Graded Cohen–Macaulay Domains and Lattice Polytopes with Short h-Vector

Abstract

Let P be a lattice polytope with the $h^{*}$-vector $(1, h^*_1, \ldots , h^*_s)$. In this note we show that if $h_s^* \le h_1^*$, then the Ehrhart ring ${\mathbb {k}}[P]$ is generated in degrees at most $s-1$ as a ${\mathbb {k}}$-algebra. In particular, if $s=2$ and $h_2^* \le h_1^*$, then P is IDP. To see this, we show the corresponding statement for semi-standard graded Cohen–Macaulay domains over algebraically closed fields.

1 Introduction

Let $R=\bigoplus _{i \in \mathbb {N}} R_i$ be a noetherian graded commutative ring. Throughout the paper, we assume that ${\mathbb {k}}:= R_0$ is a field. If $R ={\mathbb {k}}[R_1]$, that is, R is generated by $R_1$ as a ${\mathbb {k}}$-algebra, we say R is standard graded. If R is finitely generated as a ${\mathbb {k}}[R_1]$-module, we say R is semi-standard graded.

If R is a semi-standard graded ring of Krull dimension d, its Hilbert series is of the form

$$\begin{aligned}\sum _{i \in \mathbb {N}} (\dim _{{\mathbb {k}}}R_i)\cdot t^i =\frac{h_0+h_1 t + \cdots +h_st^s}{(1-t)^d}\end{aligned}$$

for some integers $h_0, h_1, \ldots , h_s$ with $\sum _{i=0}^s h_i \ne 0$ and $h_s \ne 0$. We call the vector $(h_0,h_1,\ldots , h_s)$ the h-vector of R. We always have $h_0=1$ and $\deg R=\sum _{i=0}^s h_i$. If R is Cohen–Macaulay, then $h_i \ge 0$ for all i. We have the following.

Theorem 1.1

Let R be a semi-standard graded Cohen–Macaulay domain with the h-vector $(h_0, h_1, h_2)$. Assume that ${\mathbb {k}}$ is algebraically closed. If $h_2 \le h_1$, then R is standard graded. If further $h_2 < h_1$ and ${\text {char}}{\mathbb {k}}=0$, then R is Koszul.

In fact, Theorem 2.1 states that, under the same situation, if the h-vector of R is $(h_0,h_1,\ldots , h_s)$ with $h_s \le h_1$, then R is generated in degrees at most $s-1$ as a ${\mathbb {k}}$-algebra. So the first statement of Theorem 1.1 follows. To see the latter statement, we can use [5, Thm. 5.2 (1)], since we know R is standard graded now.

An important class of semi-standard graded Cohen–Macaulay domains are the Ehrhart rings of lattice polytopes, which we now recall. Let $P \subset \mathbb {R}^d$ be a lattice polytope. Its Ehrhart ring ${\mathbb {k}}[P]$ is the monoid algebra of the monoid of lattice points in the cone $C = {{\,\mathrm{cone}\,}}{(\{1\} \times P)}\subset \mathbb {R}^{d+1}$ over P. The additional coordinate in the construction of C yields a natural grading on ${\mathbb {k}}[P]$, such that ${\mathbb {k}}[P]$ is semi-standard graded, and its Hilbert series is the Ehrhart series of P. In particular, the h-vector of ${\mathbb {k}}[P]$ is the $h^*$ -vector of P (for general information about this notion and its background, see [1]). Hence the Krull dimension $\dim {\mathbb {k}}[P]$ equals $\dim P+1$.

It is well known that ${\mathbb {k}}[P]$ is a normal domain, and by Hochster’s Theorem [12, Thm. 1], it is Cohen–Macaulay. We refer the reader to the monograph by Bruns and Gubeladze [2] for more information on Ehrhart rings. The index of the last non-zero entry of the $h^*$-vector is called the degree of P. We always have $\deg P \le \dim P$. The $h^*$-vector of P is sometimes denoted by $(h_0^*, h_1^*, \ldots ,h^*_{\dim P})$, even if $\deg P<\dim P$. In this case, $h_i^*=0$ for all $i > \deg P$. We also remark that there is no direct relation between $\deg P$ and $\deg {\mathbb {k}}[P]=\sum _{i=0}^d h_i^*$, the latter being the multiplicity of ${\mathbb {k}}[P]$, which also equals the normalized volume ${{\,\mathrm{Vol}\,}}P$ of P.

A lattice polytope P is called IDP (an abbreviation for “integer decomposition property”) if for every $k \in \mathbb {N}$ and every lattice point $p \in kP \cap \mathbb {Z}^d$, there exist k lattice points $p_1, \dotsc , p_k \in P \cap \mathbb {Z}^d$ with $p = \sum _i p_i$. Clearly, P is IDP if and only if ${\mathbb {k}}[P]$ is standard graded. Hence we obtain the following (here we do not have to assume that ${\mathbb {k}}$ is algebraically closed, since we can replace ${\mathbb {k}}[P]$ by ${\overline{{\mathbb {k}}}}[P] \cong {\overline{{\mathbb {k}}}} \otimes _{\mathbb {k}}{\mathbb {k}}[P]$):

Corollary 1.2

Let $P \subset \mathbb {R}^d$ be a lattice polytope of degree 2 with $h^*$-vector $(1, h^*_1, h^*_2)$. If $h^*_2 \le h^*_1$, then P is IDP. If further $h^*_2 < h^*_1$, the Ehrhart ring ${\mathbb {k}}[P]$ is Koszul for a filed ${\mathbb {k}}$ of characteristic 0.

Note that if $P \subset \mathbb {R}^2$ is a lattice polygon, then it has degree at most 2, and it always satisfies $h^*_2 \le h^*_1$. Therefore, the former half of this corollary can be seen as an extension of the well-known fact that lattice polygons are IDP. See also Remark 3.3 (i) below. We give an example to show that the bound in Corollary 1.2 is sharp:

Example 1.3

Let P be the 3-simplex with vertices

$$\begin{aligned}\begin{pmatrix}0\\ 0\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}, \text { and }\begin{pmatrix}1\\ 1\\ 2\end{pmatrix}.\end{aligned}$$

It is Reeve’s simplex (cf. [1, Exam. 3.22]) and its $h^*$-vector is (1, 0, 1). It is not IDP, and hence the bound $h^{*}_2 \le h^{*}_1$ is sharp.

In the latter part of the present paper, we take a more combinatorial approach. Especially, keeping Corollary 1.2 in mind, we analyze the results of [9], and give several (counter)examples of the related statements.

2 Proofs of the Main Results

Let $R=\bigoplus _{i \in \mathbb {N}} R_i$ be a noetherian graded commutative ring such that ${\mathbb {k}}:= R_0$ is an algebraically closed field. We are going to regard R as a module over $S := {{\,\mathrm{Sym}\,}}_{\mathbb {k}}R_1$. Note that S is isomorphic to the polynomial ring ${\mathbb {k}}[x_1, \ldots , x_n]$ with $n=\dim _{\mathbb {k}}R_1$. Moreover, R is standard graded if and only if R is a quotient ring of S, and R is semi-standard graded if and only if R is finitely generated as an S-module. For a finitely generated graded S-module M, $i \in \mathbb {N}$ and $j \in \mathbb {Z}$, set

$$\begin{aligned} \beta _{i,j}^S(M):= \dim _{\mathbb {k}}[{\text {Tor}}_i^S({\mathbb {k}}, M)]_j. \end{aligned}$$

In particular, $\beta _{0,j}^S(M)$ is the number of S-module generators for M in degree j.

2.1 A Bound on the Degrees of the Generators

Assume that R is semi-standard graded, and has the h-vector $(h_0, h_1, \ldots , h_s)$. The goal of this section is to obtain a bound on the degrees of the generators of R as an S-module. If R is Cohen–Macaulay, it is well known that the generators have degree at most s. Our result is a sufficient criterion when this bound can be improved by one:

Theorem 2.1

Let R be a semi-standard graded Cohen–Macaulay domain with h-vector $(h_0, h_1, h_2, \dotsc , h_s)$, and $S := {{\,\mathrm{Sym}\,}}_{\mathbb {k}}R_1$. Then it holds that

$$\begin{aligned} \beta ^S_{p,p+s}(R) = 0\quad \ \text { for }\ \ 0 \le p \le h_1 - h_s.\end{aligned}$$

In particular, if $h_s \le h_1$, then R is generated by elements of degree $\le s-1$ as an S-module, and hence as a ${\mathbb {k}}$-algebra.

Note that Theorem 1.1 amounts to the special case $s = 2$ and $p = 0$. This result and its proof have been inspired by Green’s Theorem of the Top Row, [7, Thm. 4.a.4]. For the proof of Theorem 2.1, we are going to use the following version of Green’s vanishing theorem:

Theorem 2.2

([6, Thm. 1.1]) Let $\mathfrak {p}\subset S$ be a homogeneous prime ideal, which does not contain any linear forms. Let M be a torsion-free finitely generated graded $S/\mathfrak {p}$-module and let $q \in \mathbb {Z}$ be the minimal integer such that $M_q \ne 0$. Then it holds that

$$\begin{aligned} \beta ^S_{p,p+q}(M) = 0 \quad \ \text { for }\ \ p \ge \dim _{\mathbb {k}}M_q. \end{aligned}$$

In addition, we need the following result:

Theorem 2.3

([2, Thm. 6.18], [3, Thm. 4.4.5]) Let M be a finitely generated graded Cohen–Macaulay module over $S={\mathbb {k}}[x_1, \ldots , x_n]$ with $d = \dim M$. Define $M' := {{\,\mathrm{Ext}\,}}_S^{n - d}(M, \omega _S)$. Then

• (i) $M'$ is also Cohen–Macaulay, ${{\,\mathrm{Ann}\,}}M' = {{\,\mathrm{Ann}\,}}M$, and $M'' \cong M$;

• (ii) $\beta ^S_{p,q}(M') = \beta ^S_{n - d - p, n - q}(M)$;

• (iii) $H_{M'}(t) = (-1)^{d}H_M(t^{-1})$.

Here, $\omega _S:=S(-n)$ denotes the canonical module of S, and $H_M(t)$ denotes the Hilbert series $\sum _{i\in \mathbb {Z}}(\dim _{\mathbb {k}}M_i)\cdot t^i$ of M.

Proof of Theorem 2.1

Let $d := \dim R$. By [3, Prop. 3.6.12], $\omega _R := {{\,\mathrm{Ext}\,}}_S^{n-d}(R, \omega _S)$ is a canonical module for R. Using Theorem 2.3 with $M=R$ and $M' =\omega _R$, we have

$$\begin{aligned} \beta ^S_{p,p+q}(R) = \beta ^S_{n-d-p,n-(p+q)}(\omega _R) = \beta ^S_{n-d-p,(n-d-p)+ (d-q)}(\omega _R).\end{aligned}$$

and the Hilbert series of $\omega _R$ is

$$\begin{aligned}\frac{ h_s t^{d-s}+ h_{s-1} t^{d-s+1}+ \cdots + h_0 t^d}{(1-t)^d}.\end{aligned}$$

In particular, $\omega _R$ has no elements in degrees below $d-s$ and we have that $\dim _{\mathbb {k}}(\omega _R)_{d-s} = h_s$. Now, since R is a domain and $\omega _R$ is a canonical module, it is torsion-free over R by Theorem 2.3 (i), and thus it satisfies the hypothesis of Theorem 2.2. Applying that result to $\omega _R$ (note that $q=d-s$ in the notation there) yields that

$$\begin{aligned}\beta ^S_{p,p+s}(R) = \beta ^S_{n-d-p, (n-d-p) + (d-s)}(\omega _R) = 0 \ \quad \text { if } \ \ (n-d) - p \ge h_s.\end{aligned}$$

Finally, note that $h_1 = n - d$, and the proof is completed. $\square$

Fig. 1

The poset P of Remark 2.4

Remark 2.4

Another important example of a semi-standard graded ring appearing in combinatorial commutative algebra is the face ring $A_P$ of a simplicial poset P. See [15] for details. For the simplicial poset P given in Fig. 1, we have

$$\begin{aligned}A_P \cong \frac{{\mathbb {k}}[x,y,z,u,v]}{(xz, uz, vz, uv, xy-u-v)},\end{aligned}$$

where $\deg x =\deg y=\deg z=1$ and $\deg u=\deg v=2$, and ${\widehat{0}}, w \in P$ correspond to $1, yz \in A_P$, respectively. It is easy to see that $A_P$ is a 2-dimensional Cohen–Macaulay reduced semi-standard graded ring with the h-vector (1, 1, 1), but it is not standard graded. It means that Theorem 2.1 indeed requires the assumption that R is a domain.

3 Further Discussion on Ehrhart Rings

3.1 Direct Applications of Theorem 2.1

We now apply Theorem 2.1 in the setting of Ehrhart theory. Let $P \subset \mathbb {R}^n$ be a lattice polytope. We write $M(P)\subset \mathbb {Z}^{n+1}$ for the affine monoid generated by the lattice points in $\{1\} \times P\subset \mathbb {R}^{n+1}$, and ${\widehat{M}}(P)\subset \mathbb {Z}^{n+1}$ for its integral closure inside $\mathbb {Z}^{n+1}$. Let $R={\mathbb {k}}[P]$ be the Ehrhart ring of P and ${\mathbb {k}}[R_1]$ its subalgebra generated by $R_1$. Then R and ${\mathbb {k}}[R_1]$ are the monoid algebras of the monoids ${\widehat{M}}(P)$ and $M(P)$, respectively. It is well known that ${\widehat{M}}(P)$ is generated by elements of degree at most $\min { \{ \deg P, \dim P-1 \}}$ as a module over $M(P)$ (cf. [2, Thm. 2.52]). Equivalently, $R={\mathbb {k}}[P]$ is generated by elements of at most that degree as a ${\mathbb {k}}[R_{1}]$-module, and hence in particular as a ${\mathbb {k}}$-algebra.

Since ${\mathbb {k}}[P]$ is always Cohen–Macaulay, Theorem 2.1 allows us to improve this bound under an additional assumption as follows. Clearly, this generalizes Corollary 1.2.

Corollary 3.1

Suppose $P \subset \mathbb {R}^n$ is a lattice polytope of degree s with $h^*$-vector $(1, h^*_1, \dotsc , h^*_s)$. If $h^*_s \le h^*_1$, then ${\widehat{M}}(P)$ is generated by elements of degrees $\le s - 1$ as an $M(P)$-module.

Let $P^\circ$ be the relative interior of P. The lattice points in $P^\circ$ are closely related to the canonical module of ${\mathbb {k}}[P]$ (cf. [3, Thm. 6.3.5 (b)]). In general, it holds that $h^{*}_{\dim P} \le h^{*}_1$ (because $h^{*}_{\dim P} = \#\,(P^\circ \cap \mathbb {Z}^n) \le \#\,(P \cap \mathbb {Z}^n) - (\dim P +1) = h^{*}_1$), therefore this corollary extends the above mentioned fact that ${\widehat{M}}(P)$ is generated in degrees at most $\min {\{ \deg P, \dim P-1 \}}$.

If P is IDP, then $R = {\mathbb {k}}[P]$ is the quotient ring of $S = {{\,\mathrm{Sym}\,}}_{\mathbb {k}}R_1$ by a certain prime ideal $I \subseteq S$, which is called the toric ideal of P. It is known that I is generated by polynomials of degree at most $\deg P + 1\le \dim P+ 1$ (Sturmfels, cf. [2, Cor. 7.27]), and again we can improve these bounds by one:

Corollary 3.2

Let P be an IDP lattice polytope and let $I \subset S$ be its toric ideal.

• (i) If $h^*_s \le h^*_1 - 1$, then I is generated in degrees $\le \deg P$.

• (ii) If P is not a clean simplex, then I is generated in degrees $\le \dim P$.

Recall that a clean simplex is a lattice simplex where the only lattice points on its boundary are the vertices.

Proof

   

• (i) Apply Theorem 2.1 to $R = {\mathbb {k}}[P]$ with $p = 1$.

• (ii) If I has a generator in degree $\dim P+ 1$, then it holds that $\deg P = \dim P$ by the above mentioned fact that I is generated in degrees at most $\deg P + 1\le \dim P+ 1$. Moreover, the hypothesis of part (i) needs to be violated, hence it holds that $h^{*}_{\dim P} \ge h^{*}_1$. It follows that $h^{*}_{\dim P} = h^{*}_1$, which is equivalent to P being a clean simplex. $\square$

Remark 3.3

   

• (i) Let $P \subset \mathbb {R}^2$ be a lattice polygon. Then we have $\deg P \le 2$ and P is IDP. Moreover, Koelman [13] showed that the toric ideal of P is generated by quadrics if and only if $h^*_2 < h^*_1$. Hence Corollary 3.2 (i) is an extension of one implication of his result. In particular, the result of [13] shows that the bound $h^*_2 < h^*_1$ is sharp.

• (ii) In [14], Schenck also applied the theory of Green to the study of Ehrhart rings ${\mathbb {k}}[P]$. However, the focus of [14] is different from ours. More precisely, he always assumed that ${\mathbb {k}}[P]$ is standard graded (i.e., P is IDP), and treated the case the toric ideal is generated by quadrics.

3.2 Combinatorial Proofs

Corollary 1.2 is a purely combinatorial statement, and hence one might hope for a combinatorial proof. As a first step, we prove a weak variant of Corollary 1.2 which admits an elementary proof.

We remind the reader that a lattice polytope $P \subset \mathbb {R}^n$ is called spanning [11], if $( \{1\} \times P) \cap \mathbb {Z}^{n+1}$ generates the lattice $\mathbb {Z}^{n+1}$. Every IDP polytope is spanning, but the converse is far from being true. Algebraically, for the Ehrhart ring $R={\mathbb {k}}[P]$, P is spanning if and only if the field of fractions of R coincides with that of ${\mathbb {k}}[R_{1}]$.

Proposition 3.4

Let $P \subset \mathbb {R}^n$ be a d-dimensional lattice polytope with $h^{*}$-vector $(h^{*}_0, h^{*}_1, \dotsc )$. If $h^{*}_1 + h^{*}_d \ge \sum _{i=2}^{d-1}h^{*}_i$, then P is spanning.

Proof

Let L be the lattice generated by the lattice points in P, and q the index of L in $\mathbb {Z}^n$. Further, let ${\widetilde{P}}$ be the polytope P considered in the lattice L (see [11]). We write ${\widetilde{h}}^{*}$ for the $h^{*}$-vector of ${\widetilde{P}}$. It holds that

$$\begin{aligned} \sum _{i=0}^d h^{*}_i = {{\,\mathrm{Vol}\,}}P= q {{\,\mathrm{Vol}\,}}{\widetilde{P}}= q \sum _{i=0}^d {\widetilde{h}}^{*}_i. \end{aligned}$$

(1)

Moreover, it holds that $h^{*}_1={\widetilde{h}}^{*}_1$, $h^{*}_d={\widetilde{h}}^{*}_d$, and $h^{*}_i \ge {\widetilde{h}}^{*}_i$ for $1 \le i \le d$ (see [11, Sect. 3.2]). Now (1) and the assumption that $\sum _{i=2}^{d-1} h^{*}_i \le h^{*}_1 + h^{*}_d$ imply that

$$\begin{aligned} 0 \le q \sum _{i=2}^{d-1} {\widetilde{h}}^{*}_i&=\sum _{i=2}^{d-1} h^{*}_i - (q-1)(1+h^{*}_1+h^{*}_d) \\&\le h^{*}_1 + h^{*}_d - (q-1)(1+h^{*}_1+h^{*}_d) = (2-q) (1+h^{*}_1 + h^{*}_d) -1. \end{aligned}$$

Since $1+h^{*}_1 + h^{*}_d \ge 1$, we have $q=1$, and it means that P is spanning. $\square$

In the next corollary, (i) is just a weak version of Corollary 1.2, but (ii) and (iii) are new.

Corollary 3.5

With the above notation, the following hold:

• (i) If $\deg P = 2$ and $h^{*}_1 \ge h^{*}_2$, then P is spanning.

• (ii) If $\dim P= 3$ and $h^{*}_1 + h^{*}_3 \ge h^{*}_2$, then P is spanning.

• (iii) If $\dim P = 4$, $\deg P\ge 3$, and $h^{*}_1 + h^{*}_4 \ge h^{*}_2 + h^{*}_3$, then P is spanning. In this case, it holds that $h^{*}_1 = h^{*}_2 = h^{*}_3 = h^{*}_4$.

Proof

Only the very last statement is not immediate from Proposition 3.4. If $\dim P = 4$, then $h^{*}_4 \le h^{*}_1$. By assumption and Proposition 3.4, P is spanning, and hence by [10, Thm. 1.4] it holds that $h^{*}_1 \le h^{*}_i$ for $1 \le i < \deg P$. As $\deg P\ge 3$ it holds that $h^{*}_1 \le h^{*}_2$ and $h^{*}_3 > 0$, and thus $h^{*}_1 < h^{*}_2 + h^{*}_3$. It follows that $h^{*}_4 \ne 0$, so $\deg P= 4$. Hence we have that $h^{*}_4 \le h^{*}_1 \le h^{*}_2$ and $h^{*}_1 \le h^{*}_3$. These inequalities, together with $h^{*}_1 + h^{*}_4 \ge h^{*}_2 + h^{*}_3$, imply that $h^{*}_1 = h^{*}_2 = h^{*}_3 = h^{*}_4$. $\square$

Unfortunately, these are the only cases where Proposition 3.4 can be applied, due to the following observation:

Proposition 3.6

If $d:=\dim P \ge 5$ and $\deg P \ge 3$, then $h^{*}_1 + h^{*}_d < \sum _{i=2}^{d-1} h^{*}_i$.

Proof

Set $s:=\deg P$. If $h^{*}_d=0$ (i.e., $s < d$), then we have

$$\begin{aligned} h^{*}_1 +h^{*}_d =h^{*}_1< h^{*}_0+ h^{*}_1 \le h^{*}_s+h^{*}_{s-1} \le \sum _{i=2}^{d-1} h^{*}_i, \end{aligned}$$

where the first $\le$ follows from [16, Prop. 4.1] and the second follows from the assumption that $d \ge 5$ and $s \ge 3$. If $h^{*}_d \ne 0$, we have $h^{*}_i \ge h^{*}_1$ for all $1 \le i \le d-1$ by [8, Thm. 1.1]. Hence

$$\begin{aligned} h^{*}_1+h^{*}_d \le 2 h^{*}_1 < h^{*}_2 + h^{*}_3+ h^{*}_4 \le \sum _{i=2}^{d-1} h^{*}_i. \end{aligned}$$

Here we use the assumption that $d \ge 5$. $\square$

3.3 About Polytopes of Degree 2

One can combine our Corollary 1.2 with the results of [9] to obtain the following web of implications for lattice polytopes of degree 2:

Theorem 3.7

Let $P \subset \mathbb {R}^n$ be a lattice polytope of degree 2 with $h^{*}$-vector $(1, h^{*}_1, h^{*}_2)$, and let ${\widetilde{P}}$ denote the polytope P considered as a lattice polytope inside the lattice generated by the lattice points in P. Then the following implications hold:

Here, we say that a lattice polytope P is level if its Ehrhart ring $R={\mathbb {k}}[P]$ is level, that is, its canonical module $\omega _R$ is generated in a single degree as an R-module. The levelness of P is a combinatorial property of the monoid ${\widehat{M}}(P)$ (c.f. [9, Prop. 4.3]), and does not depend on the base field ${\mathbb {k}}$.

Proof

   

• :   This is elementary.

• :   We show the contrapositive. Assume that $\deg {\widetilde{P}}= 1$. Denote the $h^*$-vector of ${\widetilde{P}}$ by ${\tilde{h}}^*$. The volume of ${\widetilde{P}}$ divides the volume of P, since the latter is normalized with respect to a finer lattice. Thus we have that

  $$\begin{aligned} (1 + {\tilde{h}}^*_1 + {\tilde{h}}^*_2) \mid (1 + h^*_1 + h^*_2). \end{aligned}$$

  On the other hand, we have that ${\tilde{h}}^*_1 = h^*_1$ and by assumption, ${\tilde{h}}^*_2 = 0$. It follows that $(1+h^*_1) \mid h^*_2$.

• $h^{*}_1 \ge h^{*}_2\ \Longrightarrow$ IDP:   This is Corollary 1.2.

• IDP $\Longrightarrow$ spanning:   This is well known (and elementary), and it does not need the assumption $\deg P \le 2$.

• spanning $\Longrightarrow ~\!\!\deg {\widetilde{P}}\ne 1$:   $\deg {\widetilde{P}}= \deg P = 2 \ne 1$.

• $\deg {\widetilde{P}}\ne 1~\!\!\Longrightarrow$ level:   Under this hypothesis, the degree of ${\widetilde{P}}$ is either 0 or 2. We distinguish those cases:

  $\deg {\widetilde{P}}= 0$::

  In this case $h^*_1 = 0$, the claim follows from [9, Lem. 2.1].

  $\deg {\widetilde{P}}= 2$::

  Let R and ${\widetilde{R}}$ denote the Ehrhart rings of P and ${\widetilde{P}}$, respectively. Then $\deg {\widetilde{R}}= 1 + h^*_1 + {\tilde{h}}^*_2 > 1 + h^*_1$ by assumption, and thus R is level by [9, Prop. 3.4].

• level $\Longrightarrow ~\!\!\deg {\widetilde{P}}\ne 1$:   This follows from more general Lemma 3.8 below.$\square$

Lemma 3.8

Let $P \subset \mathbb {R}^n$ be a lattice polytope. If P is level, then $\deg {\widetilde{P}}\ne \deg P - 1$.

Proof

Let $c(P) := \min {\{\ell \in \mathbb {Z}_{>0} :\ell P^\circ \cap \mathbb {Z}^n \ne \emptyset \}}$. It is well known that $\deg P= \dim P+ 1 - c(P)$.

We are going to use [9, Prop. 4.3], which we recall for convenience: If P is level, then for any $k \ge c(P)$ and $\alpha \in kP^\circ \cap \mathbb {Z}^n$, there exist $\beta \in c(P)P^\circ \cap \mathbb {Z}^n$ and $\gamma \in (k - c(P))P \cap \mathbb {Z}^n$ such that

$$\begin{aligned} \alpha = \beta + \gamma . \end{aligned}$$

Now, assume that $\deg {\widetilde{P}}= \deg P - 1$, and note that this implies $c({\widetilde{P}}) = c(P) + 1$. Let $L \subset \mathbb {Z}^n$ be the sublattice spanned by the lattice points in P. As $P \ne {\widetilde{P}}$, this is a proper sublattice of $\mathbb {Z}^n$. Choose $\alpha \in c({\widetilde{P}})P^\circ \cap L$. Then, if P were level, there would exist $\beta$ and $\gamma$ as above. As $\beta \in c(P)P^\circ \cap \mathbb {Z}^n$, it follows that has no interior lattice points). Further, $\gamma$ lies in $(c({\widetilde{P}}) - c(P))P = P$ and thus $\gamma \in L$. But this contradicts $\beta + \gamma = \alpha \in L$. $\square$

We provide some examples to show that all the implications are strict and that there are no other implications. In each example, the claimed properties can conveniently be verified using normaliz [4].

Example 3.9

( spanning, IDP, $h^{*}_1 \ge h^{*}_2$). Consider the 4-polytope P with vertices

$$\begin{aligned}\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix},\begin{pmatrix}1\\ 1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 1\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 0\\ 1\end{pmatrix},\begin{pmatrix}0\\ 1\\ 1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\\ 0\\ 1\end{pmatrix}, \text { and }\begin{pmatrix}0\\ 0\\ 1\\ 1\end{pmatrix}. \end{aligned}$$

Its $h^{*}$-vector is (1, 2, 5), so it satisfies , but $h^{*}_1 \ngeq h^{*}_2$. To see that it is not spanning (and thus not IDP), consider the vector

$$\begin{aligned} v := \begin{pmatrix}1\\ 1\\ 1\\ 0\end{pmatrix} = \frac{1}{2}\left( \begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix} +\begin{pmatrix}1\\ 1\\ 0\\ 0\end{pmatrix} + \begin{pmatrix}1\\ 0\\ 1\\ 0\end{pmatrix} + \begin{pmatrix}0\\ 1\\ 1\\ 0\end{pmatrix} \right) .\end{aligned}$$

It lies in $2P \cap \mathbb {Z}^4$, but the sum of its coordinates is odd, while the coordinate sum of each vertex of P is even. Hence v cannot lie in the lattice spanned by them.

Example 3.10

(IDP and spanning ). Let P be the 3-simplex with vertices

$$\begin{aligned}\begin{pmatrix}0\\ 0\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 4\\ 0\end{pmatrix}, \text { and }\begin{pmatrix}1\\ 0\\ 3\end{pmatrix}.\end{aligned}$$

It is IDP and its $h^{*}$-vector is (1, 5, 6), so $h^{*}_1+1\mid h^{*}_2$ and $h^{*}_1 \ngeq h^{*}_2$.

Example 3.11

(spanning IDP). It is well known that this implication does not hold in general. For an example with degree 2, see [2, Exe. 2.24]. This is a very-ample (and thus spanning) 3-polytope which is not IDP. Its $h^{*}$-vector is (1, 4, 5), so it has degree 2.

Example 3.12

(, spanning). Let P be the polytope of Example 1.3 with $h^{*}$-vector (1, 0, 1). In this case ${\widetilde{P}}$ is a unit simplex and thus $\deg {\widetilde{P}}= 0 \ne 1$. However, P is not spanning and it holds that $h^{*}_1+1\mid h^{*}_2$.

Example 3.13

( and IDP ). Let P be the 3-simplex with vertices

$$\begin{aligned}\begin{pmatrix}0\\ 0\\ 0\end{pmatrix},\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\begin{pmatrix}0\\ 4\\ 0\end{pmatrix}, \text { and }\begin{pmatrix}1\\ 0\\ 4\end{pmatrix}. \end{aligned}$$

Its $h^{*}$-vector is (1, 6, 9), so it satisfies , but $h^{*}_1 \ngeq h^{*}_2$. Moreover, it is IDP.