# Hodge numbers of hypergeometric data

Ling Long  Department of Mathematics, Louisiana State University, Baton Rouge,
LA 70803-4918, USA<EMAIL_ADDRESS>and  Yifan Yang  Department of Mathematics,
National Taiwan University and National Center for Theoretical Sciences,
Taipei, Taiwan 10617<EMAIL_ADDRESS>
###### Abstract.

In this paper, based on the toric hypergeometric model given in a paper by
Beukers–Cohen–Mellit, we provide two other ways to explain why the zig-zag
diagram method can be used to compute Hodge numbers for hypergeometric data
defined over $\mathbb{Q}$.

##  1\. Introduction

Let
$HD=\\{\alpha=\\{a_{1},a_{2},\cdots,a_{m}\\},\beta=\\{1,b_{2},\cdots,b_{m}\\}\\}$
with $a_{i},b_{j}\in\mathbb{Q}\cap(0,1]$ be a hypergeometric datum defined
over $\mathbb{Q}$, which means $\prod_{i=1}^{m}(X-e^{2\pi
ia_{i}})\in\mathbb{Z}[X]$ and $\prod_{i=1}^{m}(X-e^{2\pi
ib_{i}})\in\mathbb{Z}[X]$. The purpose of this note is to compute the Hodge-
Tate weights arising from the generic newton polygon of $HD$ in terms of
combinatorics via the Hodge numbers of toric varieties. They form a set of
fundamental invariants when we consider the hypergeometric Galois
representations arising from $HD$, as in [2] and [6]. Classically, each Hecke
eigenform have two important features: level and weight, say $k$. The latter
one means the Hodge-Tate weights of the corresponding Galois representation
are $\\{0,k-1\\}$.

Write

$\frac{\prod_{i=1}^{m}(X-e^{2\pi ia_{i}})}{\prod_{i=1}^{m}(X-e^{2\pi
ib_{i}})}=\frac{(X^{p_{1}}-1)\cdots(X^{p_{r}}-1)}{(X^{q_{1}}-1)\cdots(X^{q_{s}}-1)}$

where $p_{i},q_{j}\in\mathbb{N}$, $\\{p_{i}\\}\cap\\{q_{j}\\}=\emptyset$, and

(1) $\sum_{i=1}^{r}p_{i}=\sum_{j=1}^{s}q_{j}:=L(HD).$

We call $L(HD)$ the _natural length_ of $HD$. We also refer to

(2) $\Gamma(HD):=[p_{1},\cdots,p_{r},-q_{1},\cdots,-q_{s}]$

as the _hypergeometric gamma vector_ of $HD$. Here we assume $q_{s}=1$.
Further we let $n=r+s-1$ and

$M(HD):=\text{lcd}(\alpha\cup\beta)=\text{lcm}(\\{p_{i},q_{j}\\}),$

be the _level_ of $HD$, where lcd means least positive common denominator. Let
$\bf o$ be the origin of $\mathbb{Z}^{n}$ and fix a set of vectors in
$\mathbb{Z}^{n}$:

(3) $\begin{array}[]{rl}{\bf{e}}_{0}=&[1,0,\cdots,0]\\\
{\bf{e}}_{i}=&[1,0,\cdots,0,\underset{i+1}{1},0,\cdots,0],\quad 1\leq i\leq
n-1\\\
{\bf{e}}_{n}=&[1,p_{1},\cdots,p_{r-1},-q_{1},\cdots,-q_{s-1}]\end{array}$

Let $\Delta(HD)$ be the convex polyhedron with vertices $\\{{\bf
o}\\}\cup\\{{\bf{e}}_{i}\\}_{i=0}^{n}$ and $\text{Vol}(\Delta(HD))$ be its
volume. Its Hodge numbers are non-negative integers. Let $H_{\Delta(HD)}(k)$
be the multiplicity of $k$ among these Hodge numbers. They form a set of
useful invariants when we study the toric varieties parameterized by $HD$ as
follows. Given a gamma vector $\Gamma(HD)$ as above, $\lambda\in\mathbf{C}$,
there is a toric variety given by (1.1) of [3] by Beukers, Cohen and Mellit of
the form

$\quad x_{1}+\cdots+x_{r}-y_{1}-\cdots-
y_{s}=0,\quad\frac{\lambda}{C}x_{1}^{p_{1}}\cdots
x_{r}^{p_{r}}=y_{1}^{q_{1}}\cdots y_{s}^{q_{s}},\quad
C=\frac{\prod_{i=1}^{r}p_{i}^{p_{i}}}{\prod_{i=1}^{s}q_{i}^{q_{i}}}.$

Following [3], we consider the inhomogeneous model (note that $q_{s}=1$) by
letting

$x_{r}=1,\quad y_{s}=\frac{\lambda}{C}\frac{x_{1}^{p_{1}}\cdots
x_{r-1}^{p_{r-1}}}{y_{1}^{q_{1}}\cdots y_{s-1}^{q_{s-1}}}$

and hence the above variety is reduced to

(4) $f_{HD(\lambda)}=1+x_{1}+\cdots+x_{r-1}-y_{1}-\cdots-
y_{s-1}-\frac{\lambda}{C}\frac{x_{1}^{p_{1}}\cdots
x_{r-1}^{p_{r-1}}}{y_{1}^{q_{1}}\cdots y_{s-1}^{q_{s-1}}}=0.$

Example 1. $\Gamma(HD)=[3,-1,-1,-1]$ In this case, $C=27$, for any
$\lambda\neq 0,1$ $f_{HD}(\lambda)=\displaystyle
1-y_{1}-y_{2}-\frac{\lambda}{27y_{1}y_{2}}=0$ is an elliptic curve.

Given a fixed finite field $\mathbf{F}_{q}$ whose characteristic is assumed to
be coprime to $M(HD)$ throughout this paper and let
$\lambda\in\mathbf{F}_{q}$. To compute the number of solutions of
$f_{HD}(\lambda)=0$ in
$\mathbb{G}_{m}^{r+s-2}(\mathbf{F}_{q}):=(\mathbf{F}_{q}^{\times})^{r+s-2}$,
one can make use of any nontrivial additive character $\psi$ of
$\mathbf{F}_{q}$ by computing

(5) $\frac{1}{q}\sum_{\begin{subarray}{c}x_{0}\in\mathbf{F}_{q}\\\
\,\,(x_{1},\cdots,y_{s-1})\in(\mathbf{F}_{q}^{\times})^{r+s-2}\end{subarray}}\psi\left(x_{0}f_{HD}\right),$

plus contributions from the boundary. See [3] for a chosen compactification.
Below we regard $x_{0}f_{HD}(\lambda)$ as a Laurent polynomial in $n=r+s-1$
variables $x_{0},x_{1},\cdots,x_{r-1},y_{1},\cdots,y_{s-1}$. Belowe we treat
it as a vector ${\mathbb{x}}$. Using the trace map, $\psi$ can be extended to
an additive character $\psi_{r}$ for any degree-$r$ extension
$\mathbf{F}_{q^{r}}$ of $\mathbf{F}_{q}$.

To continue we first recall some more background information on exponential
sums. Assume that $f(\mathbb{x})$ is a Laurent polynomial in $n$ variables
with coefficients in $\mathbf{F}_{q}$. Let

(6)
$S_{r}(f;\mathbf{F}_{q}):=\sum_{\mathbb{x}\in\mathbb{G}_{m}^{n}(\mathbf{F}_{q^{r}})}\psi_{r}(f(\mathbb{x}))$

Then by Dwork

$Z_{q}(f;T):=\displaystyle\exp\left((-1)^{n-1}\sum_{r\geq
1}S_{r}(f;\mathbf{F}_{q})\frac{T^{r}}{r}\right)$

is a rational function of $T$. A fundamental result of Adolphson and Sperber
[1] says when $f$ is non-degenerate (see Definition 1.1 of [9]), $Z_{q}(f;T)$
is a polynomial in $\mathbb{Z}[T]$ with constant 1, whose degree is
$\operatorname{Vol}\Delta_{f}\cdot n!$ where $\Delta_{f}$ is a convex
polyhedron in $\mathbb{Q}^{n}$ whose vertices are $\mathbf{o}$ together with
the exponent vectors corresponding to the monomials of $f$ with nonzero
coefficients.

Note that for $f=x_{0}f_{HD}$ the vertex-vectors coincide with
${\bf{e}}_{0},{\bf{e}}_{1},\cdots,{\bf{e}}_{n}$. For $\lambda\neq 0,1$ in
$\mathbf{F}_{q}$, the polynomial $x_{0}f_{HD(\lambda)}$ is non-degenerate. By
induction on $n$, the corresponding Euler polynomial has degree

(7) $\operatorname{Vol}\Delta_{HD}\cdot n!=L(HD)=\sum_{j=1}^{r}p_{j}.$

Let $u_{1}(\lambda,p),\cdots,u_{L(HD)}(\lambda,p)\in\mathbb{Z}_{p}$ be the
list of reciprocal roots of the Euler polynomial in
$\overline{\mathbb{Q}}_{p}$, the algebraic closure of $\mathbb{Q}_{p}$. Let
$r_{i}(\lambda,p)=\text{ord}_{p}u_{i}(\lambda,p)$. A priori, these are non-
negative rational numbers. For $x_{0}f_{HD}$, they are integers. From these
$r_{i}(\lambda,p)$ one can compute the Newton Polygon in a way analogous to
Definition 1.2 of [9]. It was conjectured by Wan in [9] that for generic
choice of $\lambda$ and large prime $p$, the Newton polygon and the Hodge
polygon agree. This conjecture was proved by Blache in [4]. This means the
Hodge numbers that can be computed combinatorially also encode arithmetic
information of the exponential sums. They are also called Hodge-Tate weights
of the datum $HD$.

Example 2. ${\Gamma=[3,-1,-1,-1]},\lambda=2$

$Z_{p}(f_{HD}(2);T)=(1-T)(1-a_{p}(f_{54.2.a.b}){p}T+p^{{2}+1}T^{2}),\quad\forall
p>3$

where $a_{p}$ denotes the $p$th coefficient, $f_{54.2.a.b}$ is a weight-2
level 54 modular form in its L-Functions and Modular Forms Database label.
Thus for a generic $p$, its reciprocal roots in $\mathbf{C}_{p}$ have $p$-adic
orders $\\{0,1,2\\}$.

In the literature of hypergeometric functions and motives, there is an
explicit way to compute the Hodge numbers using the so-called zig-zag diagram
(see [5] by Corti and Golyshev and [8] by Roberts and Rodriguez-Villegas).
These Hodge numbers are useful for the determination of the weights of the
modular forms according to Galois representations arising from hypergeometric
motives, see for example [2, 6] for some applications. In this paper we give
two more ways to explain why:

> _The zig-zag diagram algorithm will produce the the Hodge multiplicity and
> hence Newton multiplicity numbers for $x_{0}f_{HD}$._

Our approaches below are different from [5]. They are combinatorial and
explicit.

The paper will be organized in the following manner. The main statement is
given in §2, the relation between the zig-zag diagrams and $p$-adic orders of
the hypergeometric coefficients will be recalled in §3. In §4 and §5, we give
two different proofs of the main statement.

##  2\. Main statement

Given a hypergeometric gamma vector
$\Gamma=[p_{1},\cdots,p_{r},-q_{1},\dots,-q_{s}]$, let

(8)
$HD_{\Gamma}=\left\\{\alpha=\left\\{\bigcup_{j=1}^{r}\left\\{\frac{i}{p_{j}}\right\\}_{j=1}^{p_{i}}\right\\},\quad\beta=\left\\{\bigcup_{j=1}^{s}\left\\{\frac{i}{q_{j}}\right\\}_{i=1}^{q_{j}}\right\\}\right\\}.$

Each multi-set has length $L:=L(HD)=\sum_{i=1}^{r}p_{i}$. By construction, the
multiplicity $n_{b,\beta}$ of any element $b=\frac{i}{q_{j}}<1$ in $\beta$ can
be computed by

(9) $n_{b,\beta}=\\#\\{q_{k}\mid 1\leq k\leq s,\,bq_{k}\in\mathbb{Z}\\}.$

Sort the combined elements in both $\alpha$ and $\beta$ into a new list
satisfying

(10) $\mu_{1}\leq\mu_{2}\leq\cdots\leq\mu_{2L},$

where $\mu_{2L}=1$ with the following convention:

> if $a\in\alpha\cap\beta$, then _those copies of $a$’s in $\beta$ all go
> before those $a$’s in $\alpha$_.

Further let $\mu_{0}=0$.

Given $\Gamma$, we define an arithmetic function $\Phi$ recursively as
follows:

(11)
$\Phi(0)=\Phi_{HD}(0)=r(\Gamma(HD)),\quad\Phi_{HD}=\Phi_{HD}(i)=\begin{cases}\Phi_{HD}(i-1)+1,&\text{
if }\quad\mu_{i}\in\alpha;\\\ \Phi_{HD}(i-1)-1,&\text{ if
}\quad\mu_{i}\in\beta.\end{cases}$

######  Theorem 1.

Let

(12) $HT(HD):=\\{\Phi(i)\mid\text{ if }\mu_{i}\in\beta\\}.$

They are the Hodge numbers of $HD$.

Equivalently,

######  Theorem 2.

Given a hypergeometric vector $\Gamma$ as (2), let $HD=HD_{\Gamma}$ as in (8).
Then the Hodge numbers $H_{\Delta(HD)}(k)$ of the convex polyhedron
$\Delta(HD)$ (to be defined in (24)) satisfy that

(13) $H_{\Delta(HD)}(k)=\\#\\{w\in HT(HD)\mid w=k\\}.$

Removing overlapping elements from both $\alpha,\beta$ with the same
multiplicity will result in a _derived primitive hypergeometric datum_ ,
denoted by $HD^{\text{red}}$. The zig-zag diagram for $HD^{\text{red}}$ can be
obtained by concatenating the “V” ditches caused by elements in
$\alpha\cap\beta$. Typically, we further normalize the Hodge numbers of
$HD^{\text{red}}$ so that the minimal Hodge number becomes 0.

Example 3.  Let $\Gamma=[3,-1,-1,-1]$, then
$HD_{\Gamma}=\\{\alpha=\\{\frac{1}{3},\frac{2}{3},1\\},\beta=\\{{\color[rgb]{0,0,1}1,1,1}\\}\\}.$
The sorted list is

$\frac{1}{3}\leq\frac{2}{3}\leq{{{\color[rgb]{0,0,1}1}}}\leq{{{\color[rgb]{0,0,1}1}}}\leq{{{\color[rgb]{0,0,1}1}}}\leq
1$

$HD^{\text{red}}=\\{\\{\frac{1}{3},\frac{2}{3}\\},\\{1,1\\}\\}$.

##  3\. Zig-zag diagrams and $p$-adic orders

The zig-zag diagram is related to the $p$-adic order of

(14) $\displaystyle
A_{HD}(k):=\frac{\prod_{a\in\alpha}(a)_{k}}{\prod_{b\in\beta}(b)_{k}},\quad\text{where}\quad(a)_{k}:=\frac{\Gamma(a+k)}{\Gamma(a)}=a(a+1)\cdots(a+k-1).$

It is related to the discussion in [7]. Here we recast the idea. For
$a\in\mathbb{Z}_{p}$, use $[a]_{0}$ to denote the first $p$-adic digit of $a$.
If $a=\frac{c}{d}$ where $c,d\in\mathbb{N},1\leq c<d<p$, and $p\equiv 1\mod d$
then

(15) $\left[-\frac{c}{d}\right]_{0}=\frac{cp-c}{d}=\frac{c}{d}(p-1).$

For $a\in(0,1]$ and $1\leq k<p$,

(16) $\text{ord}_{p}(a)_{k}=-\left\lfloor
a-\frac{k}{p-1}\right\rfloor=\begin{cases}0&\text{if }1\leq k\leq[-a]_{0};\\\
1&\text{if }[-a]_{0}<k<p.\end{cases}$

Let $p\equiv 1\mod M(HD)$ be a fixed prime throughout this paper. It follows
that

(17)
$0\leq[-\mu_{1}]_{0}\leq[-\mu_{2}]_{0}\leq\cdots\leq[-\mu_{2L}]_{0}\leq[-1]_{0}=p-1.$

When $k$ goes from 0 to $p-1$, $\text{ord}_{p}A_{HD}(k)$ is a step function
starting from 0, and when $k$ passes through $[-\mu_{i}]_{0}$, its value will
change by $n_{\mu_{i},\alpha}-n_{\mu_{i},\beta}$ where $n_{\mu_{i},\alpha}$
(resp. $n_{\mu_{i},\beta}$) denotes the multiplicity of $\mu_{i}$ in $\alpha$
(resp. $\beta$). We summarize the above discussion into the next Proposition.

######  Proposition 1.

For a given datum
$HD=\\{\alpha=\\{a_{1},\cdots,a_{L}\\},\beta=\\{b_{1},\cdots,b_{L}\\}\\}$
defined over $\mathbb{Q}$ where $a_{i},b_{j}\in(0,1]$, $b_{1}=1$, whose Gamma
vector is $\Gamma(HD)$, let $\mu_{1}\leq\cdots\leq\mu_{2L}=1$ be the sorted
combined list. For any prime $p\equiv 1\mod M(HD)$

(18)
$\Phi_{HD}(i)=\text{ord}_{p}A_{HD}(k)+r(\Gamma(HD)),\quad\text{if}\quad[-\mu_{i}]_{0}<k\leq[-\mu_{i+1}]_{0}.$

In particular, for each $i$ such that $\mu_{i}<\mu_{i+1}$, by (15) and (18),

(19) $\Phi_{HD}(i)=\text{ord}_{p}A_{HD}(\mu_{i+1}(p-1))+r(\Gamma(HD)).$

Next we will consider the first term on the right hand side of (18) by
recalling some useful facts. For $r\in\mathbb{R}$, let $\\{r\\}:=r-\lfloor
r\rfloor$. Then

(20) $\\{r\\}+\\{-r\\}=\begin{cases}0&r\in\mathbb{Z};\\\
1&r\notin\mathbb{Z}.\end{cases}$

Also (from the multiplication formula of Gauss sums and the Gross-Koblitz
formula), for any $m\in\mathbb{Z}_{>0}$

$\\{rm\\}+\sum_{i=1}^{m}\left\\{\frac{i}{m}\right\\}=\sum_{i=1}^{m}\left\\{r+\frac{i}{m}\right\\}.$

Equivalently,

(21) $\\{rm\\}=\sum_{i=1}^{m}\left\\{r+\frac{i}{m}\right\\}-\frac{m-1}{2}.$

For $1\leq k\leq p-1,$ from (16)

(22) $\text{ord}_{p}A_{HD}(k)=\sum_{i=1}^{L}-\left\lfloor
a_{i}-\frac{k}{p-1}\right\rfloor+\left\lfloor
b_{i}-\frac{k}{p-1}\right\rfloor.$

In particular, $\text{ord}_{p}A_{HD}(p-1)=s-r$.

######  Lemma 1.

Given a Gamma vector $\Gamma$, let $HD_{\Gamma}$ be as in (8). For $p\equiv
1\mod M(HD_{\Gamma})$, $\mu\in\alpha\cup\beta,$

(23) $\text{ord}_{p}A_{HD_{\Gamma}}(\mu(p-1))=s-r+\sum_{i=1}^{r}\\{-\mu
p_{i}\\}-\sum_{j=1}^{s}\\{-\mu q_{j}\\}.$

###### Proof.

R.H.S. $\displaystyle\overset{\eqref{eq:fr}}{=}$ $\displaystyle
s-r+\sum_{i=1}^{r}\left(\sum_{l=1}^{p_{i}}\left\\{-\mu+\frac{l}{p_{i}}\right\\}-\frac{p_{i}-1}{2}\right)-\sum_{j=1}^{s}\left(\sum_{l=1}^{q_{i}}\left\\{-\mu+\frac{l}{q_{j}}\right\\}-\frac{q_{j}-1}{2}\right)$
$\displaystyle=$
$\displaystyle\frac{s-r}{2}+\sum_{i=1}^{r}\sum_{l=1}^{p_{i}}\left(-\mu+\frac{l}{p_{i}}-\left\lfloor-\mu+\frac{l}{p_{i}}\right\rfloor\right)-\sum_{j=1}^{s}\sum_{l=1}^{q_{i}}\left(-\mu+\frac{l}{q_{j}}-\left\lfloor-\mu+\frac{l}{q_{j}}\right\rfloor\right)$
$\displaystyle=$
$\displaystyle\sum_{i=1}^{r}\sum_{k=1}^{p_{i}}-\left\lfloor\mu-\frac{k}{p_{i}}\right\rfloor+\sum_{j=1}^{s}\sum_{k=1}^{q_{j}}\left\lfloor\mu-\frac{k}{q_{i}}\right\rfloor$
$\displaystyle\overset{\eqref{eq:ordpA}}{=}$
$\displaystyle\text{ord}_{p}A_{HD_{\Gamma}}(\mu(p-1))\in\mathbb{Z}.$

∎

##  4\. Hodge numbers of convex cones

###  4.1. Background

The background for this section is from [9] by Wan. Let $\Delta$ be a
polyhedron in $\mathbb{R}^{n}$ whose vertices are points in $\mathbb{Z}^{n}$.
For each ${\bf{v}}\in\mathbb{R}^{n}$, its _depth_ , denoted by $dep({\bf{v}})$
is the sum of its negative entries. For the computation below we assume the
interior of $\Delta$ contains no lattice point (vectors in $\mathbb{Z}^{n})$
so that its denominator $1$. Consider the cone
$C(\Delta):=C(\Delta,\mathbb{R}_{\geq 0})$ obtained from rays starting from
the origin pointing to points in $\Delta$. For each lattice point ${\bf{v}}$
in this cone, the _weight_ of ${\bf{v}}$, denoted by $wt({\bf{v}})$ is given
as follows: $wt({\bf o})=0$, otherwise $wt({{\bf{v}}})$ is the non-negative
scalar $k$ such that $\bf v$ is located on one of the outer faces of the
convex polyhedron obtained from $\Delta$ re-scaled by $k$. Given
$k\in\mathbb{Z}_{\geq 0}$, let $w_{\Delta}(0)=1$ and for $k\geq 1$,

$w_{\Delta}(k)=\\#\left\\{{\bf{v}}\in C(\Delta)\cap\mathbb{Z}^{n}\mid
wt({\bf{v}})=k\right\\}.$

Then the Hodge numbers of the cone is computed by the inductive formula

(24) $H_{\Delta}(k):=\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}w_{\Delta}(k-i).$

And $H_{\Delta}(k)=0$ for $k>n$, see [9].

Now let $\Delta(HD)$ be the polyhedron with vertices $\\{{\bf
o}\\}\cup\\{{\bf{e}}_{i}\\}_{i=0}^{n}$, where ${\bf{e}}_{i}$ as in (3). It has
dimension $n$, and no interior lattice point. For any lattice point ${\bf{v}}$
of $C(\Delta)$, $wt({\bf{v}})$ is the value of its first entry.

###  4.2. Primitive elements and a generating set

Let

$G:=C(\Delta(HD))\cap\mathbb{Z}^{n}=\left\\{\sum_{i=0}^{n}c_{i}{\bf{e}}_{i}\mid
c_{i}\in\mathbb{R}_{\geq 0}\right\\}\cap\mathbb{Z}^{n},$

which is a semi-group. It is larger than
$\\{\sum_{i=0}^{n}a_{i}{\bf{e}}_{i},a_{i}\in\mathbb{Z}_{\geq 0}\\}$ as
$\\{{\bf{e}}_{i}\\}_{i=0}^{n}$ is linearly dependent. Let

$H:=C(\\{{\bf{e}}_{0},\cdots,{\bf{e}}_{n-1}\\},\mathbb{Z}_{\geq
0})=C(\\{{\bf{e}}_{0},\cdots,{\bf{e}}_{n-1}\\},\mathbb{Q}_{\geq
0})\cap\mathbb{Z}^{n}.$

Any ${\bf{v}}\in G$ is said to be _primitive_ if its depth is negative. In
this case ${\bf{v}}=\sum_{i=0}^{n}c_{i}{\bf{e}}_{i}$ where
$c_{i}\in\mathbb{Q}_{\geq 0}$, $c_{n}>0$. As ${\bf{v}}$ lies in one of the
faces of $\Delta(HD)$ re-scaled by $k$, we may assume $c_{i}=0$ for one
$i\in[0,n]$. Then $\\{{\bf{e}}_{j}\\}_{j=0,j\neq i}^{n}$ form an invertible
matrix whose determinant is either $\pm p_{j}$ or $\pm q_{j}$. For more
details, see Lemma 2 and its proof. Thus for $l\neq i$,
$c_{l}=\frac{k_{l}}{p_{j}}$ or $\frac{k_{l}}{q_{j}}$ for some positive
integers $k_{l},j$. For any $\mu\in(\alpha\cup\beta)$, let

(25) ${\bf{u}}_{\mu}:=\mu{\bf{e}}_{n}+\left\\{-\mu
p_{r}\right\\}{\bf{e}}_{0}+\sum_{i=1}^{r-1}\\{-\mu
p_{i}\\}{\bf{e}}_{i}+\sum_{j=1}^{s-1}\\{\mu q_{j}\\}{\bf{e}}_{r-1+j}.$

In particular when $\mu=1$, ${\bf{u}}_{1}={\bf{e}}_{n}$. It has weight 1 and
depth $-L$. When $\mu<1$,

(26) $\begin{split}wt({\bf{u}}_{\mu})&=\sum_{i=1}^{r}\\{-\mu
p_{i}\\}+\sum_{j=1}^{s}\\{\mu q_{j}\\}\\\
&\overset{\eqref{eq:fracsum}}{=}\sum_{i=1}^{r}\\{-\mu
p_{i}\\}-\sum_{j=1}^{s}\\{-\mu q_{j}\\}+s-\\#\\{j\mid 1\leq j\leq s,\mu
q_{j}\in\mathbb{Z}\\}\\\ &\overset{\eqref{eq:nb},\eqref{eq:p-ord-
mu(p-1)}}{=}r+\text{ord}_{p}A_{HD}(\mu(p-1))-\\#\\{\mu\in\beta\\}.\end{split}$

From construction, ${\bf{u}}_{\mu}$ is in $G$. Intuitively, it is the
“ceiling” function of $\mu{\bf{e}}_{n}$, namely the element in $G$ closest to
the vector $\mu{\bf{e}}_{n}$ with $dep({\bf{u}}_{\mu}-\mu{\bf{e}}_{n})=0$.

Assume the $i$th entry of the combined list (10) is in $\beta$, namely
$\mu_{i}\in\beta$ with multiplicity $n_{\mu_{i},\beta}$ such that all of its
other duplicates in $\beta$ have smaller indices. Thus
$\mu_{i-n_{\mu_{i},\beta}}<\mu_{i-n_{\mu_{i},\beta}+1}=\mu_{i}$. Then

(27)
$wt({\bf{u}}_{\mu_{i}})=wt({\bf{u}}_{\mu_{i-n_{\mu_{i},\beta}+1}})\overset{\eqref{eq:wt-
ceil}}{=}r+\text{ord}_{p}A_{HD}(\mu_{i-n_{\mu_{i},\beta}+1}(p-1))-n_{\mu_{i},\beta}\overset{\eqref{eq:ordpA-
Phi}}{=}\Phi(i-n_{\mu_{i},\beta})-n_{\mu_{i},\beta}\overset{\eqref{eq:Phi}}{=}\Phi(i).$

Also for $0\leq l\leq n_{\mu_{i},\beta}-1$

(28) $\Phi(i-l)\overset{\eqref{eq:Phi}}{=}\Phi(i)+l.$

We now focus on ${\bf{u}}_{\mu}$ where $\mu\in\beta$. For convenience, we let
$b_{0}=0,b_{L+1}=2$. They are not in $\beta$.

######  Lemma 2.

Assume the elements in $\beta$ are listed in a non-decreasing order, i.e.
$0=b_{0}<b_{1}\leq b_{2}\leq\cdots\leq b_{L}<b_{L+1}$.

1) For each positive $b\in\beta$, ${\bf{u}}_{b}$ is primitive.

2) For $b_{i}\neq b_{j}$ in $\beta$, ${\bf{u}}_{b_{i}}\neq{\bf{u}}_{b_{j}}$.

3) Assume $b_{i-1}<b_{i}$ and let $S_{i}=\\{1\leq j\leq s-1\mid
b_{i}q_{j}\in\mathbb{Z}\\}$, then ${\bf{u}}_{b_{i}}+\sum_{j\in
S_{i}}{\bf{e}}_{r+j}\in{\bf{u}}_{b_{i-1}}+\,H$

###### Proof.

Note that the $(r-1+j)$th entry of ${\bf{u}}_{b_{i}}$ is $-\lfloor
b_{i}q_{j}\rfloor$ and at least one of them is negative, so ${\bf{u}}_{b_{i}}$
is primitive. If $b_{i}\neq b_{j}$, by (25)
${\bf{u}}_{b_{i}}\neq{\bf{u}}_{b_{j}}$.

Each time when $i$ increases by 1, the $(r-1+j)$’s entry of ${\bf{u}}_{b_{i}}$
either stays the same or decrease by 1 depending on whether
$u_{i}q_{j}\in\mathbb{Z}$. ∎

######  Lemma 3.

For each primitive ${\bf v}\in G$, there is exists an element $b\in\beta$,
such that ${\bf v}\in{\bf{u}}_{b}+\,\langle
H,{\bf{e}}_{n}\rangle=\\{i{\bf{u}}+j{\bf{e}}_{n}\mid{\bf{u}}\in
H,i,j\in\mathbb{Z}_{\geq 0}\\}.$

###### Proof.

Assume ${\bf v}=\sum_{i=0}^{n}c_{i}{\bf{e}}_{i}$ and $c_{n}>0$. If
$c_{n}\in\mathbb{Z}$ and then we can pick $b=1$; otherwise, replace $\bf v$ by
${\bf v}-\lfloor c_{n}\rfloor{\bf{e}}_{n}$ we may assume
$c_{n}\in(0,1)\cap\mathbb{Q}$. Next pick the largest element $b\in\beta$,
satisfying $b\leq c_{n}$. Namely pick the largest value among fractions of the
form $\frac{\lfloor c_{n}q_{j}\rfloor}{q_{j}}$ in $\beta$. Up to relabelling,
we may assume $b=\frac{\lfloor c_{n}q_{1}\rfloor}{q_{1}}$. The $r-1+j$ entry
of ${\bf{u}}_{c_{n}}$ (resp. ${\bf{u}}_{\frac{\lfloor
c_{n}q_{1}\rfloor}{q_{1}}}$) is $-\lfloor c_{n}q_{j}\rfloor$ (resp.
$-\lfloor\frac{\lfloor c_{n}q_{1}\rfloor}{q_{1}}q_{j}\rfloor$). From the
choice of $b$, either $\lfloor c_{n}q_{j}\rfloor=\lfloor\frac{\lfloor
c_{n}q_{1}\rfloor}{q_{1}}q_{j}\rfloor$, or $\lfloor
c_{n}q_{j}\rfloor=\lfloor\frac{\lfloor
c_{n}q_{1}\rfloor}{q_{1}}q_{j}\rfloor+1$. In the latter case,

$\frac{\lfloor c_{n}q_{j}\rfloor}{q_{j}}=\frac{\lfloor\frac{\lfloor
c_{n}q_{1}\rfloor}{q_{1}}q_{j}\rfloor+1}{q_{j}}\leq\frac{\lfloor
c_{n}q_{1}\rfloor}{q_{1}}$

which implies $1-\left\\{\frac{\lfloor
c_{n}q_{1}\rfloor}{q_{1}}q_{j}\right\\}<0$, a contradiction. Thus
${\bf{u}}_{c_{n}}$ and ${\bf{u}}_{\frac{\lfloor c_{n}q_{1}\rfloor}{q_{1}}}$
have the same $r-1+j$ entries for $j=1,\cdots,s-1$. Also for each $2\leq i\leq
r$, the $i$th entry of ${\bf{u}}_{c_{n}}$ is greater or equal to the $i$th
entry of ${\bf{u}}_{\frac{\lfloor c_{n}q_{1}\rfloor}{q_{1}}}$. Thus ${\bf
v}\in{\bf{u}}_{b}+\,\langle H,{\bf{e}}_{n}\rangle.$ ∎

###  4.3. A set of primitive elements

We now modify the list $\\{{\bf{u}}_{b}\mid b\in\beta\\}$ by incorporating the
multiplicities of the elements to get another set consisting of distinct
elements. For each $\mu_{i}\in\beta$, let $S_{i}=\\{1\leq j\leq
s-1\mid\mu_{i}q_{j}\in\mathbb{Z}\\}$ as in Lemma 2, let
$n_{\mu_{i},\beta}=|S_{i}|$. Further we let $\displaystyle{\bf
v}_{i}:=\sum_{j\in S_{i}}{\bf{e}}_{r-1+j}$ and let ${\bf v}_{i,l}$ be the
partial summand of ${\bf v}_{i}$ consisting the first $l$ entries,
$wt({\bf{v}}_{i,l})=l$.

Like before, we assume all of its other duplicates of $\mu_{i}$ in $\beta$
appear before $\mu_{i}$. For each $1\leq l<m_{i}$,

(29) $w_{dep({\bf{u}}_{\mu_{i}})-l}:={\bf{u}}_{\mu_{i}}+{\bf{v}}_{i,l}.$

It has weight $wt({\bf{u}}_{b_{i}})+l=\Phi(i)+l=\Phi(i-l)$ by (27) and (28),
and depth $dep({\bf{u}}_{\mu_{i}})+l$.

Further we let ${\bf{w}}_{0}={\bf o}$. In conclusion, we have

######  Proposition 2.

For each $0\leq i\leq L$, $dep({\bf{w}}_{i})=-i$, and

(30) $\\{wt({\bf{w}}_{i})\\}_{i=1}^{L}=\\{\Phi(i)\mid 1\leq i\leq
2L,\mu_{i}\in\beta\\}.$

Example 4.  ${\bf{v}}=[4,4,2,-3,-6,-1]$

$\begin{array}[]{|c|l|l|c|}\hline\cr b&{\bf{u}}_{b}&{\bf{w}}_{i}\\\
\hline\cr\frac{1}{6}&[2,1,1,0,-1]&{\bf{w}}_{1}=[{{\color[rgb]{0,0,1}2}}\color[rgb]{0,0,0},1,1,0,-1]\\\
\frac{1}{3}&[2,2,2,-1,-2]&{\bf{w}}_{2}=[{{\color[rgb]{0,0,1}3}}\color[rgb]{0,0,0},2,2,0,-2]\\\
\frac{1}{2}&[1,2,2,-1,-3]&{\bf{w}}_{3}=[{{\color[rgb]{0,0,1}2}}\color[rgb]{0,0,0},2,2,-1,-2]\\\
\frac{2}{3}&[2,3,3,-2,-4]&{\bf{w}}_{4}=[{{\color[rgb]{0,0,1}1}}\color[rgb]{0,0,0},2,2,-1,-3]\\\
\frac{5}{6}&[3,3,3,-2,-4]&{\bf{w}}_{5}=[{{\color[rgb]{0,0,1}3}}\color[rgb]{0,0,0},3,3,-1,-4]\\\
1&[1,4,4,-3,-6]&{\bf{w}}_{6}=[{{\color[rgb]{0,0,1}2}}\color[rgb]{0,0,0},3,3,-2,-4]\\\
&&{\bf{w}}_{7}=[{{\color[rgb]{0,0,1}3}}\color[rgb]{0,0,0},4,4,-2,-5]\\\
&&{\bf{w}}_{8}=[{{\color[rgb]{0,0,1}2}}\color[rgb]{0,0,0},4,4,-2,-6]\\\
&&{\bf{w}}_{9}=[{{\color[rgb]{0,0,1}1}}\color[rgb]{0,0,0},4,4,-3,-6]\\\
\hline\cr\end{array}$

In this case
$HD_{\Gamma}=\\{\\{\frac{1}{4},\cdots,\frac{4}{4},\frac{1}{4},\cdots,\frac{4}{4},\frac{1}{2},1\\},\\{\frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{1}{6},\cdots,\frac{6}{6},1\\}\\}$

######  Proposition 3.

For each $i\in[0,L]$ there is an integer $s(i)\in\\{1,\cdots,n\\}$ such that

(31)
$G=\bigsqcup_{i=0}^{L}\left({\bf{w}}_{i}+C(\\{{\bf{e}}_{0},{\bf{e}}_{1},\cdots,\check{{\bf{e}}}_{s(i)},\cdots,{\bf{e}}_{n}\\},\mathbb{Z}_{\geq
0})\right),$

where ${\bf{w}}_{i}$ is defined in (29), $\check{{\bf{e}}}_{s(i)}$ means
removing ${{\bf{e}}}_{s(i)}$ from the list.

###### Proof.

We first prove that the right-hand side is a disjoint union. We let $s(0)=n$.
So
${\bf{w}}_{0}+C(\\{{\bf{e}}_{0},{\bf{e}}_{1},\cdots,\check{{\bf{e}}}_{n-1}\\},\mathbb{Z}_{\geq
0})=\,H$. Since ${\bf{w}}_{1}$ has depth $-1$, there exists a unique
${\bf{e}}_{i}$, whose index is $s(1)$ such that ${\bf{w}}_{1}+{\bf{e}}_{s(1)}$
is in $H$. Thus $H$ and
${\bf{w}}_{1}+C(\\{{\bf{e}}_{0},{\bf{e}}_{1},\cdots,\check{{\bf{e}}}_{s(1)},\cdots,{\bf{e}}_{n}\\},\mathbb{Z}_{\geq
0})$ are disjoint. Proceed by induction and assume the sets

$U_{k-1}:=\,H\bigsqcup_{i=1}^{k-1}\left({\bf{w}}_{i}+C(\\{{\bf{e}}_{0},\cdots,\check{{\bf{e}}}_{s(i)},\cdots,{\bf{e}}_{n}\\},\mathbb{Z}_{\geq
0})\right)$

is a disjoint union, as claimed. For ${\bf{w}}_{k}$, by construction, there
exists a unique $s(k)$ such that ${\bf{w}}_{k+1}+{\bf{e}}_{s(k)}$ is in
$U_{k-1}$. Thus
${\bf{w}}_{k}+C(\\{{\bf{e}}_{0},\cdots,\check{{\bf{e}}}_{s(k)},\cdots,{\bf{e}}_{n}\\},\mathbb{Z}_{\geq
0})$ is disjoint from $U_{k-1}$ and let
$U_{k}=U_{k-1}\bigsqcup({\bf{w}}_{k}+C(\\{{\bf{e}}_{0},\cdots,\check{{\bf{e}}}_{s(k)},\cdots,{\bf{e}}_{n}\\},\mathbb{Z}_{\geq
0})).$

Next we show that every primitive element ${\bf{v}}$ lies in one of the cones
with vertices ${\bf{w}}_{i}$. Taking away a positive integer multiple of
${\bf{e}}_{n}$ if necessary, we may assume the depth of ${\bf{v}}$ is larger
than $-L$. By the previous Lemma, ${\bf{v}}={\bf{u}}_{b_{i}}+{\bf{w}}$ for
some $b_{i}<1\in\beta$ and ${\bf{w}}\in\,H$. Assume now $b_{i}$ is the least
among $\beta$ such that ${\bf{v}}-{\bf{u}}_{b_{i}}$ in $H$, by 3) of Lemma 2
this means ${\bf{w}}-(\sum_{i\in S_{i}}{\bf{e}}_{r+i})\not\in\,H$ so it lies
in one of the cone
${\bf{w}}_{i}+C(\\{{\bf{e}}_{0},{\bf{e}}_{1},\cdots,\check{{\bf{e}}}_{s(i)},\cdots,{\bf{e}}_{n}\\},\mathbb{Z}_{\geq
0}).$ ∎

The previous proposition gives a partition of $G$ into lattices points located
in $L+1$ cones. Each of them is generated by a polyhedron, denoted by
$\Delta_{i}$, with vertices $\\{{\bf
o},{\bf{e}}_{0},{\bf{e}}_{1},\cdots,\check{{\bf{e}}}_{s(i)},\cdots,{\bf{e}}_{n}\\}$
whose apex located at ${\bf{w}}_{i}$.

Example 5.  For $\Gamma=[5,-2,-2,-1]$,
${\bf{w}}_{0}=[0,0,0],{\bf{w}}_{1}=[2,0,-1],s(1)=2;{\bf{w}}_{2}=[1,-1,-1],s(2)=1;{\bf{w}}_{3}=[2,-1,-2],s(3)=2;{\bf{w}}_{4}=[1,-2,-2],s(4)=1.$
In the following plot, we mark the lattices in
${\bf{w}}_{i}+C(\Delta_{i},\mathbb{Z}_{\geq 0})$ with colors green, red,
purple, blue and orange respectively.

The following identity follows from the Chu-Vandermonde’s identity that for
any $n,k\in\mathbb{Z}_{\geq 0},$

(32)
$\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}\binom{n-1+k-i}{n-1}=\begin{cases}1&\text{
if}\quad k=0;\\\ 0&\text{else}.\end{cases}$

######  Proposition 4.

For $0\leq k\leq n=r+s-1$, $H_{\Delta(HD)}(k)=\\#\\{{\bf{w}}_{i}\mid 0\leq
i\leq L-1,wt({\bf{w}}_{i})=k\\}.$

###### Proof.

When $k=0$, both hand sides take value 1. The number of weight-$k$ elements in
each ${\bf{w}}_{i}+C(\Delta_{i},\mathbb{Z}_{\geq 0})$ is the number of weight
$k-wt({\bf{w}}_{i})$ elements in $C(\Delta_{i},\mathbb{Z}_{\geq 0})$. For each
$\Delta_{i}$,
$\\{{\bf{e}}_{0},{\bf{e}}_{1},\cdots,\check{{\bf{e}}}_{s(i)},\cdots,{\bf{e}}_{n}\\}$
are linearly independent in $\mathbb{R}^{n}$ and each has weight 1. Thus

$w_{\Delta_{i}}(k):=\\#\\{{\bf{w}}\in{\bf{w}}_{i}+C(\Delta_{i},\mathbb{Z}_{\geq
0})\mid wt({\bf{w}})=k\\}=\binom{n-1+k-wt({\bf{w}}_{i})}{k-wt({\bf{w}}_{i})}.$

$\displaystyle H_{\Delta(HD)}(k)$ $\displaystyle\overset{\eqref{eq:m}}{=}$
$\displaystyle\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}w_{\Delta(HD)}(k)$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}\sum_{i=0}^{L}w_{\Delta_{i}}(k)$
$\displaystyle=$
$\displaystyle\sum_{i=0}^{L}\sum_{i=0}^{n}(-1)^{i}\binom{n}{i}\binom{n-1+k-wt({\bf{w}}_{i})}{k-wt({\bf{w}}_{i})}$
$\displaystyle\overset{\eqref{eq:eval}}{=}$
$\displaystyle\\#\\{{\bf{w}}_{i}\mid 0\leq i\leq L,wt({\bf{w}}_{i})=k\\}.$

∎

The proof of Theorem 2 follows from putting together Propositions 2 and 4.

##  5\. Another proof

Let $\Gamma=[p_{1},\cdots,p_{r},-q_{1},\cdots,-q_{s}]$ be a Gamma vector.
Below we modify the notation by letting $p_{r-1+j}=q_{j}$ for $1\leq j\leq s$
with $p_{r+s-1}=1$. Set $n=r+s-1$. Let
$\\{{\bf{e}}_{0},\ldots,{\bf{e}}_{n}\\}$ as before. Note that

$L=p_{1}+\cdots+p_{r}=q_{1}+\cdots+q_{s}$

is equivalent to the single relation satisfied by the vectors
${\bf{e}}_{i},0\leq i\leq n.$

(33)
$p_{1}{\bf{e}}_{1}+\cdots+p_{r-1}{\bf{e}}_{r-1}=p_{r}{\bf{e}}_{r}+\cdots+p_{n}{\bf{e}}_{n}.$

We let $\Delta=\Delta(HD)$ denote the $n$-dimensional polyhedron with the
origin $\mathbf{o}$ and ${\bf{e}}_{j}$, $j=0,\ldots,n$, as vertices.
Furthermore, for a nonempty subset $T$ of

$S:=\\{0,1,\ldots,n\\},$

we let $\Delta_{T}$ denote the $(|T|-1)$-dimensional polytope with vertices
${\bf{e}}_{j}$, where $j\in T$. Also, we let $C(\Delta_{T})$ denote the cone
$C(\Delta_{T}):=\\{\sum_{j\in T}a_{j}{\bf{e}}_{j}:a_{j}\geq 0\\}$. The
notation $C(\Delta)$ is similarly defined. Recall that a nonzero vector $v$ in
$C(\Delta_{T})$ is said to have _weight_ $w$ if $w$ is the smallest real
number such that $v\in w\Delta_{T}$. Since all ${\bf{e}}_{j}$ are lying on the
hyperplane $x_{1}=1$, the weight of $v$ is simply the first coordinate of $v$.
In particular, even though a vector $v$ may belong to several $C(\Delta_{T})$,
the weight of $v$ does not depend on $T$.

For a non-negative integer $k$, let $m(k;T)$ (respectively, $m(k)$) denote the
number of lattice points (points in $\mathbb{Z}^{n}$) of weight $k$ in the
cone $C(\Delta_{T})$ (respectively, $C(\Delta)$) and

(34) $g(x;T)=\sum_{k=0}^{\infty}m(k;T)x^{k}$

(respectively, $g(x)$) be the generating function of $m(k;T)$ (respectively,
$m(k)$).

Let

$S_{+}=\\{0,\ldots,r-1\\},\qquad S_{-}=\\{r,\ldots,r+s-1\\}.$

######  Lemma 1.

We have

$\begin{split}g(x)&=\sum_{k=0}^{r-1}(-1)^{r-1-k}\sum_{T\subset
S_{+},|T|=k}g(x;T\cup S_{-})\\\ &=\sum_{k=0}^{s-1}(-1)^{s-1-k}\sum_{T\subset
S_{-},|T|=k}g(x;S_{+}\cup T).\end{split}$

###### Proof.

Here we prove only the first formula; the proof of the second formula is
similar.

We claim that each nonzero vector $v$ in $C(\Delta)$ has a unique expression
$v=\sum_{j\in T}a_{j}{\bf{e}}_{j}$, $a_{j}>0$ for all $j\in T$, for some
subset $T$ of $S$ such that at least one of $0,\ldots,r-1$ is missing from
$T$.

We first prove the existence of such an expression. Assume that $v$ is a
nonzero vector in $C(\Delta)$. Then $v\in C(\Delta_{T^{\prime}})$ for some
subset $T^{\prime}$ of $S$ of cardinality $\leq n$, say, $v=\sum_{j\in
T^{\prime}}a_{j}{\bf{e}}_{j}$, $a_{j}>0$. If $S_{+}\not\subseteq T^{\prime}$,
then we are done. If $S_{+}\subseteq T^{\prime}$, then setting $c=\min_{j\in
S_{+}}a_{j}/p_{j}$ and using (33), we may write $v$ as

$v=\sum_{j\in S_{+}}(a_{j}-cp_{j}){\bf{e}}_{j}+\sum_{j\in
S_{-}}(a_{j}+cp_{j}){\bf{e}}_{j}$

(we set $a_{j}=0$ in the second sum when $j\notin T^{\prime}$). This is an
expression for $v$ with the claimed property. The uniqueness follows
immediately from the facts that any subset of $S$ of cardinality $\leq n$ is
linearly independent and that all the coefficients $b_{j}$ for $j\in S_{+}$ in
a nontrivial relation $\sum_{j\in S}b_{j}{\bf{e}}_{j}=0$ have the same sign.

Having proved the claim, for a nonzero vector $v$ we let
$\operatorname{supp}(v)$ denote the set $T$ in the claim. We also set
$\operatorname{supp}({\bf o})=\\{\\}$. We now prove the formula.

Given a nonzero vector $v$ in $\mathbb{Z}^{n}\cap C(\Delta)$, we write its
support as $\operatorname{supp}(v)=T_{+}\cup T_{-}$, where
$T_{\pm}=S_{\pm}\cap\operatorname{supp}(v)$. Let $m=|T_{+}|$. We need to show
that the vector $v$ is counted exactly once in the expression

$\sum_{k=0}^{r-1}(-1)^{r-1-k}\sum_{T\subset S_{+},|T|=k}g(x;T\cup S_{-}).$

Indeed, the number of times $v$ is counted in the expression is

$\sum_{k=m}^{r-1}(-1)^{r-1-k}\binom{r-m}{k-m}=(-1)^{r-m-1}\sum_{k=0}^{r-1-m}(-1)^{k}\binom{r-m}{k}=1.$

This proves the formula. ∎

We now determine the generating functions $g(x;T)$.

######  Lemma 2.

For $i\in S$, let $M$ be the $n\times n$ matrix whose rows are $v_{j}$, $j\in
S$, $j\neq i$. Then $|\det M|=p_{j}$.

###### Proof.

If $i=0$, after applying suitable elementary row operations, we see that $\det
M$ is equal to the determinant of the matrix formed by
${\bf{e}}_{1},\ldots,{\bf{e}}_{n-1}$ and $(1-p_{1}-\ldots-
p_{r-1}+p_{r}+\cdots+p_{n-1},0,\ldots,0)=(p_{r},0,\ldots,0)$. Thus, $|\det
M|=p_{r}$. Likewise, if $i=1,\ldots,n$, then $\det M$ is equal to the
determinant of the matrix formed by
${\bf{e}}_{0},\ldots,{\bf{e}}_{i-1},{\bf{e}}_{i+1},\ldots,{\bf{e}}_{n-1}$, and
a vector of the form $(\ast,0,\ldots,0,\pm p_{i},0,\ldots,0)$. Therefore,
$|\det M|=p_{i}$. ∎

######  Lemma 3.

Let $T$ be a proper subset of $S$. Assume that $v\in\mathbb{Z}^{n}\cap
C(\Delta_{T})$, say, $v=\sum_{j\in T}a_{j}{\bf{e}}_{j}$. Then
$a_{j}\in\frac{1}{d_{T}}\mathbb{Z}$ for all $j\in T$, where
$d_{T}=\gcd\\{p_{i}:i\in S\setminus T\\}$.

###### Proof.

For $i=1,\ldots,n+1$, let $T_{i}=S\setminus\\{i\\}$. By Lemma 2, if $T=T_{i}$
for some $i$, then for $\sum_{j\in
T_{i}}a_{j}{\bf{e}}_{j}\in\mathbb{Z}^{n}\cap C(\Delta_{T_{i}})$, we have
$a_{j}\in\frac{1}{p_{i}}\mathbb{Z}$ for all $j$ and the statement holds.

More generally, if $T$ is a proper subset of $S$ and $v=\sum_{j\in
T}a_{j}{\bf{e}}_{j}$ is a lattice point contained in $C(\Delta_{T})$, then $v$
is contained in $T_{i}$ for every $i$ in $S\setminus T$. Then by the result
above, $a_{j}\in\frac{1}{p_{i}}\mathbb{Z}$ for all $j\in T$ and all $i\in
S\setminus T$. Therefore, $a_{j}\in\frac{1}{d_{T}}\mathbb{Z}$ for all $j$. ∎

In the following, for a positive integer $d$ and an integer $m$, we let
$\\{m\\}_{d}$ denote the residue of $m$ modulo $d$. Also, for a real number
$x$, we let $\\{x\\}:=x-\lfloor x\rfloor$ as before. For a rational number
$t$, we define

$w(t):=\sum_{j\in S_{+}}\\{-p_{j}t\\}+\sum_{j\in S_{-}}\\{p_{j}t\\}.$

We remark that by the assumption that $\sum_{j\in S_{+}}p_{j}=\sum_{j\in
S_{-}}p_{j}$, the function $w(t)$ is integer-valued.

######  Lemma 4.

Let $T$ be a proper subset of $S$ and $d:=d_{T}$ be the integer defined in
Lemma 3. Then we have

$g(x;T)=\frac{f(x;T)}{(1-x)^{|T|}},$

where

(35) $f(x;T)=\sum_{c=0}^{d-1}x^{w(c/d)}.$

###### Proof.

We call an element $u$ of $\mathbb{Z}^{n}\cap C(\Delta_{T})$ to be
$T$-primitive (or simply primitive) if $u-{\bf{e}}_{j}\notin\mathbb{Z}^{n}\cap
C(\Delta)$ for any $j\in T$, so every element $v$ of $\mathbb{Z}^{n}\cap
C(\Delta_{T})$ can be written uniquely as $v=u+\sum_{j\in T}a_{j}{\bf{e}}_{j}$
for some $T$-primitive element $u$ and some non-negative integers $a_{j}$.
Then we have

$g(x;T)=\frac{1}{(1-x)^{|T|}}\sum_{u\text{ $T$-primitive}}x^{\text{weight of
}u}.$

Now we determine the $T$-primitive elements of $\mathbb{Z}^{n}\cap
C(\Delta_{T})$ and prove the lemma.

We first consider the case $n\notin T$. Since $p_{n}=1$, we have $d_{T}=1$.
Thus, $f(x;T)=1$. On the other hand, by Lemma 3, $\bf o$ is the only
$T$-primitive element, so $g(x;T)=1/(1-x)^{|T|}$ and the lemma is proved in
this case.

We next consider the case $0,n\in T$. Let $u=\sum_{j\in T}a_{j}{\bf{e}}_{j}$
be a $T$-primitive element. By Lemma 3, $da_{j}$ are integers for all $j\in
T$. Let $c:=da_{n}$. The $T$-primitive condition implies that $0\leq c\leq
d-1$. Furthermore, the assumption that $u\in\mathbb{Z}^{n}$ implies that

$\begin{cases}a_{j}+p_{j}a_{n}\in\mathbb{Z},&\text{if }j\in T_{+},j\neq 0,\\\
a_{j}-p_{j}a_{n}\in\mathbb{Z},&\text{if }j\in T_{-},\end{cases}$

and

$a_{0}+\sum_{j\in T,j\neq 0}a_{j}\in\mathbb{Z}.$

Thus,

$da_{j}=\begin{cases}\\{-p_{j}c\\}_{d},&\text{if }j\in T_{+},j\neq 0,\\\
\\{p_{j}c\\}_{d},&\text{if }j\in T_{-},\end{cases}$

and

$da_{0}=\left\\{c\sum_{j\in T_{+},j\neq 0}p_{j}-c\sum_{j\in
T_{-}}p_{j}\right\\}_{d}.$

This shows that every primitive element is uniquely determined by its
coefficient of ${\bf{e}}_{n}$. It shows also that for each integer $c$ with
$0\leq c\leq d-1$, there is a $T$-primitive element such that the coefficient
of $v_{n}$ is $c/d$. We now compute their weights.

From the computation above, we see that the weight of the $T$-primitive
element $\sum_{j\in T}a_{j}{\bf{e}}_{j}$ with $a_{n+1}=c/d$ is equal to

$\sum_{j\in T}a_{j}=\frac{1}{d}\left(\left\\{c\sum_{j\in T_{+},j\neq
1}p_{j}-c\sum_{j\in T_{-}}p_{j}\right\\}_{d}+\sum_{j\in T_{+},j\neq
0}\\{-p_{j}c\\}_{d}+\sum_{j\in T_{-}}\\{p_{j}c\\}_{d}\right).$

Observing that $d|p_{j}$ for any $j\notin T$, we may rewrite the expression
above as

$\sum_{j\in T}a_{j}=\frac{1}{d}\left(\left\\{c\sum_{j\in S_{+},j\neq
1}p_{j}-c\sum_{j\in S_{-}}p_{j}\right\\}_{d}+\sum_{j\in S_{+},j\neq
0}\\{-p_{j}c\\}_{d}+\sum_{j\in S_{-}}\\{p_{j}c\\}_{d}\right).$

Then from the assumption that $\sum_{j\in S_{+}}p_{j}=\sum_{j\in S_{-}}p_{j}$,
we see that the expression can be simplified to

$\sum_{j\in T}a_{j}=\frac{1}{d}\left(\sum_{j\in
S_{+}}\\{-p_{j}c\\}_{d}+\sum_{j\in S_{-}}\\{p_{j}c\\}_{d}\right)=w(c/d).$

Therefore,

$g(x;T)=\frac{\sum_{c=0}^{d-1}x^{w(c/d)}}{(1-x)^{|T|}}.$

This proves the lemma for the case $0,n\in T$.

Finally, assume that $n\in T$ but $0\notin T$. As before, we find that for
each integer $c$ with $0\leq c\leq d-1$, there is exactly a primitive element
$u=\sum_{j\in T}a_{j}{\bf{e}}_{j}$ with $a_{n+1}=c/d$ and

$da_{j}=\begin{cases}\\{-p_{j}c\\}_{d},&\text{if }j\in T_{+},\\\
\\{p_{j}c\\}_{d},&\text{if }j\in T_{-}.\end{cases}$

Then the weight of $u$ is equal to

$\begin{split}\sum_{j\in T}a_{j}&=\frac{1}{d}\left(\sum_{j\in
T_{+}}\\{-p_{j}c\\}_{d}+\sum_{j\in T_{-}}\\{p_{j}c\\}_{d}\right)\\\
&=\frac{1}{d}\left(\sum_{j\in S_{+}}\\{-p_{j}c\\}_{d}+\sum_{j\in
S_{-}}\\{p_{j}c\\}_{d}\right)=w(c/d).\end{split}$

Again, we have $g(x;T)=f(x;T)/(1-x)^{|T|}$ with $f(x;T)$ given by (35). This
completes the proof of the lemma. ∎

######  Corollary 1.

For a proper subset $T$ of $S$, let $f(x;T)$ be defined by (35). Then the
generating function $g(x)=\sum_{k=0}^{\infty}m(k)x^{k}$ for the number $m(k)$
of points of weight $k$ in $\mathbb{Z}^{n}\cap C(\Delta)$ is equal to
$f(x)/(1-x)^{n}$, where

(36) $\begin{split}f(x)&=\sum_{k=0}^{r-1}(-1)^{r-1-k}\sum_{T\subset
S_{+},|T|=k}f(x;T\cup S_{-})(1-x)^{r-1-k}\\\
&=\sum_{k=0}^{s-1}(-1)^{s-1-k}\sum_{T\subset S_{-},|T|=k}f(x;S_{+}\cup
T)(1-x)^{s-1-k}.\end{split}$

Consequently, the Hodge number multiplicity $H_{\Delta}(k)$ is equal to the
coefficient of $x^{k}$ in $f(x)$.

Example 6.  Let $\Gamma=[s,-1,\cdots,-1]$ with $r=1$, $p_{1}=s$, and
$q_{1}=\cdots=q_{s}=1$, so $n=s$, $S_{+}=\\{0\\}$ and
$S_{-}=\\{1,\ldots,s\\}$. This corresponds to the Dwork family. When $s=3$, as
in Example 1 and when $s=4$, $f_{HD}(\lambda)=0$ is generically a K3 surface.
According to the corollary, we have

$\sum_{k=0}^{\infty}m(k)x^{k}=\frac{f(x;S_{-})}{(1-x)^{s}}.$

It is easy to see that primitive element of $C(\Delta_{S_{-}})$ are

$\frac{c}{s}({\bf{e}}_{1}+\cdots+{\bf{e}}_{s})$

for $c=0,\ldots,s-1$. Thus,

$f(x;S_{-})=1+x+\cdots+x^{s-1}$

Thus, $H_{\Delta}(k)=1$ for $k=0,\ldots,s-1$ and $H_{\Delta}(k)=0$ for $k\geq
s$.

Example 7.  Let $\Gamma=[6,-3,-2,-1]$ with $r=1$, $s=3$, and $n=3$. We have

$\sum_{k=0}^{\infty}m(k)x^{k}=\frac{f(x;S_{-})}{(1-x)^{3}}.$

The primitive elements of $C(\Delta_{S_{-}})$ are

$\frac{1}{6}\left(\\{3c\\}_{6}{\bf{e}}_{1}+\\{2c\\}_{6}{\bf{e}}_{2}+c{\bf{e}}_{3}\right),\qquad
c=0,\ldots,5,$

whose weights are

$\begin{cases}0,&\text{if }c=0,\\\ 1,&\text{if }c=1,2,3,4,\\\ 2,&\text{if
}c=5.\end{cases}$

Thus,

$\sum_{k=0}^{\infty}m(k)x^{k}=\frac{1+4x+x^{2}}{(1-x)^{3}}.$

Example 8.  Let $\Gamma=[5,2,-6,-1]$ with $r=s=2$ and $n=3$. Then the
polynomial $f(x)$ in $\sum_{k=0}^{\infty}m(k)x^{k}=f(x)/(1-x)^{3}$ is

$f(x)=f(x;\\{1,2,3\\})+f(x;\\{1,2,4\\})-f(x;\\{1,2\\})(1-x).$

By Lemma 4, $f(x;\\{1,2,3\\})=f(x;\\{1,2\\})=1$, and

$f(x;\\{1,2,4\\})=\sum_{c=0}^{5}x^{w(c;\\{1,2,4\\})}=1+3x+2x^{2}.$

Thus,

$f(x)=1+(1+3x+2x^{2})-(1-x)=1+4x+2x^{2}.$

We now deduce an alternative expression for $f(x)$, from which we will see
that it is equal to the Hodge polynomial from the zigzag algorithm. Before
that, let us remark that since $w(t)$ is periodic of period $1$, the
polynomial $f(x;T)$ in (35) can also be written as

$f(x;T)=\sum_{c=1}^{d}x^{w(c/d)}.$

We let $\mathscr{T}$ denote the set

$\mathscr{T}:=\\{t\in\mathbb{Q}:0<t\leq 1,~{}p_{j}t\in\mathbb{Z}\text{ for
some }j\in S_{+}\\}.$

For $t\in\mathscr{T}$, we let

$T(t):=\\{j\in S_{+}:p_{j}t\in\mathbb{Z}\\},\qquad m(t):=|T(t)|.$

That is,

$\mathscr{T}=\bigcup_{j\in S_{+}}\left\\{\frac{k}{p_{j}}:k\in\mathbb{Z},1\leq
k\leq p_{j}\right\\},$

$T(t)$ is the set of indices $j$ such that $t$ is contained in the set
$\\{k/p_{j}:k\in\mathbb{Z},1\leq k\leq p_{j}\\}$.

Recall from (36) that

$f(x)=\sum_{k=0}^{r-1}(-1)^{r-1-k}\sum_{T\subset S_{+},|T|=k}f(x;T\cup
S_{-})(1-x)^{r-1-k},$

which is equal to

$\sum_{c/d\in\mathscr{T},(c,d)=1}x^{w(c/d)}\sum_{T\subsetneq S_{+},d|d_{T\cup
S_{-}}}(x-1)^{r-1-|T|}.$

Now for $T\subset S_{+}$, we have $d|d_{T\cup S_{-}}$ if and only if
$T^{\prime}:=S\setminus(T\cup S_{-})$ satisfies $T^{\prime}\subseteq T(c/d)$.
Thus,

(37)
$\begin{split}f(x)&=\sum_{t\in\mathscr{T}}x^{w(t)}\sum_{T^{\prime}\subseteq
T(t),T^{\prime}\neq\emptyset}(x-1)^{|T^{\prime}|-1}\\\
&=\sum_{t\in\mathscr{T}}x^{w(t)}\sum_{j=1}^{m(t)}\binom{m(t)}{j}(x-1)^{j}\\\
&=\sum_{t\in\mathscr{T}}x^{w(t)}(1+x+\cdots+x^{m(t)-1}).\end{split}$

This means that for each fraction $t$ in $\\{k/p_{j}:1\leq k\leq p_{j}\\}$,
$j\in S_{+}$, if we let $m(t)$ denote the number of times it appears in the
sets, then $t$ contributes $m(t)$ terms continually to $f(x)$, starting from
the exponent $w(t)$ and ending at the exponent $w(t)+m(t)-1$. Now we have

(38) $\begin{split}w(t)&=\sum_{j\in S_{+}}(-p_{j}t-\lfloor-
p_{j}t\rfloor)+\sum_{j\in S_{-}}(p_{j}t-\lfloor p_{j}t\rfloor)\\\
&=-\sum_{j\in S_{+}}\lfloor-p_{j}t\rfloor-\sum_{j\in S_{-}}\lfloor
p_{j}t\rfloor.\end{split}$

Using

$\lfloor-
p_{j}t\rfloor=-\\#\left\\{k\in\mathbb{Z}:0<\frac{k}{p_{j}}<t\right\\}-1$

and

$\lfloor p_{j}t\rfloor=\\#\left\\{k\in\mathbb{Z}:0<\frac{k}{p_{j}}\leq
t\right\\},$

we find that

(39) $w(t)=\sum_{j\in
S_{+}}\\#\left\\{k\in\mathbb{Z}:0<\frac{k}{p_{j}}<t\right\\}-\sum_{j\in
S_{-}}\\#\left\\{k\in\mathbb{Z}:0<\frac{k}{p_{j}}\leq t\right\\}+r.$

From this, we see that $f(x)$ is precisely equal to the polynomial from the
zigzag algorithm.

## Acknowledgements

Long is supported in part by the Simons Foundation grant #MP-TSM-00002492 and
the LSU Micheal F. and Roberta Nesbit McDonald Professorship. She is grateful
for the opportunity to visit Academia Sinica in Taiwan in 2023 during when
this paper was written. Yang is supported by Grant 112-2115-M-002-002 of
National Science and Technology Council of Taiwan (R.O.C.).

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