Universal conditions on $h^*$-vectors of lattice simplices

TL;DR

This paper establishes universal conditions on the $h^*$-vectors of lattice simplices, proving existence results under certain vanishing conditions and providing examples to show the necessity of these conditions.

Contribution

It introduces new conditions on $h^*$-vectors of lattice simplices and proves existence results, advancing understanding of their structure.

Findings

Existence of lattice simplices with prescribed $h^*$-vector segments.

Necessary conditions for vanishing $h^*$-vector entries.

Examples demonstrating the sharpness of the conditions.

Abstract

In this paper, we will prove that given a lattice simplex with its $h^*$-polynomial $\sum_{i \geq 0}h_i^*t^i$, if $h_{k+1}^*=\cdots=h_{2k}^*=0$ holds, then there exists a lattice simplex of degree $k$ whose $h^*$-polynomial coincides with $\sum_{i=0}^k h_i^*t^i$. Moreover, we will present the examples showing that the condition $h_{k+1}^*=h_{k+2}^*=\cdots=h_{2k-1}^*=0$ is necessary.

Equations

$$1+\sum _{n=1}^{\infty }|nP\cap {\mathbb{Z}}^d|t^n=\frac{{\sum }_{i=0}^d{h}_i^{\ast }{t}^i}{(1-t{)}^{d+1}},$$

$$h_1^{\ast }\ge h_d^{\ast }.$$

$${\Lambda }_{\Delta }:=\left\{(r_1,\dots ,r_{d+1}):0\le r_j<1\text{ for }\forall j\in [d+1],\sum _{i=1}^{d+1}{r}_i{\mathbf{v}}_i\in {\mathbb{Z}}^d,\sum _{i=1}^{d+1}{r}_i\in \mathbb{Z}\right\}.$$

$$\alpha +\beta =(\{{\alpha }_1+{\beta }_1\},\dots ,\{{\alpha }_{d+1}+{\beta }_{d+1}\})\in {\Lambda }_{\Delta }$$

$${\Lambda }_{\Delta }^{(h)}=\{\alpha \in {\Lambda }_{\Delta }:\mathrm{ht}(\alpha )=h\}.$$

$$h_{\Delta }^{\ast }(t)=\sum _{\alpha \in {\Lambda }_{\Delta }}{t}^{\mathrm{ht}(\alpha )}.$$

$$|\alpha |=\mathrm{ht}(\alpha )+\mathrm{ht}(-\alpha ).$$

$$\mathrm{ht}(\alpha )+\mathrm{ht}(-\alpha )=\sum _{i\in \mathrm{supp}(\alpha )}{\alpha }_i+\sum _{i\in \mathrm{supp}(\alpha )}(1-{\alpha }_i)=|\alpha |.$$

$$\mathrm{ht}(-\alpha )=|\alpha |-\mathrm{ht}(\alpha )\ge k+1.$$

$$\mathrm{ht}(j\alpha )=\sum _{i=1}^{d+1}\{j{\alpha }_i\}\le \sum _{i=1}^{d+1}\{(j-1){\alpha }_i\}+\sum _{i=1}^{d+1}{\alpha }_i=\mathrm{ht}((j-1)\alpha )+h.$$

$$2k+1\le \mathrm{ht}(-\alpha )=|\alpha |-\mathrm{ht}(\alpha )\le |\alpha |-2k-1,$$

$$P\star Q:=\mathrm{conv}(\{(0,\mathbf{x},0_e):\mathbf{x}\in P\}\cup \{(1,0_d,\mathbf{y}):\mathbf{y}\in Q\})\subset {\mathbb{R}}^{d+e+1},$$

$$h_{P\star Q}^{\ast }(t)=h_P^{\ast }(t)h_Q^{\ast }(t).$$

$$\mathrm{conv}(0,{\mathbf{e}}_1,\dots ,{\mathbf{e}}_4,{\mathbf{e}}_1+4{\mathbf{e}}_2+7{\mathbf{e}}_3+8{\mathbf{e}}_4+9{\mathbf{e}}_5\})\subset {\mathbb{R}}^5.$$

Full text

The author is partially supported by JSPS Grant-in-Aid for Young Scientists (B) $\sharp$17K14177.

1. Introduction

A lattice polytope is a convex polytope all of its vertices belong to the standard lattice ${\mathbb{Z}}^{d}$. Let $P\subset{\mathbb{R}}^{d}$ be a lattice polytope $P$ of dimension $d$. Given $n\in{\mathbb{Z}}_{>0}$, we consider the number of lattice points $|nP\cap{\mathbb{Z}}^{d}|$ contained in the $n$-th dilation of $P$. Ehrhart [4] proved that $|nP\cap{\mathbb{Z}}^{d}|$ can be expressed by a polynomial in $n$ of degree $d$ for each $n$, denoted by $E_{P}(n)$. This polynomial $E_{P}(n)$ is called the Ehrhart polynomial of $P$. Moreover, the generating function of $|nP\cap{\mathbb{Z}}^{d}|$ becomes the rational function which is of the form

[TABLE]

where each $h_{i}^{*}$ is an integer. The sequence of integers $h^{*}(P)=(h_{0}^{*},h_{1}^{*},\ldots,h_{d}^{*})$ is called the $h^{*}$-vector (or $\delta$-vector) of $P$, and the polynomial $h_{P}^{*}(t)=h_{0}^{*}+h_{1}^{*}t+\cdots+h_{d}^{*}t^{d}$ is called the $h^{*}$-polynomial (or $\delta$-polynomial) of $P$. We call the degree of the $h^{*}$-polynomial of $P$ the degree of $P$, denoted by $\deg(P)$. Note that the degree of a lattice polytope is at most the dimension of polytope. This implies that lattice polytopes having degree at most $d$ are a kind of generalization of lattice polytopes of dimension $d$.

On the $h^{*}$-vector $h^{*}(P)=(h_{0}^{*},\ldots,h_{d}^{*})$ of a lattice polytope $P$ of dimension $d$, the following facts are well know.

• •

We have $\displaystyle E_{P}(n)=\sum_{i=0}^{d}h_{i}^{*}\binom{n+d-i}{d}$.

• •

Some of $h_{i}^{*}$ has a simple description in terms of $P$, e.g., $h_{0}^{*}=1$, $h_{1}^{*}=|P\cap{\mathbb{Z}}^{d}|-(d+1)$ and $h_{i}^{*}=|(d+1-i)P^{\circ}\cap{\mathbb{Z}}^{d}|$ for each $\deg(P)\leq i\leq d$, where $P^{\circ}$ denotes the relative interior of $P$. In particular, $h_{d}^{*}=|P^{\circ}\cap{\mathbb{Z}}^{d}|$. Hence, we have

[TABLE]

• •

The leading coefficient of $E_{P}(n)$ is equal to $\sum_{i=0}^{d}h_{i}^{*}/d!$ and this coincides with the relative volume of $P$ ([3, Corollaries 3.20, 3.21]). The sum $\sum_{i=0}^{d}h_{i}^{*}$ of $h^{*}$-vector is called the normalized volume of $P$.

• •

Each $h_{i}^{*}$ is nonnegative ([13]).

• •

If $h_{d}^{*}>0$, then we have $h_{i}^{*}\geq h_{1}^{*}$ for each $1\leq i\leq d-1$ ([6]).

For more details on Ehrhart polynomials and $h^{*}$-vectors, consult e.g., [3].

One of the most important problems in Ehrhart theory is the characterization of the polynomials that are the Ehrhart polynomials of some lattice polytopes. Since the Ehrhart polynomial and the $h^{*}$-vector are equivalent, i.e., we can know another one once we know the one, we can rephrase this problem as follows:

Question 1.1** (cf. [2, Question 1.1]).**

Characterize the sequences $(h_{0}^{*},h_{1}^{*},\ldots,h_{d}^{*})$ of nonnegative integers that are the $h^{*}$-vectors of some lattice polytope of dimension $d$.

It is easy to see that $(1,a)$ is always the $h^{*}$-vector of some lattice polytope of dimension $1$ for any nonnnegative integer $a$. The case of $d=2$ is highly non-trivial, but the following is known:

Theorem 1.2** ([12]).**

Let $P$ be a lattice polytope $P$ of dimension $2$ with its $h^{*}$-vector $h^{*}(P)=(1,h_{1}^{*},h_{2}^{*})$. Then one of the following conditions is satisfied:

1. (1)

$h^{*}_{2}=0$; 2. (2)

$h_{2}^{*}\leq h^{*}_{1}\leq 3h^{*}_{2}+3$; 3. (3)

$h^{*}_{1}=7$* and $h^{*}_{2}=1$.*

We can see that the converse of Theorem 1.2 is also true, i.e., we can construct a lattice polytope of dimension $2$ whose $h^{*}$-vector is equal to $(1,h_{1}^{*},h_{2}^{*})$ for each $(h_{1}^{*},h_{2}^{*})$ which satisfies one of the conditions (1), (2) and (3) in Theorem 1.2.

The case $d\geq 3$ is widely open, while some necessary inequalities have been recently conjectured in [1, Conjecture 8.7] in the case of $d=3$.

As a generalization of Theorem 1.2, the following is also known:

Theorem 1.3** ([14, Theorem 2]).**

Let $P$ be a lattice polytope of degree at most two with its $h^{*}$-polynomial $h^{*}_{P}(t)=1+h^{*}_{1}t+h^{*}_{2}t^{2}$. Then one of the following conditions is satisfied:

1. (1)

$h^{*}_{2}=0$; 2. (2)

$h^{*}_{1}\leq 3h^{*}_{2}+3$; 3. (3)

$h^{*}_{1}=7$* and $h^{*}_{2}=1$.*

It is proved in [5, Proposition 1.10] that the converse of Theorem 1.3 is also true. Note that the inequality $h_{1}^{*}\geq h_{2}^{*}$ of Theorem 1.2 comes from (1.1), while it does not appear in Theorem 1.3 since the dimension may be greater than two.

Moreover, as a further generalization of Theorem 1.3, the following is proved:

Theorem 1.4** ([2, Theorem 1.4]).**

Let $P$ be a lattice polytope with its $h^{*}$-polynomial $h_{P}^{*}(t)=\sum_{i\geq 0}h_{i}^{*}t^{i}$. Assume $h^{*}_{3}=0$. Then one of the following conditions is satisfied:

1. (1)

$h^{*}_{2}=0$; 2. (2)

$h^{*}_{1}\leq 3h^{*}_{2}+3$; 3. (3)

$h^{*}_{1}=7$* and $h^{*}_{2}=1$.*

In [2], the inequalities (1), (2) and (3) are called universal because those are valid independently of both the dimension and the degree of lattice polytopes. A proof of Theorem 1.4 is based on the idea of spanning polytopes developed in [11].

The motivation to organize this paper is to find a new kind of universal inequalities or universal conditions.

Definition 1.5**.**

We say that a sequence $(a_{0},a_{1},\ldots,a_{k})$ of nonnegative integers satisfies degree $k$ condition if there exists a lattice polytope $P$ of degree at most $k$ such that $h_{P}^{*}(t)=\sum_{i=0}^{k}a_{i}t^{i}$.

For any lattice polytope $P$ with its $h^{*}$-vector $(h_{0}^{*},h_{1}^{*},\ldots,h_{d}^{*})$, we see that $(h_{0}^{*},h_{1}^{*})$ satisfies degree $1$ conditions. Moreover, Theorem 1.4 says that the assumption “$h_{3}^{*}=0$” implies degree $2$ condition of $(h_{0}^{*},h_{1}^{*},h_{2}^{*})$ of any $h^{*}$-vector $(h_{0}^{*},h_{1}^{*},\ldots,h_{d}^{*})$ of lattice polytopes. Note that the assumption $h_{3}^{*}=0$ is necessary for degree $2$ condition. See [2, Example 1.5].

Hence, it is natural to think of the assumptions which imply degree $k$ condition for $(h_{0}^{*},h_{1}^{*},\ldots,h_{k}^{*})$ of any $h^{*}$-vector $(h_{0}^{*},h_{1}^{*},\ldots,h_{d}^{*})$ of lattice polytopes in the case $k\geq 3$. The main result of this paper is the following:

Theorem 1.6**.**

Let $\Delta$ be a lattice simplex with its $h^{*}$-polynomial $h_{\Delta}^{*}(t)=\sum_{i\geq 0}h_{i}^{*}t^{i}$. Assume that $h_{k+1}^{*}=h_{k+2}^{*}=\cdots=h_{2k}^{*}=0$ for some $k\geq 3$. Then there exists a face $\Delta^{\prime}$ of $\Delta$ such that $h^{*}_{\Delta^{\prime}}(t)=\sum_{i=0}^{k}h_{i}^{*}t^{i}$.

As an immediate corollary, we obtain the following:

Corollary 1.7**.**

Let $\Delta$ be a lattice simplex with $h_{\Delta}^{*}(t)=\sum_{i\geq 0}h_{i}^{*}t^{i}$. Assume that $h_{k+1}^{*}=h_{k+2}^{*}=\cdots=h_{2k}^{*}=0$ for some $k\geq 3$. Then $(h_{0}^{*},h_{1}^{*},\ldots,h_{k}^{*})$ satisfies degree $k$ condition.

A brief organization of this paper is as follows. First, in Section 2, we prepare a material which we will use in the proof of the main results. Next, in Section 3, we give a proof of Theorem 1.6. Finally, in Section 4, we show that the condition $h_{k+1}^{*}=h_{k+2}^{*}=\cdots=h_{2k-1}^{*}=0$ of Theorem 1.6 is necessary for degree $k$ condition. See Proposition 4.3.

2. Preliminary

First of all, we prepare some tools for the computation of $h^{*}$-vectors of lattice simplices.

Let $\Delta\subset{\mathbb{R}}^{d}$ be a lattice simplex of dimension $d$ and let ${\bf v}_{1},\ldots,{\bf v}_{d+1}\in{\mathbb{Z}}^{d}$ be the vertices of $\Delta$. We use the usual notation $[d+1]:=\{1,\ldots,d+1\}$. We define

[TABLE]

We see that $\Lambda_{\Delta}$ is a finite abelian group by its addition

[TABLE]

for $\alpha,\beta\in\Lambda_{\Delta}$, where $\{r\}=r-\lfloor r\rfloor$ for $r\in{\mathbb{R}}$. Note that ${\bf 0}=(0,\ldots,0)\in\Lambda_{\Delta}$ and $-\alpha=(\{1-\alpha_{1}\},\ldots,\{1-\alpha_{d+1}\})\in\Lambda_{\Delta}$ for $\alpha\in\Lambda_{\Delta}$.

We collect the notation concerning $\Lambda_{\Delta}$ which we will use.

• •

For $\alpha\in\Lambda_{\Delta}$, let ${\mathrm{ht}}(\alpha)=\sum_{i=1}^{d+1}\alpha_{i}$.

• •

For $h=0,1,\ldots,d$, let

[TABLE]

Note that $\Lambda_{\Delta}=\bigsqcup_{h=0}^{d}\Lambda_{\Delta}^{(h)}$. The $h^{*}$-polynomial of $\Delta$ can be computed as follows:

[TABLE]

See [3, Corollary 3.11].

• •

For $\alpha\in\Lambda_{\Delta}$, let ${\mathrm{supp}}(\alpha)=\{i\in[d+1]:\alpha_{i}>0\}$ and let $|\alpha|=|{\mathrm{supp}}(\alpha)|$.

• •

For $\Gamma\subset\Lambda_{\Delta}$, let ${\mathrm{supp}}(\Gamma)=\bigcup_{\alpha\in\Gamma}{\mathrm{supp}}(\alpha)$.

We observe that for any $\alpha\in\Lambda_{\Delta}$, we have

[TABLE]

In fact,

[TABLE]

3. A proof of Theorem 1.6

Fix an integer $k$ with $k\geq 3$. In what follows, we consider a lattice simplex $\Delta$ with its $h^{*}$-polynomial $\sum_{i\geq 0}h_{i}^{*}t^{i}$ satisfying $h_{k+1}^{*}=\cdots=h_{2k}^{*}=0$. Namely, we have $\Lambda_{\Delta}^{(k+1)}=\cdots=\Lambda_{\Delta}^{(2k)}=\emptyset$. Our goal is to find a face $\Delta^{\prime}$ of $\Delta$ such that $h_{\Delta^{\prime}}^{*}(t)=\sum_{i=0}^{k}h_{i}^{*}t^{i}$.

Lemma 3.1**.**

For any $\alpha\in\Lambda_{\Delta}^{(h)}$ and $0\leq h\leq k$, we have $|\alpha|\leq k+h$.

Proof.

Take $\alpha\in\Lambda_{\Delta}^{(h)}$ with $0\leq h\leq k$. Then ${\mathrm{ht}}(\alpha)=h$.

Suppose that $|\alpha|\geq k+h+1$. Then it follows from (2.2) that

[TABLE]

Hence ${\mathrm{ht}}(-\alpha)\geq 2k+1$ by our assumption. Let $m$ denote the order of $\alpha$, i.e., $m\alpha={\bf 0}$. Note that $-\alpha=(m-1)\alpha$. Take any $1\leq j\leq m-1$. Then we see that

[TABLE]

On the one hand, ${\mathrm{ht}}(\alpha)=h\leq k$. On the other hand, ${\mathrm{ht}}((m-1)\alpha)\geq 2k+1$. This means together with ${\mathrm{ht}}(j\alpha)\leq{\mathrm{ht}}((j-1)\alpha)+k$ that there is $1<j^{\prime}<m-1$ with $k+1\leq{\mathrm{ht}}(j^{\prime}\alpha)\leq 2k$, a contradiction. ∎

Lemma 3.2**.**

Let $\Lambda^{\prime}=\{\alpha\in\Lambda_{\Delta}:{\mathrm{ht}}(\alpha)\leq k\}$. Then the following statements hold:

• (a)

$\Lambda^{\prime}$* is a subgroup of $\Lambda_{\Delta}$;*

• (b)

$|{\mathrm{supp}}(\Lambda^{\prime})|\leq 4k-1$.

Proof.

(a) In general, one sees that ${\mathrm{ht}}(\alpha+\beta)\leq{\mathrm{ht}}(\alpha)+{\mathrm{ht}}(\beta)$ for $\alpha,\beta\in\Lambda_{\Delta}$. Since ${\mathrm{ht}}(\alpha+\beta)\leq 2k$ for any $\alpha,\beta\in\Lambda^{\prime}$, one has ${\mathrm{ht}}(\alpha+\beta)\leq k$. Thus, $\alpha+\beta\in\Lambda^{\prime}$. Moreover, we see from Lemma 3.1 that one has ${\mathrm{ht}}(-\alpha)=|\alpha|-{\mathrm{ht}}(\alpha)\leq k$ for $\alpha\in\Lambda^{\prime}$. Hence, $-\alpha\in\Lambda^{\prime}$. Clearly, ${\bf 0}\in\Lambda^{\prime}$ since ${\mathrm{ht}}({\bf 0})=0$. Therefore, $\Lambda^{\prime}$ is a subgroup of $\Lambda_{\Delta}$.

(b) Let $s=\max\{{\mathrm{ht}}(\alpha):\alpha\in\Lambda^{\prime}\}$. Then $s\leq k$. By [10, Lemma 2.2 and Proposition 2.3 (a)], we obtain that $|{\mathrm{supp}}(\Lambda^{\prime})|\leq 2(2s)-1\leq 4k-1$. ∎

Now we are ready to give a proof of Theorem 1.6.

Proof of Theorem 1.6.

Work with the same notation as in Lemma 3.2. Let $\Delta^{\prime}={\mathrm{conv}}(\{v_{i}:i\in{\mathrm{supp}}(\Lambda^{\prime})\})$. Then $\Delta^{\prime}$ is a face of $\Delta$. Note that ${\mathrm{supp}}(\Lambda_{\Delta^{\prime}})={\mathrm{supp}}(\Lambda^{\prime})\subset[d+1]$ by the definition. In what follows, we prove that $h_{\Delta^{\prime}}^{*}(t)=\sum_{i=0}^{k}h_{i}^{*}t^{i}$. Our goal is to show that $\Lambda_{\Delta^{\prime}}=\Lambda^{\prime}$. Clearly, we have $\Lambda^{\prime}\subset\Lambda_{\Delta^{\prime}}$, where we regard $\Lambda_{\Delta^{\prime}}$ as a subgroup of $\Lambda_{\Delta}$. We prove that for any $\alpha\in\Lambda_{\Delta}\setminus\Lambda^{\prime}$, we have $\alpha\not\in\Lambda_{\Delta^{\prime}}$.

Let $\alpha\in\Lambda_{\Delta}\setminus\Lambda^{\prime}$. Then ${\mathrm{ht}}(\alpha)\geq 2k+1$. If ${\mathrm{ht}}(-\alpha)\leq k$, then ${\mathrm{ht}}(\alpha)\leq k$ by Lemma 3.2 (a), a contradiction. Thus, ${\mathrm{ht}}(-\alpha)\geq 2k+1$. From (2.2), we see that

[TABLE]

i.e., $|\alpha|\geq 4k+2$. Now, Lemma 3.2 (b) says that $|\gamma|\leq 4k-1$ for any $\gamma\in\Lambda^{\prime}$. Since $|\alpha|\geq 4k+2>4k-1$, one sees that ${\mathrm{supp}}(\alpha)\not\subset{\mathrm{supp}}(\Lambda^{\prime})$. Since ${\mathrm{supp}}(\Lambda^{\prime})={\mathrm{supp}}(\Lambda_{\Delta^{\prime}})$, we conclude that $\alpha\not\in\Lambda_{\Delta^{\prime}}$, as required. ∎

4. Examples showing that $h_{k+1}^{*}=h_{k+2}^{*}=\cdots=h_{2k-1}^{*}=0$ is necessary

We will provide the example which shows that Theorem 1.6 is not true if we drop the assumption $h_{j}^{*}=0$ for each $k+1\leq j\leq 2k-1$ (Proposition 4.3).

We recall a useful construction of lattice polytopes. Let $P\subset{\mathbb{R}}^{d}$ and $Q\subset{\mathbb{R}}^{e}$ be lattice polytopes. We define the join of $P$ and $Q$:

[TABLE]

where ${\bf 0}_{d}$ (resp. ${\bf 0}_{e}$) denotes the origin of ${\mathbb{R}}^{d}$ (resp. ${\mathbb{R}}^{e}$). Note that $P\star Q$ is a simplex if and only if so are $P$ and $Q$. It is known from [5, Lemma 1.3] that

[TABLE]

Lemma 4.1**.**

For any positive integers $a,b,k,\ell$, there exists a lattice simplex $\Delta$ with its $h^{*}$-polynomial $1+at^{k}+bt^{\ell}+(a+b)t^{k+\ell}$.

Proof.

For positive integers $c$ and $m$, let $\Delta_{c,m}$ be a lattice simplex with its $h^{*}$-polynomial $1+ct^{m}$. Such a simplex is given in e.g., [3, Example 3.22]. A required simplex can be obtained by $\Delta_{a,k}\star\Delta_{b,\ell}$. ∎

Lemma 4.2**.**

For any prime number $p$ with $p\geq 5$ and integers $i,j$ with $2\leq i<j$, there exists no lattice polytope whose $h^{*}$-polynomial is $1+t^{i}+(p-2)t^{j}$.

Proof.

Suppose that there is a lattice polytope $P$ with $h^{*}_{P}(t)=1+t^{i}+(p-2)t^{j}$. Since the linear term of $h_{P}^{*}(t)$ vanishes, $P$ must be a simplex. Now, [9, Theorem 1.1 (a)] says that the $h^{*}$-polynomial with a prime normalized volume should satisfy $h_{i+1}^{*}=h_{s-i}^{*}$ for any $i=0,1,\ldots,s$, where $s$ is the degree of lattice polytope. However, since $p\geq 5$, this never happens, a contradiction. ∎

Proposition 4.3**.**

For any $k\geq 3$ and $k+1\leq j\leq 2k-1$, there exists a lattice simplex $\Delta$ with $h^{*}_{\Delta}(t)=\sum_{i\geq 0}h_{i}^{*}t^{i}$ such that

• •

$h_{k+1}^{*}=\cdots=h_{j-1}^{*}=h_{j+1}^{*}=\cdots=h_{2k-1}^{*}=0$,

• •

$h_{j}^{*}\neq 0$* and*

• •

$(h_{0}^{*},\ldots,h_{k}^{*})$* does not satisfy degree $k$ condition.*

Proof.

When $j\geq k+2$, let $a=j-k$ and $b=k$. Note that $2\leq a\leq k-1$. Then Lemma 4.1 says that there exists a lattice simplex $\Delta$ with its $h^{*}$-polynomial $1+t^{a}+(p-2)t^{b}+(p-1)t^{a+b}=1+t^{j-k}+(p-2)t^{k}+(p-1)t^{j}$, where we let $p$ be a prime number with $p\geq 5$. Therefore, by Lemma 4.2, this $h^{*}$-polynomial enjoys the required properties.

In the case $j=k+1$ and $k\geq 4$, there exists a lattice simplex $\Delta$ with its $h^{*}$-polynomial $1+t^{2}+(p-2)t^{k-1}+(p-1)t^{k+1}$ by Lemma 4.1, where we let $p$ be a prime number with $p\geq 5$, and this $h^{*}$-polynomial enjoys the required properties by Lemma 4.2.

In the case $j=k+1$ and $k=3$, consider the lattice simplex

[TABLE]

Then the $h^{*}$-polynomial of this simplex is equal to $1+2t^{2}+4t^{3}+2t^{4}$. Since $(1,0,2,4)$ does not satisfy degree $3$ condition by Lemma 4.2, we conclude that this $h^{*}$-polynomial enjoys the required properties. ∎

Proposition 4.3 says that we cannot drop the assumption $h_{j}^{*}=0$ in Corollary 1.7 (Theorem 1.6, too) for any $k+1\leq j\leq 2k-1$.

Remark 4.4**.**

There exists a lattice simplex $\Delta$ of dimension $3k-1$ such that $h_{\Delta}^{*}(t)=1+t^{k}+t^{2k}$ for any $k\geq 2$. In fact, we may set $\Delta=\mathrm{conv}(\{{\bf 0},{\mathbb{e}}_{1},\ldots,{\mathbb{e}}_{d-1},2({\mathbb{e}}_{1}+\cdots+{\mathbb{e}}_{d-1})+3{\mathbb{e}}_{d})\})$, where $d=3k-1$ and ${\mathbb{e}}_{1},\ldots,{\mathbb{e}}_{d}$ denote the unit vectors of ${\mathbb{R}}^{d}$. Moreover, we see from [7, Section 4.2] that this is the only lattice simplex of dimension $3k-1$ up to unimodular equivalence whose $h^{*}$-polynomial is $1+t^{k}+t^{2k}$. On the other hand, we see that this $h^{*}$-vector is shifted symmetric (see [8]), i.e., $h_{i+1}^{*}=h_{d-i}^{*}$ for each $0\leq i\leq d-1$. Thus, the $h^{*}$-polynomial of every face of $\Delta$ is $1$ by [8, Theorem 2.1].

Therefore, Theorem 1.6 is not true under the assumption $h_{k+1}^{*}=\cdots=h_{2k-1}^{*}=0$, i.e., the assumption $h_{2k}^{*}=0$ is necessary.

However, Corollary 1.7 might be still true even if $h_{2k}^{*}\neq 0$. In fact, $1+t^{k}$ is a possible $h^{*}$-polynomial of a lattice simplex $\Delta_{1,k}$.

5. Future Questions

Finally, we suggest two questions:

Question 5.1**.**

Let $P$ be a lattice polytope with $h_{P}^{*}(t)=\sum_{i\geq 0}h_{i}^{*}t^{i}$. Assume that $h_{k+1}^{*}=h_{k+2}^{*}=\cdots=h_{2k}^{*}=0$. Then does $(h_{0}^{*},h_{1}^{*},\ldots,h_{k}^{*})$ satisfy degree $k$ condition?

Question 5.2**.**

Let $\Delta$ be a lattice simplex with $h_{\Delta}^{*}(t)=\sum_{i\geq 0}h_{i}^{*}t^{i}$. Assume that $h_{k+1}^{*}=h_{k+2}^{*}=\cdots=h_{2k-1}^{*}=0$. Then does $(h_{0}^{*},h_{1}^{*},\ldots,h_{k}^{*})$ satisfy degree $k$ condition? Is this also true for polytopes?