=c_{1}(\mathbb{L}_{i})$

These elements in $R^{1}(\overline{\mathcal{M}}_{g,n})$ are usually referred
to as $\psi$-classes. They play a central role in the tautological
intersection theory of $\overline{\mathcal{M}}_{g,n}$ for a multitude of
reasons. In particular, all tautological intersection numbers can be reduced
to intersection numbers only involving $\psi$-classes, that is, intersection
numbers of the form

$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}:=\int_{\overline{\mathcal{M}}_{g,n}}\psi_{1}^{d_{1}}\ldots\psi_{n}^{d_{n}}$

The main theorem of this paper shows that these intersection numbers are
evaluations of _Ehrhart polynomials_ of partial polytopal complexes. An
Ehrhart polynomial is a counting function for integer lattice points of
dilations of an integral polytope. More precisely, given an integral
$d$-polytope $P\subset\mathbb{R}^{d}$, its Ehrhart polynomial is

$L_{P}(g):=\left|gP\cap\mathbb{Z}^{d}\right|$

where $g$ is a positive integer, and $gP$ is the $g^{th}$ dilate of $P$. One
can extend the notion of an Ehrhart polynomial to more general polyhedral
objects, in particular, to _partial polytopal complexes_ , which are disjoint
unions of open polytopes (see Section 2 below).

Here is our main theorem:

######  Theorem 1.

Let $n\geq 1$, $\vec{d}:=(d_{1},\ldots,d_{n})\in\mathbb{Z}^{n}_{\geq 0}$.
Define $|\vec{d}|:=\sum_{i=1}^{n}d_{i}$,
$C(\vec{d}):=\prod_{i=1}^{n}(2d_{i}+1)!!$, and
$m(\vec{d})=m:=\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$. Then there
exists an integral partial polytopal complex $P_{\vec{d}}$ of dimension
$|\vec{d}|$ and volume $\text{vol}\left(P_{\vec{d}}\right)=6^{|\vec{d}|}$ such
that

$24^{g+m}(g+m)!C(\vec{d})\int_{\overline{\mathcal{M}}_{g+m,n+1}}\psi_{1}^{d_{1}}\ldots\psi_{n}^{d_{n}}\psi_{n+1}^{3(g+m)-2+n-|\vec{d}|}=\\#\\{\text{integer
lattice points in $gP_{\vec{d}}$}\\}$

where $gP_{\vec{d}}$ is the $g^{th}$ dilate of $P_{\vec{d}}$.

The statement of Theorem 1 has some idiosyncratic notation, and might seem a
bit opaque, so let us take a moment to explain how one should think about what
Theorem 1 actually says.

Suppose you fix an integer vector
$\vec{d}=(d_{1},\ldots,d_{n})\in\mathbb{Z}_{\geq 0}^{n}$ which corresponds to
a monomial of $\psi$-classes $\psi_{1}^{d_{1}}\ldots\psi_{n}^{d_{n}}$.
Consider the family of intersection numbers

$\left\\{\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g}\right\\}_{g.d_{n+1}\geq
0}$

Since $\vec{d}$ is fixed, in order for
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g}$ to be nonzero,
$d_{n+1}$ must be

$3g-3+(n+1)-|\vec{d}|=3g-2+n-|\vec{d}|$

which explains the exponent of the last insertion in Theorem 1. Furthermore,
notice that there exists a smallest genus $g$ such that
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g}\not=0$. This
genus is the smallest genus $g$ such that the exponent on the last insertion
is nonnegative, that is, the smallest genus $g$ such that

$3g-2+n-|\vec{d}|\geq 0$

which is precisely

$\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil$

Consequently, we see that
$m(\vec{d}):=\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$ is designed to
be an appropriate shift of the genera, in that it ensures the following
equivalence:

$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g+m(\vec{d})}\not=0\iff
g\geq 1$

The statement of Theorem 1 then says that there exists a partial polytopal
complex $P_{\vec{d}}$ that only depends on $\vec{d}$, such that

$24^{g+m(\vec{d})}(g+m(\vec{d}))!C(\vec{d})\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g+m(\vec{d})}=\\#\\{\text{integer
lattice points in $gP_{\vec{d}}$}\\}$ (1)

Phrased in this way, we see that the smallest genus in which
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g+m(\vec{d})}\not=0$
corresponds to the first dilate of $P_{\vec{d}}$, the next smallest genus will
correspond to the 2nd dilate of $P_{\vec{d}}$, and so on. Of course, one could
easily rewrite the equation in Theorem 1 as

$24^{g}g!C(\vec{d})\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g}=\\#\\{\text{integer
lattices points in $(g-m(\vec{d}))P_{\vec{d}}$}\\}$ (2)

However, notice that Equation 2 is only valid for
$g\geq\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil$. Both ways of phrasing
Theorem 1 i.e. Equation 1 and Equation 2, are equivalent. However, Equation 1
emphasizes the role of the partial polytopal complex and its dilates, while
Equation 2 emphasizes the role of the intersection numbers
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g}$. In the
present paper, we have chosen the former.

In addition to the connection with lattice-point counting in partial polytopal
complexes, Theorem 1 implies that, as a formal consequence of Ehrhart theory,
the quantity

$24^{g}g!\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{3g-2+n-|\vec{d}|}\right>_{g}$

is a _polynomial_ in $g$, whose leading coefficient is

$\frac{6^{|\vec{d}|}}{C(\vec{d})}$

As far as the author is aware, this observation has not been fleshed out in
the literature. We hope that this observation of polynomiality will contribute
to the development of faster algorithms to compute
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$. It may also contribute to a
different understanding of the large-genus asymptotics of
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$.

###  1.2 Strategy of Proof

The proof of Theorem 1 is in Section 4. There are three main components that
comprise the argument.

Let $\vec{d}\in\mathbb{Z}^{n}_{\geq 0}$, and define

$L_{\vec{d}}(g):=24^{g}g!C(\vec{d})\left<\tau_{\vec{d}}\tau_{3g-2+n-|\vec{d}|}\right>_{g}$

A priori, this is an arbitrary family of intersection numbers parametrized by
genera $g$. However, we prove that $L_{\vec{d}}(g)$ is an _integer-valued
polynomial_ in $g$ whose leading coefficient is $6^{|\vec{d}|}$. In order to
prove this, we use the fact that $L_{\vec{d}}(g)$ can be recursively computed
by the _String Equation_ , the _Dilaton Equation_ , and the (higher) _Virasoro
constraints_. We then prove that the property of being an integer-valued
polynomial with leading coefficient $6^{|\vec{d}|}$ is a property that remains
invariant under all three of these recursive operations. This concludes the
first component of the argument.

Once we know that $L_{\vec{d}}(g)$ is an integer-valued polynomial, we can
consider the shifted integer-valued polynomial $L_{\vec{d}}(g+m(\vec{d}))$
(see the paragraphs directly following the statement of Theorem 1 for an
explanation of the shift $m(\vec{d})$). We then expand
$L_{\vec{d}}(g+m(\vec{d}))$ in the binomial basis $\\{{g-1\choose k}\\}$. The
choice of the binomial basis $\\{{g-1\choose k}\\}$ is not an arbitrary
choice: in the field of Ehrhart theory, such an expansion of an Ehrhart
polynomial computes the $f^{*}$-vector of the polynomial, which, under certain
assumptions, gives one information about the geometry of the corresponding
polytope (see Section 2 below). We prove that the $f^{*}$-vector of
$L_{\vec{d}}(g+m(\vec{d}))$ is nonnegative. The strategy for this component of
the argument is the same as in the previous one: we show that nonnegativitiy
of the $f^{*}$-vector is a property that remains invariant throughout all
recursive procedures that compute $L_{\vec{d}}(g+m(\vec{d}))$.

Once we know that the $f^{*}$-vector of $L_{\vec{d}}(g+m(\vec{d}))$ is
nonnegative, we apply the following classification theorem of Breuer: _a
polynomial $P(g)$ of degree $d$ is the Ehrhart polynomial of a partial
polytopal complex of dimension $d$ if and only if the $f^{*}$-vector of $P(g)$
is integral and nonnegative_.

###  1.3 Outline of Paper

Our intention is to make the paper readable to combinatorialists from Ehrhart
theory and algebraic geometers interested in $\overline{\mathcal{M}}_{g,n}$.
Consequently, we spend a good portion of the paper explaining fundamental
ideas and results from both fields. However, we keep technical details to a
minimum, and we refer the reader to sources in the literature when necessary.

In Section 2, we discuss standard results and basic notions from Ehrhart
theory. The main goal of Section 2 is to define what one means when one refers
to the ‘Ehrhart polynomial of a partial polytopal complex’. If one is already
familiar with what this means, it is safe to skip this section.

In Section 3, we recall results concerning tautological intersection numbers,
especially the ones needed for this paper. The goal of this section to show
how one computes the tautological intersection number
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$ recursively using the
String/Dilaton equation and the Virasoro constraints.

In Section 4 we present the proof of Theorem 1. The main computational idea is
to view the Virasoro constraints as a recursion for integer-valued polynomials
in general, and Ehrhart polynomials in particular.

The paper ends with the computation of a few examples (see (Section 5), along
with an outline for future work (Section 6).

###  1.4 Ackowledgments

The author is grateful to Renzo Cavalieri for his comments and suggestions on
an earlier draft. Many thanks to David and Majka Phillips for their
hospitality while the author was making edits.

This work was supported by the Cluster of Excellence _Mathematics Münster_ and
the CRC 1442 _Geometry: Deformations and Rigidity_. “Funded by the Deutsche
Forschungsgemeinschaft (DFG, German Research Foundation) - Project-I 427320536
- SFB 1442, as well as under Germany’s Excellence Strategy EXC 2044390685587,
Mathematics Münster: Dynamics - Geometry - Structure”

##  2 Ehrhart Theory

The purpose of this section is to define the Ehrhart polynomial of a partial
polytopal complex. Pedagogically, it seems natural to first begin with a
discussion on polytopes, which will provide the necessary background to
discuss partial polytopal complexes.

For more details concerning the Ehrhart theory of convex integral
$d$-polytopes, we recommend ([BR07], Chapter 3). For the most part, the
standard techniques and ideas of Ehrhart theory in the context of polytopes
extend to the context of partial polytopal complexes. For an exposition on
partial polytopal complexes that aligns well with the purposes of this paper,
see ([Bre12], Section 2).

###  2.1 Ehrhart Polynomials of Convex Integral $d$-Polytopes

Let $P\subset\mathbb{R}^{d}$ be an integral convex polytope, and let
$v_{1},\ldots,v_{n}\in\mathbb{Z}^{d}$ be the vertices of $P$,

$P=\text{Conv}(v_{1},\ldots,v_{n})\subset\mathbb{R}^{d}$

Throughout, we always assume that $P$ is _full-dimensional_ , that is, $P$ is
a $d$-polytope. For an integer $g\geq 1$, the $g^{th}$-dilate of $P$, denoted
$gP$, is defined to be

$gP:=\text{Conv}(gv_{1},\ldots,gv_{n})=\\{gp:p\in P\\}$

Ehrhart theory is chiefly concerned with the task of counting _lattice points_
in $gP$, i.e. understanding the function

$L_{P}(g):g\mapsto|gP\cap\mathbb{Z}^{d}|$

Here is the main theorem of Ehrhart theory:

######  Theorem 2 (Ehrhart’s Theorem).

$L_{P}(g)$ is a rational polynomial in $g$ of degree $d$.

We call $L_{P}(g)$ the _Ehrhart polynomial_ of $P$. Theorem 2 is originally
due to Eugène Ehrhart. For a proof, see ([BR07], Theorem 3.8).

An important family of polytopes are the $d$-dimensional _simplices_. An
integral _$d$ -simplex_ is a $d$-dimensional polytope
$\Delta\subset\mathbb{R}^{d}$ that is the convex hull of $d+1$ affinely
independent integral points. The _standard $d$-simplex_ is given by
$\Delta_{d}:=\text{Conv}(\vec{0},e_{1},\ldots,e_{d})\subseteq\mathbb{R}^{d}$.
An _open $d$-simplex_ is the relative interior of a $d$-simplex.

######  Example 1.

Let $P\subset\mathbb{R}^{2}$ be the standard $2$-simplex,

$P=\text{Conv}(\vec{0},e_{1},e_{2})\subset\mathbb{R}^{2}$

By Ehrhart’s theorem, we know there exists rational numbers $a_{0},a_{1}$, and
$a_{2}$ such that

$L_{P}(g)=a_{2}g^{2}+a_{1}g+a_{0}$

Counting lattice points by hand (see Figure 1), we see that
$L_{P}(1)=3,L_{P}(2)=6$, and $L_{P}(3)=10$. These equations suffice to
determine the coefficients $a_{0},a_{1}$, and $a_{2}$, and we obtain

$L_{P}(g)=\frac{1}{2}g^{2}+\frac{3}{2}g+1$

$L_{P}(1)=3$$L_{P}(2)=6$$L_{P}(3)=10$ Figure 1: As is in Example 1, we can
compute $L_{P}(g)$ by polynomial interpolation

We need a slight generalization of integral $d$-polytopes. Instead of dealing
with one polytope at a time, it is possible to take a collection of polytopes
and glue them along their faces, albeit in a compatible way. We call these
objects _polytopal complexes_ (see also [Zie12], Definition 5.1):

######  Definition 1.

A _polytopal complex_ is a finite collection $K$ of polytopes that satisfies
the following three properties:

  1. 1.

The empty polytope is in $K$

  2. 2.

$P\in K,f\subseteq P$ is a face of P $\implies f\in K$

  3. 3.

$P,Q\in K\implies P\cap Q\in K$, and $P\cap Q$ is a face of both $P$ and $Q$

The elements of $K$ are called the _faces_ of $K$. The dimension of $K$ is the
maximum dimension of the faces of $K$.

Ehrhart’s theorem still holds in the context of polytopal complexes, that is,
the counting function $L_{K}(g)=\\#\\{\text{integer lattice points in
$gK$}\\}$ is a rational polynomial in $g$ of degree $d$.

Let $P,Q\subseteq\mathbb{R}^{d}$ be integral $d$-polytopes. We say $P$ and $Q$
are _lattice equivalent_ if there exists an affine isomorphism
$\phi:\mathbb{Z}^{d}\rightarrow\mathbb{Z}^{d}$ sending the vertices of $P$ to
the vertices of $Q$. A polytopal complex $K$ is a _simplicial complex_ if
every face of $K$ is a simplex. A _triangulation_ of an integral polytope $P$
is a simplicial complex whose support is $P$. We say that a triangulation is
_unimodular_ if the simplices in the corresponding simplicial complex are
lattice equivalent to the standard simplex.

Unimodular triangulations are useful in Ehrhart theory due to the following
result:

######  Theorem 3.

Let $(P,K)$ be a unimodular triangulation of an integral $d$-polytope $P$.
Define

$f_{i}^{*}:=\\#\\{\text{$i$-dimensional open simplices in $K$}\\}$

Then the Ehrhart polynomial of $P$ has the following form:

$L_{P}(g):=\sum_{k=0}^{d}f_{k}^{*}{g-1\choose k}$

We define the _$f^{*}$ -vector_ of $P$ to be the vector of integers
$(f_{0}^{*},\ldots,f_{d}^{*})$.

######  Example 2.

Recall the polytope $P$ given in Example 3,
$P=\text{Conv}(\vec{0},e_{1},e_{2})$. This is just the standard two-
dimensional simplex. The simplex itself provides a unimodular triangulation.
Upon inspection we see that its $f^{*}$-vector is $(3,3,1)$. Therefore,

$\displaystyle L_{P}(g)$ $\displaystyle=3{g-1\choose 0}+3{g-1\choose
1}+{g-1\choose 2}$ $\displaystyle=3+3(g-1)+\frac{1}{2}(g-1)(g-2)$
$\displaystyle=3g+\frac{1}{2}(g^{2}-3g+2)$
$\displaystyle=\frac{1}{2}g^{2}+\frac{3}{2}g+1$

as expected.

Not every integral polytope $P$ admits a unimodular triangulation. However, it
always makes sense to talk about the $f^{*}$-vector of an integral polytope.
The reason is as follows. Suppose $L(g)\in\mathbb{Q}[g]$ is a polynomial of
degree $d$. The set $\\{{g-1\choose k}\\}_{k=0}^{d}$ forms $\mathbb{Q}$-basis
for the vector space of all rational polynomials of degree $d$. Therefore,
there exists a unique vector $(f_{0}^{*},\ldots,f_{d}^{*})\in\mathbb{Q}^{d+1}$
such that $L(g)=\sum_{k=0}^{d}f_{k}^{*}{g-1\choose k}$. The $f^{*}$-vector of
an integral polytope $P$ is the unique vector $(f_{0}^{*},\ldots
f_{d}^{*})\in\mathbb{Q}^{d+1}$ such that
$L_{P}(g)=\sum_{k=0}^{d}f_{k}^{*}{g-1\choose k}$.

However, notice that $L_{P}(g)$ is actually an _integer-valued polynomial_ :

######  Definition 2.

An _integer-valued polynomial_ $L(g)\in\mathbb{Q}[g]$ is a polynomial such
that $L(\mathbb{N})\subseteq\mathbb{Z}$.

It turns out that the $f^{*}$-vector of an integer-valued polynomial is always
integral. Therefore, we can make the following definition:

######  Definition 3.

Let $P$ be a convex integral $d$-polytope. The $f^{*}$-vector of $P$ is the
unique integer tuple $(f_{0}^{*},\ldots,f_{d}^{*})\in\mathbb{Z}^{d+1}$ such
that

$L_{P}(g)=\sum_{k=0}^{d}f_{k}^{*}{g-1\choose k}$

###  2.2 Ehrhart Polynomials of Partial Polytopal Complexes

Generalizing even further, we need to consider open $d$-polytopes. An _open
$d$ polytope_ is the relative interior of an integral $d$-polytope. This
brings us to the generalization of polytopal complexes that we need:

######  Definition 4.

An _integral partial polytopal complex_ $K$ is the disjoint union of a finite
collection of open integral polytopes. The elements of $K$ are called the
faces of $K$. The dimension $d$ of $K$ is the maximum dimension of the faces
of $K$. The Ehrhart polynomial of $K$, denoted $L_{K}(g)$, is the sum of the
Ehrhart polynomials of each face of $K$.

######  Remark 1.

Notice that an integral partial polytopal complex is a generalization of an
integral polytopal complex. This follows from the observation that the support
of an integral polytopal complex is the disjoint union of the relative
interiors of all of its faces. Thus, every polytopal complex is neccessarily a
partial polytopal complex. Intuitively, one can think of this generalization
as simply allowing oneself to ‘excise’ or ‘throw away’ faces of any polytope
$P\in K$ in a polytopal complex.

######  Definition 5.

Let $K$ be a partial polytopal complex of dimension $d$. A _triangulation_
$\mathcal{T}$ of $K$ is a disjoint union of open simplices whose support is
$K$. The triangulation $\mathcal{T}$ is _unimodular_ if the closure of each
open simplex in $\mathcal{T}$ is lattice equivalent to the standard simplex.
The $f^{*}$-vector of $\mathcal{T}$ is $(f_{0}^{*},\ldots f_{d}^{*})$, where
$f_{i}^{*}:=\\#\\{\text{$i$-dimensional open simplices in $\mathcal{T}$}\\}$.

As in the case of polytopal complexes, if one can find a unimodular
triangulation of a partial polytopal complex, this immediately gives its
Ehrhart polynomial:

######  Theorem 4.

Let $K$ be a partial polytopal complex of dimension $d$ and let $\mathcal{T}$
be a unimodular triangulation of $K$. Then

$L_{K}(g)=\sum_{i=0}^{d}f_{i}^{*}{g-1\choose i}$

where $(f_{0}^{*},\ldots,f_{d}^{*})$ is the $f^{*}$-vector of $\mathcal{T}$.

Even if a partial polytopal complex does not admit a unimodular triangulation,
it still makes sense to talk about its $f^{*}$-vector. Furthermore, Ehrhart
polynomials of partial polytopal complexes are completely classified by their
$f^{*}$-vector due to the following result of Breuer:

######  Theorem 5 ([Bre12], Theorem 2).

Let $P(g)$ be an integer-valued polynomial of degree $d$. Then $P(g)$ is the
Ehrhart polynomial of an integral partial polytopal complex if and only if the
$f^{*}$-vector $(f_{0}^{*},\ldots,f_{d}^{*})$ of $P(g)$ is non-negative i.e.
$f_{i}^{*}\geq 0$ for all $0\leq i\leq d$.

###  2.3 Useful Properties of Ehrhart Polynomials

We recall properties of Ehrhart polynomials of partial polytopal complexes
that will be useful in the proof of Theorem 1.

######  Lemma 1.

Let $K$ be a partial polytopal complex of dimension $d$, and let $L_{K}(g)$ be
its Ehrhart polynomial. Then, for any $k\geq 0$, there exists a partial
polytopal complex $K^{\prime}$ such that $L_{K}(g+k)=L_{K^{\prime}}(g)$ is the
Ehrhart polynomial of $K^{\prime}$.

###### Proof.

This is a direct application of Theorem 5 and the basic binomial identity
${g\choose k}={g-1\choose k}+{g-1\choose k-1}$. Indeed, by Theorem 5, there
exists a non-negative $f^{*}$-vector $(f_{0}^{*},\ldots,f_{d}^{*})$ such that

$L_{K}(g)=\sum_{k=0}^{d}f_{k}^{*}{g-1\choose k}$

and therefore,

$\displaystyle L_{K}(g+1)$ $\displaystyle=\sum_{k=0}^{d}f_{k}^{*}{g\choose k}$
$\displaystyle=f_{0}^{*}{g-1\choose 0}+\sum_{k=1}^{d}f_{k}^{*}{g\choose k}$
$\displaystyle=f_{0}^{*}{g-1\choose
0}+\sum_{k=1}^{d}f_{k}^{*}\left({g-1\choose k}+{g-1\choose k-1}\right)$

By theorem 5, $L_{K}(g+1)$ is the Ehrhart polynomial of some partial polytopal
complex $K^{\prime}$. Therefore, the desired result is true for $k=1$, and we
proceed by induction. ∎

The next property concerns the cartesian products of partial polytopal
complexes. To get a better understanding of how this works, it makes sense to
first start with taking products of polytopes, and then generalize.

######  Definition 6.

Let $P\subset\mathbb{R}^{d_{1}}$ and $Q\subset\mathbb{R}^{d_{2}}$ be integral
polytopes of dimension $d_{1}$ and $d_{2}$, respectively. The _cartesian
product_ , or simply _product_ of the polytopes $P$ and $Q$ is the
$(d_{1}+d_{2})$-dimensional integral polytope denoted by

$\displaystyle P\times Q$
$\displaystyle:=\\{(p,q)\in\mathbb{R}^{d_{1}}\times\mathbb{R}^{d_{2}}=\mathbb{R}^{d_{1}+d_{2}}:p\in
P,q\in Q\\}$ $\displaystyle=\text{Conv}\\{(p,q)\in\mathbb{R}^{d_{1}+d_{1}}:p\
\text{a vertex of $P$},\ q\ \text{a vertex of $Q$}\\}$

The product of two open polytopes is defined similarly.

The operation of taking products of (open) polytopes plays well with Ehrhart
polynomials in that, if $P$ and $Q$ are integral polytopes,

$L_{P}(g)\times L_{Q}(g)=L_{P\times Q}(g)$

Furthermore, we can also define the product of two partial polytopal complexes
$K_{1}$ and $K_{2}$ in a way that also plays well with taking their Ehrhart
polynomials.

######  Definition 7.

Let $K_{1}$ and $K_{2}$ be partial polytopal complexes. The product
$K_{1}\times K_{2}$ is the partial polytopal complex defined by

$\displaystyle K_{1}\times K_{2}$ $\displaystyle:=\\{(p,q)\in P\times Q:P\in
K_{1},Q\in K_{2}\\}$ $\displaystyle=\coprod_{P\in K_{1},Q\in K_{2}}(P\times
Q)$

######  Lemma 2.

Let $K_{1}$ and $K_{2}$ be partial polytopal complexes, and let $L_{K_{1}}(g)$
and $L_{K_{2}}(g)$ be their Ehrhart polynomials, respectively. Then

$L_{K_{1}}(g)\times L_{K_{2}}(g)=L_{K_{1}\times K_{2}}(g)$

###### Proof.

The Ehrhart polynomial of $K_{1}\times K_{2}$ is, by definition, the sum of
the Ehrhart polynomials of the faces of $K_{1}\times K_{2}$. But the faces of
$K_{1}\times K_{2}$ are the open polytopes $\\{P\times Q:P\in K_{1},Q\in
K_{2}\\}$. Therefore

$\displaystyle L_{K_{1}\times K_{2}}(g)$ $\displaystyle=\sum_{P\in K_{1},Q\in
K_{2}}L_{P\times Q}(g)$ $\displaystyle=\sum_{P\in K_{1},Q\in
K_{2}}L_{P}(g)\times L_{Q}(g)$ $\displaystyle=\left(\sum_{P\in
K_{1}}L_{P}(g)\right)\left(\sum_{Q\in K_{2}}L_{Q}(g)\right)$
$\displaystyle=L_{K_{1}}(g)\times L_{K_{2}}(g)$

∎

######  Lemma 3.

Let $K$ be a partial polytopal complex, and let $L_{K}(g)$ be its Ehrhart
polynomial. Then the leading coefficient of $L_{K}(g)$ is the (Euclidean)
volume of $K$.

###### Proof.

In the case that $K$ is an integral $d$-polytope, the statement is classical
(see [BR07], Corollary 3.20). The result then follows from the observation
that the volume of an integral $d$-polytope is the same as the volume of its
relative interior. ∎

##  3 Tautological Intersection Numbers on $\overline{\mathcal{M}}_{g,n}$

In this section, we introduce $\overline{\mathcal{M}}_{g,n}$, its tautological
ring $R^{*}(\overline{\mathcal{M}}_{g,n})$, and all of the necessary results
that are required to compute intersection numbers of the form

$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}:=\int_{\overline{\mathcal{M}}_{g,n}}\psi_{1}^{d_{1}}\ldots\psi_{n}^{d_{n}}$

For readers who would like a nice introduction to the tautological
intersection theory of $\overline{\mathcal{M}}_{g,n}$, especially one that
focuses on computation, we would recommend [Zvo14] and [Pan15].

###  3.1 The Moduli of Stable Curves

Let $(g,n)$ be a pair of nonnegative integers such that $2g-2+n>0$, and let
$(C,p_{1},\ldots,p_{n})$ be an at-worst nodal curve of genus $g$, along with
$n$ smooth marked points $p_{1},\ldots,p_{n}$. We say that
$(C,p_{1},\ldots,p_{n})$ is _stable_ if the automorphism group of
$(C,p_{1},\ldots,p_{n})$ is finite. Alternatively, we say
$(C,p_{1},\ldots,p_{n})$ is stable if it satisfies the following two
conditions:

  1. 1.

If $C_{0}\subseteq C$ is a rational irreducible component, then $C_{0}$ is
incident to at least 3 ‘special points’, that is, nodes or marked points

  2. 2.

If $C_{1}\subseteq C$ is an elliptic irreducible component, then $C_{1}$ is
incident to at least 1 ‘special point’

We denote by $\overline{\mathcal{M}}_{g,n}$ the moduli space of stable
$n$-pointed genus $g$ curves. It is a smooth Deligne-Mumford stack of
dimension $\text{dim}(\overline{\mathcal{M}}_{g,n})=3g-3+n$. The universal
curve is denoted

$\pi_{n+1}:\overline{\mathcal{M}}_{g,n+1}\rightarrow\overline{\mathcal{M}}_{g,n}$

We often call $\pi_{n+1}$ the _forgetful morphism_ : it sends a point
$[C,p_{1},\ldots,p_{n+1}]$ to $[C,p_{1},\ldots,p_{n}]$, that is, it ‘forgets’
the last marked point. However, in order for this map to be well defined, one
must contract any rational components that are unstable. Similarly, we define
$\pi_{i}:\overline{\mathcal{M}}_{g,n+1}\rightarrow\overline{\mathcal{M}}_{g,n}$
to be the morphism that forgets the $i^{th}$ marked point.

Sitting inside $\overline{\mathcal{M}}_{g,n}$ is a dense open locus of curves

$\mathcal{M}_{g,n}\subset\overline{\mathcal{M}}_{g,n}$

parametrizing smooth $n$-pointed genus $g$ curves. The _boundary_ of
$\overline{\mathcal{M}}_{g,n}$, which is the complement of $\mathcal{M}_{g,n}$
in $\overline{\mathcal{M}}_{g,n}$, parametrizes nodal stable curves. The
boundary of $\overline{\mathcal{M}}_{g,n}$ has a _stratification_ , and each
stratum parametrizes curves of a fixed topological type.

######  Example 3.

Consider the moduli space $\overline{\mathcal{M}}_{2,4}$. As a running
example, we consider two strata
$\Gamma_{2}\subset\Gamma_{1}\subset\overline{\mathcal{M}}_{2,4}$.

Let $\Gamma_{1}\subset\overline{\mathcal{M}}_{2,4}$ be the stratum
parametrizing curves of the following topological type: a generic curve
$(C,p_{1},p_{2},p_{3},p_{4})\in\Gamma_{1}$ is a curve $C=C_{1}\cup C_{1}$,
where $C_{1}$ and $C_{2}$ are stable curves of genus 1 attached via a node,
$\\{p_{1},p_{2}\\}\subset C_{1}$, and $\\{p_{3},p_{4}\\}\subset C_{2}$. Below
is a drawing of a generic curve in $\Gamma_{1}$ as a two-dimensional nodal
topological surface:

$p_{1}$$p_{2}$$p_{3}$$p_{4}$

Similarly, let
$\Gamma_{2}\subset\Gamma_{1}\subset\overline{\mathcal{M}}_{2,4}$ be the
stratum parametrizing curves of the following topological type: a generic
curve $(C,p_{1},p_{2},p_{3},p_{4})\in\Gamma_{2}$ is a curve
$C=C_{1}\cup\tilde{C_{2}}$, where $C_{1}$ is a stable curve of genus 1,
$\tilde{C_{2}}$ is a nodal curve of genus 1, $\\{p_{1},p_{2}\\}\subset C_{1}$,
and $\\{p_{3},p_{4}\\}\subset\tilde{C_{2}}$. Below is a drawing of a generic
curve in $\Gamma_{2}$ as a two-dimensional nodal topological surface:

$p_{1}$$p_{2}$$p_{3}$$p_{4}$

The boundary strata in $\overline{\mathcal{M}}_{g,n}$ can be indexed by
_stable graphs_.

######  Definition 8 ([Pan15], Section 5.2).

A stable graph $\Gamma$ is the data

$\Gamma=(V,H,L,g:V\rightarrow\mathbb{Z}_{\geq 0},v:H\rightarrow
V,\iota:H\rightarrow H)$

where,

  1. 1.

$V$ is a vertex set, and $g:V\rightarrow\mathbb{Z}_{\geq 0}$ is the genus
assignment

  2. 2.

$H$ is a set of half-edges, $v:H\rightarrow V$ is a vertex assignment i.e.
indicates which vertex each half-edge is incident to, and $\iota:H\rightarrow
H$ is an involution that indicates when two half edges are glued together

  3. 3.

$E$ is an edge set, determined by the $2$-cycles of $\iota$

  4. 4.

$L$ is a set of legs, determined by the fixed points of $\iota$; it is in
bijection with the set of markings $\\{1,\ldots,n\\}$

  5. 5.

The data $(V,E)$ defines a connected graph.

  6. 6.

For each vertex $v\in V$, $2g(v)-2+n(v)>0$, where $n(v)$ is the valence of
$\Gamma$ at v including both edges and legs

######  Example 4.

The stratum $\Gamma_{1}\subset\overline{\mathcal{M}}_{2,4}$ from Example 3
corresponds to the following stable graph:

$1$$2$$3$$4$$1$$1$

The above stable graph consists of two vertices, both with genus assignment
$1$. Each vertex is incident to three half edges. The only $2$-cycle of the
involution $\iota$ corresponds to the two half edges that glue together to
form the edge connecting both vertices. The set of legs $L$ corresponds to the
half-edges labelled $1,2,3$, and $4$.

The stratum $\Gamma_{2}\subset\overline{\mathcal{M}}_{2,4}$ from Example 3
corresponds to the following stable graph:

$1$$2$$3$$4$$1$$0$

Notice that for this stable graph, the vertex with genus assignment $0$ has a
self-edge, corresponding to the self-node.

Due to the dictionary between stable graphs and boundary strata, we
unambiguously refer to a boundary stratum by its stable graph. In the case of
our running example, $\Gamma_{1}$ will mean ‘the stable graph corresponding to
the boundary stratum $\Gamma_{1}$’. Similarly, $\Gamma_{2}$ will mean ‘the
stable graph corresponding to the boundary stratum $\Gamma_{2}$.

Every stable graph $\Gamma$ corresponds to a product of moduli spaces:

$\overline{\mathcal{M}}_{\Gamma}:=\prod_{v\in
V}\overline{\mathcal{M}}_{g(v),n(v)}$

For instance, in our running example, we have

$\displaystyle\overline{\mathcal{M}}_{\Gamma_{1}}$
$\displaystyle=\overline{\mathcal{M}}_{1,3}\times\overline{\mathcal{M}}_{1,3}$
$\displaystyle\overline{\mathcal{M}}_{\Gamma_{2}}$
$\displaystyle=\overline{\mathcal{M}}_{1,3}\times\overline{\mathcal{M}}_{0,5}$

For every stable graph $\Gamma$, there exists a canonical morphism

$\xi_{\Gamma}:\overline{\mathcal{M}}_{\Gamma}\rightarrow\overline{\mathcal{M}}_{g,n}$

whose image is the boundary stratum corresponding to $\Gamma$ (see [Pan15],
Section 5.2 for details).

###  3.2 The Tautological Ring

######  Definition 9.

The _tautological ring_ $R^{*}(\overline{\mathcal{M}}_{g,n})\subset
A^{*}(\overline{\mathcal{M}}_{g,n})$ is the smallest $\mathbb{Q}$-subalgebra
of $A^{*}(\overline{\mathcal{M}}_{g,n})$ closed under pushforwards of the
morphisms $\pi_{i}$ and $\xi_{\Gamma}$. Elements in
$R^{*}(\overline{\mathcal{M}}_{g,n})$ are called tautological classes.

It turns out that many important cycles in
$A^{*}(\overline{\mathcal{M}}_{g,n})$ are tautological, and this has prompted
an intensive investigation of this ring in recent times. However, it is
difficult to gain access to the tautological ring using only its definition.
Fortunately, there is a nice result due to Graber and Pandharipande that gives
an explicit additive generating set for $R^{*}(\overline{\mathcal{M}}_{g,n})$
(see Theorem 6 below). In order to state this result, we need to define two
types of Chow classes in $A^{*}(\overline{\mathcal{M}}_{g,n})$, the
$\psi$-classes, and the $\kappa$-classes.

######  Definition 10.

For $1\leq i\leq n$, let $\mathbb{L}_{i}$ the line bundle on
$\overline{\mathcal{M}}_{g,n}$ whose fiber over a point
$[C,p_{1},\ldots,p_{n}]\in\overline{\mathcal{M}}_{g,n}$ is $T_{p_{i}}^{*}C$,
the cotangent space to $C$ at the $i^{th}$ marked point. For $1\leq i\leq n$
and $0\leq m\leq 3g-3+n$, define

$\displaystyle\psi_{i}$ $\displaystyle:=c_{1}(\mathbb{L}_{i})\in
A^{1}(\overline{\mathcal{M}}_{g,n})$ $\displaystyle\kappa_{m}$
$\displaystyle:={\pi_{n+1}}_{*}\left(\psi_{n+1}^{m+1}\right)\in
A^{m}(\overline{\mathcal{M}}_{g,n})$

######  Theorem 6 ([GP03], Proposition 11).

The set

$\left\\{{\xi_{\Gamma}}_{*}\left(\prod_{v\in V}\theta_{v}\right)\right\\}$

where $\Gamma$ is a stable graph, and $\theta_{v}$ is a monomial of
$\psi$-classes and $\kappa$-classes on $\overline{\mathcal{M}}_{g(v),n(v)}$,
forms an additive generating set of $R^{*}(\overline{\mathcal{M}}_{g,n})$.

A direct consequence of Theorem 6 is that, any tautological intersection
number can be reduced to intersection numbers involving only $\psi$-classes
and $\kappa$-classes. However, one can do even better than this: tautological
intersection numbers involving $\kappa$-classes can be reduced to intersection
numbers only involving $\psi$-classes. This is due to the following result:

######  Proposition 1 ([Zvo14], Corollary 3.23).

Let $Q$ be a polynomial in the variables
$\kappa_{m},\psi_{1},\ldots,\psi_{n}$, and let $\tilde{Q}$ be the polynomial
obtained from Q by the substitution
$\kappa_{i}\mapsto\kappa_{i}-\psi_{n+1}^{i}$. Then

$\int_{\overline{\mathcal{M}}_{g,n}}\kappa_{m}Q=\int_{\overline{\mathcal{M}}_{g,n+1}}\psi_{n+1}^{m+1}\tilde{Q}$

###  3.3 Computing $\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$

Since all tautological intersection numbers reduce to intersection numbers
only involving $\psi$-classes, we can restrict our attention to the rational
numbers

$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}:=\int_{\overline{\mathcal{M}}_{g,n}}\psi_{1}^{d_{1}}\ldots\psi_{n}^{d_{n}}$
(3)

It now remains to compute these numbers. When $g=0$, there is a closed form
expression in terms of multinomial coefficients:

$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{0}={n-3\choose d_{1}\ldots
d_{n}}$ (4)

When a $\psi$-class at only one marked point occurs, there is also a closed
form expression:

######  Lemma 4.

For $n\geq 1$,

$\left<\tau_{3g-3+n}\tau_{0}^{n-1}\right>_{g}=\frac{1}{24^{g}g!}$

###### Proof.

The proof can be found in ([Koc01] Section 3.3, Example 3.3.5) ∎

For $g>0$, there are various recursions that completely determine
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$. First, we recall the
_String Equation_ and the _Dilaton Equation_ :

$\text{\emph{The String Equation:}}\
\left<\tau_{d_{1}},\ldots,\tau_{d_{n}}\tau_{0}\right>_{g}=\sum_{i=1}^{n}\left<\tau_{d_{1}}\ldots\tau_{d_{i}-1}\ldots\tau_{d_{n}}\right>_{g}$

$\text{\emph{The Dilaton Equation:}}\
\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{1}\right>_{g}=(2g-2+n)\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$

When one wants to compute $\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$,
and $d_{i}\leq 1$ for all $i$, the String Equation and Dilaton Equation will
suffice. However, if $d_{i}\geq 2$ for some $i$, one needs the so-called
_Virasoro constraints_ :

######  Theorem 7 (Virasoro Constraints).

Let $m\geq 1$. Then

$\displaystyle(2m+3)!!\left<\tau_{m+1}\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}=$
$\displaystyle\sum_{i=1}^{n}\frac{(2d_{i}+1+2m)!!}{(2d_{i}-1)!!}\left<\tau_{d_{1}}\ldots\tau_{d_{i}+m}\ldots\tau_{d_{n}}\right>$
$\displaystyle+\frac{1}{2}\sum_{a+b=m-1}(2a+1)!!(2b+1)!!\left<\tau_{a}\tau_{b}\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g-1}$
$\displaystyle+\frac{1}{2}\sum_{\begin{subarray}{c}a+b=m-1\\\ I\amalg
J=\\{1,\ldots,n\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!(2b+1)!!\left<\tau_{a}\prod_{i\in
I}\tau_{d_{i}}\right>_{g_{1}}\left<\tau_{b}\prod_{i\in
J}\tau_{d_{i}}\right>_{g_{2}}$

###### Proof.

See [Zvo14], Section 4.2 ∎

So, in summary, with the closed from expression in genus zero (Equation 4),
the String Equation, the Dilaton Equation, the closed formula for intersection
numbers with $\psi$-classes supported at one marked point (Lemma 4), and the
Virasoro constraints (Theorem 7), one can compute any intersection number
$\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\right>_{g}$.

##  4 Proof of Theorem 1

Throughout this section, we use the following shorthand notation:

  * •

$\displaystyle\vec{d}:=(d_{1},\ldots,d_{n})\in\mathbb{Z}^{n}_{\geq 0}$

  * •

$\displaystyle|\vec{d}|:=\sum_{i=1}^{n}d_{i}$

  * •

$\displaystyle C(\vec{d}):=\prod_{i=1}^{n}(2d_{i}+1)!!$

  * •

$\tau_{\vec{d}}:=\tau_{d_{1}}\ldots\tau_{d_{n}}$

  * •

$d_{n+1}=d_{n+1}(g,|\vec{d}|):=3g-2+n-|\vec{d}|$

  * •

$\displaystyle
L_{\vec{d}}(g):=24^{g}g!C(\vec{d})\left<\tau_{\vec{d}}\tau_{d_{n+1}}\right>_{g}$

  * •

$m=m(\vec{d}):=\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$

In the case that $n=0$, $\vec{d}$ is the empty vector, which we denote by
$\vec{d}=\varnothing$. We define

$C(\varnothing):=1$

By Lemma 4, we have

$L_{\varnothing}(g)=24^{g}g!\left<\tau_{3g-2}\tau_{0}^{n}\right>_{g}=1$

The proof of Theorem 1 consists of three parts. We begin by showing that

$L_{\vec{d}}(g):=24^{g}g!C(\vec{d})\left<\tau_{\vec{d}}\tau_{3g-2+n-|\vec{d}|}\right>_{g}$

is an integer-valued polynomial in $g$ whose leading coefficient is
$6^{|\vec{d}|}$. This is proven by carefully examining the recursions that
determine $L_{\vec{d}}(g)$, and making sure that the property of being an
integer-valued polynomial with leading coefficient $6^{|\vec{d}|}$ is
preserved under these recursions.

Once this is established, we can consider the shifted polynomial

$L_{\vec{d}}(g+m(\vec{d}))=24^{g+m(\vec{d})}(g+m(\vec{d}))!C(\vec{d})\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{3(g+m(\vec{d}))-2+n-|\vec{d}|}\right>_{g+m(\vec{d})}$

(see the explanation directly following the statement of Theorem 1 for a
justification as to why $m(\vec{d})$ is a natural shift of the genera) The
second part of the proof is to show that $L_{\vec{d}}(g+m)$ has a nonnegative
$f^{*}$-vector. The strategy is the same as in the first part, that is, we
make sure that the property of having a nonnegative $f^{*}$-vector is
preserved under the recursions that determine $L_{\vec{d}}(g+m)$.

The final part of the proof is to apply Breuer’s theorem (Theorem 5), which
ensures that one can always find a partial polytopal complex $P_{\vec{d}}$
whose Ehrhart polynomial is $L_{\vec{d}}(g+m)$.

###  4.1 $L_{\vec{d}}(g)$ Is An Integer-Valued Polynomial

Consider the intersection number

$L_{\vec{d}}(g):=24^{g}g!C(\vec{d})\left<\tau_{d_{1}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g}$

The String Equation and the Dilaton Equation implies:

######  Lemma 5 (String and Dilaton Equation for $L_{\vec{d}}(g)$).

The String Equation and the Dilaton Equation, respectively, imply that

$\displaystyle L_{\vec{d}\cup\\{0\\}}(g)$
$\displaystyle=\left(\sum_{i=1}^{n}(2d_{i}+1)L_{\vec{d}\setminus\\{d_{i}\\}\cup\\{d_{i}-1\\}}(g)\right)+L_{\vec{d}}(g)$
$\displaystyle L_{\vec{d}\cup\\{1\\}}(g)$
$\displaystyle=(6g-3+3n)L_{\vec{d}}(g)$

###### Proof.

Indeed, we have

$\displaystyle L_{\vec{d}\cup\\{0\\}}(g)$
$\displaystyle=24^{g}g!C(\vec{d})\left<\tau_{\vec{d}}\tau_{0}\tau_{3g-2+(n+1)-|\vec{d}|}\right>_{g}$
$\displaystyle=24^{g}g!C(\vec{d})\left[\sum_{i=1}^{n}\left<\tau_{d_{1}}\ldots\tau_{d_{i}-1}\ldots\tau_{d_{n}}\tau_{3g-2+n-(|\vec{d}|)-1}\right>_{g}+\left<\tau_{\vec{d}}\tau_{3g-2+n-|\vec{d}|}\right>_{g}\right]$
$\displaystyle=\left(\sum_{i=1}^{n}24^{g}g!(2d_{i}+1)C(d_{1},\ldots,d_{i}-1,\ldots
d_{n})\left<\tau_{d_{1}}\ldots\tau_{d_{i}-1}\ldots\tau_{d_{n}}\tau_{3g-2+n-(|\vec{d}|-1)}\right>_{g}\right)$
$\displaystyle+24^{g}g!C(\vec{d})\left<\tau_{\vec{d}}\tau_{3g-2+n-|\vec{d}|}\right>_{g}$
$\displaystyle=\left(\sum_{i=1}^{n}(2d_{i}+1)L_{\vec{d}\setminus\\{d_{i}\\}\cup\\{d_{i}-1}(g)\right)+L_{\vec{d}}(g)$

Furthermore,

$\displaystyle L_{\vec{d}\cup\\{1\\}}(g)$
$\displaystyle=24^{g}g!C(\vec{d}\cup\\{1\\})\left<\tau_{\vec{d}}\tau_{1}\tau_{3g-2+(n+1)-(|\vec{d}|+1)}\right>_{g}$
$\displaystyle=24^{g}g!C(\vec{d})3!!(2g-2+(n+1))\left<\tau_{\vec{d}}\tau_{3g-2+n-|\vec{d}|}\right>_{g}$
$\displaystyle=3(2g-1+n)L_{\vec{d}}(g)$
$\displaystyle=(6g-3+3n)L_{\vec{d}}(g)$

∎

Now consider the case that there exists some $d_{i}$, say $d_{1}$, such that
$d_{1}\geq 2$. We can use the Virasoro constraints (Theorem 7) to evaluate
$L_{\vec{d}}(g)$ by letting $d_{1}$ play the role of $`m+1^{\prime}$ , so that
$m=d_{1}-1$. Multiplying through by $24^{g}g!C(\vec{d})$, and dividing by
$(2(d_{1}-1)+3)!!=(2d_{1}+1)!!$, the Virasoro constraints tell us that

$L_{\vec{d}}(g)=\frac{1}{(2d_{1}+1)!!}24^{g}g!C(\vec{d})\left(T_{1}(\vec{d})+T_{2}(\vec{d})+T_{3}(\vec{d})+T_{4}(\vec{d})\right)$
(5)

where

$\displaystyle T_{1}(\vec{d})$
$\displaystyle:=\sum_{i=2}^{n}\frac{(2d_{i}+2(d_{1}-1)+1)!!}{(2d_{i}-1)!!}\left<\tau_{d_{2}}\ldots\tau_{d_{i}+d_{1}-1}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g}$
$\displaystyle T_{2}(\vec{d})$
$\displaystyle:=\frac{(2(3g-2+n-|\vec{d}|)+2(d_{1}-1)+1)!!}{(2(3g-2+n-|\vec{d}|)-1)!!}\left<\tau_{d_{2}}\ldots\tau_{d_{n}}\tau_{d_{n+1}+d_{1}-1}\right>_{g}$
$\displaystyle T_{3}(\vec{d})$
$\displaystyle:=\frac{1}{2}\sum_{a+b=d_{1}-2}(2a+1)!!(2b+1)!!\left<\tau_{a}\tau_{b}\tau_{d_{2}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g-1}$
$\displaystyle T_{4}(\vec{d})$
$\displaystyle:=\frac{1}{2}\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n+1\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!(2b+1)!!\left<\tau_{a}\tau_{I}\right>_{g_{1}}\left<\tau_{b}\tau_{J}\right>_{g_{2}}$

In the term $T_{4}(\vec{d})$, we are using the shorthand notation

$I\amalg J=\\{2,\ldots,n+1\\}\implies\tau_{I}:=\prod_{i\in I}\tau_{d_{i}}\
\tau_{J}:=\prod_{i\in J}\tau_{d_{i}}$

Our goal now is express the right hand side of Equation 5 in terms of
$L_{\vec{\ell}}(g)$, where $|\vec{\ell}|<|\vec{d}|$. Let’s begin with
$T_{1}(\vec{d})$. First, define

$\vec{d}(i):=(d_{2},d_{3},\ldots,d_{i-1},d_{i}+d_{1}-1,d_{i+1},\ldots d_{n})$

Then

$\displaystyle C(\vec{d}(i))$
$\displaystyle=\frac{C(\vec{d})}{(2d_{1}+1)!!}\cdot\frac{(2(d_{i}+d_{1}-1)+1)!!}{(2d_{i}+1)!!}$
$\displaystyle=\frac{C(\vec{d})}{(2d_{1}+1)!!}\cdot\frac{(2d_{i}+(2d_{1}-1))!!}{(2d_{i}+1)!!}$

which implies that

$(2d_{i}+1)C(\vec{d}(i))=\frac{C(\vec{d})}{(2d_{1}+1)!!}\cdot\frac{(2d_{i}+(2d_{1}-1))!!}{(2d_{i}-1)!!}$

Therefore, we have

$\frac{C(\vec{d})}{(2d_{1}+1)!!}24^{g}g!T_{1}(\vec{d})=\sum_{i=2}^{n}(2d_{i}+1)L_{\vec{d}(i)}(g)$
(6)

For the term involving $T_{2}(\vec{d})$, we have

$\displaystyle\frac{(2(3g-2+n-|\vec{d}|)+2(d_{1}-1)+1)!!}{(2(3g-2+n-|\vec{d}|)-1)!!}$
$\displaystyle=\frac{(6g-4+2n-2|\vec{d}|+(2d_{1}-1))!!}{(6g-4+2n-2|\vec{d}|-1)!!}$
$\displaystyle=\prod_{k=1}^{d_{1}}(6g-4+2n-2|\vec{d}|+(2k-1))$

and therefore,

$\frac{C(\vec{d})}{(2d_{1}+1)!!}24^{g}g!T_{2}(\vec{d})=\left(\prod_{k=1}^{d_{1}}(6g-4+2n-2|\vec{d}|+(2k-1))\right)L_{\vec{d}\setminus\\{d_{1}\\}}(g)$
(7)

For the term involving $T_{3}(\vec{d})$,

$\displaystyle\frac{1}{(2d_{1}+1)!!}24^{g}g!C(\vec{d})T_{3}(\vec{d})=24g\left(\frac{1}{2}\sum_{a+b=d_{1}-2}24^{g-1}(g-1)!C(\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\})\left<\tau_{a}\tau_{b}\tau_{d_{2}}\ldots\tau_{d_{n}}\tau_{d_{n+1}}\right>_{g-1}\right)$

The expression $d_{n+1}=3g-2+n-|\vec{d}|$ can be written as

$3(g-1)-2+(n+1)-(|\vec{d}|-d_{1}+a+b)$

and therefore,

$\frac{1}{(2d_{1}+1)!!}24^{g}g!C(\vec{d})T_{3}(\vec{d})=12g\sum_{a+b=d_{1}-2}L_{\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}}(g-1)$
(8)

Finally, we consider the term containing $T_{4}(\vec{d})$. Using the symmetry
of the summation, we can get rid of the factor of $\frac{1}{2}$:

$\displaystyle T_{4}=\frac{1}{2}$
$\displaystyle\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n+1\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!(2b+1)!!\left<\tau_{a}\tau_{I}\right>_{g_{1}}\left<\tau_{b}\tau_{J}\right>_{g_{2}}$
$\displaystyle=\frac{1}{2}$
$\displaystyle\left(\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!(2b+1)!!\left<\tau_{a}\tau_{I}\tau_{d_{n+1}}\right>_{g_{1}}\left<\tau_{b}\tau_{J}\right>_{g_{2}}\right.$
$\displaystyle\left.+\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!(2b+1)!!\left<\tau_{a}\tau_{I}\right>_{g_{1}}\left<\tau_{b}\tau_{J}\tau_{d_{n+1}}\right>_{g_{2}}\right)$
$\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!(2b+1)!!\left<\tau_{a}\tau_{I}\right>_{g_{1}}\left<\tau_{b}\tau_{J}\tau_{d_{n+1}}\right>_{g_{2}}$

For any partition $I\amalg J=\\{2,\ldots,n\\}$, and for any pair
$(g_{1},g_{2})$ such that $g_{1}+g_{2}=g$, we have

$\displaystyle\frac{C(\vec{d})}{(2d_{1}+1)!!}$
$\displaystyle=C(\vec{d}_{I})C(\vec{d}_{J})$ $\displaystyle 24^{g}$
$\displaystyle=24^{g_{1}}24^{g_{2}}$ $\displaystyle g!$
$\displaystyle=g_{1}!g_{2}!{g\choose g_{1}}$

Therefore,

$\displaystyle\frac{C(\vec{d})}{(2d_{1}+1)!!}24^{g}g!T_{4}(\vec{d})$
$\displaystyle=\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!24^{g_{1}}g_{1}!C(\vec{d}_{I})\left<\tau_{a}\tau_{I}\right>_{g_{1}}(2b+1)!!24^{g_{2}}g_{2}!C(\vec{d}_{J})\left<\tau_{b}\tau_{J}\tau_{d_{n+1}}\right>_{g_{2}}{g\choose
g_{1}}$ $\displaystyle=\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\\\
g_{1}+g_{2}=g\end{subarray}}(2a+1)!!L_{\vec{d}_{I}}(g_{1})L_{\vec{d}_{J}\cup\\{b\\}}(g_{2}){g\choose
g_{1}}$

Now, notice that, for any given pairs $(a,b)$ and $(I,J)$ such that
$a+b=d_{1}-2$ and $I\amalg J=\\{2,\ldots,n\\}$, there exists _at most_ one
pair $(g_{1},g_{2})=(g_{1},g-g_{1})$ such that $L_{\vec{d}_{I}}(g_{1})$ and
$L_{\vec{d}_{J}}(g_{2})$ are simultaneously non-zero. In fact, it is a simple
computation to determine $g_{1}$ (and consequently $g_{2}=g-g_{1}$) as a
function of $I$ and $a$:

$3g_{1}-3+1+|I|=a+|\vec{d}_{I}|\implies
g_{1}=\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)$

######  Remark 2.

Implicit in our explanation is the assumption that
$\left<\tau_{a}\tau_{I}\right>_{\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)}$ is zero
whenever $\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)$ is not a positive integer.
Geometrically, this is just the simple statement that we are excluding the
possibility of fractional genera.

This is all to say that

$\frac{C(\vec{d})}{(2d_{1}+1)!!}24^{g}g!T_{4}(\vec{d})=\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\
I\amalg
J=\\{2,\ldots,n\\}\end{subarray}}(2a+1)!!L_{\vec{d}_{I}}\left(g_{1}(I,a)\right)L_{\vec{d}_{J}}\left(g-g_{1}(I,a)\right){g\choose
g_{1}(I,a)}$ (9)

where $g_{1}(I,a):=\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)$. Putting all of these
calculations together, we obtain the following recursion:

######  Proposition 2 (Virasoro Constraints for $L_{\vec{d}}(g)$).

Let $\vec{d}:=(d_{1},\ldots,d_{n})\in\mathbb{Z}^{n}_{\geq 0}$ where $d_{1}\geq
2$. We have

$\displaystyle L_{\vec{d}}(g)$
$\displaystyle=\sum_{i=2}^{n}(2d_{i}+1)L_{\vec{d}(i)}(g)+\left(\prod_{k=1}^{d_{1}}(6g-4+2n-2|\vec{d}|+(2k-1))\right)L_{\vec{d}\setminus\\{d_{1}\\}}(g)$
$\displaystyle+12g\sum_{a+b=d_{1}-2}L_{\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}}(g-1)$
$\displaystyle+\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\end{subarray}}(2a+1)!!L_{\vec{d}_{I}}(g_{1})L_{\vec{d}_{J}\cup\\{b\\}}(g-g_{1}){g\choose
g_{1}}$

where, in the last summation on the right hand side,

$\displaystyle g_{1}$
$\displaystyle:=\frac{1}{3}\left(a+|\vec{d}_{I}|+2-|I|\right)$ $\displaystyle
L_{\vec{d}_{I}}(g_{1})$
$\displaystyle:=\begin{cases}L_{\vec{d}_{I}}\left(\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)\right),\quad
a+|\vec{d}_{I}|+2-|I|\equiv 0\ \text{mod}\ 3\\\
0,\quad\text{otherwise}\end{cases}$

###### Proof.

This is the combination of Equations 5, 6, 7, 8, 9. ∎

We have the following direct corollary of Proposition 2:

######  Corollary 1.

The intersection number $L_{\vec{d}}(g)$ is an integer-valued polynomial in
$g$ of degree $|\vec{d}|$ with leading coefficient $6^{|\vec{d}|}$.

###### Proof.

We first establish a few base cases. By the Dilaton Equation, we have

$\displaystyle L_{(1)}(g)$
$\displaystyle=24^{g}g!C((1))\left<\tau_{1}\tau_{3g-2+1-1}\right>_{g}$
$\displaystyle=24^{g}g!(3!!)\left<\tau_{1}\tau_{3g-2}\right>_{g}$
$\displaystyle=(2g-2+1)24^{g}g!(3!!)\left<\tau_{3g-2}\right>_{g}$
$\displaystyle=6g-3$

$\displaystyle L_{(1,1)}(g)$ $\displaystyle=(6g-3+3)L_{(1)}(g)$
$\displaystyle=(6g)(6g-3)$

Using Proposition 2, we have

$\displaystyle L_{(2)}(g)$ $\displaystyle=(6g-4+2-4+1)(6g-4+2-4+3)+12g$
$\displaystyle=(6g-5)(6g-3)+12g$ $\displaystyle=36g^{2}-48g+15+12g$
$\displaystyle=36g^{2}-36g+15$

Thus, if $|\vec{d}|\leq 2$, the desired result is satisfied.

By way of induction, suppose there exists an integer $d>1$ such that
$L_{\vec{\ell}}(g)$ is an integer valued polynomial of degree $|\vec{\ell}|$
with leading coefficient $6^{|\vec{\ell}|}$ whenever $|\vec{\ell}|<d$. Let
$\vec{d}\in\mathbb{Z}^{n}_{\geq 0}$ be an integer vector such that
$|\vec{d}|=d$. Without loss of generality, by Lemma 5, assume $d_{i}\geq 2$
for all $i$. By Proposition 2, $L_{\vec{d}}(g)$ is a sum of four terms. What
we need to check is that all four terms sum to an integer-valued polynomial of
degree $d$. The first term is

$\sum_{i=2}^{n}(2d_{i}+1)L_{\vec{d}(i)}(g)$

Since $|\vec{d}(i)|=|\vec{d}|-d_{1}-d_{i}+(d_{i}+d_{1}-1)=d-1<d$, by the
induction hypothesis, this term is an integer-valued polynomial in $g$ of
degree $|\vec{d}(i)|=|\vec{d}|-1$

The second term is

$\left(\prod_{k=1}^{d_{1}}(6g-4+2n-2|\vec{d}|+(2k-1))\right)L_{\vec{d}\setminus\\{d_{1}\\}}(g)$

By the induction hypothesis, this is an integer-valued polynomial in $g$ of
degree $|\vec{d}\setminus\\{d_{1}\\}|+d_{1}=d$.

The third term is

$12g\sum_{a+b=d_{1}-2}L_{\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}}(g-1)$

By the induction hypothesis, this is an integer-valued polynomial in $g$ of
degree $1+|\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}|=1+d-d_{1}+d_{1}-2=d-1<d$

Finally, the fourth term is

$\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\end{subarray}}(2a+1)!!L_{\vec{d}_{I}}(g_{1})L_{\vec{d}_{J}\cup\\{b\\}}(g-g_{1}){g\choose
g_{1}}$

The binomial term in the sum, ${g\choose g_{1}}$, is an integer-valued
polynomial in $g$ of degree $g_{1}=\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)$.
Therefore, by the induction hypothesis, each summand is an integer-valued
polynomial in $g$ of degree

$\displaystyle g_{1}+|\vec{d}_{J}|+b$
$\displaystyle=\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)+|\vec{d}_{J}|+b$
$\displaystyle=\left(\frac{a}{3}+b\right)+\left(\frac{|\vec{d}_{I}|}{3}+|\vec{d}_{J}|\right)+\left(\frac{2}{3}-\frac{|I|}{3}\right)$
$\displaystyle\leq\left(a+b\right)+\left(|\vec{d}_{I}|+|\vec{d}_{J}|\right)+\frac{2}{3}$
$\displaystyle=d_{1}-2+|\vec{d}|-d_{1}+\frac{2}{3}$
$\displaystyle=|\vec{d}|-\frac{4}{3}$ $\displaystyle<d$

Thus, it follows that $L_{\vec{d}}(g)$ is an integer-valued polynomial of
degree $|\vec{d}|=d$. Its leading coefficient comes from the contribution of
the fourth term, which is

$6^{d_{1}}6^{|\vec{d}\setminus\\{d_{1}\\}|}=6^{|\vec{d}|}$

as desired. ∎

###  4.2 The $f^{*}$-Vector of $L_{\vec{d}}(g+m(\vec{d}))$ Is Nonnegative

Now we need a second corollary that says the $f^{*}$-vector of
$L_{\vec{d}}(g+m)$ is nonnegative. However, before we prove that corollary, we
need the following lemma:

######  Lemma 6.

Let $\vec{d}=(d_{1},\ldots,d_{n})\in\mathbb{Z}^{n}_{\geq 2}$ be an integer
vector, and define
$m=m(\vec{d}):=\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$. Then the
polynomial

$\prod_{k=1}^{d_{1}}(6(g+m)-4+2n-2|\vec{d}|+(2k-1))$

has nonnegative $f^{*}$-vector.

###### Proof.

For $1\leq k\leq d_{1}$, consider the linear polynomial

$L_{k}(g):=6(g+m)-4+2n-2|\vec{d}|+(2k-1)$

Suppose we compute the $f^{*}$-vector of this linear polynomial, i.e. we find
integers $f_{0}^{*}$ and $f_{1}^{*}$ such that

$L_{k}(g)=f_{0}^{*}{g-1\choose 0}+f_{1}^{*}{g-1\choose 1}$

This means $f_{0}^{*}=L_{k}(1)$ and $f_{1}^{*}=\text{leading coefficient of
$L_{k}(g)$}=6$. So all we need to check is whether $f_{0}^{*}=L_{k}(1)$ is
always nonnegative.

In the case that $2-n+|\vec{d}|\equiv 0\ \text{mod}\
3\implies\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil=\frac{2-n+|\vec{d}|}{3}$,
we have

$\displaystyle f_{0}^{*}$ $\displaystyle=L_{k}(1)$
$\displaystyle=6\left(1+\frac{2-n+|\vec{d}|}{3}-1\right)-4+2n-2|\vec{d}|+(2k-1)$
$\displaystyle=2k-1$

Similar calculations show that

$\displaystyle 2-n+|\vec{d}|\equiv 1\ \text{mod}\
3\implies\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil=\frac{2-n+|\vec{d}|}{3}+\frac{2}{3}\implies
f_{0}^{*}=L_{k}(1)=2k+3$ $\displaystyle 2-n+|\vec{d}|\equiv 2\ \text{mod}\
3\implies\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil=\frac{2-n+|\vec{d}|}{3}+\frac{1}{3}\implies
f_{0}^{*}=L_{k}(1)=2k+1$

Thus, the $f^{*}$-vector of $L_{k}(g)$ is nonnegative, and by Lemma 2, the
product $\prod_{k=1}^{d_{1}}L_{k}(g)$ also has nonnegative $f^{*}$-vector, as
desired.

∎

######  Corollary 2.

For any vector $\vec{d}\in\mathbb{Z}_{\geq 0}^{n}$, define
$m=m(\vec{d}):=\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$. Then the
$f^{*}$-vector of $L_{\vec{d}}(g+m)$ is nonnegative.

###### Proof.

For the base cases, we have

$\displaystyle m((0))=0,\ L_{(0)}(g+0)=1=1{g-1\choose 0}$ $\displaystyle
m((1))=0,\ L_{(1)}(g+0)=6g-3=3{g-1\choose 0}+6{g-1\choose 1}$ $\displaystyle
m((1,1))=0,\ L_{(1,1)}(g+0)=18{g-1\choose 0}+90{g-1\choose 1}+72{g-1\choose
2}$ $\displaystyle m((2))=0,\ L_{(2)}=15{g-1\choose 0}+72{g-1\choose
1}+72{g-1\choose 2}$

By way of induction, suppose that $L_{\vec{\ell}}(g+m(\vec{\ell}))$ has a
nonnegative $f^{*}$-vector for all vectors $\vec{\ell}$ where $|\vec{\ell}|<d$
for some positive integer $d>1$. Let $\vec{d}$ be an integer vector such that
$|\vec{d}|=d$. By Lemma 5, without loss of generality, assume $d_{i}\geq 2$
for all $i$. This means we can use Proposition 2 to compute
$L_{\vec{d}}(g+m(\vec{d}))$. All that remains is to check that all four terms
that arise in Proposition 2 add up to a polynomial with a nonnegative
$f^{*}$vector.

When we use Proposition 2 to compute $L_{\vec{d}}(g+m(\vec{d}))$, the
contribution coming from the first term is

$\sum_{i=2}^{n}(2d_{i}+1)L_{\vec{d}(i)}(g+m(\vec{d}))$

Since $|\vec{d}(i)|=|\vec{d}|-1$, by the induction hypothesis, the integer-
valued polynomial

$L_{\vec{d}(i)}(g+m(\vec{d}(i)))$

has a non-negative $f^{*}$-vector. However, notice that

$\displaystyle m(\vec{d}(i))$
$\displaystyle=\left\lceil\frac{2-(n-1)+(|\vec{d}|-1)}{3}\right\rceil-1$
$\displaystyle=\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$
$\displaystyle=m(\vec{d})$

Therefore, $L_{\vec{d}(i)}(g+m(\vec{d}))=L_{\vec{d}(i)}(g+m(\vec{d}(i)))$, so
the contribution coming from the first term has nonnegative $f^{*}$-vector.

Now consider the contribution coming from the second term,

$\left(\prod_{k=1}^{d_{1}}(6(g+m(\vec{d})-4+2n-2|\vec{d}|+(2k-1)\right)L_{\vec{d}\setminus\\{d_{1}\\}}(g+m(\vec{d}))$

By Lemma 6, the product of linear terms is an integer-valued polynomial with
non-negative $f^{*}$-vector. Furthermore, since

$\displaystyle m(\vec{d}\setminus\\{d_{1}\\})$
$\displaystyle=\left\lceil\frac{2-(n-1)+(|\vec{d}|-d_{1})}{3}\right\rceil-1$
$\displaystyle=\left\lceil\frac{2-n+|\vec{d}|-(d_{1}-1)}{3}\right\rceil-1$
$\displaystyle\leq\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$
$\displaystyle=m(\vec{d})$

then by the induction hypothesis and Lemma 1, it follows that
$L_{\vec{d}\setminus\\{d_{1}\\}}(g+m(\vec{d}))$ is an integer-valued
polynomial with nonnegative $f^{*}$-vector. Therefore, by Lemma 2, the
contribution of the second term is an integer-valued polynomial with
nonnegative $f^{*}$-vector.

Now consider the contribution coming from the third term,

$12(g+m(\vec{d}))\sum_{a+b=d_{1}-2}L_{\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}}(g+m(\vec{d})-1)$

Since
$|\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}|=|\vec{d}|-d_{1}+(d_{1}-2)=|\vec{d}|-2$,
by the induction hypothesis,

$L_{\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}}(g+m(\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}))$

is an integer-valued polynomial with nonnegative $f^{*}$-vector. However,
notice that

$\displaystyle m(\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\})$
$\displaystyle=\left\lceil\frac{2-(n+1)+|\vec{d}|-2}{3}\right\rceil-1$
$\displaystyle=\left\lceil\frac{2-n+|\vec{d}|-3}{3}\right\rceil-1$
$\displaystyle<\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1$
$\displaystyle=m(\vec{d})$

Therefore, by Lemma 1,
$L_{\vec{d}\setminus\\{d_{1}\\}\cup\\{a,b\\}}(g+m(\vec{d})-1)$ has a
nonnegative $f^{*}$-vector. Since the $f^{*}$-vector of the polynomial $12g$
is $(12,12)$, using Lemma 1 again, we see that $12(g+m(\vec{d}))$ has a
nonnegative $f^{*}$-vector. By Lemma 2, the total contribution coming from the
third term is an integer-valued polynomial with nonnegative $f^{*}$-vector.

Finally, consider the contribution coming from the fourth term,

$\sum_{\begin{subarray}{c}a+b=d_{1}-2\\\ I\amalg
J=\\{2,\ldots,n\\}\end{subarray}}(2a+1)!!L_{\vec{d}_{I}}(g_{1})L_{\vec{d}_{J}\cup\\{b\\}}(g+m(\vec{d})-g_{1}){g+m(\vec{d})\choose
g_{1}}$

Recall that

$L_{\vec{d}_{I}}(g_{1})=\begin{cases}L_{\vec{d}_{I}}\left(\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)\right)\quad
a+|\vec{d}_{I}|+2-|I|\equiv 0\ \text{mod}\ 3\\\
0\quad\text{otherwise}\end{cases}$

so we can assume that $a+|\vec{d}_{I}|+2-|I|\equiv 0\ \text{mod}\ 3$. Now,
notice that, for any pairs $(a,b)$ and $(I,J)$ such that $a+b=d_{1}-2$,
$I\amalg J=\\{2,\ldots,n\\}$, we have

$\displaystyle(2-n+|\vec{d}|)$
$\displaystyle-(a+|\vec{d}_{I}|+2-|I|)-(2-(|J|+1)+|\vec{d}_{J}|+b)$
$\displaystyle=-1+(|I|+|J|-n)+(|\vec{d}|-|\vec{d}_{I}|-|\vec{d}_{J}|)-(a+b)$
$\displaystyle=-1-1+d_{1}-(d_{1}-2)$ $\displaystyle=0$

But since $a+|\vec{d}_{I}|+2-|I|\equiv 0\ \text{mod}\ 3$, it follows that

$2-n+|\vec{d}|\equiv 2-(|J|+1)+|\vec{d}_{J}|+b\ \text{mod}\ 3$

In particular, there exists an integer $0\leq k\leq 2$ such that

$\displaystyle\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil$
$\displaystyle=\frac{2-n+|\vec{d}|}{3}+\frac{k}{3}$
$\displaystyle\left\lceil\frac{2-(|J|+1)+|\vec{d}_{J}|+b}{3}\right\rceil$
$\displaystyle=\frac{2-(|J|+1)+|\vec{d}_{J}|+b}{3}+\frac{k}{3}$

and thus,

$\displaystyle(m(\vec{d})-g_{1})-m(\vec{d}_{J}\cup\\{b\\})$
$\displaystyle=\left(\left\lceil\frac{2-n+|\vec{d}|}{3}\right\rceil-1-\frac{1}{3}(a+|\vec{d}_{I}|+2-|I|)\right)-\left(\left\lceil\frac{2-(|J|+1)+|\vec{d}_{J}|+b}{3}\right\rceil-1\right)$
$\displaystyle=\frac{1}{3}\left((2-n+|\vec{d}|)-(a+|\vec{d}_{I}|+2-|I|)-(2-(|J|+1)+|\vec{d}_{J}|+b)+k-k\right)$
$\displaystyle=0$

It follows that $m(\vec{d})-g_{1}=m(\vec{d}_{J}\cup\\{b\\})$, so

$L_{\vec{d}_{J}\cup\\{b\\}}(g+m(\vec{d})-g_{1})=L_{\vec{d}_{J}\cup\\{b\\}}(g+m(\vec{d}_{J}\cup\\{b\\}))$

Therefore, since
$|\vec{d}_{J}\cup\\{b\\}|=|\vec{d}_{J}|+b\leq|\vec{d}|-d_{1}+d_{1}-2=|\vec{d}|-2<|\vec{d}|$,
by the induction hypothesis, $L_{\vec{d}_{J}\cup\\{b\\}}(g+m(\vec{d})-g_{1})$
is an integer-valued polynomial with nonnegative $f^{*}$-vector. Furthermore,
since the $f^{*}$-vector of ${g+m(\vec{d})\choose g_{1}}$ is nonnegative (by
Lemma 1), the contribution of the fourth term is a sum of products of
polynomials with nonnegative $f^{*}$-vector. By Theorem 2, the total
contribution of the fourth term is a polynomial with nonnegative
$f^{*}$-vector.

In summary, we have shown that $L_{\vec{d}}(g+m(\vec{d}))$ is a sum of
integer-valued polynomials with nonnegative $f^{*}$-vector, and therefore
$L_{\vec{d}}(g+m(\vec{d})$ has nonnegative $f^{*}$-vector, as desired.

∎

###  4.3 Putting the Pieces Together

###### Proof of Theorem 1.

By Corollary 1 and Corollary 2, we know that

$L_{\vec{d}}(g+m)=24^{g+m}(g+m)!C(\vec{d})\left<\tau_{\vec{d}}\tau_{3(g+m)-2+n-d}\right>_{g+m}$

is an integer-valued polynomial in $g$ of degree $|\vec{d}|$ with nonnegative
$f^{*}$-vector. By Breuer’s theorem (Theorem 5), there exists a partial
polytopal complex $P_{\vec{d}}$ of dimension $|\vec{d}|$ such that

$L_{\vec{d}}(g+m)=\\#\\{\text{integer lattice points in $gP_{\vec{d}}$}\\}$

By Theorem 3, the volume of $P_{\vec{d}}$ is $6^{|\vec{d}|}$.

∎

##  5 Examples

In this section, we compute some examples of the Ehrhart polynomials
$L_{\vec{d}}(g+m)$, along with their corresponding partial polytopal
complexes. When $|\vec{d}|\leq 2$, we show that the corresponding partial
polytopal complexes can be presented as _inside-out polytopes_ , a type of
partial polytopal complex first studied by Beck and Zaslavsky [BZ06].

######  Definition 11.

Let $P\subseteq\mathbb{R}^{d}$ be a full dimensional integral $d$-polytope,
and let $\mathcal{H}$ be a _hyperplane arrangement_ , that is, a finite
collection of hyperplanes in $\mathbb{R}^{d}$. An _inside-out polytope_ is any
set of the form

$P\setminus\left(\bigcup_{H\in\mathcal{H}}H\right)$

The reasoning behind the name is that, one should think of the hyperplanes as
dissecting the polytope $P$ into various regions, and the various hyperplanes
serve as ‘boundaries’ that lie on the ‘inside’ of the polytope.

###  5.1 $|\vec{d}|=1$

We’ve already computed the Ehrhart polynomial

$L_{(1)}(g+m)=L_{1}(g)=6g-3=3{g-1\choose 0}+6{g-1\choose 1}=3\left[{g-1\choose
0}+2{g-1\choose 1}\right]$

Define $P_{(1)}$ as the inside-out polytope

$P_{(1)}=[-3,3]\setminus\\{\pm 2,\pm 3\\}$

which can be visualized as

$P_{(1)}=$$-3$$-2$$-1$$0$$1$$2$$3$

The inside-out polytope $P_{(1)}$ has a unimodular triangulation given by

$\mathcal{T}=\\{0\\}\amalg\\{-1\\}\amalg\\{1\\}\amalg(-3,-2)\amalg(-2,-1)\amalg(-1,0)\amalg(0,1)\amalg(1,2)\amalg(2,3)$

The $f^{*}$-vector of this triangulation is $f^{*}=(3,6)$, and by Theorem 4,

$L_{(1)}(g+m)=\left|gP_{(1)}\cap\mathbb{Z}\right|$

Alternatively, we can consider the inside-out polytope

$\widetilde{P_{(1)}}=[-1,1]\setminus\\{\pm 1\\}$

which can be visualized as

$\widetilde{P_{(1)}}=$$-1$$0$$1$

The unimodular triangulation given by
$\mathcal{T}=\\{0\\}\amalg(-1,0)\amalg(0,1)$ has support $P_{(1)}$, and has
$f^{*}$-vector $(1,2)$. Therefore, by Theorem 4

$L_{(1)}(g)=3\left|g\widetilde{P_{(1)}}\cap\mathbb{Z}\right|$

###  5.2 $|\vec{d}|=2$

There are two vectors to consider, $(1,1)$ and $(2)$. We have computed the
polynomial $L_{(1,1)}(g+m)=L_{(1,1)}(g)$ previously in the proof of Corollary
2,

$\displaystyle L_{(1,1)}(g+m)=L_{(1,1)}(g)$ $\displaystyle=18{g-1\choose
0}+90{g-1\choose 1}+72{g-1\choose 2}$ $\displaystyle=18\left[{g-1\choose
0}+5{g-1\choose 1}+4{g-1\choose 2}\right]$

Consider the inside-out polytope
$P_{(1,1)}=\left([-3,3]\times[-3,3]\right)\setminus\mathcal{H}_{(1,1)}\subset\mathbb{R}^{2}$,
where $\mathcal{H}_{(1,1)}$ is the hyperplane arrangement

$\mathcal{H}_{(1,1)}:=\\{x_{2}=3,x_{2}=2,x_{2}=1,x_{2}=0,x_{1}=3\\}$

Here is a visualization of $P_{(1,1)}$:

$(3,3)$$(-3,3)$$(3,-3)$$(-3,-3)$

$P_{(1,1)}$ admits a unimodular triangulation as suggested below:

$(3,3)$$(-3,3)$$(3,-3)$$(-3,-3)$

The $f^{*}$-vector of this triangulation is $f^{*}=(18,90,72)$. Therefore, by
Theorem 4,

$L_{(1,1)}(g+m)=\left|gP_{(1,1)}\cap\mathbb{Z}^{2}\right|$

Alternatively, consider the inside-out polytope
$\widetilde{P_{(1,1)}}:=\left([0,2]\times[0,1]\right)\setminus\widetilde{\mathcal{H}}_{(1,1)}\subset\mathbb{R}^{2}$,
where $\widetilde{\mathcal{H}}_{(1,1)}$ is the hyperplane arrangement given by

$\widetilde{\mathcal{H}}_{(1,1)}:=\\{x_{2}=1,x_{1}=1,x_{1}=2\\}$

Here is a visualization of $\widetilde{P_{(1,1)}}$:

$(0,0)$

The claim is that

$L_{(1,1)}(g)=18\left|g\left(P_{(1,1)}\right)\cap\mathbb{Z}^{2}\right|$

Indeed, $P_{(1,1)}$ admits a unimodular triangulation as suggested below:

Since the $f^{*}$-vector of this triangulation is $(1,5,4)$, the claim follows
from Theorem 4.

Now consider the vector $\vec{d}=(2)$. We have already computed
$L_{(2)}(g+m)=L_{(2)}(g)$ in the proof of Corollary 2:

$\displaystyle L_{(2)}(g+m)=L_{(2)}(g)$ $\displaystyle=15{g-1\choose
0}+72{g-1\choose 1}+72{g-1\choose 2}$ $\displaystyle=3\left[5{g-1\choose
0}+24{g-1\choose 1}+24{g-1\choose 2}\right]$

Consider the inside-out polytope given by

$P_{(2)}:=\left([-3,3]\times[-3,3]\right)\setminus\mathcal{H}_{(2)}\subset\mathbb{R}^{2}$

where the hyperplane arrangement $\mathcal{H}_{(2)}$ is given by

$\mathcal{H}_{(2)}:=\\{x_{1}=\pm 3,x_{2}=\pm 3,x_{2}=\pm 2,x_{1}\pm x_{2}=\pm
4,x_{1}\pm x_{2}=\pm 5\\}$

Here is a visualization of $P_{(2)}$:

$P_{(2)}$ admits a unimodular triangulation as suggested below:

The $f^{*}$ vector of this triangulation is $f^{*}=(15,72,72)$, so by Theorem
4,

$L_{(2)}(g+m)=\left|.gP_{(2)}\cap\mathbb{Z}^{2}\right|$

Alternatively, consider the inside-out polytope
$\widetilde{P_{(2)}}:=([-3,3]\times[-1,1])\setminus\widetilde{\mathcal{H}}_{(2)}$,
where $\widetilde{\mathcal{H}}_{(2)}$ is the hyperplane arrangement given by

$\widetilde{\mathcal{H}}_{(2)}=\\{x_{1}=\pm 1,x_{2}=\pm 3,x_{1}\pm x_{2}=\pm
3\\}$

Here is a visualization of $\widetilde{P_{(2)}}$:

$(3,1)$$(-3,1)$$(-3,-1)$$(3,-1)$

The claim is that

$L_{(2)}(g+m)=3\left|g\widetilde{P_{(2)}}\cap\mathbb{Z}^{2}\right|$

Indeed, $\widetilde{P_{(2)}}$ admits a unimodular triangulation as suggested
below:

The $f^{*}$-vector of this triangulation is $(5,24,24)$, so the claim follows
from Theorem 4.

##  6 What Next?

There are many basic open questions that seem natural in light of Theorem 1
and the computations provided in Section 5.

######  Question 1.

Let $\vec{d}\in\mathbb{Z}^{n}_{\geq 0}$, and let
$f^{*}=(f_{0}^{*},\ldots,f_{|\vec{d}|}^{*})$ be the $f^{*}$-vector of
$L_{\vec{d}}(g+m)$. If
$n(\vec{d}):=\text{gcd}(f_{0}^{*},\ldots,f_{|\vec{d}|}^{*})$, is it always
possible to find an inside-out polytope $P_{\vec{d}}$ of dimension $|\vec{d}|$
such that

$L_{\vec{d}}(g+m(\vec{d}))=n(\vec{d})L_{P_{\vec{d}}}(g)\ ?$

######  Question 2.

Besides $\psi$-classes, one can consider other tautological classes,

$\lambda_{i},\kappa_{i},\delta_{j,k}\in R^{*}(\overline{\mathcal{M}}_{g,n})$

Does an Ehrhart phenomenon still occur if we allow for the insertions of these
classes as well?

######  Question 3.

The space $\overline{\mathcal{M}}_{g,n}$ can be viewed as the moduli stack of
stable maps to a point. So one way to generalize is to consider the stack of
stable maps to a smooth projective variety $X$,

$\overline{\mathcal{M}}_{g,n}(X,d)$

and try to play the same game: does an Ehrhart phenomenon still occur when
considering descendent integrals on $\overline{\mathcal{M}}_{g,n}(X,d)$?

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