Separation in Simply Linked Neighbourly 4-Polytopes

TL;DR

This paper investigates the separation problem in convex geometry, verifying a conjecture about the maximum number of hyperplanes needed to separate a point from all faces in a specific class of 4-polytopes.

Contribution

It proves the conjecture for simply linked neighbourly 4-polytopes, advancing understanding of hyperplane separation in convex polytopes.

Findings

Verified the conjecture s(O,K) ≤ 2^d for simply linked neighbourly 4-polytopes

Established bounds on hyperplane separation in specific convex polytopes

Contributed to the theory of convex polytope separation problems

Abstract

The Separation Problem asks for the minimum number s(O,K) of hyperplanes required to strictly separate any interior point O of a convex body K from all faces of K. The Conjecture is s(O,K) is at most 2 to the power d in real d-space , and we verify this for the class of simply linked neighbourly 4-polytopes.

Figures

Equations

$$[Y_1,Y_2,\cdots ,Y_k]=conv(Y_1\cup Y_2\cup ...\cup Y_k)$$

$$P_{m+1}\supset P_{m}and\mathcal{U}(P_{m+1})\backslash \mathcal{U}(P_m)\ne \varnothing$$

$$P_m=[x_m,x_{m-1},\cdots ,x_1]and\mathcal{U}(P_{m+1})\backslash \mathcal{U}(P_m)\ne \varnothing$$

$$Y_a=\{y_1,y_2,\cdots ,y_a\},z\in \mathcal{V}(Q)\backslash Y_a,$$

$$Z_t=\{z|⟨v,y_1,y_t,y_{t+1}⟩strictlyseparates{y}_{2}andz\}$$

$$Z_t^{\prime }=\{z|⟨v,y_2,y_t,y_{t+1}⟩strictlyseparates{y}_{1}andz\}$$

$$w_s\to w_{s-1}\to \cdots \to w_1.$$

$$⟨x_a,x_b,x_c,x_s⟩\cap [x_g,x_h]=\varnothing =⟨x_a,x_b,x_c,x_q⟩\cap [x_g,x_h].$$

$$Z_t^{-}=\{y_3,\cdots ,y_{t-1}\}\cup Z_3\cup \cdots \cup Z_{t-1},$$

$$Z_t^{+}=\{y_{t+2},\cdots ,y_a\}\cup Z_{t+1}\cup \cdots \cup Z_{a-1}$$

$$\mathcal{V}(P_m)={\hat{Y}}_b\cup {\hat{Z}}_3\cup \cdots \cup {\hat{Z}}_{b-1}=\{{\hat{y}}_1,{\hat{y}}_2,{\hat{y}}_k,{\hat{y}}_{k+1}\}\cup {\hat{Z}}_k^{-}\cup {\hat{Z}}_k\cup {\hat{Z}}_k^{+}$$

$$\mathcal{V}(P_m)={\bar{Y}}_c\cup {\bar{Z}}_3\cup \cdots \cup {\bar{Z}}_{c-1}=\{{\bar{y}}_1,{\bar{y}}_2,{\bar{y}}_i,{\bar{y}}_{i+1}\}\cup {\bar{Z}}_i^{-}\cup {\bar{Z}}_i\cup {\bar{Z}}_i^{+}$$

$$O\in [x_1,x_3,x_5]\cap [x_2,x_4,x_6].$$

Full text

 

2010 MSC: 52B11, 52B05.

keywords: Gohberg-Markus-Hadwiger covering number, neighbourly polytope, separation problem

1 INTRODUCTION

We recall that the Separation Problem is the polar version of the Gohberg-Markus-Hadwiger Covering Problem for convex bodies, and refer to [2], [6] and [9] for an overview of the topic.

For convex d-polytopes $P$, the Conjecture has been verified in the case that $P$ is cyclic or a type of neighbourly 4-polytope( totally-sewn or with at most ten vertices). We refer to [3] for an overview of these results.

In the following, we assume that $P$ is a neighbourly 4-dimensional polytope in $\bbbr^{4}$. Then $P$ is convex and any two distinct vertices determine an edge of $P$. We refer to [8] and [12] for the basic geometric and combinatorial properties of $P$.

With formal definitions to follow; we note only that cyclic polytopes are neighbourly and totally-sewn, and that totally-sewn $P$ are linked. Thus, we verify the Conjecture for a new class of $P$.

As for organization: Section 2 contains definitions and conventions. In Section 3, we examine the inner structure of $P$. In Section 4, we determine some separation properties of P. We introduce simply linked $P$ and present our separation results in Section 5 and 6.

2 DEFINITIONS

Let $Y$ be a set of points in $\bbbr^{d}$. Then $conv\ Y$ and aff $Y$ denote, respectively, the convex hull and the affine hull of $Y$. For sets $Y_{1},Y_{2},\cdots,Y_{k}$, let

[TABLE]

and $\langle Y_{1},Y_{2},\cdots,Y_{k}\rangle=$ aff $(Y_{1}\cup Y_{2}\cup\cdots\cup Y_{k})$. For a point $y$, let $[y]=[\{y\}]$ and $\langle y\rangle=\langle\{y\}\rangle$.

Let $Q\in\bbbr^{d}$denote a (convex) d-polytope with $\mathcal{V}(Q),\mathcal{E}(Q)$ and $\mathcal{F}(Q)$ denoting, respectively, its sets of vertices, edges and facets. For $x\in\mathcal{V}(Q),\ Q/_{x}$ denotes the vertex figure of Q at x. For $E=[x,y]\in\mathcal{E}(Q),\ Q/_{E}$ denotes the quotient polytope $(Q/_{y})/_{x}$. We note that $Q/_{E}$ is a $(d-2)$-polytope.

Let $d=4$. As a simplification, we assume always that $Q/_{x}$ is contained in a hyperplane $H\subset\bbbr^{4}$ that strictly separates $x$ from each $y\in\mathcal{V}(Q)/\{x\}$, and denote $H\cap[x,y]=H\cap\langle x,y\rangle$ also by $y$. Then of importance here are the following:

2.1

For $y_{i}\in\mathcal{V}(Q)/\{x\}$; a plane $\langle y_{1},y_{2},y_{3}\rangle$ separates $y_{4}$ and $y_{5}$ in $\langle Q/_{x}\rangle$ if, and only if, the hyperplane $\langle x,y_{1},y_{2},y_{3}\rangle$ separates $y_{4}$ and $y_{5}$ in $\bbbr^{4}$, and

2.2

For $y_{i}\in\mathcal{V}(Q)/E$; a line $\langle z_{1},z_{2}\rangle$ separates $z_{3}$ and $z_{4}$ in $\langle Q/_{E}\rangle$ if, and only if the hyperplane$\langle E,z_{1},z_{2}\rangle$ separates $z_{3}$ and $z_{4}$ in $\bbbr^{4}$.

Let $S\subset\bbbr^{3}$ be a 3-polytope with $s\geq 4$ vertices. Then $S$ is stacked if either $s=4$ or $S$ is the convex hull of a stacked 3-polytope with $s-1$ vertices and a point in $\bbbr^{3}$ that is beyond exactly one facet of $S$.

Let $S$ be stacked, $\{x,y,z\}\subset\mathcal{V}(S)$ and $C=[x,y,z]$ be a triangle. We say that C is a cut of $S$ if $\mathcal{E}(C)\subset\mathcal{E}(S)$ but $C\notin\mathcal{F}(S)$. All the cuts of $S$ decompose $S$ into components, each of which is a 3-simplex. We note that $|\mathcal{V}(S)|=s$ yields that $S$ has $s-4$ cuts and $s-3$ components.

Let $\mathcal{N}_{m}$ denote the family of combinatorially distinct neighbourly 4-polytopes with $m\geq 5$ vertices, $P\in\mathcal{N}_{m+1},x\in\mathcal{V}(P)$ and $Q=[\mathcal{V}(P)\backslash\{x\}]$. We note that $Q\in\mathcal{N}_{m}$.

The relevance of stacked 3-polytopes here is the following result in [1]:

2.3

$P/_{x}$* is a stacked 3-polytope with $m$ vertices; furthermore, $[y_{1},y_{2},y_{3},y_{4}]$ is a component of $P/_{x}$ if, and only if, $[y_{1},y_{2},y_{3},y_{4}]\in\mathcal{F}(Q)\backslash\mathcal{F}(P)$. Hence, $x$ is beyond exactly $m-3$ facets of $Q$.*

Next, let $E=[x,y]\in\mathcal{E}(P)$. Then $E$ is a universal edge of $P$ if $[E,z]$ is a 2-face of $P$ for each $z\in\mathcal{V}(P)/\{x,y\}$. Let $\mathcal{U}(P)$ denote the set of universal edges of $P$. We observe from [12] and [13] that

2.4

$E=[x,y]\in\mathcal{U}(P)$* if, and only if, x and y lies on the same side of every hyperplane determined by the vertices of $P$. From the same sources; if $|\mathcal{V}(P)\geq 7$ then any vertex of $P$ is on at most two members of $\mathcal{U}(P)$, and $|\mathcal{U}(P)|\leq|\mathcal{V}(P)|$.*

We recall that a cyclic 4-polytope $C_{m}$ with $m$ vertices is combinatorially equivalent to the convex hull of $m$ points on the moment curve in $\bbbr^{4}$. From [7], [8] and [12], we note that $C_{m}\in\mathcal{N}_{m}$, $\mathcal{N}_{6}=\{C_{6}\}$, $|\mathcal{U}(C_{6})|=9$, $\mathcal{N}_{7}=\{C_{7}\}$, $|\mathcal{U}(C_{m})|=m$ for $m\geq 7$, and any 4-subpolytope of $C_{m}$ is again cyclic. For $m\geq 6$, there is a natural ordering (Gale’s Evenness Condition) of $\mathcal{V}(P_{m})$ that corresponds to the order of appearance of equivalent points on the moment curve.

Let $m\geq 8$. Most of our knowledge about members of $\mathcal{N}_{m}$ is based upon various construction techniques: given $Q\in\mathcal{N}_{m-1}$, find a point $\bar{x}\in\bbbr^{4}/Q$ such that $\bar{Q}=[Q,\bar{x}]\in\mathcal{N}_{m}$. It is noteworhy that, at present, known constructions such as Shemer Sewing, Extended Sewing and Gale Sewing(cf. [12], [10] and [11]) yield that $\mathcal{U}(\bar{Q})\backslash\mathcal{U}(Q)\neq\emptyset$. We introduce a class of polytopes to reflect this fact.

Let $n\geq 7$ and $P_{n}\in\mathcal{N}_{n}$. We say that $P_{n}$ is linked if for $m=n-1,\cdots,6$, there is a $P_{m}\in\mathcal{N}_{m}$ with the property that

[TABLE]

We say that $P_{n}$ is linked under the (vertex) array $x_{n}>x_{n-1}>\cdots>x_{1}$ if for $m=n-1,\cdots,6$,

[TABLE]

For $x_{t}\in\{x_{7},\cdots,x_{n}\}$ and $x_{r}\in\{x_{1},\cdots,x_{t-1}\}$, we say that $x_{t}$ is linked to $x_{r}\ (x_{t}\rightarrow x_{r})$ if $[x_{t},x_{r}]\in\mathcal{U}(P_{t})$ and $[x_{t},x_{j}]\notin\mathcal{U}(P_{t})$ for $j>max\{6,r\}$.

By way of clarification for requiring that $t\geq 7$; we note that

2.5

$P_{6}$* is cyclic and there are disjoint three element subsets $Y$ and $Z$ of $\mathcal{V}(P_{6})$ such that $\mathcal{U}(P_{6})=\{[y,z]|y\in Y\ and\ z\in Z\}$. Thus, there is no meaningful labeling of a greatest or a least vertex of $P_{6}$*

3 THE INNER STRUCTURE OF $P$

Let $v\in\mathcal{V}(P),\ Q\subset P,\ v\notin Q,\ Q\in\mathcal{N}_{m}$ and $R=[Q,v]$. We recall that

$R^{*}=R/_{v}$ is a stacked 3-polytope and that $y\in\mathcal{V}(Q)$ denotes also $\{y^{*}\}=\langle v,y\rangle\cap\langle R^{*}\rangle$. We describe $R^{*}$.

Let

[TABLE]

[TABLE]

and

[TABLE]

From 2.1, we have that

• $\bullet$

$Z_{t}\neq\emptyset(Z^{\prime}_{t}\neq\emptyset)$ if and only if, $[y_{1},y_{t},y_{t+1}]\ ([y_{2},y_{t},y_{t+1}])$ is a cut of $R^{*}$.

Hence, we have a generic description of $R^{*}$; cf. the Schlegel diagram in Figure 1.

Next, we observe from 2.3 and 2.1 that

• $\bullet$

$\langle y_{1},y_{2},y_{t},y_{t+1}\rangle$ separates $v$ and $Q$ for $t=3,\cdots,a-1$, and

• $\bullet$

$\langle v,y_{1},y_{2},y_{t}\rangle$ separates $Z_{r}\cup Z^{\prime}_{r}(r<t)$ and $Z_{s}\cap Z^{\prime}_{s}(s\geq t)$ for $t=4,\cdots,a-1.$

From 2.2, we depict these separation properties with respect to $Q/_{[y_{1},y_{2}]}$ and $R/_{[y_{1},y_{2}]}$ in Figure 2.

REMARKS Let $R^{*}=R/_{v}$ be labeled as above.

3.1

If $a=m$ then $\mathcal{F}(Q)\backslash\mathcal{F}(R)=\{[y_{1},y_{2},y_{t},y_{t+1}]|\ t=3,\cdots,a-1\},\ [y_{1},y_{2}]\in\mathcal{U}(Q)$ and $\{[v,y_{1}],[v,y_{2}]\}\subset\mathcal{U}(R).$

There is such a labeling of $R^{*}$ if $R$ is cyclic or if $R$ is constructed by a Shemer sewing of $v$ onto $Q$.

3.2

If $w\in\mathcal{V}(P)\backslash\mathcal{V}(Q)$ and $[w,v]\in\mathcal{U}([Q,v,w])$ then $\mathcal{F}(Q)\backslash\mathcal{F}([Q,w])=\mathcal{F}(Q)\backslash\mathcal{F}(R)$ by 2.4.

3.3

Let $3<t<a$. If $\langle v,y_{1},y_{2},y_{t}\rangle$ strictly separates vertices $p_{t}$ and $s_{t}$ of $Q$ then $[p_{t},s_{t}]$ is not an edge of $R^{*}$, and $[v,p_{t},s_{t}]$ is not a face of $R$.

We note from Figure 2 that under the hypotheses of 3.3, each hyperplane through $\langle y_{1},y_{2},y_{t}\rangle$ strictly separates some two of $v$, $p_{t}$ and $s_{t}$. Thus, the following is the more general result; cf. [5].

3.4

If $\{x_{a},x_{b},x_{c},x_{e},x_{f},x_{g}\}$ is a set of six vertices of $P$ and each hyperplane of $\bbbr^{4}$ though $\{x_{a},x_{b},x_{c}\}$ strictly separates two of $x_{e},x_{f}$ and $x_{g}$, then $[x_{a},x_{b},x_{c}]$ and $[x_{e},x_{f},x_{g}]$ are not faces of $P$.

4 GENERIC SEPARATION PROPERTIES OF P

Let $P\in\mathcal{N}_{m},m\geq 6$, and $O$ be an interior point of $P$. We determine hyperplanes $H\in\bbbr^{4}$ that strictly separate $O$ from facets of $P$. As a simplification, we determine $H$ that do not contain $O$. We consider first $F\in\mathcal{F}(P)$ that either are contained in a subpolytope $Q$ such that $O\notin int\ Q$ or have a common vertex $w$.

Lemma A

( cf. [4] ) Let $O\in bd(Q)$. Then $O$ is strictly separated from any $F\in\mathcal{F}(P)\cap\mathcal{F}(Q)$ by one of at most three hyperplanes.

Lemma B

Let $w\in\mathcal{V}(P),R\in\mathcal{N}_{m-1},P=[R,w]$ and $F\in\mathcal{F}(P)$ such that $w\in F$. Then $O$ is strictly separated from any such $F$ by one of at most four(six) hyperplanes in case $O$ is (is not) an interior point of $R$.

Proof

Since $P^{*}=P/_{w}$ is stacked and $O\in int\ P$, it follows that $O^{*}\in\langle w,O\rangle\cap P^{*}$ is in a component $A^{*}=[x^{*},y^{*},z^{*},v^{*}]$ of $P^{*}$. If $O^{*}\in relint\ A^{*}$ then $O$ is separated from $F$ by one of $\langle w,x,y,z\rangle,\ \langle w,x,y,v\rangle,\ \langle w,x,z,v\rangle$ and $\langle w,y,z,v\rangle$.

Let $O^{*}\in B^{*}=[x^{*},y^{*},z^{*}]$, say. Then $B^{*}$ is a cut of $P^{*},\ O\in\langle w,x,y,z\rangle$ and there are subpolytopes $P^{\prime}$ and $P^{\prime\prime}$ of $P$ such that $P^{\prime}\cap P^{\prime\prime}=[w,x,y,z],[P^{\prime},P^{\prime\prime}]=P$ and (since $w\in F$) either $F\subset P^{\prime}$ or $F\subset P^{\prime\prime}$.

We recall from $2.3$ that $[x,y,z,v]\in\mathcal{F}(R)\backslash\mathcal{F}(P)$. If $O\in int\ R$ then it is clear that $O\notin[w,x,y,z]$; that is, $O\notin P^{\prime}\cup P^{\prime\prime}$ and $O$ is separated from $F$ by one of two hyperplanes. If $O\in[w,x,y,z]$ then $O\in bd(P^{\prime})\cap bd(P^{\prime\prime})$ and we apply LEMMA A. ∎

REMARKS Let $Q$ be a subpolytope of $P$ such that $O\notin int\ Q$.

4.1

If $Q\in\mathcal{N}_{m-1}$ then $O$ is strictly separated from any $F\in\mathcal{F}(P)$ by one of at most nine( three from LEMMA A, six from LEMMA B) hyperplanes.

4.2

If $Q\in\mathcal{N}_{m-3}$ then $O$ is strictly separated from any $F\in\mathcal{F}(P)$ by one of at most fifteen hyperplanes.

For $4.2$, we apply $LEMMA\ B$ under the assumption that $O$ is an interior point of any $Q^{\prime}\in\mathcal{N}_{m-1}$ such that $Q^{\prime}\subset P$

5 SIMPLY LINKED P

Let $n\geq 7$ and $P=P_{n}\in\mathcal{N}_{n}$ be linked under the array $x_{n}>x_{n-1}>\cdots>x_{1}$.

Let $\mathcal{W}=\{w_{s},w_{s-1},\cdots,w_{1}\}$ be an s element subset of $\mathcal{V}(P)$ with the induced array $w_{s}>w_{s-1}>\cdots>w_{1}$ in the case $s>1$. Then $\mathcal{W}$ is a chain if either $s=1$ or

[TABLE]

For $x_{k}\in\mathcal{V}(P)$, let $\mathcal{V}^{k}$ denote the union of all chains of $P$ with $x_{k}$ as the least vertex.

Finally, we say that $P_{n}$ is simply linked if for $k=7,\cdots,n:$

• $\bullet$

$\mathcal{V}^{k}$ is a chain, and

• $\bullet$

for disjoint chains $\mathcal{V}^{c}$ and $\mathcal{V}^{d}$, there are $x_{i}\neq x_{j}$ in $\mathcal{V}(P_{6})$ such that $x_{c}\rightarrow x_{i},x_{d}\rightarrow x_{j}$ and $[x_{i},x_{j}]\notin\mathcal{U}(P_{6})$.

Henceforth, we assume that $P_{n}$ is simply linked. Then it follows from $2.5$ that $\{x_{7},\cdots,x_{n}\}$ is the union of at most three pairwise disjoint maximal chains.

Lemma C

*Let $6\leq m<n,x_{m}<x_{t},x_{k}<x_{t}$ and $x_{t}\notin\mathcal{V}^{k}.$

C.1 $H\cap[\mathcal{V}^{t}]=\emptyset$ for any hyperplane $H$ spanned from $\{x_{1},\cdots,x_{m}\}$.

C.2 Let $H_{h}=\langle x_{a},x_{b},x_{c},x_{h}\rangle$ be a hyperplane with $\{x_{a},x_{b},x_{c}\}\subset\{x_{1},\cdots,x_{m}\}$ and $H_{h}\cap\mathcal{V}^{k}=\{x_{h}\}$. Then $H_{h}\cap[\mathcal{V}^{t}]=\emptyset$.

C.3 Let $x_{t}\rightarrow x_{j},x_{j}\notin\{x_{a},x_{b},x_{c}\}$ and $H_{h}$ be defined as above. Then $H_{h}\cap[\mathcal{V}^{j}]=\emptyset$.*

Proof

Since $P$ is simplicial, it follows from $H\cap[\mathcal{V}^{t}]\neq\emptyset$ that $H$ strictly separates some $x_{v}$ and $x_{u}$ in the chain $\mathcal{V}^{t}$ such that $x_{v}\rightarrow x_{u}$. Then $[x_{v},x_{u}]\in\mathcal{U}(P_{v})$ and $P_{m}\subset P_{t}\subset P_{v}$ yield a contradiction by $2.4$.

As above, $H_{h}\cap[\mathcal{V}^{t}]\neq\emptyset$ implies that $H_{h}$ strictly separates some $x_{s}$ and $x_{q}$ in $\mathcal{V}^{t}$ such that $x_{s}\rightarrow x_{q}$. Thus, C.1 yields that $x_{s}<x_{h}$ and $x_{t}\in P_{s}\subset P_{h}$. From $x_{h}\in\mathcal{V}^{k}$ and $x_{k}<x_{t}<x_{h}$, there is an $x_{g}\in\mathcal{V}^{k}$ such that $x_{h}\rightarrow x_{g}$. Then $[x_{g},x_{h}]\in\mathcal{U}(P_{h}),x_{m}<x_{t}<x_{h},\mathcal{V}^{k}\cap\{x_{a},x_{b},x_{c}\}=\emptyset$ and $2.4$ yield that in the pencil of hyperplanes containing $\langle x_{a},x_{b},x_{c}\rangle:$

[TABLE]

Hence, $\langle x_{a},x_{b},x_{c},x_{g}\rangle$also strictly separates $x_{s}$ and $x_{q}$, and $x_{t}\in P_{s}\subset P_{g}\subset P_{h}.$ It now follows from $x_{h}\rightarrow x_{g}\rightarrow\cdots\rightarrow x_{k}$ that $x_{t}\in P_{s}\subset P_{k}\subset P_{h};$ a contradiction.

We note that $\mathcal{V}^{j}=\{x_{j}\}\cup\mathcal{V}^{t}$ and that if $H_{h}\cap[\mathcal{V}^{j}]\neq\emptyset$ then $H_{h}$ strictly separates $x_{t}$ and $x_{j}$ by C.2, and $x_{t}<x_{h}$ by C.1. We now argue on above and obtain a contradiction. ∎

REMARKS We recall that $P_{m}=[x_{m},x_{m-1},\cdots,x_{1}]$ for $m=n,\cdots,6.$ Let $P_{5}$ denote any 4-subpolytope of $P_{6}$. In view of $2.5$,

5.1

there is a labeling of $\mathcal{V}(P_{6})$, which we may denote by $x_{1},x_{2}\cdots,x_{6}$, such that

• $\bullet$

$P_{6}$* satisfies Gale’s Evenness Condition with $x_{1}<x_{2}<\cdots<x_{6},Y=\{x_{1},x_{3},x_{5}\},Z=\{x_{2},x_{4},x_{6}\}$*

• $\bullet$

$P_{5}=[x_{1},x_{2},\cdots,x_{5}]$, and

• $\bullet$

any hyperplane through $\langle Y\rangle$ strictly separates two elements of $Z$.

We recall that $P=P_{n}$ is simply linked under $x_{n}>x_{n-1}>\cdots>x_{1}$ and $P_{m}=[x_{m},\cdots,x_{1}]$ for $m\geq 5$. Let $O$ be an interior point of $P,6\leq m\leq n-1$ and $O\in P_{m}\backslash P_{m-1}$. We note that a vertex of $[x_{m+1},\cdots,x_{n}]$ is linked to a vertex of $P_{m}$.

With $v=x_{w}>x_{m},Q=P_{m}$ and $R=[Q,v]$, we label $Q$ and $R^{*}=R/_{v}$

as in Section 3 so that $x_{w}\rightarrow y_{1}$ (hence, each $Z^{\prime}_{t}$ is empty) and $\langle x_{w},O\rangle\cap[y_{1},y_{2},y_{t},y_{t+1}]\neq\emptyset$ for some $3\leq t\leq a-1$. We let $T=[y_{1},y_{2},y_{t},y_{t+1}]$,

[TABLE]

and note that

$T\in\mathcal{F}(Q)\backslash\mathcal{F}(R)$ and $\mathcal{V}(P_{m})=Y_{a}\cup Z_{3}\cup\cdots\cup Z_{a-1}=\mathcal{V}(T)\cup Z_{t}^{-}\cup Z_{t}\cup Z_{t}^{+}$.

From the Schlegel diagram of $R^{*}$ on $[y_{1},y_{2},y_{a}]$ in Figure 1, we readily obtain diagrams of $R^{*}$ on 2-faces containing $[y_{1},y_{t}]$ or $[y_{2},y_{t}]$. In Figure 3, 4 and 5, we depict associated polygons $R/_{[y_{1},y_{2}]},R/_{[y_{1},y_{t}]}$ and $R/_{[y_{2},y_{t}]}$ that include $[\mathcal{V}^{w}]$(as per 3.2 and C.1) and hyperplanes $H_{1},H_{2},H_{3},H_{4}$ and $H_{5}$ that separate $O$ and $[\mathcal{V}^{w}]$. We note that each of $H_{2},H_{3},H_{4}$ and $H_{5}$ intersects and supports $[\mathcal{V}^{w}]$. For $i=1,\cdots,5$, let $H_{i}^{-}$ and $H_{i}^{+}$ denote the open half-spaces of $\bbbr^{4}$ determined by $H_{i}$ with $\mathcal{V}^{w}\subset H_{i}\cup H_{i}^{+}$.

REMARKS Let $F\in\mathcal{F}(P)$ and assume by 4.2 that $m\leq n-3$. From $\langle x_{w},O\rangle\cap T\neq\emptyset$, we have the following:

5.2

O is separated from all F with a common vertex by one of at most four hyperplanes; cf. LEMMA B.

5.3

$O\in[x_{m},P_{m-1}]\backslash P_{m-1}$* is separated from any $F\in\mathcal{F}(P_{m})$ by one of at most five hyperplanes.*

5.4

If $F\cap\mathcal{V}^{w}\neq\emptyset$ then $F$ intersects at most one of $Z_{t}^{-},Z_{t}$ and $Z_{t}^{+}$; cf. 3.4.

5.5

If $O\notin T$ then $O$ is separated from any $F$ such that $F\cap\mathcal{V}^{w}\neq\emptyset$ and $\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w}$ by one of $H_{2},H_{3},H_{4}$, and $H_{1}$ or $H_{5}$ in the case $F\cap(Z_{t}^{-}\cup Z_{t}\cup Z_{t}^{+})=\emptyset$.

5.6

Let $x_{m}\notin T$. Then $O$ is separated from any $F$ such that $\mathcal{V}(F)\subset\mathcal{V}(P_{m-1})\cup\mathcal{V}^{w}$ by one hyperplane.

Regarding 5.6; let $\langle x_{w},O\rangle\cap T=\{p_{1}\}$. Then $\langle x_{w},O\rangle\cap bd(P_{m-1})=\{p_{1},p_{2}\}$ and $\langle x_{w},O\rangle\cap bd(P_{m})=\{p_{1},p_{3}\}$ with $p_{2}\in[p_{1},p_{3}]$. From $O\in[x_{m},P_{m-1}]\backslash P_{m}$, we obtain that $O\in[p_{2},p_{3}]$. It is now clear that there is an $F^{\prime}\in\mathcal{F}(P_{m-1})$ such that $p_{2}\in F^{\prime}$ and $\langle F^{\prime}\rangle$ separates $O$ from $F$ with $\mathcal{V}(F)\subset\mathcal{V}(P_{m-1})\cup\mathcal{V}^{w}$.∎

Since $P$ is simply linked and $m\leq n-3$, we consider the case that $\{x_{m+1},\cdots,x_{n}\}$ is the union of mutually disjoint chains $\mathcal{V}^{w},\mathcal{V}^{s}$ and $\mathcal{V}^{r}$ with $x_{w}\rightarrow y_{1}\in P_{m},x_{s}\rightarrow\hat{y}_{1}\in P_{m}$ and $x_{r}\rightarrow\bar{y}_{1}=x_{m}$. Then with labelings analogous to the one for $Q=P_{m}$ and $R^{*}[P_{m},x_{w}]/_{x_{w}}$;

[TABLE]

corresponds to $[P_{m},x_{s}]/_{x_{s}}$ and

[TABLE]

corresponds to $[P_{m},x_{r}]/_{x_{r}}$.

We simplify the notation and let $u_{j}=\hat{y}_{j},U_{j}=\hat{Z}_{j},r_{j}=\bar{y}_{j}$ and $V_{j}=\bar{Z}_{j}$.

With reference to Figure 3, 4 and 5, we assume that $\{T,L,I\}\subset\mathcal{F}(P_{m})$ and that

• $\bullet$

$\langle x_{w},O\rangle\cap T\neq\emptyset$ with $T=[y_{1},y_{2},y_{t},y_{t+1}]\notin\mathcal{F}([P_{m},x_{w}])$,

• $\bullet$

$\langle x_{s},O\rangle\cap K\neq\emptyset$ with $K=[u_{1},u_{2},u_{k},u_{k+1}]\notin\mathcal{F}([P_{m},x_{s}])$,

• $\bullet$

$\langle x_{r},O\rangle\cap I\neq\emptyset$ with $I=[v_{1},v_{2},v_{i},v_{i+1}]\notin\mathcal{F}([P_{m},x_{r}])$,

• $\bullet$

$O$ is separated from $[\mathcal{V}^{s}]([\mathcal{V}^{r}])$ by $\hat{H}_{1},\cdots,\hat{H}_{5}(\bar{H}_{1},\cdots,\bar{H}_{5})$ and

• $\bullet$

$\mathcal{V}^{s}\subset\hat{H}_{j}\cup\hat{H}_{j}^{+}(\mathcal{V}^{r}\subset\bar{H}_{j}\cup\bar{H}_{j}^{+})$ for $j=1,\cdots,5$.

REMARK We refer to Figure 4, and consider any hyperplane $H^{\prime}$ through $\langle y_{1},y_{t},y_{t+1}\rangle$ in the case that $x_{s}\rightarrow u_{1}=y_{2}$. Then $[x_{s},y_{2}]\in\mathcal{U}(P_{s})$, and it follows from LEMMA C that if $H^{\prime}\cap Z_{t}\neq\emptyset$, then $H^{\prime}\cap[y_{2},x_{s}]=\emptyset,H^{\prime}\cap[\mathcal{V}^{s}]=\emptyset$ and $H^{\prime}$ strictly separates $[\mathcal{V}^{s}]$ and $[\mathcal{V}^{w}]$. The following now follows from 3.4:

5.7

If $u_{1}=y_{2}$ and $F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{w}$ then $F\cap Z_{t}=\emptyset$.

Lemma D

Let $F\in\mathcal{F}(P)$. Then $F$ intersects at most two of $\mathcal{V}^{2},\mathcal{V}^{4}$ and $\mathcal{V}^{6}$, and at most two of $\mathcal{V}^{w},\mathcal{V}^{s}$ and $\mathcal{V}^{r}$.

Proof

The existence of $\mathcal{V}^{w},\mathcal{V}^{s}$ and $\mathcal{V}^{r}$ imply that $\{x_{7},\cdots,x_{n}\}$ is the union of pairwise disjoint chains $\mathcal{V}^{e},\mathcal{V}^{f}$ and $\mathcal{V}^{g}$, say. Since $P_{6}$ is cyclic with $x_{1}<x_{2}<\cdots<x_{6}$, we may assume by 2.5 and 5.1 that $x_{e}\rightarrow x_{2},x_{f}\rightarrow x_{4}$ and $x_{g}\rightarrow x_{6}$.

From 5.1 and LEMMA C, we obtain that any hyperplane $H$ through $\langle x_{1},x_{3},x_{5}\rangle$ strictly separates two of $\mathcal{V}^{2},\mathcal{V}^{4}$ and $\mathcal{V}^{6}$. Hence, no face of $P$ intersects each of $\mathcal{V}^{2},\mathcal{V}^{4}$ and $\mathcal{V}^{6}$ by 3.4. ∎

REMARK We refer to Figure 3, 4 and 5, and consider a $v\in\mathcal{V}(P)$ with the property that $v\in H_{1}^{-},v\notin H_{2}^{+}\cup H_{3}^{+}\cup H_{4}^{+}$ and no hyperplane spanned from $v,y_{1},y_{2},y_{t},y_{t+1}$ intersects $[\mathcal{V}^{w}]$.

Then $\langle x_{w},y_{1},y_{2},y_{t}\rangle$ separates $v$ and $Z_{t}^{-},\langle x_{w},y_{1},y_{2},y_{t+1}\rangle$ separates $v$ and $Z_{t}^{+}$, and $\langle x_{w},y_{1},y_{t},y_{t+1}\rangle$ separates $v$ and $Z_{t}$. From $[P_{m},x_{w}]/_{[x_{w},y_{1}]}$, it now follows that $[x_{w},y_{1},v]$ is not a 2-face of $[P_{m},x_{w},v]$ or $[x_{w},y_{1},y_{2},y_{t},y_{t+1},v]$. Since the latter polytope is cyclic, we obtain from 5.1 that

5.8

$\langle v,y_{2},y_{t},y_{t+1}\rangle$* strictly separates $x_{w}$ and $y_{1}$.*

Lemma E

*Let $x_{w}<x_{s},F\in\mathcal{F}(P)$ and $F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}$. Then $O$ is separated from any such F by

E.1 at most three hyperplanes $(\hat{H}_{2},\hat{H}_{3},\hat{H}_{4})$ in the case $x_{w}\in\hat{H}_{1}^{-}$ and $O\notin bd(K)$,

E.2 one hyperplane $(H_{i},2\leq i\leq 4)$ in the case $u_{1}\notin T$ and $O\notin bd(T)$,

E.3 one hyperplane $(H_{i},2\leq i\leq 5)$ in the case $x_{w}\in\hat{H}_{i}^{+},u_{1}\in T,x_{s}\in H_{1}^{-}$ and $O\notin bd(T)$, and

E.4 at most two hyperplanes from $H_{2},H_{3},\hat{H}_{2},\hat{H}_{3}$ in the case $x_{w}\in\hat{H}_{1}^{+},x_{s}\in H_{1}^{+},O\notin bd(K)\cup bd(T)$ and either $T\neq K$ , or $T=K$ and $F\cap H_{1}^{-}\neq\emptyset\neq F\cap\hat{H}_{2}^{-}$.*

Proof

We refer to Figure 3, 4 and 5, and the analogous figures with $\mathcal{V}^{s},K,U_{j}=\hat{Z}_{j}$ and $\hat{H}_{j}$ for the location of $O$, and note $\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w}\cup\mathcal{V^{s}}$ by LEMMA D.

E.1 Let $x_{w}\in\hat{H}_{1}^{-}$. Then $\mathcal{V}^{w}\subset\hat{H}_{1}^{-}$ by C.1, and either $(\mathcal{V}(F)\cap\hat{H}_{1}^{-})\cap\hat{H}_{j}^{+}\neq\emptyset$ for some $j\in\{2,3,4\}$ or there is a $v\in\mathcal{V}^{w}$ such that $v\notin\hat{H}_{2}^{+}\cup\hat{H}_{3}^{+}\cup\hat{H}_{4}^{+}$. In case of the former, $O$ is separated from $F$ by $\hat{H}_{j}$; cf. 5.4. In case of the latter, it follows from $x_{w}<x_{s}$ and C.3 that $\langle v,u_{2},u_{k},u_{k+1}\cap[\mathcal{V}^{s}\cup\{u_{1}\}]=\emptyset$; a contradiction of 5.8.

E.2 Let $u_{1}\notin T$. Then $x_{s}\rightarrow u_{1}\in P_{m}$ yields that $u_{1}\in Z_{t}^{-}\cup Z_{t}\cup Z_{t}^{+}\subset H_{2}^{+}\cup H_{3}^{+}\cup H_{4}^{+}$ and $\mathcal{V}^{s}\subset H_{1}^{-}$. Now $x_{w}<x_{s}$ and C.1 yield that if $u_{1}\subset H_{j}^{+}$ then $\mathcal{V}^{s}\subset H_{j}^{+}$ and $O$ is separated from $F$ by $H_{j}$.

E.3 Let $x_{w}\in\hat{H}_{1}^{+}$ and $u_{1}\in T$. Then $u_{1}\in\{y_{2},y_{y},y_{t+1}\},y_{1}\in\{u_{2},u_{k},u_{k+1}\}$ and may assume that $u_{1}=y_{2}$ and $y_{1}=u_{2}$. From 5.7, we obtain that $F\cap Z_{t}=\emptyset=F\cap U_{k}$.

Let $x_{s}\in H_{1}^{-}$. Then $\langle T\rangle=H_{1}\neq\hat{H}_{1}=\langle K\rangle$ and $T\neq K$; cf. Figure 6 with $x_{s}\in H_{3}^{+}$, say, and $\mathcal{V}^{s}\subset H_{1}^{-}\cap H_{3}^{+}$. We consider the hyperplanes through $\langle y_{1},y_{2},y_{t+1}\rangle$ and obtain from 3.4 that $F\cap Z_{t}^{-}=\emptyset$. Hence, $O$ is separated from $F$ by $H_{3}$ in this case.

E.4 Let $x_{w}\in\hat{H}_{1}^{+},x_{s}\in H_{1}^{+}$ and $O\notin bd(K)\cup bd(T)$. Then $u_{1}\in T$, we assume that $(y_{1},y_{2})=(u_{2},u_{1})$ and note that $F\cap(Z_{t}\cup U_{k})=\emptyset$.

If $T\neq K$ then we may assume also that $\mathcal{V}^{s}\subset H_{1}^{+}\cup H_{3}^{+}$ as in Figure 6. If there is an $F^{\prime}\in\mathcal{F}(P)$ such that $O$ is not separated from $F^{\prime}$ by $H_{3}$ then $F^{\prime}\cap Z_{t}^{+}=\emptyset$ by 5.4. From $F^{\prime}\cap Z_{t}=\emptyset$, it now follows that $F^{\prime}\cap(Z_{t}^{-}\cup\{y_{t}\})\neq\emptyset$.

Let $Z_{t}^{-}(s)=\{z\in Z_{t}^{-}|\langle y_{1},y_{2},y_{t+1},z\rangle\ does\ not\ separate\ x_{w}\ and\ x_{s}\}$ . We apply C.1 and 3.4, and obtain that $F^{\prime}\cap Z_{t}^{-}\subset Z_{t}^{-}(s)$. Thus, either $O$ is separated from $F^{\prime}$ by $H_{1}$ (and so $\hat{H}_{2}$), or there is an $F^{\prime}$ such that $F^{\prime}\cap Z_{t}^{-}\neq\emptyset$. In the latter case, it is easy to check (cf. Figure 6) that $O$ is separated from any such $F^{\prime}$ by $H_{2}$ or $\hat{H}_{2}$.

Finally, let $T=K$ with $(y_{1},y_{2},y_{t},y_{t+1})=(u_{2},u_{1},u_{k},u_{k+1})$ and, say, $\mathcal{V}^{s}\subset H_{3}^{+}$. We argue now as above that if $O$ is not separated from $F^{\prime}\in\mathcal{F}(P)$ by $H_{3}$ or $H_{1}=\hat{H}_{1}$ then $O$ is separated from $F^{\prime}$ by $H_{2}$ or $\hat{H}_{2}$. ∎

REMARK We observe that under the hypotheses of LEMMA E, it follows from LEMMA A that

5.9

If $O\in bd(k)\cup bd(T)$ and $H$ is a separating hyperplane through $O$ then we may replace $H$ by three strictly separating hyperplanes.

6 SEPARATION RESULTS

With O$\in$intP, let $s(O)$ denote the minimum number of hyperplanes required to strictly separate $O$ from any facet of $P$. We prove that $s(O)\leq 16$ under the assumption that $P=P_{n}$ is simply linked under the array $x_{n}>x_{n-1}>\cdots>x_{1},P_{m}=[x_{m},\cdots,x_{2},x_{1}]$ for $m=n-1,\cdots,5$ and $\mathcal{V}(P)=\{x_{1},x_{3},x_{5}\}\cup\mathcal{V}^{2}\cup\mathcal{V}^{4}\cup\mathcal{V}^{6}$.

6.1

We consider first the case of $O\in P_{m}\backslash P_{m-1}$ for some $6\leq m\leq n$. As noted in Sections 4 and 5, we may assume that $m\leq n-3$ and that $\{x_{m+1},\cdots,x_{n}\}$ is the union of non-empty chains $\mathcal{V}^{r},\mathcal{V}^{s}$ and $\mathcal{V}^{w}$ described in Section 5.

Our arguments are based upon

• $\bullet$

the location of $x_{m}$ with respect to $T$ and $K$,

• $\bullet$

the order of $x_{r}$ with respect to $x_{w}<x_{s}$, and

• $\bullet$

the location of $O$ with respect to $T,K$ and $I$.

For each location of $O$, we present the separation result

• $\bullet$

$\{k\}$: property: rationale

to indicate that at most $k$ separating hyperplanes suffice for $F\in\mathcal{F}(P)$ with the indicated property due to the specified reasons. $\{-\}$ indicates that the separating hyperplanes for this case have already counted.

I. $x_{m}\notin T\cup K$

Then $T\cup K\subset P_{m-1}$ and $O\notin T\cup K$. From $x_{m}\in H_{1}^{-}\cap\hat{H}_{1}^{-},x_{r}\rightarrow x_{m}$ and 2.4,we have that $x_{r}\in H_{1}^{-}\cap\hat{H}_{1}^{-}$. Next, LEMMA D and its proof yield that any $F$ intersects at most two of $\mathcal{V}^{w},\mathcal{V}^{r}$ and $\mathcal{V}^{s}$, and that $[v_{1},u_{1},y_{1}]$ is not a 2-face of $P_{m}$. Hence , $x_{m}\in I$ implies that $\{u_{1},y_{1}\}\not\subset I$.

I.1 $O\notin bd(I)$

We apply our Lemmas and Remarks. Then

• $\bullet$

$\{4\}:x_{m}\in F:5.2$

• $\bullet$

$\{1\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m-1})\cup\mathcal{V}^{w}:5.6$

• $\bullet$

$\{1\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m-1})\cup\mathcal{V}^{s}:5.6$

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:E.1$, and

• $\bullet$

$\{4\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset:5.5$ with $\bar{H}_{2},\bar{H}_{3},\bar{H}_{4},\bar{H}_{5}$(as $v_{1}=x_{m}\notin F$).

It remains to consider $F$ that inersect $\mathcal{V}^{r}$ and $\mathcal{V}^{w}\cup\mathcal{V}^{s}$. Here, we apply $\{u_{1},y_{1}\}\not\subset I$ and LEMMA E with relabeling as necessary

I. 1.1 $x_{r}<x_{w}<x_{s}$

As $x_{w}$ and $x_{s}$ are interchangeable with respect to $x_{r}$, we assume that $u_{1}\notin I$,

say. Then

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{r}<x_{s},u_{1}\notin I$ and E.2 with $\bar{H}_{2},\bar{H}_{3}$ and $\bar{H}_{4}$ already counted, and

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{w}:x_{r}<x_{w},x_{r}\in H_{1}^{-}$ and E.1.

I. 1.2 $x_{w}<x_{r}<x_{s}$

If $u_{1}\notin I$ then one case is above, and

• $\bullet$

$\{1\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{w}:x_{w}<x_{r},v_{1}=x_{m}\notin T$ and E.2

If $u_{1}\in I$ and $y_{1}\notin I$ then

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{w}:x_{w}<x_{r},x_{w}\in\bar{H}_{1}^{-}$ and E.1, and

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{r}<x_{s},x_{r}\in\hat{H}_{1}^{-}$ and E.1 .

I. 1.3 $x_{w}<x_{s}<x_{r}$

Then $v_{1}=x_{m}\notin T\cup K$ and E.2 yield $\{1\}$ for $F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}$, and $\{1\}$ for $F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{r}$.

I.2 $O\in bd(I)$

We recall that $\langle x_{w},O\rangle\cap T\neq\emptyset$ and $O\notin T$. Hence, $H_{1}=\langle T\rangle$ strictly separates $O$ and $x_{w}$, and $x_{w}$ is necessarily beneath any facet of $P_{m}$ that contains $O$. Thus $x_{w}\in\bar{H}_{1}^{-}$ and, similarly,$x_{s}\in\bar{H}_{1}^{-}$; whence $\mathcal{V}^{w}\cup\mathcal{V}^{s}\subset\bar{H}_{1}^{-}$.Since $v_{1}=x_{m}$ implies that $\bar{H}_{1}\cap\bar{H}_{5}=[v_{2},v_{i},v_{i+1}]\subset bd(P_{m-1})$, it follows from $O\in\bar{H}_{1}\backslash P_{m-1}$ that $O\notin[v_{2},v_{i},v_{i+1}]$. From these observations, we have that

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{r}=\emptyset:(LEMMA)A$.

• $\bullet$

$\{8\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset$ :5.5, A and 5.9 with $\bar{H}_{1},\bar{H}_{2},\bar{H}_{3},\bar{H}_{4}$ as separating hyperplanes and $O\in\bar{H}_{1}\cap\bar{H}_{j}$ for some $j\in\{2,3,4\}$. We apply LEMMA A and replace $\bar{H}_{1}$ and $\bar{H}_{j}$ as per 5.9. We indicate these eight hyperplanes by $2\bar{H}_{i}+3+3$.

We now argue as in I.1.1, I.1.2 and I.3 with 5.9 applied for $\bar{H}_{1}$ and $\bar{H}_{j}$, and obtain the same counts. Thus, $s(O)\leq 16$ in each of these cases.

II. $x_{m}\in K$ and $x_{m}\notin T$.

Then $T\subset P_{m-1},O\notin T,x_{r}\in H_{1}^{-}$ and we let $x_{m}=u_{2}$. We have again that $\{y_{1},u_{1}\}\not\subset I$; and from $\{v_{1},u_{1}\}\subset K$, it follows that $y_{1}\notin K$ and $x_{w}\in\hat{H}_{1}^{-}$. We note that $x_{m}=u_{2}$ yields that $\hat{H}_{4}$ separates $O$ from any $F$ with $x_{m}\notin F$ and $\mathcal{V}(F)\subset\mathcal{V}^{s}\cup\{u_{1},u_{2},u_{k},u_{k+1}\}$.

II.1 $O\notin bd(I)\cup bd(K)$

Similarly to I.1, we obtain

• $\bullet$

$\{4\}:x_{m}\in F:5.2$,

• $\bullet$

$\{1\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m-1})\cup\mathcal{V}^{s}:5.6$,

• $\bullet$

$\{3\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{s},F\cap\mathcal{V}^{s}\neq\emptyset:5.5$ with $\hat{H}_{2},\hat{H}_{3},\hat{H}_{4}$,

• $\bullet$

$\{4\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset:5.5$ with $\bar{H}_{2},\bar{H}_{3},\bar{H}_{4},\bar{H}_{5}$ and

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s},x_{w}\in\hat{H}_{1}^{-}$, and E.1.

II.1.1 $x_{r}<x_{w}<x_{s}$

If $u_{1}\notin I$ then we recall that $x_{r}\in H_{1}^{-}$ and argue as in I.1. If $y_{1}\notin I$ then

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{w}:x_{r}<x_{w}$ and E.2.

For $F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}$; we obtain from $x_{r}<x_{s}$ and $u_{1}\in I$ that one of E.1, E.3 or E.4 is applicable. We note that E.1 and E.3 yield $\{-\}$, and E.4 yields either $\{-\}$ or $\{3\}$ with $O\in I=K$ and 5.9 applied to $\hat{H}_{1}=\bar{H}_{1}$.

Henceforth, as a simplification, we list only “worst case scenario”results. In that regard, it is noteworthy that the assertion of E.4 is the same if $x_{s}$ and $x_{w}$ are interchanged.

II.1.2 $x_{w}<x_{r}<x_{s}$ or $x_{w}<x_{s}<x_{r}$

Then as worst case scenarios, we have

• $\bullet$

$\{1\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}:x_{m}=v_{1}\notin T$ and E.2, and

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{r}:$ E.4, 5.9 with $O\in\hat{H}_{1}=\bar{H}_{1}$.

II.2 $O\in bd(I)$

We note that as in I.2; $x_{w}\in\bar{H}_{1}^{-}$ and

• $\bullet$

$\{8\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cap\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset:2\bar{H}_{i}+3+3$ with $O\in\bar{H}_{1}\cap\bar{H}_{j}$ for some $j\in\{2,3,4\}$.

If $x_{s}\in\bar{H}_{1}^{-}$ then $\mathcal{V}^{w}\cup\mathcal{V}^{s}\subset\bar{H}_{1}^{-}$,

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{r}=\emptyset$: LEMMA A

and, as worst case scenario, E.2 and $\{y_{1},u_{1}\}\not\subset I$ yield $x_{r}<x_{w}<x_{s}$ and $u_{1}\in I$. Then

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{w}:y_{1}\notin I$ and E.2, and

• $\bullet$

$\{5\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{r}\in\hat{H}_{1}^{-}$, E.1, 5.9 with $O\in\hat{H}_{i}$ for some $i\in\{2,3,4\}$

Let $x_{s}\in\bar{H}_{1}^{+}$. Then $u_{1}\in I$ with $u_{1}=v_{2}$, say, and $\langle x_{s},O\rangle\cap\ bd(P_{m})\subset I.$ Hence, we choose $K=I$ with $u_{2}=v_{1},u_{k}=v_{i}$ and $u_{k+1}=v_{i+1}$. Then $\hat{H}_{1}=\bar{H}_{1}$ and

• $\bullet$

$\{5\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{s},F\cup\mathcal{V}^{s}\neq\emptyset:2\hat{H}_{i}+3$ with $O\in\bar{H}_{1}\cap\hat{H}_{j}$ and $\{\hat{H}_{j},\hat{H}_{i},\hat{H}_{i}\}=\{\hat{H}_{2},\hat{H}_{3},\hat{H}_{4}\}$.

From $\{y_{1},u_{1}\}\not\subset I$, we obtain that $y_{1}\notin I$ and $x_{w}\in\bar{H}_{1}^{-}=\hat{H}_{1}^{-}$ and

• $\bullet$

$\{3\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w}$: LEMMA A,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s},x_{w}\in\hat{H}_{1}^{-}$ and E.1,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}$ : either $x_{w}<x_{r}$ and E.1, or $x_{r}<x_{w},y_{1}\notin I$ and E.2, and

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}$: E.4 with $\bar{H}_{1}=\hat{H}_{1},\bar{H}_{2},\bar{H}_{3},\hat{H}_{2},\hat{H}_{3}$.

II.3 $O\in bd(K)$ and $O\notin bd(I)$

We recall that $x_{w}\in\hat{H}_{1}^{-}$, and note that $I\neq K$ implies that $x_{r}\in\hat{H}_{1}^{-}$. Then

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{s}=\emptyset$: LEMMA A,

• $\bullet$

$\{8\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{s},F\cap\mathcal{V}^{s}\neq\emptyset:2\hat{H}_{i}+3+3$ with $O\in\hat{H}_{1}\cap\hat{H}_{j}$ for some $j=\{2,3,4\}$,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s},x_{w}\in\hat{H}_{1}^{-}$ and E.1, and as worst case scenario,

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{r}:x_{s}<x_{r},x_{s}\in\bar{H}_{1}^{-}$ and E.1.

III. $x_{m}\in T$ and $x_{m}\notin K$.

As $\{y_{1},v_{1}=x_{m}\}\subset$T and K$\subset$Pm-1, we have that $u_{1}\in H_{1}^{-},\mathcal{V}^{s}\subset H_{1}^{-},\mathcal{V}^{r}\subset\hat{H}_{1}^{-}$ and $O\notin K$. We let $v_{1}=y_{2}$ and note that $H_{4}$ separates $O$ from any $F$ with $x_{m}\notin F\subset[\mathcal{V}^{w}\cup\{y_{1},y_{t},y_{t+1}\}]$.

III.1. $O\notin bd(I)\cup bd(T)$

As in II.1, we obtain that

• $\bullet$

$\{4\}:x_{m}\in F$: 5.2,

• $\bullet$

$\{1\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m-1})\cup\mathcal{V}^{s}$: 5.6,

• $\bullet$

$\{3\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w},F\cap\mathcal{V}^{w}\neq\emptyset$: 5.5 with $H_{2},H_{3},H_{4}$,

• $\bullet$

$\{4\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset$: 5.5 with $\bar{H}_{2},\bar{H}_{3},\bar{H}_{4},\bar{H}_{5}$, and

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s},u_{1}\neq T$ and E.2.

We note that our repetitive arguments are dependent upon Lemmas A and E, and $\{u_{1},y_{1}\}\not\subset I$. Also that we present only worst case scenarios.

If $x_{r}<x_{s}$ then with $u_{1}\in I$ and $y_{1}\notin I$, we have

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{r}\in\hat{H}_{1}^{-}$ with E.1, and

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{w}$: either $x_{r}<x_{w}$ with E.2, or $x_{w}<x_{r}$ with E.1.

Let $x_{w}<x_{s}<x_{r}$. By E.1, we may assume that $\{x_{w},x_{s}\}\not\subset\bar{H}_{1}^{-}$. With $x_{s}\in\bar{H}_{1}^{+}$ and $x_{r}\in\hat{H}_{1}$, we have $u_{1}\in I,y_{1}\notin I,x_{w}\in\bar{H}_{1}^{-}$,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}$: E.1, and

• $\bullet$

$\{1\}:F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{r}$ :E.2 or E.3.

With $x_{w}\in\bar{H}_{1}^{+}$, we have $x_{s}\in\bar{H}_{1}^{-}$,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{r}$ : E.1, and

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}$ : E.4 and 5.9 with $O\in T=I$.

III.2 $O\in bd(I)$.

We note as in I.2 that $x_{s}\in\bar{H}_{1}^{-}$ follows from $O\notin K$. Next, we obtain the same separating hyperplanes for $F$ with $F\cap\mathcal{V}^{r}\neq\emptyset$ and $\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r}$ as in II.2, and with $x_{w}$ and $x_{s}$ interchanged, the corresponding worst case scenario for $x_{w}\in\bar{H}_{1}^{-}$.

Let $x_{w}\in\bar{H}_{1}^{+}$. Then we choose $T=I$ and, similarly to II.2, obtain that

• $\bullet$

$\{5\}:\mathcal{V}(P)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w},F\cap\mathcal{V}^{w}\neq\emptyset:2H_{i}+3$ with $H_{1}=\bar{H_{1}}$, and $O\in H_{1}\cap H_{j}$ for some $j\in\{2,3,4\}$,

• $\bullet$

$\{3\}:\mathcal{V}(P)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{s}$: LEMMA A,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s},u_{1}\notin T$ and E.2,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}$: either $x_{r}<x_{s},u_{1}\notin I$ and E.2, or $x_{s}<x_{r},x_{s}\in\bar{H}_{1}^{-}$ and E.1, and

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}$: E.4.

III.3 $O\in bd(T)$ and $O\notin bd(I)$.

Then $I\neq T$ and $x_{r}\in H_{1}^{-}$. We recall that $y_{1}\notin T$and $x_{s}\in H_{1}^{-}$. Hence,

• $\bullet$

$\{8\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w},F\cap\mathcal{V}^{w}\neq\emptyset:2H_{i}+3+3$ with $O\in H_{1}\cap H_{j}$ and $\{H_{j},H_{i},H_{i}\}=\{H_{2},H_{3},H_{4}\}$.

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{w}=\emptyset$: LEMMA A,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s},u_{1}\notin T$ and E.2, and as worst case scenario,

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}:x_{w}<x_{r},x_{w}\in\bar{H}_{1}^{-}$ and E.1.

IV $x_{m}\in T\cap K$.

We let $v_{1}=x_{m}=y_{2}=u_{2}$, and note that $\{v_{1},y_{1}\}\subset T$ implies that $u_{1}\notin T$ and $x_{s}\in H_{1}^{-}$; and $\{v_{1},u_{1}\}\subset K$ implies that $y_{1}\notin K$ and $x_{w}\in\hat{H}_{1}^{-}$.

IV.1 $O\notin bd(K)\cup bd(T)\cup bd(I)$.

We recall that $O\in P_{m}\backslash P_{m-1}$. Then

• $\bullet$

$\{5\}:x_{m}\in F$ or $F\subset P_{m}$: 5.3,

• $\bullet$

$\{3\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w}\neq\emptyset,y_{2}=x_{m}\notin F$: 5.5 with $H_{2},H_{3}$, and $H_{4}$ for $F\cap Z_{t}\neq\emptyset$ or $F\cap(Z_{t}^{-}\cup Z_{t}\cup Z_{t}^{+})=\emptyset$,

• $\bullet$

$\{3\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{s},F\cap\mathcal{V}^{s}\neq\emptyset,u_{2}=x_{m}\notin F$: 5.5 with $\hat{H}_{2},\hat{H}_{3},\hat{H}_{4}$,

• $\bullet$

$\{4\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset$: 5.5 with $\bar{H}_{2},\bar{H}_{3},\bar{H}_{4},\bar{H}_{5}$, and

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s},x_{w}\in\hat{H}_{1}^{-}$ and E.1.

We observe that for $x_{r}$ and $x_{s}:$ E.1 and E.2 yield $\{-\}$ for $F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}$, and E.3 and E.4 yield $u_{1}\in I,v_{1}\in K$ and the worst case scenario

• $\bullet$

$\{1\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}$ : either $x_{s}<x_{r}$ with $\hat{H}_{5}$, or $x_{r}<x_{s}$ with $\bar{H}_{1}=\hat{H}_{1}$.

The corresponding observation for $x_{r}$ and $x_{w}$, and $\{u_{1},y_{1}\}\not\subset I$, now yield

• $\bullet$

$\{1\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap(\mathcal{V}^{w}\cup\mathcal{V}^{s})$.

IV.2 $O\in bd(K)$.

Then $O\notin P_{m-1}$ and $u_{2}=x_{m}$ imply that $O\notin[u_{1},u_{k},u_{k+1}]$ and

• $\bullet$

$\{8\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{s},F\cap\mathcal{V}^{s}\neq\emptyset:2\hat{H}_{i}+3+3$ with $O\in\hat{H}_{1}\cap\hat{H}_{j}$ and $\{\hat{H}_{j},\hat{H}_{i},\hat{H}_{i}\}=\{\hat{H}_{2},\hat{H}_{3},\hat{H}_{4}\}$

We recall that $x_{w}\in\hat{H}_{1}^{-}$. If $x_{r}\in\hat{H}_{1}^{-}$ then

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{s}=\emptyset$: LEMMA A,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s}$ and E.1, and

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{r}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{s}<x_{r}$ and either $x_{s}\in\bar{H}_{1}^{-}$ and E.1, or $x_{s}\in\bar{H}_{1}^{+},v_{1}\in K,x_{r}\in\hat{H}_{1}^{-}$ and E.3, 5.9 with $O\in\hat{H}_{5}$.

Let $x_{r}\in\hat{H}_{1}^{+}$. Then we choose $I=K$ with $(v_{1},v_{2},v_{i},v_{i+1})=(u_{2},u_{1},u_{k},u_{k+1})$, and note that $y_{1}\in I=K$ and $x_{w}\in\bar{H}_{1}^{-}=\hat{H}_{1}^{-}$. Now

• $\bullet$

$\{3\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w}$: LEMMA A with $\hat{H}_{1}^{-}$,

• $\bullet$

$\{5\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset:2\bar{H}_{i}+3$ with $O\in\hat{H}_{1}\cap\bar{H}_{j}$ and $\{\bar{H}_{j},\bar{H}_{i},\bar{H}_{i}\}=\{\bar{H}_{2},\bar{H}_{3},\bar{H}_{4}\}$,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s}$ and E.1,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}$: either $x_{w}<x_{r}$ and E.1, or $x_{r}<x_{w}$ and E.2, and

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{r}$: E.4.

IV.3 $O\in bd(T)$

We argue as in IV.2 with

• $\bullet$

$\{8\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{w},F\cap\mathcal{V}^{w}\neq\emptyset:2H_{i}+3+3$,

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{s}:x_{w}<x_{s}$ and E.2, and the cases $x_{r}\in H_{1}^{-}$ and $x_{r}\in H_{1}^{+}$.

IV.4 $O\in bd(I)$ and $O\notin bd(K)\cup bd(T)$

Since we choose $K=I(T=I)$ if $x_{s}\in\bar{H}_{1}^{+}(x_{w}\in\bar{H}_{1}^{+})$, we may assume that $\{x_{w},x_{s}\}\subset\bar{H}_{1}^{-}$. Then

• $\bullet$

$\{8\}:\mathcal{V}(F)\subset\mathcal{V}(P_{m})\cup\mathcal{V}^{r},F\cap\mathcal{V}^{r}\neq\emptyset:2\bar{H}_{1}+3+3$ with $O\in\bar{H}_{1}\cap\bar{H}_{j}$ and $\{\bar{H}_{j},\bar{H}_{i},\bar{H}_{i}\}=\{\bar{H}_{2},\bar{H}v,\bar{H}_{4}\}$, and

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{r}=\emptyset$: LEMMA A

If $u_{1}\notin I$ ,then

• $\bullet$

$\{-\}:F\cap\mathcal{V}^{s}\neq\emptyset\neq F\cap\mathcal{V}^{r}$: either $x_{r}<x_{s}$ and E.2, or $x_{s}<x_{r}$ and E.1, and

• $\bullet$

$\{3\}:F\cap\mathcal{V}^{w}\neq\emptyset\neq F\cap\mathcal{V}^{r}:x_{r}<x_{w}$ and E.1.

Let $u_{1}\in I$. Then $y_{1}\notin I$ and we argue as above with $x_{s}$ and $x_{w}$ interchanged.∎

6.2

It remains to determine $s(O)$ in the case that $O\in P_{6}$ and (in view of 5.1) $O$ is contained in every 4-subpolytope of $P_{6}$. Since $P_{6}$ satisfies Gale’s Evenness Condition with $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}<x_{6}$ and $O$ is not contained in any facet of $P_{6}$, it follows that

[TABLE]

From LEMMA D and its proof, we have

• $\bullet$

$\mathcal{V}(P)=\{x_{1},x_{3},x_{5}\}\cup\mathcal{V}^{2}\cup\mathcal{V}^{4}\cup\mathcal{V}^{6}$,

• $\bullet$

any hyperplane through $\langle x_{1},x_{3},x_{5}\rangle$ intersects at most two of $\mathcal{V}^{2},\mathcal{V}^{4}$ and $\mathcal{V}^{6}$, and

• $\bullet$

any $F\in\mathcal{F}(P)$ intersects at most two of $\mathcal{V}^{2},\mathcal{V}^{4}$ and $\mathcal{V}^{6}$.

Let $\mathcal{W}^{ij}=[\{x_{1},x_{3},x_{5}\}\cup\mathcal{V}^{i}\cup\mathcal{V}^{j}],i\neq j$ in $\{2,4,6\}$. Then $[x_{1},x_{3},x_{5}]\subset bd(\mathcal{W}^{ij})$, any $F\in\mathcal{F}(P)$ is contained is some $\mathcal{W}^{ij}$, and $s(O)\leq 9$ by LEMMA A. ∎

We conclude with the observation that any linked $P$ with $|\mathcal{V}(P)|\leq 11$ is simply linked, and the problem: Is every linked $P$ also simply linked?