The numbers of edges of 5-polytopes with a given number of vertices

TL;DR

This paper characterizes the possible pairs of the number of vertices and edges in 5-dimensional convex polytopes, extending known results from lower dimensions and confirming recent independent findings.

Contribution

It provides a complete characterization of the first two entries of the $f$-vector for 5-polytopes, filling a gap in the combinatorial understanding of higher-dimensional polytopes.

Findings

Characterization of vertex-edge pairs in 5-polytopes.

Extension of Steinitz and Grünbaum's results to 5 dimensions.

Independent confirmation of the characterization by other researchers.

Abstract

A basic combinatorial invariant of a convex polytope $P$ is its $f$-vector $f(P)=(f_0,f_1,\dots,f_{\dim P-1})$, where $f_i$ is the number of $i$-dimensional faces of $P$. Steinitz characterized all possible $f$-vectors of $3$-polytopes and Gr\"unbaum characterized the pairs given by the first two entries of the $f$-vectors of $4$-polytopes. In this paper, we characterize the pairs given by the first two entries of the $f$-vectors of $5$-polytopes. The same result was also proved by Pineda-Villavicencio, Ugon and Yost independently.

Equations

$$\big{\{}(f_{i}(P),f_{j}(P)):P\mbox{ is a $4$-polytope}\big{\}}.$$

$$\mathcal{E}^{d}=\{(f_{0}(P),f_{1}(P)):\mbox{ $P$ is a $d$-polytope}\}.$$

$${\mathcal{E}}^3=\{(v,e):\frac{3}{2}v\le e\le 3v-6\}.$$

$$\frac{d}{2}{f}_0(P)\le f_1(P)\le \left(\genfrac{}{}{0ex}{}{f_0(P)}{2}\right).$$

$$\mathcal{E}^{4}=\left\{(v,e):2v\leq e\leq{v\choose 2}\right\}\setminus\big{\{}(6,12),(7,14),(8,17),(10,20)\big{\}}.$$

$${\mathcal{E}}^5=\left\{(v,e):\frac{5}{2}v\le e\le \left(\genfrac{}{}{0ex}{}{v}{2}\right)\right\}\setminus (L\cup G).$$

$$f_0(Q)=f_0(P)+1\text{mbox}and{f}_1(Q)=f_0(P)+f_1(P).$$

$$\mathcal{E}^{5}\supset\left\{(v,e):3v-3\leq e\leq{v\choose 2}\right\}\setminus\big{\{}(7,18),(8,21),(9,25),(11,30)\big{\}}.$$

$$X=\left\{(v,e):\frac{5}{2}v\le e\le \left(\genfrac{}{}{0ex}{}{v}{2}\right)\right\}\setminus (L\cup G)$$

$$X_k=\{(v,e)\in X:v=k\}$$

$$f_0(\mathrm{tr}(P,v))=f_0(P)+4\text{mbox}and{f}_1(\mathrm{tr}(P,v))=f_1(P)+10.$$

$$X_k=\left\{(k,e):\frac{5}{2}k\le e\le \left(\genfrac{}{}{0ex}{}{k}{2}\right)\right\}\setminus \left\{\left(k,\lfloor \frac{5}{2}k+1\rfloor \right)\right\}.$$

$$f_0(Q)=k+4\text{mbox}and{f}_1(Q)=f_1(P)+10$$

$${\mathcal{E}}^5$$

$$\displaystyle=\left\{\big{(}(k+4),e^{\prime}\big{)}:\frac{5}{2}(k+4)\leq e^{\prime}\leq 3(k+4)-3\right\}\setminus\left\{\left(k+4,\left\lfloor\frac{5}{2}(k+4)+1\right\rfloor\right)\right\}.$$

$${\mathcal{E}}^5\supset X_6\cup X_7\cup X_8\cup X_9\cup X_{12}\cup X_{13}\cup X_{17}.$$

$${\mathcal{E}}^5\supset \{(12,30),(12,32),(13,34),(17,44),(17,45),(17,46),(17,47)\}.$$

$$f_0(C^{\prime \prime })=f_0(C^{\prime })+1=f_0(C^{\ast })+5=17$$

$$f_1(C^{\prime \prime })=f_1(C^{\prime })+5=f_1(C^{\ast })+15=45.$$

$$\partial\Gamma=\langle H\in\Gamma:\mbox{ $H$ is a $3$-face of $\Gamma$ contained in exactly one facet of $\Gamma$}\rangle.$$

$$f_2(G)\le f_3(P)-1=6.$$

$$h_i(\Delta )=\sum _{j=0}^i(-1{)}^{i-j}(\genfrac{}{}{0ex}{}{d-j}{i-j})f_{j-1}(\Delta )$$

$$(h_0,h_0+h_1-h_d,h_0+h_1+h_2-h_d-h_{d-1},\cdots ).$$

$$\sum _{k=1}^{13}\mathrm{deg}{v}_k=2f_1(P)=70$$

$$f_0(F^{\ast })=f_3(F)=8\text{mbox}and{f}_1(F^{\ast })=f_2(F)=\frac{1}{2}(5\times 2+4\times 6)=17.$$

$$B=\langle G:G\mbox{ is a simplex facet of $P^{*}$}\rangle.$$

$$F_1=\mathrm{conv}(v_1,v_2,u_1,u_2,u_3,u_4),$$

$$F_2=\mathrm{conv}(v_2,v_3,u_1,u_2,u_3,u_4),$$

$$⋮$$

$$F_5=\mathrm{conv}(v_5,v_1,u_1,u_2,u_3,u_4)$$

$$(1,5,8,5,1)$$

Full text

The second author was partially supported by KAKENHI16K05102.

1. Introduction

The study of $f$-vectors of convex polytopes is one of the central research topic in convex geometry. We call a $d$-dimensional convex polytope a $d$-polytope. For a convex polytope (or a polyhedral complex) $P$, we write $f_{i}(P)$ for the number of $i$-dimensional faces of $P$. The $f$-vector of a $d$-polytope $P$ is the vector $f(P)=(f_{0}(P),f_{1}(P),\dots,f_{d-1}(P))$. In 1906, Steinitz characterized all possible $f$-vectors of $3$-polytopes (see [Gr, §10.3]). While a characterization of $f$-vectors of $4$-polytopes is a big open problem in convex geometry, for any $0\leq i<j\leq 3$, the following set was characterized by Grünbaum [Gr], Barnette [Ba] and Barnette–Reay [BR] (see also [BL, Theorem 3.9])

[TABLE]

Moreover, Sjöberg and Ziegler [SZ] recently characterize all possible values of the pairs $(f_{0},f_{03})$ of flag face numbers of $4$-polytopes. In this paper, we characterize all possible $(f_{0},f_{1})$ pairs of $5$-polytopes.

Let

[TABLE]

The set $\mathcal{E}^{3}$ was determined by Steinitz in 1906 who shows that

[TABLE]

Note that, by Euler’s relation, this actually determines all possible $f$-vectors of $3$-polytopes. In higher dimensions, it is easy to see that any $d$-polytope $P$ satisfies

[TABLE]

Indeed, the first inequality follows since $f_{1}(P)$ equals to $\frac{1}{2}$ times the sum of degrees of the vertices of $P$ and since each vertex of $P$ has degree $\geq d$. Grünbaum [Gr, §10.4] proved that the inequality (1) characterizes $\mathcal{E}^{4}$, with four exceptions. More precisely, he proved the following statement.

Theorem 1.1** (Grünbaum).**

[TABLE]

In dimension $5$, the situation is more complicated. The set $\mathcal{E}^{5}$ is close to the set of integer points satisfying (1), but there are not only a finite list of exceptions but also an infinite family of exceptions. Indeed, our main result is the following.

Theorem 1.2**.**

Let $L=\{(v,\lfloor\frac{5}{2}v+1\rfloor):v\geq 7\}$ and $G=\{(8,20),(9,25),(13,35)\}$. Then

[TABLE]

Here $\lfloor a\rfloor$ denotes the integer part of a rational number $a$. Note that it is not hard to see $(8,20)\not\in\mathcal{E}^{5}$ since a $5$-polytope $P$ with $f_{1}(P)=\frac{5}{2}f_{0}(P)$ must be a simple polytope. Also, $(9,25)\not\in\mathcal{E}^{5}$ was proved in [PUY1] recently.

The following table illustrates the shape of $\mathcal{E}^{5}$.

In the table, black dots represent points in $\mathcal{E}^{5}$, white circles and triangles represent points in $L$ and $G$ respectively. For example, on the line $v=9$, $(9,23)\in L$ is presented by a white circle, $(9,25)\in G$ is presented by a triangle, and the possible numbers of edges are $24,26,27,\dots,36$.

Theorem 1.2 was also independently proved by Pineda-Villavicencio, Ugon and Yost [PUY2] by a different method.

It would be interesting to determine $\mathcal{E}^{d}$ for $d\geq 6$, and more generally to characterize the set \{(f_{i}(P),f_{j}(P)):\mbox{P$is a$d-polytope}\} for any $0\leq i<j<d$. About the latter problem, Sjöberg and Ziegler [SZ] recently study the case when $i=0$ and $j=d-1$.

2. sufficiency

In this section, we prove the sufficiency part of Theorem 1.2. If a polytope $Q$ is the pyramid over a polytope $P$, then we have

[TABLE]

This simple fact and Theorem 1.1 prove the next lemma.

Lemma 2.1**.**

[TABLE]

Let $P$ be a $d$-polytope. The degree $\deg v$ of a vertex $v$ of $P$ is the number of edges of $P$ that contain $v$. We say that a vertex $v$ is simple if $\deg v=d$. Let $V(P)$ be the vertex set of $P$.

Lemma 2.2**.**

If $P$ is a $5$-polytope such that $f_{1}(P)\leq 3f_{0}(P)-1$, then $P$ has a simple vertex.

Proof.

Observe $\deg v\geq 5$ for any $v\in V(P)$. Since $\frac{1}{2}\sum_{v\in V(P)}\deg v=f_{1}(P)=3f_{0}(P)-1$, there must exist a vertex $v\in V(P)$ of degree $<6$. ∎

Let

[TABLE]

be the right-hand side of Theorem 1.2, and let

[TABLE]

for $k\in\mathbb{Z}$. We want to prove $\mathcal{E}^{5}\supset X_{k}$ for all $k\geq 6$. To prove this, we use truncations. For a $5$-polytope $P$ and its vertex $v\in V(P)$, we write $\mathrm{tr}(P,v)$ for a polytope obtained from $P$ by truncating the vertex $v$. If $v$ is simple, then

[TABLE]

Lemma 2.3**.**

For $k\geq 6$ with $k\not\in\{8,9,13\}$, if $\mathcal{E}^{5}\supset X_{k}$ then $\mathcal{E}^{5}\supset X_{k+4}$.

Proof.

Since $k\not\in\{8,9,13\}$,

[TABLE]

By Lemma 2.2, for any $5$-polytope $P$ with $f_{0}(P)=k$ and $f_{1}(P)\leq 3k-1$, we can make a $5$-polytope $Q$ with

[TABLE]

by truncating a simple vertex from $P$. Since $\mathcal{E}^{5}\supset X_{k}$, this implies

[TABLE]

The above inclusion and Lemma 2.1 prove the desired statement. ∎

Now, we prove the main result of this section. For a convex polytope $P$, we write $P^{*}$ for its dual polytope. In the rest of the paper, if a face of a convex polytope is a simplex, then we call it a simplex face. A face which is not a simplex is called a non-simplex face.

Theorem 2.4**.**

$\mathcal{E}^{5}\supset X$.

Proof.

By Lemma 2.3, it is enough to show that

[TABLE]

Let $\varphi(v)=3v-3$. By Lemma 2.1, $(v,e)\in\mathcal{E}^{5}$ if $\varphi(v)\leq e\leq{v\choose 2}$ and $(v,e)\not\in\{(7,18),(8,21),(9,25),(11,30)\}$. Observe $\varphi(6)=15,\varphi(7)=18,\varphi(8)=21,\varphi(9)=24,\varphi(12)=33,\varphi(13)=36,\varphi(17)=48$. Then, to prove (2), what we must prove is

[TABLE]

(See also Table 1.) Note that this observation says $\mathcal{E}^{5}\supset X_{k}$ for $k\leq 9$.

Let $C$ be the cyclic $5$-polytope with $7$ vertices. Then $f(C)=(7,21,34,30,12)$ (see [Br, §18]). Hence $f_{0}(C^{*})=12$ and $f_{1}(C^{*})=30$, and therefore $(12,30)\in\mathcal{E}^{5}$. Let $C^{\prime}$ be the polytope obtained from $C^{*}$ by truncating its vertex. Note that every vertex of $C^{*}$ is simple. Since a truncation of a simple vertex creates a simplex facet, $C^{\prime}$ contains a simplex facet $F$. Let $C^{\prime\prime}$ be the polytope obtained from $C^{\prime}$ by adding a pyramid over $F$. Then

[TABLE]

and

[TABLE]

Hence $(17,45)\in\mathcal{E}^{5}$. We already see $(8,22),(9,24)\in\mathcal{E}^{5}$. Then, using truncations of simple vertices and Lemma 2.2, we have $(12,32),(13,34),(17,44)\in\mathcal{E}^{5}$. Also, since $(13,36),(13,37)\in\mathcal{E}^{5}$, by the same argument we have $(17,46),(17,47)\in\mathcal{E}^{5}$. These complete the proof of the theorem. ∎

3. Necessity

In this section, we prove the necessity part of Theorem 1.2. We first show that any element in $L$ is not contained in $\mathcal{E}^{5}$. We introduce some lemmas which we need. The following fact appears in [Gr, §6.1] (see also [Zi, Problem 6.8]).

Lemma 3.1**.**

There are exactly four combinatorially different $4$-polytopes with $6$ facets. They are

• ($P_{A}$)

Pyramid over a square pyramid;

• ($P_{B}$)

Pyramid over a triangular prism;

• ($P_{C}$)

A polytope obtained from a $4$-simplex by truncating its vertex;

• ($P_{D}$)

Product of two triangles.

Here are Schlegel diagrams and a list of facets of $P_{A},P_{B},P_{C}$ and $P_{D}$.

P_{A}$$P_{B}$$P_{C}$$P_{D}

[TABLE]

Recall that a convex polytope $P$ is said to be simplicial if all its proper faces are simplices. A simplicial $k$-sphere is a simplicial complex whose geometric realization is homeomorphic to the $k$-sphere. The boundary complex of a simplicial $d$-polytope is a simplicial $(d-1)$-sphere. The next statement easily follows from the Lower Bound Theorem (see [Ka]) and the Upper Bound Theorem (see [St, Corollary II.3.5]) for simplicial spheres.

Lemma 3.2**.**

Let $\Delta$ be a simplicial $3$-sphere.

• (i)

$f_{3}(\Delta)\neq 6,7,10$.

• (ii)

If $f_{3}(\Delta)=9$, then $\Delta$ is neighbourly, that is, every pair of vertices of $\Delta$ are connected by an edge.

Proof.

By the Lower Bound Theorem and the Upper Bound Theorem, we have

• •

if $f_{0}(\Delta)=5$, then $5\leq f_{3}(\Delta)\leq 5$;

• •

if $f_{0}(\Delta)=6$, then $8\leq f_{3}(\Delta)\leq 9$;

• •

if $f_{0}(\Delta)\geq 7$, then $11\leq f_{3}(\Delta)$.

These clearly imply (i). The statement (ii) follows from the fact that if the number of facets of a simplicial $(d-1)$-sphere equals to the bound in the Upper Bound Theorem, then it must be neighbourly (see e.g. the proof of [Br, Theorem 18.1]). ∎

We now prove that any element in $L$ is not contained in $\mathcal{E}^{5}$.

Proposition 3.3**.**

If $P$ is a $5$-polytope, then $f_{1}(P)\neq\lfloor\frac{5}{2}f_{0}(P)+1\rfloor$.

Proof.

Suppose to the contrary that $f_{1}(P)=\lfloor\frac{5}{2}f_{0}(P)+1\rfloor$. We first consider the case when $f_{0}(P)$ is odd. Then, since $\sum_{v\in V(P)}\deg v=2f_{1}(P)=5f_{0}(P)+1$, $P$ has one vertex having degree $6$ and all other vertices have degree $5$. Then $P^{*}$ has one facet $F$ with $f_{3}(F)=6$ and all other facets of $P^{*}$ are simplices. However this implies that the $4$-polytope $F$ must be simplicial, which contradicts Lemma 3.2(i).

Next, we consider the case when $f_{0}(P)$ is even. In this case, $\sum_{v\in V(P)}\deg v=2f_{1}(P)=5f_{0}(P)+2$, so one of the following two cases occurs:

• (a)

$P^{*}$ has one facet $F$ with $f_{3}(P)=7$ and all other facets of $P^{*}$ are simplices;

• (b)

$P^{*}$ has two facets $F$ and $G$ with $f_{3}(F)=f_{3}(G)=6$ and all other facets of $P^{*}$ are simplices.

Since there are no simplicial $4$-polytope with $7$ facets by Lemma 2.2(i), the case (a) cannot occur. Also, if the case (b) occurs, then $F$ and $G$ can have at most one non-simplex facet. However, Lemma 3.1 says that any $4$-polytope with $6$ facets have at least two non-simplex facets. ∎

Next, we show that any element of $G$ is not contained in $\mathcal{E}^{5}$. Let $\phi(v,d)=\frac{1}{2}dv+\frac{1}{2}(v-d-1)(2d-v)$. The following result was proved by Pineda-Villavicencio, Ugon and Yost [PUY1, Theorems 6 and 19].

Theorem 3.4**.**

Let $P$ be a $d$-polytope.

• (i)

If $f_{0}(P)\leq 2d$, then $f_{1}(P)\geq\phi(f_{0}(P),d)$.

• (ii)

If $d\geq 4$, then $(f_{0}(P),f_{1}(P))\neq(d+4,\phi(d+4,d)+1).$

By considering the special case when $d=5$ of the above theorem, we obtain the following.

Corollary 3.5**.**

$(8,20),(9,25)\not\in\mathcal{E}^{5}$.

By Proposition 3.3 and Corollary 3.5, to prove Theorem 1.2, we only need to prove $(13,35)\not\in\mathcal{E}^{5}$. We will prove this in the rest of this paper.

Let $P$ be a $5$-polytope. For faces $F_{1},F_{2},\dots,F_{k}$ of $P$, we write $\langle F_{1},\dots,F_{k}\rangle$ for the polyhedral complex generated by $F_{1},\dots,F_{k}$. Let $\{G_{1},\dots,G_{l}\}$ be a subset of the set of facets of $P$. Then any $3$-face of $\Gamma=\langle G_{1},\dots,G_{l}\rangle$ is contained in at most two facets of $\Gamma$. We write

[TABLE]

We often use the following trivial observation: If $\{G_{1},\dots,G_{l}\}$ is the set of non-simplex facets of $P$, then $\partial\langle G_{1},\dots,G_{l}\rangle$ is a simplicial complex.

We say that a $d$-polytope $P$ is almost simplicial if all facets of $P$ except for one facet are simplices. (We consider that simplicial polytopes are not almost simplicial.) The next lemma can be checked by using a complete list of $4$-polytopes with at most $8$ vertices (see [FMM]), but we write its proof for completeness.

Lemma 3.6**.**

Let $P$ be a $4$-polytope.

• (i)

Suppose that $P$ is almost simplicial and $f_{3}(P)=7$. Then $P$ is the pyramid over a triangular bipyramid.

• (ii)

Suppose that $P$ is almost simplicial and $f_{3}(P)=8$. Then $P$ does not contain a triangular bipyramid as a facet.

• (iii)

Suppose that $f_{3}(P)=7$ and $P$ has exactly two non-simplex facets $F$ and $G$. Then none of $F$ and $G$ are square pyramids.

Proof.

(i) Let $F$ be the unique non-simplex facet of $P$. Clearly, $F$ is simplicial and $f_{2}(F)\leq f_{3}(P)-1=6$ since, for each $2$-face of $F$, there is a unique $3$-face of $P$ that contains it other than $F$. Since a $3$-simplex and a triangular bipyramid are the only simplicial $3$-polytopes having at most $6$ facets, $F$ is a triangular bipyramid. Then, since $P$ has $7$ facets, $P$ must be the pyramid over $F$.

(ii) Let $F$ be the unique non-simplex facet of $P$. If $F$ is a triangular bipyramid, then by subdividing $F$ into two tetrahedra without introducing edges, one obtains a simplicial $3$-sphere $\Delta$ with $9$ facets. Since $F$ is a triangular bipyramid, there are two vertices $u$ and $v$ of $F$ such that $u$ and $v$ are not connected by an edge in $F$. These vertices are not connected by an edge in $\Delta$ by the construction of $\Delta$, which contradicts Lemma 3.2(ii) saying that $\Delta$ must be neighbourly.

(iii) Suppose to the contrary that $F$ is a square pyramid. We claim that $G$ is also a square pyramid. Indeed, since $\partial\langle F,G\rangle$ is a simplicial complex, $G$ contains exactly one non-simplex facet, and this facet must be a square and equals to $F\cap G$. Also,

[TABLE]

Let $\Delta$ be a simplicial $2$-sphere obtained from $G$ by subdividing the square $F\cap G$ into two triangles. Then $f_{2}(\Delta)\leq 7$, but since the number of $2$-faces of a simplicial $2$-sphere is even, $f_{2}(\Delta)=6$ and therefore $f_{2}(G)=5$. This forces that $G$ is a square pyramid.

Let $F=\mathrm{conv}(v_{1},a,b,c,d)$ and $G=\mathrm{conv}(v_{2},a,b,c,d)$, where $\mathrm{conv}(v_{1},\dots,v_{k})$ denotes the convex hull of points $v_{1},\dots,v_{k}$. Note that $\mathrm{conv}(a,b,c,d)$ is a square. We assume that $\mathrm{conv}(a,c)$ and $\mathrm{conv}(b,d)$ are non-edges of $P$. By subdividing each $F$ and $G$ into two tetrahedra by adding an edge $\mathrm{conv}(a,c)$, we can make a simplicial $3$-sphere $\Gamma$ with $f_{3}(\Gamma)=f_{3}(P)+2=9$. By the construction of $\Gamma$, $\mathrm{conv}(b,d)$ is not an edge of $\Gamma$, but this contradicts Lemma 3.2(ii). ∎

We also recall some known results on $h$-vectors of simplicial balls and their boundaries. For a simplicial complex $\Delta$ of dimension $d-1$, its $h$-vector $h(\Delta)=(h_{0}(\Delta),h_{1}(\Delta),\dots,h_{d}(\Delta))$ is defined by

[TABLE]

where $f_{-1}(\Delta)=1$. A simplicial $d$-ball is a simplicial complex whose geometric carrier is homeomorphic to a $d$-dimensional ball. The following facts are known. See [St, Chapter II and Problem 12].

Lemma 3.7**.**

Let $\Delta$ be a simplicial $d$-ball and $(h_{0},h_{1},\dots,h_{d+1})$ its $h$-vector. Then

• (i)

$h_{0}=1$* and $h_{d+1}=0$.*

• (ii)

$h_{0}+\cdots+h_{d+1}=f_{d}(\Delta)$.

• (iii)

$h_{i}\geq 0$* for all $i$.*

• (iv)

The $h$-vector of the boundary complex of $\Delta$ is

[TABLE]

We now complete the proof of Theorem 1.2.

Theorem 3.8**.**

$(13,35)\not\in\mathcal{E}^{5}$.

Proof.

Suppose to the contrary that there is a $5$-polytope $P$ such that $f_{0}(P)=13$ and $f_{1}(P)=35$. Let $v_{1},\dots,v_{13}$ be the vertices of $P$ with $\deg v_{1}\geq\cdots\geq\deg v_{13}$ and let $D=(\deg v_{1},\deg v_{2},\dots,\deg v_{13})$. Since

[TABLE]

and $\deg v_{k}\geq 5$ for each $k$, $D$ must be one of the following.

• (1)

$D=(10,5,\dots,5)$;

• (2)

$D=(9,6,5,\dots,5)$;

• (3)

$D=(8,7,5,\dots,5)$;

• (4)

$D=(8,6,6,5,\dots,5)$;

• (5)

$D=(7,7,6,5,\dots,5)$;

• (6)

$D=(7,6,6,6,5,\dots,5)$;

• (7)

$D=(6,6,6,6,6,5,\dots,5)$.

Below we will show a contradiction for each case.

(1) Suppose $D=(10,5,\dots,5)$. Then $P^{*}$ has a $4$-face $F$ with $f_{3}(F)=10$. This $F$ must be a simplicial $4$-polytope since all other facets of $P^{*}$ are simplices, which contradicts Lemma 3.2(i) saying that there are no simplicial $4$-polytopes with $10$ facets.

(2) Suppose $D=(9,6,5,\dots,5)$. $P^{*}$ has only two non-simplex $4$-faces $F$ and $G$. These $4$-faces can have at most one non-simplex $3$-face. By the assumption on $D$, $F$ or $G$ must have $6$ facets. This contradicts Lemma 3.1 saying that any $4$-polytope with $6$ facets has at least two non-simplex facets.

(3) Suppose $D=(8,7,5,\dots,5)$. Let $F$ and $G$ be the $4$-faces of $P^{*}$ with $f_{3}(F)=8$ and $f_{3}(G)=7$. Then $G$ is not simplicial by Lemma 3.2(i) and therefore $F$ is also not simplicial. Hence $F$ and $G$ are almost simplicial and $F\cap G$ is a $3$-polytope which is not a simplex. Since $f_{3}(F)=8$ and $f_{3}(G)=7$, this contradicts Lemma 3.6(i) and (ii).

(4) Suppose $D=(8,6,6,5,\dots,5)$. Let $F,G$ and $G^{\prime}$ be $4$-faces of $P^{*}$ with $f_{3}(F)=8$ and $f_{3}(G)=f_{3}(G^{\prime})=6$. Since $\partial\langle F,G,G^{\prime}\rangle$ is a simplicial complex, the polytopes $F,G$ and $G^{\prime}$ have at most two non-simplex $3$-faces. By Lemma 3.1, $G$ and $G^{\prime}$ must be the polytope $P_{A}$, and $F$ must have $2$ square pyramids as its $3$-faces. Thus, the $4$-polytope $F$ has $6$ tetrahedra and $2$ square pyramids as its facets. This implies

[TABLE]

However, this contradicts Theorem 1.1 saying that $(8,17)\not\in\mathcal{E}^{4}$.

(5) Suppose $D=(7,7,6,5,\dots,5)$. Let $F,F^{\prime}$ and $G$ be $4$-faces of $P^{*}$ with $f_{3}(F)=f_{3}(F^{\prime})=7$ and $F(G)=6$. Since all other $4$-faces of $P^{*}$ are simplices, $F,F^{\prime}$ and $G$ have at most two non-simplex $3$-faces. Then, by Lemma 3.1, $G$ must be the polytope $P_{A}$, and $F$ and $F^{\prime}$ have a square pyramid as its facets. By Lemma 3.6(i), $F$ and $F^{\prime}$ are not almost simplicial. Hence $F$ and $F^{\prime}$ have exactly two non-simplex facets, but this contradicts Lemma 3.6(iii).

(6) Suppose $D=(7,6,6,6,5,\dots,5)$. Let $F,G,G^{\prime}$ and $G^{\prime\prime}$ be $4$-facets of $P^{*}$ with $f_{3}(F)=7$ and $f_{3}(G)=f_{3}(G^{\prime})=f_{3}(G^{\prime\prime})=6$. Since all other $4$-faces of $P^{*}$ are simplices, each of $F,G,G^{\prime}$ and $G^{\prime\prime}$ can have at most three non-simplex $3$-faces. By Lemma 3.1, $G,G^{\prime}$ and $G^{\prime\prime}$ must be the polytope $P_{A}$ and have exactly two non-simplex $3$-faces. Since $\partial\langle F,G,G^{\prime},G^{\prime\prime}\rangle$ is a simplicial complex, one of the following situations must occur:

• (a)

$F$ is a simplicial polytope;

• (b)

$F$ has exactly two square pyramids as its facets and all other facets are simplices.

However, (a) cannot occur by Lemma 3.2(i) and (b) cannot occur by Lemma 3.6(iii).

(7) Suppose $D=(6,6,6,6,6,5,\dots,5)$. Let $F_{1},\dots,F_{5}$ be the $4$-faces of $P^{*}$ with $f_{3}(F_{1})=\cdots=f_{3}(F_{5})=6$. Observing that each of $F_{1},\dots,F_{5}$ must be one of $P_{A},P_{B},P_{C}$ and $P_{D}$ in Lemma 3.1. It is not hard to see that one of the following situations must occur;

• (a)

All the $F_{1},\dots,F_{5}$ are $P_{A}$;

• (b)

All the $F_{1},\dots,F_{5}$ are $P_{C}$;

• (c)

$F_{1}$ and $F_{2}$ are $P_{B}$. $F_{3},F_{4}$ and $F_{5}$ are $P_{A}$.

We first show that (a) cannot occur. We may assume that $F_{i}\cap F_{i+1}$ is a square pyramid for $i=1,2,\dots,5$, where $F_{6}=F_{0}$. Since $P_{A}$ has only one square as its $2$-faces, $K=F_{1}\cap F_{2}\cap\cdots\cap F_{5}$ must be this square, and the polyhedral complex generated by the facets of $P^{*}$ that do not contain $K$ is a shellable simplicial $4$-ball by a line shelling [Zi, §8.2]. Let $B$ be this ball. Clearly,

[TABLE]

Hence $B$ has $13-5=8$ facets. Also, since each $F_{i}\cap F_{i+1}$ is a pyramid over the square $K$, the faces $F_{1},\dots,F_{5}$ can be written as

[TABLE]

[TABLE]

[TABLE]

[TABLE]

with $K=\mathrm{conv}(u_{1},\dots,u_{4})$. Then it follows that $\Gamma=\partial\langle F_{1},\dots,F_{5}\rangle$ is the join of the $5$-cycle and the $4$-cycle, and its $h$-vector is

[TABLE]

since its entries coincide with the coefficients of the polynomial $(1+3t+t^{2})(1+2t+t^{2})$. Let $h(B)=(1,h_{1},h_{2},h_{3},h_{4},0)$. Lemmas 3.7 and 3.8 say

[TABLE]

Then it is easy to see $h(B)=(1,4,3,0,0,0)$. Let $G_{1},\dots,G_{8}$ be a shelling of $B$. Let

[TABLE]

and $S_{j}=V(G_{j})\setminus R_{j}$, where $V(G_{j})$ is the vertex set of $G_{j}$. Then

[TABLE]

for all $i$ (see [Zi, §8.3]). Since $h_{3}=h_{4}=0$, we have $|S_{j}|=5-|R_{j}|\geq 3$ for all $j$. By the definition of a shelling, $\mathrm{conv}(S_{8})$ is a missing face of $\partial B$, that is, $\mathrm{conv}(S_{8})$ is not a face of $\partial B$ but any its proper face is a face of $\partial B$. Thus $\partial B$ has a missing face of dimension $\geq 2$. However, the join of two cycles of length $\geq 4$ does not have any missing face of dimension $\geq 2$, and $\partial B=\partial\langle F_{1},\dots,F_{5}\rangle$ is the join of the $5$-cycle and the $4$-cycle, a contradiction.

We next prove that (b) cannot occur. Observe that $P_{C}$ has $4$ non-simplex facets. Since $\partial\langle F_{1},\dots,F_{5}\rangle$ is a simplicial complex, each $F_{i}\cap F_{j}$ must be a triangular prism. Then it is easy to see that we can write

[TABLE]

where each $\mathrm{conv}(x_{i},x_{j},x_{k},y_{i},y_{j},y_{k})$ is a triangular prism with triangles $\mathrm{conv}(x_{i},x_{j},x_{k})$ and $\mathrm{conv}(y_{i},y_{j},y_{k})$ (we assume that each $\mathrm{conv}(x_{k},y_{k})$ is an edge of $P^{*}$). Using this formula, one conclude that

[TABLE]

is the disjoint union of two copies of the boundary of a $4$-simplex.

If $\mathrm{conv}(x_{1},\dots,x_{5})$ is not a face of $P^{*}$, then, for each $S\subset\{x_{1},\dots,x_{5}\}$ with $|S|=4$, there is a unique $4$-face $G_{S}\not\in\{F_{1},\dots,F_{5}\}$ of $P^{*}$ that contains $\mathrm{conv}(S)$. This implies that, since $P^{*}$ has only 8 facets other than $F_{1},\dots,F_{5}$, either $\mathrm{conv}(x_{1},\dots,x_{5})$ or $\mathrm{conv}(y_{1},\dots,y_{5})$ must be a face of $P^{*}$ (otherwise $P^{*}$ has at least 10 facets other than $F_{1},\dots,F_{5}$). We assume that $\mathrm{conv}(x_{1},\dots,x_{5})$ is a face of $P^{*}$. Let $G_{1},\dots,G_{8}$ be the simplex $4$-faces of $P^{*}$ and assume $G_{1}=\mathrm{conv}(x_{1},\dots,x_{5})$. Then $\Gamma=\langle G_{2},\dots,G_{8}\rangle$ is a pseudomanifold with

[TABLE]

Let $n_{i}$ be the number of interior vertices of $\Gamma$. By the Lower Bound Theorem for pseudomanifolds with boundary [Fo] (see also [Ta, Theorem 1.2]), $\Gamma$ must have at least $5+4n_{i}-4$ facets. Since $\Gamma$ only has $7$ facets, we have $n_{i}\leq 1$. However, this implies that $\Gamma$ is either the $4$-simplex or the cone over the boundary of the $4$-simplex, contradicting the fact that $\Gamma$ has $7$ facets.

We finally prove that (C) cannot occur. Since $\partial\langle F_{1},\dots,F_{5}\rangle$ is a simplicial complex, $F_{1}\cap F_{2}$ must be a triangular prism and $F_{i}\cap F_{j}$ is a square pyramid for all $i\in\{1,2\}$ and $j\in\{3,4,5\}$. It is not hard to see that $F_{1},\dots,F_{5}$ can be written as

[TABLE]

where $\mathrm{conv}(x,y,z,x^{\prime},y^{\prime},z^{\prime})$ is a triangular pyramid with triangles $\mathrm{conv}(x,y,z)$ and $\mathrm{conv}(x^{\prime},y^{\prime},z^{\prime})$ (we assume that $\mathrm{conv}(x,x^{\prime}),\mathrm{conv}(y,y^{\prime})$ and $\mathrm{conv}(z,z^{\prime})$ are edges). A routine computation shows

[TABLE]

is the union of two copies of the boundary of a $4$-simplex intersections in the edge $\mathrm{conv}(v_{1},v_{2})$. Then the exactly same argument as in the case (b) works, namely, one case show that either $\mathrm{conv}(v_{1},v_{2},x,y,z)$ or $\mathrm{conv}(v_{1},v_{2},x^{\prime},y^{\prime},z^{\prime})$ must be a face of $P^{*}$ and conclude a contradiction by the Lower Bound Theorem for pseudomanifolds with boundary. ∎