opology of complements of skeleton

Given a polytopal complex $X$, we examine the topological complement of its $k$-skeleton. We construct a long exact sequence relating the homologies of the skeleton complements and links of faces in $X$, and using this long exact sequence, we obtain characterisations of CohenMacaulay and Leray complexes, stacked balls, and neighbourly spheres in terms of their skeleton complements. We also apply these results to CAT(0) cubical complexes, and find new similarities between such a complex and an associated simplicial complex, the crossing complex.

Introduction {#sec:introduction}
============

Polytopal complexes are important objects in topology and combinatorics, which include simplicial complexes, cubical complexes and polytopes. Their geometric realisations provide examples of a wide array of topological spaces, and much research has gone into studying the interplay between their combinatorial and topological aspects.

One important feature of a polytopal complex is its *$k$-skeleton*, the set of faces of the complex of dimension less than or equal to $k$. Skeletons act as a framework which the high-dimensional faces are attached to, so studying the structure of a skeleton can reveal a lot about a complex. For example, @art:Bayer surveys many results about objects that can be reconstructed from their $k$-skeletons for certain values of $k$. Skeletons also play an important role in defining cellular homology, and in important topological results such as Poincaré duality.

In this paper, however, we aim to approach this topic from the other direction, starting from the higher-dimensional faces instead of the low-dimensional ones. We define the *$k$th co-skeleton* of a complex to be the set of faces of dimension *higher* than $k$. We discover that there is a strong relationship between co-skeletons and *links*: a link of a face in a polytopal complex captures the local structure of the complex around that face, so in a sense, the co-skeletons give us a "global" summary of the "local" information of the complex.

of this paper sets out the basic definitions we will use. In \[sec:general\], we examine some initial facts about co-skeletons of arbitrary complexes, and their connections with various forms of duality (see \[thm:polytope-duality,thm:Alexander-duality-for-coskeletons\]). In \[sec:LES\], we construct a long exact sequence relating the $k$th and $(k-1)$th co-skeletons and the links of $k$-dimensional faces (\[thm:LES\]). In \[sec:families\], we use this long exact sequence to study some families of complexes defined by properties of links — specifically, Cohen–Macaulay complexes, Leray complexes, and stacked balls — and give characterisations of these families in terms of the homology of their co-skeletons (\[thm:CM-characterisation,thm:Leray-characterisation,thm:stacked-characterisation\]). Finally, \[sec:CAT(0)\] examines cubical complexes, particularly CAT(0) cubical complexes: we examine the "crossing complex" defined by @art:Rowlands-CAT0, and show that a CAT(0) cubical complex has one of the properties discussed in \[sec:families\] if and only if its crossing complex shares that property (\[thm:CAT(0)-property-iff-crossing\]).

Acknowledgments {#acknowledgments .unnumbered}
---------------

Many thanks to Isabella Novik for her multitudinous helpful suggestions on drafts. Thanks also to Martina Juhnke-Kubitzke for her insightful comments.

The author was partially supported by a graduate fellowship from NSF grant DMS-1953815.

Preliminaries {#sec:preliminaries}
=============

We begin with some definitions. Readers familiar with polytopal complexes and topology may skip most of this section, but beware that we give slightly non-standard definitions in a couple of places, specifically for geometric realisations and links.

Polytopal complexes
-------------------

A *polytopal complex* $X$ is a collection of polytopes in Euclidean space $\mathbb R^N$ with the following properties:

-   If $\sigma$ is in $X$ and $\tau$ is a face of $\sigma$, then $\tau$ is in $X$, and

-   If $\sigma$ and $\sigma'$ are polytopes in $X$, then $\sigma \cap \sigma'$ is a face of each (possibly the empty face).

See \[fig:running-example\] for a small example. In this paper, we will only consider finite polytopal complexes.

\[scale=0.8\]

(v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v1)–(v5)–(v6)–cycle; (v1)–(v2)–(v7)–(v6)–cycle;

iin <span>1,...,7</span> at (vi) ;

If every polytope in $X$ is a simplex, then we say that $\Delta \coloneqq X$ is a *(geometric) simplicial complex*. If every polytope is a cube (that is, a polytope combinatorially equivalent to $[0,1]^i$ for some dimension $i \geq 0$), then ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}\coloneqq X$ is a *cubical complex*.

An *abstract simplicial complex* $\Delta$ is a poset isomorphic to the poset of faces of a geometric simplicial complex, ordered by inclusion. Equivalently, it is a collection of subsets of some finite set, with the property that $\sigma \in \Delta$ and $\tau \subseteq \sigma$ implies $\tau \in \Delta$. The dimension of a face $\sigma$ is $\dim \sigma \coloneq \# \sigma - 1$.

In any of these types of complex, faces of dimension $0$ and $1$ are called *vertices* and *edges* respectively, and a maximal face (by inclusion) is called a *facet*. If all facets of a complex have the same dimension, the complex is *pure*. The dimension of the complex is the maximum dimension of its faces. The number of faces of dimension $k$ in a complex $X$ is denoted $f_k(X)$.

If $\sigma$ is a polytope, $\abs{\sigma}$ will denote its relative interior. If $S$ is a collection of faces in a polytopal complex, then their *geometric realisation* $\abs{S}$ is the union of the relative interiors of the polytopes in $S$. This is slightly different from the usual definition of a geometric realisation — if $\sigma$ is in $S$, we do not include the boundary of $\sigma$ in the geometric realisation $\abs{S}$ unless those boundary faces are also part of $S$, unlike other definitions. We will take care to distinguish between $S$ as a set of polytopes and $\abs{S}$ as a topological space.

Suppose $Y$ is a subset of a polytopal complex $X$. We write $X \setminus Y$ to denote the difference of sets, so $X \setminus Y$ is the set of faces of $X$ that are not in $Y$ (which is not generally a polytopal complex). If $Y$ is a polytopal subcomplex of $X$, then $X - Y$ will denote the polytopal complex consisting of all faces of $X$ that do not intersect any faces of $Y$. We will reserve "$-$" to denote this "combinatorial" deletion, and use "$\setminus$" to denote deletion of sets or topological spaces. Note that $\abs{X \setminus Y} = \abs{X} \setminus \abs{Y}$, but the space $\abs{X - Y}$ is not in general the same; see \[fig:deletion\], for an example. However, these spaces are sometimes related by the following lemma:

\[<span>[@book:Munkres-EAT Lemma 70.1]</span>\] \[thm:deletion-homotopy-equivalence\] If $\Delta$ is a simplicial complex and $\Lambda$ is an induced subcomplex (in other words, every face of $\Delta$ whose vertices are contained in the vertex set of $\Lambda$ is a face of $\Lambda$), then $\abs{\Delta - \Lambda}$ and $\abs{\Delta} \setminus \abs{\Lambda}$ are homotopy equivalent.

<span>0.3</span>

\[yscale=0.87, baseline=-1.4cm\]

/in <span>-0.5/1, 0.5/1, 1.5/1, 0/0, 1/0, 2/0, 1.5/-1, 2.5/-1</span> (vi) at (,);

//in <span>1/2/4, 2/4/5, 2/3/5, 3/5/6, 5/6/7, 6/7/8</span> (v) – (v) – (v) – cycle;

iin <span>1,...,8</span> at (vi) ;

;

;

<span>0.3</span>

\[yscale=0.87, baseline=-1.4cm\]

/in <span>-0.5/1, 0.5/1, 1.5/1, 0/0, 1/0, 2/0, 1.5/-1, 2.5/-1</span> (vi) at (,);

(v3) – (v5) – (v6) – cycle; //in <span>1/2/4, 2/4/5, 2/3/5</span> (v) – (v) – (v) – cycle;

(v3) – (v6); (v5) – (v6);

iin <span>1,...,4</span> at (vi) ; iin <span>5,...,6</span> at (vi) ;

<span>0.3</span>

\[yscale=0.87, baseline=-1.4cm\]

/in <span>-0.5/1, 0.5/1, 1.5/1, 0/0, 1/0, 2/0, 1.5/-1, 2.5/-1</span> (vi) at (,);

//in <span>1/2/4</span> (v) – (v) – (v) – cycle;

(v2) – (v3);

iin <span>1,...,4</span> at (vi) ;

Given a face $\sigma$ of $X$, the *(open) star* of $\sigma$ is the set of faces $$\operatorname{star}_X \sigma \coloneqq \{ \tau \in X : \sigma \subseteq \tau \}.$$ The geometric realisation of a star is always contractible, if $\sigma \neq {\varnothing}$. The *link* of $\sigma$ is the set $$\begin{aligned}
\operatorname{link}_X \sigma & \coloneqq \{ \tau \in X : \sigma \subset \tau, \ \tau \neq \sigma \} \\
& = \operatorname{star}_X \sigma \setminus \{\sigma\}.\end{aligned}$$ We will sometimes simply write "$\operatorname{star}\sigma$" and "$\operatorname{link}\sigma$" if the space $X$ is clear from context. Note that if $\Delta$ is an abstract simplicial complex, the usual definition of a link is slightly different: we will refer to the usual definition as the "simplicial link", defined by $$\operatorname{s-link}_\Delta \sigma \coloneq \{ \tau \in \Delta : \sigma \cup \tau \in \Delta, \ \sigma \cap \tau = {\varnothing}\}.$$ Although our link and the simplicial link are not the same, they are homotopy equivalent.

If $X$ is a polytopal complex, its *barycentric subdivision* is the abstract simplicial complex $\operatorname{bary}(X)$ which has one vertex $v_\sigma$ for each non-empty face $\sigma$ of $X$, and a set $\{ v_{\sigma_1}, \dotsc, v_{\sigma_m} \}$ forms a face of $\operatorname{bary}(X)$ whenever $\{\sigma_1, \dotsc, \sigma_m\}$ is a chain in the poset of faces of $X$ ordered by inclusion (that is, $\sigma_1 \subset \dotsb \subset \sigma_m$, up to reordering). If each vertex $v_\sigma$ is positioned at the barycentre of the polytope $\sigma$, then the geometric realisation of $\operatorname{bary}(X)$ is exactly the geometric realisation of $X$, as a topological space. See \[fig:bary(X)\] for an example.

\[scale=0.8\]

(v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v1)–(v5)–(v6)–cycle; (v1)–(v2)–(v7)–(v6)–cycle;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[vertex\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[vertex\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[vertex\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[vertex\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

iin <span>1,...,5,12,23,34,45,15</span> (vi)–(v12345); iin <span>1,5,6,15,16,56</span> (vi)–(v156); iin <span>1,2,6,7,12,27,67,16</span> (vi)–(v1267);

iin <span>1,...,7</span> at (vi) ;

Given a polytopal complex $X$, the set of $k$-dimensional faces of $X$ will be denoted $X_k$. The *$k$-skeleton* of $X$ is the subcomplex $$\operatorname{Skel}_k X \coloneqq \{ \sigma \in X : \dim \sigma \leq k \}.$$ The central definition in this paper is the *$k$th co-skeleton* of $X$, which is the complement of the the $k$-skeleton: $$\begin{aligned}
\operatorname{Skel}^c_k X & \coloneqq X \setminus \operatorname{Skel}_k X \\
& = \{ \sigma \in X : \dim \sigma > k \}.\end{aligned}$$ We will mostly be interested in the topological properties of $\abs{\operatorname{Skel}^c_k X} = \abs{X} \setminus \abs{\operatorname{Skel}_k X}$. Note that the co-skeleton is not in general a polytopal complex, except in two special cases: $\operatorname{Skel}^c_{-1} X$ is $X$ itself (modulo the empty face, which makes no difference to the topology), and if $X$ is $d$-dimensional, $\operatorname{Skel}^c_d X$ is the polytopal complex with no faces.

Topology
--------

We will use the notation $A \simeq B$ to indicate that the spaces $A$ and $B$ are homotopy equivalent.

We assume that the reader is familiar with homology and cohomology; for background, refer to @book:Hatcher [@book:Munkres-EAT]. While this paper does focus on cell complexes, many spaces we consider are not themselves cell complexes, so $H_i(A)$ will denote singular homology, with coefficients in $R$ where $R$ is a field or $\mathbb Z$. Reduced homology is denoted by ${\widetilde}H_i(A)$. We use the convention that ${\widetilde}H_{-1}({\varnothing}) = R$. Analogous statements apply to cohomology, denoted $H^i(A)$.

For reference, here are two important theorems from algebraic topology that we will use repeatedly.

\[Mayer–Vietoris, <span>[@book:Hatcher p. 149], [@book:Munkres-EAT Theorem 33.1]</span>\] \[thm:Mayer-Vietoris\] If $A$ and $B$ are open subsets of a topological space, then there is a long exact sequence: $$\begin{tikzcd}[column sep=small]
        \dotsb \rar & H_i(A \cap B) \rar & H_i(A) \oplus H_i(B) \rar & H_i(A \cup B) \rar & H_{i-1}(A \cap B) \rar & \dotsb.
    \end{tikzcd}$$ If $A \cap B \neq {\varnothing}$, then we may replace these unreduced homology groups with reduced homology throughout.

\[Nerve theorem, <span>[@book:Hatcher Corollay 4G.3], [@art:Bjorner Theorem 10.7]</span>\] \[thm:nerve\] Suppose $\mathcal U = \{U_1, \dotsc, U_n\}$ is a family of open sets whose union is a paracompact space (e.g. any subspace of $\mathbb R^n$), or a family of closed sets whose union is a triangulable space. Suppose further that for every index set $I \subseteq \{1, \dotsc, n\}$, the intersection $\bigcap_{i \in I} U_i$ is either contractible or empty.

Construct a simplicial complex $N(\mathcal U)$ (called the "nerve" of $\mathcal U$) where the vertex set is $\{1, \dotsc, n\}$ and the set $I$ forms a face whenever $\bigcap_{i \in I} U_i$ is non-empty. Then $$\bigcup_{i=1}^n U_i \simeq \abs{N(\mathcal U)}.$$

First results about co-skeletons {#sec:general}
================================

Let us begin to study the co-skeleton $\operatorname{Skel}^c_k X$.

This first lemma tells us that while $\operatorname{Skel}^c_k X$ is not itself a polytopal complex, it is homotopy equivalent to one. See \[fig:coskel-simeq-bary\].

\[thm:coskel-simeq-bary\] $\abs{\operatorname{Skel}^c_k X} \simeq \abs{\operatorname{bary}(X) - \operatorname{bary}(\operatorname{Skel}_k X)}$.

<span>0.4</span>

\[scale=0.8\]

(v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v1)–(v5)–(v6)–cycle; (v1)–(v2)–(v7)–(v6)–cycle;

iin <span>1,...,7</span> at (vi) ;

<span>0.4</span>

\[scale=0.8\]

(v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

(v2)–(v3)–(v4)–(v5)–(v6)–(v7)–cycle–(v1)–(v5) (v1)–(v6); i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[vertex\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[vertex\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[vertex\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[vertex\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

in <span>12,23,34,45,15</span> (v)–(v12345); in <span>15,16,56</span> (v)–(v156); in <span>12,27,67,16</span> (v)–(v1267);

By definition, $$\begin{aligned}
    \abs{\operatorname{Skel}^c_k X} & = \abs{X} \setminus \abs{\operatorname{Skel}_k X} \\
    & = \abs{\operatorname{bary}(X)} \setminus \abs{\operatorname{bary}(\operatorname{Skel}_k X)}.\end{aligned}$$ Now, $\operatorname{bary}(\operatorname{Skel}_k X)$ is an induced subcomplex of the simplicial complex $\operatorname{bary}(X)$: if $v_{\sigma_1}, \dotsc, v_{\sigma_m}$ are vertices of $\operatorname{bary}(\operatorname{Skel}_k X)$ that form a face of $\operatorname{bary}(X)$, then the faces $\sigma_1, \dotsc, \sigma_m$ form a chain in the face poset of $X$, so they still form a chain in the face poset of $\operatorname{Skel}_k X$. Therefore, we may invoke \[thm:deletion-homotopy-equivalence\]: $$\begin{aligned}
    \abs{\operatorname{bary}(X)} \setminus \abs{\operatorname{bary}(\operatorname{Skel}_k X)} & \simeq \abs{\operatorname{bary}(X) - \operatorname{bary}(\operatorname{Skel}_k X)}. \qedhere\end{aligned}$$

\[thm:coskel-dimension\] $H_i(\operatorname{Skel}^c_k \Delta) = 0$ when $i > d-k-1$.

The vertices of $\operatorname{bary}(X) - \operatorname{bary}(\operatorname{Skel}_k X)$ correspond to faces $\sigma$ of $X$ of dimensions between $k+1$ and $d$; therefore, the largest possible face of $\operatorname{bary}(X) - \operatorname{bary}(\operatorname{Skel}_k X)$ has $d-k$ vertices, so its dimension is $d-k-1$.

The name "co-skeleton" was chosen to suggest "complement", but also duality. Let us illustrate why.

\[thm:polytope-duality\] Suppose $P$ is a $(d+1)$-dimensional polytope. Then $$\abs{\operatorname{Skel}^c_k \partial P} \simeq \abs{\operatorname{Skel}_{d-k-1} \partial P^*},$$ where $P^*$ is the polar dual polytope to $P$.

By \[thm:coskel-simeq-bary\], $$\abs{\operatorname{Skel}^c_k \partial P} \simeq \abs{\operatorname{bary}(\partial P) - \operatorname{bary}(\operatorname{Skel}_k \partial P)}.$$ The faces of $\operatorname{bary}(\partial P) - \operatorname{bary}(\operatorname{Skel}_k \partial P)$ are the chains in the poset of faces of $\partial P$ of dimension greater than $k$. But the poset of non-empty faces of $\partial P$ is isomorphic to the poset of non-empty faces of $\partial P^*$, flipped upside down. Under this flip, the faces of $\partial P$ of dimension greater than $k$ become faces of $\partial P^*$ of dimension less than or equal to $d-k-1$. Therefore, $\operatorname{bary}(\partial P) - \operatorname{bary}(\operatorname{Skel}_k \partial P) = \operatorname{bary}(\operatorname{Skel}_{d-k-1} \partial P^*)$, so $$\begin{aligned}
    \abs{\operatorname{bary}(\partial P) - \operatorname{bary}(\operatorname{Skel}_k \partial P)} & = \abs{\operatorname{bary}(\operatorname{Skel}_{d-k-1} \partial P^*)} \\
    & = \abs{\operatorname{Skel}_{d-k-1} \partial P^*}. \qedhere\end{aligned}$$

Note that the fact that $P$ is a polytope is not essential to this proposition, just that $P$ has an associated "dual cell structure". In fact, if $X$ is a $d$-dimensional homology manifold, the space $\abs{\operatorname{bary}(X) - \operatorname{bary}(\operatorname{Skel}_k X)}$ is exactly the $(d-k-1)$-skeleton of the dual cell structure used in some proofs of Poincaré duality — for example, @book:Munkres-EAT [§64] calls this space the "dual $(d-k-1)$-skeleton" of the manifold.

A related duality result is the Alexander duality theorem, which has direct implications for co-skeletons.

\[Alexander duality, <span>[@book:Munkres-EAT §71, particularly Theorem 71.1 and Exercise 4]</span>\] \[thm:Alexander-duality\] Suppose $X$ is a polytopal complex where $\dim \Delta = d$ and $\abs{X}$ is homeomorphic to a sphere, and suppose $Y$ is a proper, non-empty subcomplex of $X$. Then $${\widetilde}H^i(\abs{Y}) \cong {\widetilde}H_{d-i-1}(\abs{X} \setminus \abs{Y}) \qquad \text{and} \qquad {\widetilde}H_i(\abs{Y}) \cong {\widetilde}H^{d-i-1}(\abs{X} \setminus \abs{Y}).$$

Taking $Y = \operatorname{Skel}_k X$ gives us this corollary:

\[thm:Alexander-duality-for-coskeletons\] If $X$ is a $d$-dimensional polytopal complex where $\abs{X}$ is homeomorphic to a sphere, then for $k = 0, \dotsc, d-1$, $${\widetilde}H^i(\abs{\operatorname{Skel}_k X}) \cong {\widetilde}H_{d-i-1}(\abs{\operatorname{Skel}^c_k X}) \qquad \text{and} \qquad {\widetilde}H_i(\abs{\operatorname{Skel}_k X}) \cong {\widetilde}H^{d-i-1}(\abs{\operatorname{Skel}^c_k X}).$$

For a first example of an application for co-skeletons, let us consider neighbourly simplicial spheres. Suppose $\Delta$ is a simplicial sphere (that is, a simplicial complex where $\abs{\Delta}$ is homeomorphic to a sphere) with $n$ vertices. Then $\Delta$ is *$t$-neighbourly* if every set of $t$ vertices is the vertex set of a face, or equivalently, if its $(t-1)$-skeleton is isomorphic to the $(t-1)$-skeleton of a simplex with $n$ vertices. For example, the boundaries of $d$-dimensional cyclic polytopes are $\lfloor d/2 \rfloor$-neighbourly. A major reason why neighbourly spheres have been studied is their connection with the Upper Bound Conjecture (see e.g. @art:Alon-Kalai).

The following lemma will let us characterise skeletons of simplices by their homology. (It follows without much difficulty from @art:Bjorner-Kalai [Theorem 3.1] and @art:Kalai-shifting [Equation (3.6)], but we also present a self-contained proof.)

\[thm:skel-simplex-maximises-homology\] Suppose $\Delta$ is a simplicial complex with $n$ vertices, $n \geq 1$, and $\dim \Delta = k < n$. Then $$\operatorname{rank}{\widetilde}H_k(\abs{\Delta}; \mathbb Z) \leq \binom{n-1}{k+1},$$ with equality if and only if $\Delta$ is isomorphic to the $k$-skeleton of a simplex with $n$ vertices. If the coefficient ring is a field instead of $\mathbb Z$, the same holds with rank replaced by dimension.

For simplicity of notation, assume the coefficient ring is $\mathbb Z$; the argument when the coefficient ring is a field is similar.

We will prove this by induction on $n$. For the base case of $n = 1$, there is only one simplicial complex with $1$ vertex: it is the $0$-skeleton of the simplex on $1$ vertex, and its homology matches the formula above with $k = 0$.

Now, for the inductive step, label the vertices of $\Delta$ as $\{1,\dotsc,n\}$, and define the following two open sets: $$\begin{aligned}
A & \coloneq \abs{\operatorname{star}n}, \\
B & \coloneq \abs{\Delta} \setminus \{n\}.\end{aligned}$$ Then $A \cup B$ is $\abs{\Delta}$, and $A \cap B$ is $\abs{\operatorname{star}n} \setminus \{n\} = \abs{\operatorname{link}n}$. The Mayer–Vietoris theorem (\[thm:Mayer-Vietoris\]) then gives us the following long exact sequence: $$\begin{tikzcd}
\dotsb \rar & {\widetilde}H_i(\abs{\operatorname{link}n}) \rar & {\widetilde}H_i(\abs{\operatorname{star}n}) \oplus {\widetilde}H_i(\abs{\Delta} \setminus \{n\}) \rar & {\widetilde}H_i(\abs{\Delta}) \rar & \dotsb.
\end{tikzcd}$$ Since $\abs{\operatorname{star}n}$ is contractible, ${\widetilde}H_i(\abs{\operatorname{star}n})$ is zero. Now, consider the following segment of the exact sequence: $$\begin{tikzcd}
{\widetilde}H_k(\abs{\Delta} \setminus \{n\}) \rar & {\widetilde}H_k(\abs{\Delta}) \rar & {\widetilde}H_{k-1}(\abs{\operatorname{link}n}).
\end{tikzcd}$$ By \[thm:deletion-homotopy-equivalence\], $\abs{\Delta} \setminus \{n\}$ is homotopy equivalent to $\abs{\Delta - n}$, which is a simplicial complex with $n-1$ vertices and dimension less than or equal to $k$, so by induction, $\operatorname{rank}{\widetilde}H_k(\abs{\Delta} \setminus \{n\}) \leq \binom{n-2}{k+1}$. (Note that if the dimension of $\Delta - n$ is less than $k$, its degree $k$ homology must be zero.) Similarly, $\operatorname{link}n$ is homotopy equivalent to the simplicial link of $n$, which is a simplicial complex of dimension at most $k-1$ and with at most $n-1$ vertices, so by induction, $\operatorname{rank}{\widetilde}H_{k-1}(\abs{\operatorname{link}n}) \leq \binom{n-2}{k}$. Therefore, $$\operatorname{rank}{\widetilde}H_k(\abs{\Delta}) \leq \binom{n-2}{k+1} + \binom{n-2}{k} = \binom{n-1}{k+1}.$$ To obtain equality for $\operatorname{rank}{\widetilde}H_k(\abs{\Delta})$, we need equality to hold for both $\operatorname{rank}{\widetilde}H_k(\abs{\Delta - n})$ and $\operatorname{rank}{\widetilde}H_{k-1}(\abs{\operatorname{link}n})$, so by induction, both $\Delta - n$ and $\operatorname{s-link}n$ need to be skeletons of simplices of the appropriate sizes. Therefore, $\Delta$ itself can only be a $k$-skeleton of a simplex on $n$ vertices.

Finally, note that the homology of a $j$-skeleton of a simplex is concentrated in degree $j$, since in all degrees less than $j$ the simplicial chain complex agrees with that of a full simplex, which is contractible. Therefore, if $\Delta$ is the $k$-skeleton of a simplex, the Mayer–Vietoris exact sequence is zero outside the segment above, so the desired equality does indeed hold.

The previous lemma tells us that "being a skeleton of a simplex" is a homological property, if dimension and number of vertices are fixed. Combining this with \[thm:Alexander-duality-for-coskeletons\], we obtain the following characterisations of neighbourly spheres in terms of their skeletons and their co-skeletons:

\[thm:neighbourly-characterisation\] Suppose $\Delta$ is a $d$-dimensional simplicial sphere with $n$ vertices. Then for $t = 1, \dotsc, d$, the following are equivalent:

-   $\Delta$ is $t$-neighbourly,

-   $\operatorname{rank}{\widetilde}H_{t-1}(\abs{\operatorname{Skel}_{t-1} \Delta}; \mathbb Z) = \binom{n-1}{t}$,

-   $\operatorname{rank}{\widetilde}H^{d-t}(\abs{\operatorname{Skel}^c_{t-1} \Delta}; \mathbb Z) = \binom{n-1}{t}$.

By definition, $\Delta$ is $t$-neighbourly if and only if $\operatorname{Skel}_{t-1} \Delta$ is isomorphic to the $(t-1)$-skeleton of a simplex on $n$ vertices. says that this occurs if and only if $\operatorname{rank}{\widetilde}H_{t-1}(\abs{\operatorname{Skel}_{t-1} \Delta}; \mathbb Z) = \binom{n-1}{t}$. And \[thm:Alexander-duality-for-coskeletons\] says that $$\begin{aligned}
{\widetilde}H_{t-1}(\abs{\operatorname{Skel}_{t-1} \Delta}; \mathbb Z) & \cong {\widetilde}H^{d-t}(\abs{\operatorname{Skel}^c_{t-1} \Delta}; \mathbb Z). \qedhere\end{aligned}$$

The key long exact sequence {#sec:LES}
===========================

We now reach the main tool of this paper, which we will make much use of in the upcoming sections.

\[thm:LES\] If $X$ is a $d$-dimensional polytopal complex and $0 \leq k \leq d$, then there is the following long exact sequence: $$\begin{tikzcd}[column sep=small]
        \dotsb \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_i(\abs{\operatorname{link}_X \sigma}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_{i-1}(\abs{\operatorname{link}_X \sigma}) \rar & \dotsb.
    \end{tikzcd}$$

We will begin with the case $k = d$. The link of any $d$-dimensional face is empty, as is $\operatorname{Skel}^c_d X$. The $(d-1)$th co-skeleton $\operatorname{Skel}^c_{d-1} X$ consists only of the interiors of the $d$-dimensional faces, so it retracts onto $f_d(X)$ points. Therefore, the non-zero part of the claimed long exact sequence looks like this: $$\begin{tikzcd}
0 \rar & {\widetilde}H_0(\operatorname{Skel}^c_{d-1}) \arrow[d, phantom, "\cong" rotate=-90] \rar & \bigoplus_{\sigma \in X_d} {\widetilde}H_{-1}(\operatorname{link}\sigma) \arrow[d, phantom, "\cong" rotate=-90] \rar & {\widetilde}H_{-1}(\operatorname{Skel}^c_d X) \arrow[d, phantom, "\cong" rotate=-90] \rar & 0. \\
& R^{f_d(X) - 1} & R^{f_d(X)} & R
\end{tikzcd}$$ These homology groups do indeed form an exact sequence, with appropriate maps. (We won't need to know what the maps are in the rest of this paper, just that they exist.)

For the remainder of this proof, we will assume that $k < d$. We will use the Mayer–Vietoris theorem. Define the sets $A$ and $B$ by: $$\begin{aligned}
A & \coloneqq \bigcup_{\sigma \in X_k} \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma}, \\
B & \coloneqq \abs{\operatorname{Skel}^c_k X} = \abs{X} \setminus \abs{\operatorname{Skel}_k X}.\end{aligned}$$ See \[fig:Mayer-Vietoris-k=0,fig:Mayer-Vietoris-k=1\] for an example. Both $A$ and $B$ are open in $\abs{X}$: $A$ is a union of open sets, and $B$ is the complement of a closed set.

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> <span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; </span> (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

i///in <span>1/12/15/12345, 2/12/23/12345, 3/23/34/12345, 4/34/45/12345, 5/45/15/12345, 1/15/16/156, 5/15/56/156, 6/16/56/156, 1/12/16/1267, 2/12/27/1267, 7/27/67/1267, 6/16/67/1267</span> <span> (v-i) at ($(v\ix)!0.2!(v\i)$); (v-i) at ($(v\iy)!0.2!(v\i)$); (v-i) at ($(v\f)!0.2!(v\i)$); (v-i) – (vi) – (v-i) – (v-i) – cycle; (v-i) – (vi) – (v-i); (v-i) – (v-i) – (v-i); </span>

iin <span>1,...,7</span> at (vi) ;

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v1)–(v5)–(v6)–cycle; (v1)–(v2)–(v7)–(v6)–cycle;

iin <span>1,...,7</span> at (vi) ;

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> <span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; </span> (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

i///in <span>1/12/15/12345, 2/12/23/12345, 3/23/34/12345, 4/34/45/12345, 5/45/15/12345, 1/15/16/156, 5/15/56/156, 6/16/56/156, 1/12/16/1267, 2/12/27/1267, 7/27/67/1267, 6/16/67/1267</span> <span> (v-i) at ($(v\ix)!0.2!(v\i)$); (v-i) at ($(v\iy)!0.2!(v\i)$); (v-i) at ($(v\f)!0.2!(v\i)$); (v-i) – (vi) – (v-i) – (v-i) – cycle; (v-i) – (vi) – (v-i); (v-i) – (v-i) – (v-i); </span>

iin <span>1,...,7</span> at (vi) ;

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v1)–(v5)–(v6)–cycle; (v1)–(v2)–(v7)–(v6)–cycle;

iin <span>1,...,7</span> at (vi) ;

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> <span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; </span> (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

//in <span>1/2/12345, 2/3/12345, 3/4/12345, 4/5/12345, 1/5/12345, 1/5/156, 1/6/156, 5/6/156, 1/2/1267, 2/7/1267, 6/7/1267, 1/6/1267</span> <span> (v-) \[coordinate\] at ($(v\x)!0.2!(v\x\y)$) ; (v-) \[coordinate\] at ($(v\y)!0.2!(v\x\y)$) ; (v-) \[coordinate\] at ($(v\f)!0.2!(v\x\y)$) ; (v-) – (v) – (v-) – (v-) – cycle; (v-) – (v) node \[vertex\] – (v-); (v-) – (v-) – (v-); </span>

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

i/in <span>1/12345, 2/12345, 3/12345, 4/12345, 5/12345, 1/156, 5/156, 6/156, 1/1267, 2/1267, 6/1267, 7/1267</span> (vi-) at ($(v\i)!0.2!(v\f)$);

(v1-12345)–(v2-12345)–(v3-12345)–(v4-12345)–(v5-12345)–cycle; (v1-156)–(v5-156)–(v6-156)–cycle; (v1-1267)–(v2-1267)–(v7-1267)–(v6-1267)–cycle;

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> <span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; </span> (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

//in <span>1/2/12345, 2/3/12345, 3/4/12345, 4/5/12345, 1/5/12345, 1/5/156, 1/6/156, 5/6/156, 1/2/1267, 2/7/1267, 6/7/1267, 1/6/1267</span> <span> (v--) \[coordinate\] at ($(v\x)!0.2!(v\x\y)!0.1!(v\f)$) ; (v--) \[coordinate\] at ($(v\y)!0.2!(v\x\y)!0.1!(v\f)$) ; (v-) \[coordinate\] at ($(v\f)!0.2!(v\x\y)$) ; (v--) – (v--) – (v-) – cycle; (v--) – (v-) – (v--) – cycle; </span>

<span>0.35</span>

\[scale=0.8\] (v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v1)–(v5)–(v6)–cycle; (v1)–(v2)–(v7)–(v6)–cycle;

iin <span>1,...,7</span> at (vi) ;

First, we claim that the union in $A$ is in fact a *disjoint* union. If two stars, say $\operatorname{star}_{\operatorname{bary}(X)} v_\sigma$ and $\operatorname{star}_{\operatorname{bary}(X)} v_\tau$, were to intersect, they would have a face of $\operatorname{bary}(X)$ in common, and by the definition of an open star, this face would need to contain both $v_\sigma$ and $v_\tau$. But faces in $\operatorname{bary}(X)$ correspond to chains in the face poset of $X$; therefore, no face in $\operatorname{bary}(X)$ can contain both $v_\sigma$ and $v_\tau$, since $\sigma$ and $\tau$ are both $k$-dimensional and are thus incomparable in the face poset. So $A$ is a disjoint union of stars. Since these stars are open, each one thus forms a connected component of $A$.

Next, let us consider $A \cap B$. Since $\operatorname{star}_{\operatorname{bary}(X)} v_\sigma$ does not intersect any faces of $X$ of dimension less than $k$, and the only $k$-dimensional face it intersects is $\sigma$, we conclude that $$A \cap B = \bigsqcup_{\sigma \in X_k} \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}.$$ Now for $A \cup B$. Suppose $\tau$ is an arbitrary face of $X$. If $\dim \tau \leq k-1$, then $\abs \tau$ is disjoint from both $A$ and $B$; if $\dim \tau = k$, then $\abs \tau$ is a subset of $\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\tau}$ hence $\abs \tau \subseteq A$, and if $\dim \tau \geq k+1$ then $\abs \tau$ is a subset of $B$. Therefore, $A \cup B$ consists of the relative interiors of all faces of dimension $k$ and greater; that is, $$A \cup B = \abs{\operatorname{Skel}^c_{k-1} X}.$$

We can now apply the Mayer–Vietoris theorem, with unreduced homology. This gives us the following long exact sequence: <span>$$\begin{tikzcd}
        \dotsb \rar & H_i \bigg( \bigsqcup_{\sigma \in X_k} \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma} \bigg) \rar & H_i \bigg( \bigsqcup_{\sigma \in X_k} \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \bigg) \oplus H_i(\abs{\operatorname{Skel}^c_k X}) \rar & H_i(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \dotsb.
    \end{tikzcd}$$</span> Since unreduced homology of a disconnected space can be decomposed as a direct sum over components, we can simplify this long exact sequence: <span>$$\begin{tikzcd}[column sep=small]
        \dotsb \rar & \bigoplus_{\sigma \in X_k} H_i \big( \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma} \big) \rar & \bigg( \bigoplus_{\sigma \in X_k} H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma}) \bigg) \oplus H_i(\abs{\operatorname{Skel}^c_k X}) \rar & H_i(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \dotsb.
    \end{tikzcd}$$</span>

While this is a perfectly good long exact sequence, there are some steps we can do to "clean it up" and make it more convenient for later proofs. The first step is to convert from unreduced to reduced homology, which affects this sequence in degrees $0$ and $-1$. For each face $\sigma \in X_k$, there are two cases to consider: either $\sigma$ is not a facet, or it is.

If $\sigma$ is not a facet of $X$, then $\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}$ is non-empty, so $H_0(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma})$ is non-zero. The generators of a degree $0$ homology group are homotopy equivalence classes of maps from the $0$-simplex (i.e., a point) to the space in question; therefore, since the first map in our long exact sequence is induced by the inclusion of topological spaces, the map sends each generator of $H_0(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma})$ to a generator of $H_0(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma}) \cong R$. Therefore, since homology groups in degree $0$ are always free, we may remove one summand of $R$ from each of these two homology groups. The result is that the unreduced homology group $H_0(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma})$ becomes reduced, and since $\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma}$ is contractible, its homology is zero outside degree $0$, so we can omit the $H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma})$ term. This holds for every $k$-face $\sigma$ that is not a facet.

What if $\sigma$ is a facet? In this case, $\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} = \abs{\sigma}$, so $\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}$ is empty, hence its degree $0$ homology is zero in both the unreduced and the reduced form. However, its degree $(-1)$ reduced homology becomes non-zero. Therefore, we can do the following two actions:

-   First, the map from $H_0(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma}) \cong R$ to $H_0(\abs{\operatorname{Skel}^c_{k-1} X})$ takes generators to generators, so we can remove one summand of $R$ from each, by the same reasoning as above.

-   Second, we can add a copy of the sequence $\dotsb \to 0 \to R \to R \to 0 \to \dotsb$ to our sequence, adding one summand of $R$ to each of $H_0(\abs{\operatorname{Skel}^c_{k-1} X})$ and $H_{-1}(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma})$.

The result is the following:

-   The $H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma})$ term is removed (since it is zero when $i \neq 0$),

-   No net change occurs to $H_0(\abs{\operatorname{Skel}^c_{k-1} X})$, and

-   The $H_{-1}(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma})$ term turns from $0$ to $R$, effectively converting from unreduced to reduced homology.

There is one more change to make in converting from unreduced to reduced homology: the map from $H_0(\abs{\operatorname{Skel}^c_k X})$ to $H_0(\abs{\operatorname{Skel}^c_{k-1} X})$ is induced by the inclusion of non-empty topological spaces (since $k < d$), so by a similar argument, we may replace both of these homology groups with reduced homology.

Our long exact sequence then becomes: $$\begin{tikzcd}[column sep=small]
        \dotsb \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_i \big( \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma} \big) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \dotsb.
    \end{tikzcd}$$

There is one last thing to do to this long exact sequence: simplifying ${\widetilde}H_i \big( \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma} \big)$. To do this, let us call on the Mayer–Vietoris theorem again. Assume that $\sigma$ is not a facet, so $\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}$ is non-empty. Define two new spaces: $$\begin{aligned}
A' & \coloneqq \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma}, \\
B' & \coloneqq \abs{\operatorname{star}_X \sigma} \setminus \abs{\sigma} = \abs{\operatorname{link}_X \sigma}\end{aligned}$$

<span>0.3</span>

\[scale=0.8\]

(v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v1)–(v5)–(v6)–cycle; (v1)–(v2)–(v7)–(v6)–cycle;

iin <span>1,...,7</span> at (vi) ;

at (v3) ;

(v6)–(v7) node \[coordinate, midway, pin=right:$\sigma_2$\] ;

at (v156) ;

<span>0.6</span>

\[scale=0.8\]

(v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) at ($(v\i)!0.5!(v\j)$); (v12345) at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$); (v156) at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$); (v1267) at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$);

(v3)–(v23)–(v12345)–(v34)–cycle; (v6)–(v1267)–(v7)–cycle; (v1)–(v5)–(v6)–cycle;

(v1)–(v2)–(v3)–(v4)–(v5)–cycle; (v5)–(v6)–(v1); (v2)–(v7)–(v6);

iin <span>1,...,5,12,23,34,45,15</span> (vi)–(v12345); iin <span>1,5,6,15,16,56</span> (vi)–(v156); iin <span>1,2,6,7,12,27,67,16</span> (vi)–(v1267);

(v23)–(v3)–(v34); (v23)–(v12345)–(v34); (v6)–(v7); (v6)–(v1267)–(v7); (v1)–(v5)–(v6)–cycle;

at (v3) ; at (v67) ; at (v156) ;

<span>0.6</span>

\[scale=0.8\]

(v1) \[coordinate\] at (0,0) ; (v2) \[coordinate\] at (100:1.8) ; (v3) \[coordinate\] at (-1.9,0.8) ; (v4) \[coordinate\] at (-1.8,-0.9) ; (v5) \[coordinate\] at (-0.1,-2.0) ; (v6) \[coordinate\] at (10:1.8) ; (v7) \[coordinate\] at ($(v6) + (100:1.8)$) ;

i/in <span>1/2, 2/3, 3/4, 4/5, 1/5, 1/6, 5/6, 2/7, 6/7</span> (vi) \[coordinate\] at ($(v\i)!0.5!(v\j)$) ; (v12345) \[coordinate\] at ($1/5*(v1)+1/5*(v2)+1/5*(v3)+1/5*(v4)+1/5*(v5)$) ; (v156) \[coordinate\] at ($1/3*(v1)+1/3*(v5)+1/3*(v6)$) ; (v1267) \[coordinate\] at ($1/4*(v1)+1/4*(v2)+1/4*(v6)+1/4*(v7)$) ;

i/in <span>1/12345, 2/12345, 3/12345, 4/12345, 5/12345, 12/12345, 23/12345, 34/12345, 45/12345, 15/12345, 1/156, 5/156, 6/156, 15/156, 16/156, 56/156, 1/1267, 2/1267, 6/1267, 7/1267, 12/1267, 27/1267, 67/1267, 16/1267</span> (vi-) at ($(v\i)!0.2!(v\f)$);

(v1-12345)–(v2-12345)–(v3-12345)–(v4-12345)–(v5-12345)–cycle; (v1-156)–(v5-156)–(v6-156)–cycle; (v1-1267)–(v2-1267)–(v7-1267)–(v6-1267)–cycle;

(v2-12345)–(v3-12345)–(v4-12345); (v4-12345)–(v5-12345)–(v1-12345)–(v2-12345); (v6-1267)–(v7-1267); (v6-1267)–(v1-1267)–(v2-1267)–(v7-1267); (v1-156)–(v5-156)–(v6-156)–cycle;

at (v3-12345) ; at (v67-1267) ; at (v156) ;

Since $\abs{\sigma} \subseteq \abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \subseteq \abs{\operatorname{star}_X \sigma}$ — for example, see \[fig:LES-stars\] — the intersection $A' \cap B'$ is $\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}$, and the union $A' \cup B'$ is simply $\abs{\operatorname{star}_X \sigma}$. Therefore, the Mayer–Vietoris long exact sequence with reduced homology for $A'$ and $B'$ is the following: $$\begin{gathered}
    \begin{tikzcd}[ampersand replacement=\&]
        \dotsb \rar \& {\widetilde}H_{i+1}(\abs{\operatorname{star}_X \sigma}) \rar \& {\widetilde}H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma})
    \end{tikzcd} \\
    \begin{tikzcd}[ampersand replacement=\&]
        {} \rar \& {\widetilde}H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma}) \oplus {\widetilde}H_i(\abs{\operatorname{star}_X \sigma} \setminus \abs{\sigma}) \rar \& {\widetilde}H_i(\abs{\operatorname{star}_X \sigma}) \rar \& \dotsb.
    \end{tikzcd}\end{gathered}$$ But stars of non-empty faces are contractible, so this becomes: $$\begin{tikzcd}
        \dotsb \rar & 0 \rar & {\widetilde}H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}) \rar & {\widetilde}H_i(\abs{\operatorname{star}_X \sigma} \setminus \abs{\sigma}) \rar & 0 \rar & \dotsb.
    \end{tikzcd}$$ Therefore, ${\widetilde}H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}) \cong {\widetilde}H_i(\abs{\operatorname{star}_X \sigma} \setminus \abs{\sigma}) = {\widetilde}H_i(\abs{\operatorname{link}_X \sigma})$ for all non-facets $\sigma$. Finally, note that if $\sigma$ is a facet, this isomorphism still holds, since all spaces involved are empty.

Hence we obtain the desired long exact sequence.

Complexes characterised by link conditions {#sec:families}
==========================================

Now that we have a long exact sequence relating co-skeletons and links, in this section we will consider some particular families of polytopal complexes that are characterised by links of faces having zero homology in some degrees.

Cohen–Macaulay complexes
------------------------

A pure, $d$-dimensional polytopal complex $X$ is *Cohen–Macaulay* if for every face $\sigma$, including $\sigma = {\varnothing}$, and every $i < d - \dim \sigma - 1$, the homology ${\widetilde}H_i(\abs{\operatorname{link}_\Delta \sigma})$ is $0$. If $\Delta \coloneqq X$ is a simplicial complex, this condition is equivalent to the Stanley–Reisner ring of $\Delta$ being a Cohen–Macaulay ring [@art:Bjorner p. 1855].

\[thm:CM-topological\] Cohen–Macaulayness is also a topological condition: $X$ is Cohen–Macaulay if and only if ${\widetilde}H_i(\abs{X})$ and ${\widetilde}H_i(\abs{X}, \abs{X} \setminus \{p\})$ are both $0$ for all $i < \dim X$ and all points $p \in \abs{X}$. This was proved by Munkres [@art:Munkres-TRC Corollary 3.4] for simplicial complexes, but a similar proof applies to polytopal complexes as well, with the following modified version of @art:Munkres-TRC [Lemma 3.3]:

If $X$ is a polytopal complex, $\sigma$ a face of $X$ and $p$ a point in the relative interior of $\sigma$, then ${\widetilde}H_i(\abs{\operatorname{link}_X \sigma}) \cong {\widetilde}H_{i+\dim \sigma}(\abs{\operatorname{link}_{\operatorname{bary}(X)} v_\sigma}) \cong {\widetilde}H_{i + \dim \sigma + 1}(\abs{X}, \abs{X} \setminus \{p\})$.

We saw in the proof of \[thm:LES\] that $$\begin{aligned}
{\widetilde}H_i(\abs{\operatorname{link}_X \sigma}) & \cong {\widetilde}H_i(\abs{\operatorname{star}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\sigma}) \\
& = {\widetilde}H_i \big( \abs{\operatorname{link}_{\operatorname{bary}(X)} v_\sigma} \setminus (\abs{\sigma} \setminus \{v_\sigma\}) \big) \\
& \cong {\widetilde}H_i(\abs{\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\operatorname{bary}(\partial \sigma)}),\end{aligned}$$ where "$\operatorname{s-link}$" denotes the simplicial link. Since $\operatorname{bary}(\partial \sigma)$ is an induced subcomplex of the simplicial complex $\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma$, \[thm:deletion-homotopy-equivalence\] says that $$\begin{aligned}
\abs{\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma} \setminus \abs{\operatorname{bary}(\partial \sigma)} & \simeq \abs{\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma - \operatorname{bary}(\partial \sigma)}.\end{aligned}$$ It is not hard to check that $$\begin{aligned}
\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma & = \big( \operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma - \operatorname{bary}(\partial \sigma) \big) * \operatorname{bary}(\partial \sigma),\end{aligned}$$ where $*$ denotes simplicial join [@book:Munkres-EAT §62], so @book:Munkres-EAT [Theorem 62.5] implies that $$\begin{aligned}
{\widetilde}H_i(\abs{\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma - \operatorname{bary}(\partial \sigma)}) & \cong {\widetilde}H_{i+\dim \sigma}(\abs{\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma}).\end{aligned}$$ Finally, since $\operatorname{bary}(X)$ is a simplicial complex, we can apply @art:Munkres-TRC [Lemma 3.3] to conclude that $$\begin{aligned}
{\widetilde}H_{i+\dim \sigma}(\abs{\operatorname{s-link}_{\operatorname{bary}(X)} v_\sigma}) & \cong {\widetilde}H_{i+\dim \sigma + 1}(\abs{X}, \abs{X} \setminus \{v_\sigma\}),\end{aligned}$$ and note that although $v_\sigma$ is typically defined to be the barycentre of $\sigma$, it can be chosen to be any point in the relative interior of $\sigma$.

The following theorem allows us to determine whether a complex is Cohen–Macaulay by considering its co-skeletons.

\[thm:CM-characterisation\] A $d$-dimensional polytopal complex $X$ is Cohen–Macaulay if and only if it is pure and ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) = 0$ for all $i < d-k-1$ and all $k = -1, \dotsc, d$.

Note that this theorem combined with \[thm:coskel-dimension\] means that the homology of $\abs{\operatorname{Skel}^c_k X}$ is entirely concentrated in degree $d-k-1$.

\[Proof of \[thm:CM-characterisation\]\] To begin with, assume $X$ is Cohen–Macaulay. We will prove that ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) = 0$ for $i < d-k-1$ by induction on $k$.

For the base case, when $k = -1$, we have $${\widetilde}H_i(\abs{\operatorname{Skel}^c_{-1} X}) = {\widetilde}H_i(\abs{X}) = {\widetilde}H_i(\abs{\operatorname{link}_X {\varnothing}}),$$ which is zero for all $i < d$ since $X$ is Cohen–Macaulay.

Now, suppose $k > -1$, and consider this part of the long exact sequence from \[thm:LES\]: $$\begin{tikzcd}
\bigoplus_{\sigma \in X_k} {\widetilde}H_i(\abs{\operatorname{link}\sigma}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} X}).
\end{tikzcd}$$ Since $X$ is Cohen–Macaulay and $i < d-k-1$, we know that ${\widetilde}H_i(\abs{\operatorname{link}\sigma})$ is $0$. Also, since $i$ is less than $d-k-1$, it is certainly less than $d-(k-1)-1$, so by induction, ${\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} X})$ is $0$. Therefore, ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X})$ must be $0$ too, and we are done with this direction of the proof.

Now, for the reverse direction, assume that ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) = 0$ for all $i < d-k-1$ and all $k = -1, \dotsc, d$. Consider this part of the long exact sequence for $k \geq 0$: $$\begin{tikzcd}
{\widetilde}H_{i+1}(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_i(\abs{\operatorname{link}\sigma}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_k X})
\end{tikzcd}$$ By our assumption, ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) = 0$. Also, since $i < d-k-1$, we have $i+1 < d - (k-1) - 1$, so our assumption tells us that ${\widetilde}H_{i+1}(\abs{\operatorname{Skel}^c_{k-1} X}) = 0$ too. Therefore, $\bigoplus_{\sigma \in X_k} {\widetilde}H_i(\abs{\operatorname{link}\sigma})$ is $0$, so ${\widetilde}H_i(\abs{\operatorname{link}\sigma})$ must be $0$ for all $i < d-k-1 = d- \dim \sigma - 1$.

This is true for all faces $\sigma$ of dimension $k \geq 0$, which leaves the case $\sigma = {\varnothing}$. In this case, $${\widetilde}H_i(\abs{\operatorname{link}\sigma}) = {\widetilde}H_i(\abs{X}) = {\widetilde}H_i(\abs{\operatorname{Skel}^c_{-1} X}) = 0.$$ Therefore, $X$ is Cohen–Macaulay.

This theorem has potential computational applications. Naïvely, to check whether some complex $X$ is Cohen–Macaulay, one must compute the homology groups of the link of every face of $X$ — if $\dim X = d$, there are at least $2^{d+1}$ faces. But with this result, one only needs to compute homologies of the $k$th co-skeletons of $X$ for the $d+2$ values of $k$ between $-1$ and $d$, or equivalently (by \[thm:coskel-simeq-bary\]) the simplicial complexes $\operatorname{bary}(X) - \operatorname{bary}(\operatorname{Skel}_k X)$. The tradeoff is that these simplicial complexes are much larger and more complicated than the links of faces: for instance, the number of vertices of $\operatorname{bary}(X) - \operatorname{bary}(\operatorname{Skel}_k X)$ is at least on the order of $2^d$.

We saw in \[thm:neighbourly-characterisation\] that neighbourly spheres are characterised by the homology groups of either their skeletons or their co-skeletons. However, for Cohen–Macaulay complexes, the homology groups of skeletons are not sufficient for a characterisation: for example, \[fig:CM-not-characterised-by-skeletons\] shows two complexes, one Cohen–Macaulay and one not, whose $k$-skeletons have isomorphic $i$th homology groups for all $k$ and $i$.

<span>0.4</span>

(v0) at (0,0); iin <span>1,...,6</span> (vi) at (60\*i:1);

(v0)–(v1)–(v2)–cycle; (v0)–(v2)–(v3)–cycle; (v0)–(v3)–(v4)–cycle; (v0)–(v4)–(v5)–cycle; (v0)–(v5)–(v6)–cycle; (v0)–(v6)–(v1)–cycle;

at (v0); iin <span>1,...,6</span> at (vi) ;

<span>0.4</span>

(v0) at (0,0); (v1) at (1,0); (v2) at ($(v1)+(60:1)$); (v3) at ($(v1)+(-60:1)$); (v4) at (-1,0); (v5) at ($(v4)+(120:1)$); (v6) at ($(v4)+(-120:1)$);

(v1)–(v0)–(v2)–cycle; (v1)–(v2)–(v3)–cycle; (v1)–(v3)–(v0)–cycle; (v4)–(v0)–(v5)–cycle; (v4)–(v5)–(v6)–cycle; (v4)–(v6)–(v0)–cycle;

at (v0); iin <span>1,...,6</span> at (vi) ;

says that if $X$ is Cohen–Macaulay, almost all of the long exact sequence from \[thm:LES\] is $0$. The remaining non-zero terms are captured in the following corollary:

\[thm:CM-SES\] If $X$ is a Cohen–Macaulay polytopal complex and $k = 0, \dotsc, d$, we have the following short exact sequence: $$\begin{tikzcd}
0 \rar & {\widetilde}H_{d-k}(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_{d-k-1}(\abs{\operatorname{link}\sigma}) \rar & {\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_k X}) \rar & 0.
\end{tikzcd}$$

Take the following segment of the long exact sequence from \[thm:LES\]: <span>$$\begin{tikzcd}
{\widetilde}H_{d-k}(\abs{\operatorname{Skel}^c_k X}) \rar & {\widetilde}H_{d-k}(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_{d-k-1}(\abs{\operatorname{link}\sigma}) \rar & {\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_k X}) \rar & {\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_{k-1} X}).
\end{tikzcd}$$</span> The first term, ${\widetilde}H_{d-k}(\abs{\operatorname{Skel}^c_k X})$, is $0$ by \[thm:coskel-dimension\]. The last term, ${\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_{k-1} X})$, is $0$ by \[thm:CM-characterisation\].

We can combine these short exact sequences into a long exact sequence, using the following homological algebra fact:

\[thm:splicing\] Suppose we have two exact sequences: $$\begin{tikzcd}
\dotsb \rar & A_2 \arrow[r, "a_2"] & A_1 \arrow[r, "a_1"] & C \rar & 0
\end{tikzcd}
\qquad \text{and} \qquad
\begin{tikzcd}
0 \rar & C \arrow[r, "b_0"] & B_{-1} \arrow[r, "b_{-1}"] & B_{-2} \rar & \dotsb.
\end{tikzcd}$$ Then the horizontal sequence in the following diagram is exact: $$\begin{tikzpicture}[commutative diagrams/every diagram, column sep=1.0cm, row sep=1.0cm]

\matrix[matrix of math nodes, name=m, commutative diagrams/every cell] {
& & 0 & 0 \\
\dotsb & A_2 & A_1 & B_{-1} & B_{-2} & \dotsb. \\
};
\node (c) at ($(m-2-3)!0.5!(m-1-4)$) {$C$};

\path[commutative diagrams/.cd, every arrow, every label]
    (m-1-3) edge (c)
    (c) edge (m-1-4)
        edge node {$b_0$} (m-2-4)
    (m-2-1) edge (m-2-2)
    (m-2-2) edge node {$a_2$} (m-2-3)
    (m-2-3) edge node {$a_1$} (c)
        edge[commutative diagrams/dashed] node [swap] {$b_0 \circ a_1$} (m-2-4)
    (m-2-4) edge node {$b_{-1}$} (m-2-5)
    (m-2-5) edge (m-2-6);

\end{tikzpicture}$$

Since $a_1$ is surjective, $\operatorname{im}(b_0 \circ a_1) = \operatorname{im}(b_0) = \ker(b_{-1})$. Since $b_0$ is injective, $\ker(b_0 \circ a_1) = \ker a_1 = \operatorname{im}a_2$.

\[thm:CM-LES\] If $X$ is Cohen–Macaulay, we have the following long exact sequence: $$\begin{tikzcd}
0 \rar & {\widetilde}H_d(\abs{X}) \rar & \bigoplus_{\sigma \in X_0} {\widetilde}H_{d-1}(\abs{\operatorname{link}\sigma}) \rar & \bigoplus_{\sigma \in X_1} {\widetilde}H_{d-2}(\abs{\operatorname{link}\sigma}) \rar & \dotsb \rar & \bigoplus_{\sigma \in X_d} {\widetilde}H_{-1}(\abs{\operatorname{link}\sigma}) \rar & R \rar & 0.
\end{tikzcd}$$

Use \[thm:splicing\] to stitch together the short exact sequences from \[thm:CM-SES\]. The start of the sequence looks like this: <span>$$\begin{tikzpicture}[commutative diagrams/every diagram, column sep=0.6cm, row sep=1.6cm]

\matrix[matrix of math nodes, name=m, commutative diagrams/every cell] {
& & 0 & 0 \\
0 & {\widetilde}H_d(\abs{\operatorname{Skel}^c_{-1} X}) & \bigoplus_{\sigma \in X_0} {\widetilde}H_{d-1}(\abs{\operatorname{link}\sigma}) & \bigoplus_{\sigma \in X_1} {\widetilde}H_{d-2}(\abs{\operatorname{link}\sigma}) & \bigoplus_{\sigma \in X_2} {\widetilde}H_{d-3}(\abs{\operatorname{link}\sigma}) & \dotsb \\
& & & 0 & 0 \\
};
\node (hd1skel) at ($(m-2-3)!0.5!(m-1-4)$) {${\widetilde}H_{d-1}(\abs{\operatorname{Skel}^c_0 X})$};
\node (hd2skel) at ($(m-3-4)!0.5!(m-2-5)$) {${\widetilde}H_{d-2}(\abs{\operatorname{Skel}^c_1 X})$};

\path[commutative diagrams/.cd, every arrow, every label]
    (m-1-3) edge (hd1skel)
    (hd1skel) edge (m-1-4)
        edge (m-2-4)
    (m-2-1) edge (m-2-2)
    (m-2-2) edge (m-2-3)
    (m-2-3) edge (hd1skel)
        edge[commutative diagrams/dashed] (m-2-4)
    (m-2-4) edge[commutative diagrams/dashed] (m-2-5)
        edge (hd2skel)
    (m-2-5) edge[commutative diagrams/dashed] (m-2-6)
    (hd2skel) edge (m-2-5)
        edge (m-3-5)
    (m-3-4) edge (hd2skel);

\end{tikzpicture}$$</span> And the end like this: <span>$$\begin{tikzpicture}[commutative diagrams/every diagram, column sep=0.6cm, row sep=1.6cm]

\matrix[matrix of math nodes, name=m, commutative diagrams/every cell] {
& 0 & 0 \\
\dotsb & \bigoplus_{\sigma \in X_{d-2}} {\widetilde}H_1(\abs{\operatorname{link}\sigma}) & \bigoplus_{\sigma \in X_{d-1}} {\widetilde}H_0(\abs{\operatorname{link}\sigma}) & \bigoplus_{\sigma \in X_d} {\widetilde}H_{-1}(\abs{\operatorname{link}\sigma}) & {\widetilde}H_{-1}(\abs{\operatorname{Skel}^c_d X}) & 0 \\
& & 0 & 0 \\
};
\node (h1skel) at ($(m-2-2)!0.5!(m-1-3)$) {${\widetilde}H_1(\abs{\operatorname{Skel}^c_{d-2} X})$};
\node (h0skel) at ($(m-3-3)!0.5!(m-2-4)$) {${\widetilde}H_0(\abs{\operatorname{Skel}^c_{d-1} X})$};

\path[commutative diagrams/.cd, every arrow, every label]
    (m-1-2) edge (h1skel)
    (h1skel) edge (m-1-3)
        edge (m-2-3)
    (m-2-1) edge[commutative diagrams/dashed] (m-2-2)
    (m-2-2) edge[commutative diagrams/dashed] (m-2-3)
        edge (h1skel)
    (m-2-3) edge[commutative diagrams/dashed] (m-2-4)
        edge (h0skel)
    (m-2-4) edge (m-2-5)
    (m-2-5) edge (m-2-6)
    (h0skel) edge (m-2-4)
        edge (m-3-4)
    (m-3-3) edge (h0skel);

\end{tikzpicture}$$</span>

Note that this exact sequence is reminiscent of the "partition complex", a chain complex defined by @art:Adiprasito-Yashfe [Definition 25].

If the definition of a Cohen–Macaulay complex is modified to allow ${\widetilde}H_i(\abs{\operatorname{link}_X \sigma})$ to be non-zero when $\sigma = {\varnothing}$, we get the definition of a *Buchsbaum* complex. Buchsbaum complexes can also be characterised by homological properties of their co-skeletons, although the statement is not as simple as for Cohen–Macaulay complexes in \[thm:CM-characterisation\]: a complex $X$ is Buchsbaum if and only if the map ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) \to {\widetilde}H_i(\abs{X})$ induced by the inclusion of topological spaces is an isomorphism for all $i < d-k-1$ and a surjection for $i = d-k-1$, for all $k = -1, \dotsc, d$. This follows fairly easily from the long exact sequence in \[thm:LES\], although we leave the details to the reader.

Leray complexes
---------------

A simplicial complex $\Delta$ is *$r$-Leray* if every induced subcomplex $\Lambda$ of $\Delta$ has ${\widetilde}H_i(\abs{\Lambda}) = 0$ for $i \geq r$. For example, if $\Delta$ is the nerve of a family of convex open subsets of $\mathbb R^r$, then it follows from the nerve theorem (\[thm:nerve\]) that $\Delta$ is $r$-Leray (see @art:Kalai-Meshulam).

Equivalently, $\Delta$ is $r$-Leray if the homology ${\widetilde}H_i(\abs{\operatorname{link}_\Delta \sigma})$ is $0$ for every face $\sigma \in \Delta$ (including the empty face) and every $i \geq r$ [@art:Kalai-Meshulam Proposition 3.1]. We will use this condition to generalise $r$-Leray-ness to polytopal complexes: we will say a complex $X$ is $r$-Leray if ${\widetilde}H_i(\abs{\operatorname{link}_X \sigma}) = 0$ for all faces $\sigma$ (including $\sigma = {\varnothing}$) and all $i \geq r$. Note that this condition is not equivalent to the condition on induced subcomplexes in the non-simplicial case: for example, if $X$ is a square, every non-empty link is contractible so $X$ is $0$-Leray, but the subcomplex induced by a pair of diagonally opposite vertices is not connected.

The following theorem provides a characterisation of $r$-Leray complexes in terms of their co-skeletons.

\[thm:Leray-characterisation\] A polytopal complex $X$ is $r$-Leray if and only if ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) = 0$ for all $i \geq r$ and all $k = -1, \dotsc, d$.

For the forward direction, assume $X$ is $r$-Leray and let $i \geq r$. When $k = -1$, $${\widetilde}H_i(\abs{\operatorname{Skel}^c_{-1} X}) = {\widetilde}H_i(\abs{X}) = {\widetilde}H_i(\abs{\operatorname{link}_X {\varnothing}}),$$ which is $0$, since $X$ is $r$-Leray.

For $k \geq 0$, consider the following subsequence of the long exact sequence from \[thm:LES\]: $$\begin{tikzcd}[column sep=small]
\bigoplus_{\sigma \in X_k} {\widetilde}H_i(\abs{\operatorname{link}\sigma}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} X}).
\end{tikzcd}$$ Since $X$ is $r$-Leray, $\bigoplus_{\sigma \in X_k} {\widetilde}H_i(\abs{\operatorname{link}\sigma})$ is $0$. Therefore, ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X})$ injects into ${\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} X})$. Since this is true for all $k = 0, \dotsc, d$, we get a series of injections: $$\begin{tikzcd}[column sep=small]
{\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) \arrow[r, hook] & {\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} X}) \arrow[r, hook] & {\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-2} X}) \arrow[r, hook] & \dotsb \arrow[r, hook] & {\widetilde}H_i(\abs{\operatorname{Skel}^c_0 X}) \arrow[r, hook] & {\widetilde}H_i(\abs{\operatorname{Skel}^c_{-1} X}).
\end{tikzcd}$$ But we saw above that ${\widetilde}H_i(\abs{\operatorname{Skel}^c_{-1} X}) = 0$. Therefore, ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X})$ injects into $0$, so it must itself be $0$, for all $k = 0, \dotsc, d$.

For the reverse direction, assume that ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}) = 0$ for all $k = -1, \dotsc, d$ and all $i \geq r$. When $k \geq 0$, consider this part of the long exact sequence: $$\begin{tikzcd}
{\widetilde}H_{i+1}(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_i(\abs{\operatorname{link}\sigma}) \rar & {\widetilde}H_i(\abs{\operatorname{Skel}^c_k X}).
\end{tikzcd}$$ By our assumptions, the two outer terms here are both $0$, hence the middle term is $0$ as well, so ${\widetilde}H_i(\abs{\operatorname{link}\sigma}) = 0$ for all $k$-faces $\sigma$. This works for all $k = 0, \dotsc, d$, which only leaves $k = -1$: the only $(-1)$-dimensional face is ${\varnothing}$, and $${\widetilde}H_i(\abs{\operatorname{link}_X {\varnothing}}) = {\widetilde}H_i(\abs{X}) = {\widetilde}H_i(\abs{\operatorname{Skel}^c_{-1} X})$$ which we assumed to be $0$.

Therefore, ${\widetilde}H_i(\abs{\operatorname{link}_X \sigma}) = 0$ for all faces $\sigma$ and all $i \geq r$, so $X$ is $r$-Leray.

Stacked balls
-------------

A *homology sphere* is a polytopal complex $X$ of dimension $d$ where for every face $\sigma$ including $\sigma = {\varnothing}$, $${\widetilde}H_i(\abs{\operatorname{link}\sigma}) = \begin{cases*} 0 & if $i \neq d - \dim \sigma - 1$ \\ R & if $i = d - \dim \sigma - 1$. \end{cases*}$$ A *homology manifold* is defined similarly but only considering $\sigma \neq {\varnothing}$. A *homology ball* of dimension $d$ is a complex $X$ where:

-   ${\widetilde}H_i(\abs{X}) = 0$ for all $i$,

-   for every non-empty face $\sigma$, $${\widetilde}H_i(\abs{\operatorname{link}\sigma}) = \begin{cases*} 0 & if $i \neq d - \dim \sigma - 1$ \\ \text{$R$ or $0$} & if $i = d - \dim \sigma - 1$, \end{cases*}$$ and

-   the set of faces $\sigma$ with ${\widetilde}H_{d-\dim \sigma - 1}(\abs{\operatorname{link}\sigma}) = 0$ forms a subcomplex of $X$ that is a $(d-1)$-dimensional homology sphere.

If ${\widetilde}H_{d-\dim \sigma - 1}(\abs{\operatorname{link}\sigma})$ is $0$, $\sigma$ is called a *boundary face*, and if this homology is $R$, $\sigma$ is an *interior face*.

Suppose $X$ is a homology ball of dimension $d$. If every face of $X$ of dimension less than or equal to $d - s - 1$ is a boundary face, then $X$ is said to be *$s$-stacked*. For example, if $X$ is $1$-stacked, then the interior faces must all have dimension $d$ or $d-1$; this condition is sometimes simply called "stacked" for simplicial complexes (e.g. @art:Kalai-rigidity), or "capped" for cubical complexes (e.g. @art:Blind-Blind). Simplicial $s$-stacked balls are well studied due to their connection to the Lower Bound Conjecture — see e.g. @art:Murai-Nevo.

Just like Cohen–Macaulay complexes, we can characterise $s$-stacked balls using their co-skeletons.

\[thm:stacked-characterisation\] Suppose $X$ is a homology ball with dimension $d$. Then $X$ is $s$-stacked if and only if ${\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_k X}) = 0$ for all $k \leq d-s-1$ (or equivalently ${\widetilde}H_j(\abs{\operatorname{Skel}^c_{d-j-1} X}) = 0$ for all $j \geq s$, where $j = d-k-1$).

By definition, a $k$-face $\sigma$ is a boundary face if and only if ${\widetilde}H_{d-k-1}(\abs{\operatorname{link}\sigma})$ is $0$. Therefore, $X$ is $s$-stacked if and only if $$\bigoplus_{\sigma \in X_k} {\widetilde}H_{d-k-1}(\abs{\operatorname{link}\sigma}) = 0$$ for all $k \leq d-s-1$.

We could apply this fact directly to the long exact sequence from \[thm:LES\]; however, since $X$ is a homology ball, it is Cohen–Macaulay, so we can take a shortcut by using the short exact sequence from \[thm:CM-SES\]: $$\begin{tikzcd}
0 \rar & {\widetilde}H_{d-k}(\abs{\operatorname{Skel}^c_{k-1} X}) \rar & \bigoplus_{\sigma \in X_k} {\widetilde}H_{d-k-1}(\abs{\operatorname{link}\sigma}) \rar & {\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_k X}) \rar & 0.
\end{tikzcd}$$ The inner term of this short exact sequence is $0$ for $k \leq d-s-1$ if and only if both outer terms are $0$ for the same range of $k$, which is equivalent to the claimed condition.

A closely related notion is an *$s$-stacked sphere*, which is a complex that is the boundary of some $s$-stacked ball. Unfortunately, $s$-stacked spheres cannot be distinguished by topological features of their co-skeletons, at least for simplicial complexes. For example, the two spheres shown in \[fig:stacked-spheres\] have homeomorphic $k$th co-skeletons for all $k$, but only one of the spheres is $1$-stacked as a simplicial complex.

<span>0.4</span>

\[scale=1.7\]

(v1) at (-1,0.8); (v2) at (-0.5,0); (v3) at (0,0.2); (v4) at (0,0.9); (v5) at (0.5,0); (v6) at (1,0.8);

(v1)–(v2)–(v4)–cycle; (v2)–(v4)–(v5)–cycle; (v4)–(v5)–(v6)–cycle; (v1)–(v2)–(v3)–cycle; (v1)–(v3)–(v4)–cycle; (v2)–(v3)–(v5)–cycle; (v3)–(v4)–(v6)–cycle; (v3)–(v5)–(v6)–cycle;

iin <span>1,...,6</span> at (vi) ;

<span>0.4</span>

\[3D, scale=1.2\]

(v1+) at (1,0,0); (v1-) at (-1,0,0); (v2+) at (0,1,0); (v2-) at (0,-1,0); (v3+) at (0,0,1); (v3-) at (0,0,-1);

(v1+)–(v2+)–(v3-) – cycle; (v1-)–(v2+)–(v3-) – cycle; (v1+)–(v2+)–(v3+) – cycle; (v1+)–(v2-)–(v3-) – cycle; (v1-)–(v2+)–(v3+) – cycle; (v1+)–(v2-)–(v3+) – cycle; (v1-)–(v2-)–(v3-) – cycle; (v1-)–(v2-)–(v3+) – cycle; at (v1+) ; at (v1-) ; at (v2+) ; at (v2-) ; at (v3+) ; at (v3-) ;

CAT(0) cubical complexes {#sec:CAT(0)}
========================

One of the key features of a cubical complex is its *hyperplanes*. If we associate an $r$-dimensional cube with the space $[0,1]^r$ in $\mathbb R^r$, then the *$i$th hyperplane* of that cube is the subspace where $x_i = \frac{1}{2}$ — see \[fig:iterated-hyperplanes-1st\]. A hyperplane in a cubical complex ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is a maximally connected cubical complex obtained by glueing together hyperplanes of its component cubes where they meet along faces — see \[fig:hyperplanes\]. (We should note that in general, issues can arise when a hyperplane "intersects itself", in which case the hyperplane is not a true polytopal complex by our definitions. However, the focus of this section will be on CAT(0) cubical complexes, in which hyperplanes behave nicely — see e.g. \[thm:CAT(0)-hyperplanes-from-cube-embedding\] — so we will not dwell on this issue.)

<span>0.2</span>

\[3D, baseline=0\]

(0,0,1)–(1,0,1)–(1,1,1)–(0,1,1)–cycle; (0,0,0)–(1,0,0)–(1,0,1)–(0,0,1)–cycle; (0,0,0)–(0,1,0)–(0,1,1)–(0,0,1)–cycle;

<span>0.2</span>

\[3D, face/.append style=<span>opacity=1</span>, baseline=0\]

(1,0,0) – (1,1,0) – (0,1,0) (1,1,0) – (1,1,1);

(0.5,0.5,0) – (0.5,1,0) – (0.5,1,0.5) – (0.5,0.5,0.5) – cycle; (0,0.5,0) – (0.5,0.5,0) – (0.5,0.5,0.5) – (0,0.5,0.5) – cycle; (0.5,0,0) – (0.5,0.5,0) – (0.5,0.5,0.5) – (0.5,0,0.5) – cycle; (0.5,0,0.5) – (1,0,0.5) – (1,0.5,0.5) – (0.5,0.5,0.5) – cycle; (0.5,0.5,0.5) – (1,0.5,0.5) – (1,0.5,1) – (0.5,0.5,1) – cycle; (0.5,0.5,0.5) – (0.5,1,0.5) – (0.5,1,1) – (0.5,0.5,1) – cycle; (0,0.5,0.5) – (0,1,0.5) – (0.5,1,0.5) – (0.5,0.5,0.5) – cycle; (0,0,0.5) – (0.5,0,0.5) – (0.5,0.5,0.5) – (0,0.5,0.5) – cycle; (0,0.5,0.5) – (0.5,0.5,0.5) – (0.5,0.5,1) – (0,0.5,1) – cycle; (0.5,0,0.5) – (0.5,0.5,0.5) – (0.5,0.5,1) – (0.5,0,1) – cycle;

(0,0,0) – (1,0,0) – (1,0,1) – (0,0,1) – (0,1,1) – (0,1,0) – (0,0,0) – (0,0,1) (1,0,1) – (1,1,1) – (0,1,1);

<span>0.2</span>

\[3D, baseline=0\]

(1,0,0) – (1,1,0) – (0,1,0) (1,1,0) – (1,1,1);

(0.5,0.5,0)–(0.5,0.5,1); (0.5,0,0.5)–(0.5,1,0.5); (0,0.5,0.5)–(1,0.5,0.5);

(0,0,0) – (1,0,0) – (1,0,1) – (0,0,1) – (0,1,1) – (0,1,0) – (0,0,0) – (0,0,1) (1,0,1) – (1,1,1) – (0,1,1);

<span>0.2</span>

\[3D, baseline=0\]

(1,0,0) – (1,1,0) – (0,1,0) (1,1,0) – (1,1,1);

at (0.5,0.5,0.5) ;

(0,0,0) – (1,0,0) – (1,0,1) – (0,0,1) – (0,1,1) – (0,1,0) – (0,0,0) – (0,0,1) (1,0,1) – (1,1,1) – (0,1,1);

<span>0.2</span>

\[3D, scale=0.7, baseline=0\]

(0,0,1) – (1,0,1) – (1,1,1) – (0,1,1) – cycle; (1,0,1) – (2,0,1) – (2,1,1) – (1,1,1) – cycle; (1,-1,1) – (2,-1,1) – (2,0,1) – (1,0,1) – cycle; (0,0,0) – (0,1,0) – (0,1,1) – (0,0,1) – cycle; (0,0,0) – (1,0,0) – (1,0,1) – (0,0,1) – cycle; (1,-1,0) – (1,0,0) – (1,0,1) – (1,-1,1) – cycle; (1,-1,0) – (2,-1,0) – (2,-1,1) – (1,-1,1) – cycle; (2,-1,1) – (2,0,1) – (3,0,1) – (3,-1,1) – cycle; (2,-1,0) – (2,-1,1) – (3,-1,1) – (3,-1,0) – cycle;

<span>0.7</span>

\[3D, baseline=0\]

;

Since each hyperplane of a cubical complex is itself a cubical complex, it also has its own hyperplanes. Therefore, we will define an *iterated hyperplane* of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ to be a hyperplane of a hyperplane of …of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$. We will say the original hyperplanes are the "$1$st iterated hyperplanes", the hyperplanes of the hyperplanes are the "$2$nd iterated hyperplanes", and so on. See . As these figures show, an iterated hyperplane can sometimes be obtained in more than one way, when considered as a subspace of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ — for example, each line segment in \[fig:iterated-hyperplanes-2nd\] is a hyperplane of two different squares in \[fig:iterated-hyperplanes-1st\].

The following proposition reveals the connection between co-skeletons and iterated hyperplanes.

\[thm:cubical-coskeleton-simeq-iterated-hyperplanes\] Suppose ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is a cubical complex. Then $\abs{\operatorname{bary}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}) - \operatorname{bary}(\operatorname{Skel}_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})}$ is equal to the union of the $(k+1)$th iterated hyperplanes of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, considered as subspaces of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$.

For example, compare \[fig:iterated-hyperplanes-1st\] and \[fig:bary-hyperplanes\]. Note that in combination with \[thm:coskel-simeq-bary\], this proposition tells us that $\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$ is homotopy equivalent to the union of the $(k+1)$th iterated hyperplanes.

\[Proof of \[thm:cubical-coskeleton-simeq-iterated-hyperplanes\]\] We will show that the intersections of $\abs{\operatorname{bary}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}) - \operatorname{bary}(\operatorname{Skel}_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})}$ and of the union of the $(k+1)$th iterated hyperplanes with each cube of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ are the same.

\[3D, face/.append style=<span>opacity=1</span>, scale=1.8\]

(v00\*) at (0,0,0.5); (v01\*) at (0,1,0.5); (v10\*) at (1,0,0.5); (v11\*) at (1,1,0.5); (v0\*0) at (0,0.5,0); (v0\*1) at (0,0.5,1); (v1\*0) at (1,0.5,0); (v1\*1) at (1,0.5,1); (v\*00) at (0.5,0,0); (v\*01) at (0.5,0,1); (v\*10) at (0.5,1,0); (v\*11) at (0.5,1,1); (v0\*\*) at (0,0.5,0.5); (v1\*\*) at (1,0.5,0.5); (v\*0\*) at (0.5,0,0.5); (v\*1\*) at (0.5,1,0.5); (v\*\*0) at (0.5,0.5,0); (v\*\*1) at (0.5,0.5,1); (v\*\*\*) at (0.5,0.5,0.5);

(1,0,0) – (1,1,0) – (0,1,0) (1,1,0) – (1,1,1);

/in <span>\*10/\*\*0, \*10/\*1\*, 1\*0/1\*\*, 1\*0/\*\*0, 11\*/1\*\*, 11\*/\*1\*</span> (v)–(v)–(v\*\*\*)–cycle; in <span>\*10, 1\*0, 11\*</span> at (v) ;

/in <span>\*00/\*\*0, \*00/\*0\*, 0\*0/\*\*0, 0\*0/0\*\*, 01\*/\*1\*, 01\*/0\*\*, \*11/\*\*1, \*11/\*1\*, 1\*1/\*\*1, 1\*1/1\*\*, 10\*/\*0\*, 10\*/1\*\*</span> (v)–(v)–(v\*\*\*)–cycle; in <span>\*\*0, \*1\*, 1\*\*</span> at (v) ;

/in <span>00\*/\*0\*, 00\*/0\*\*, 0\*1/\*\*1, 0\*1/0\*\*, \*01/\*\*1, \*01/\*0\*</span> (v)–(v)–(v\*\*\*)–cycle; (0,0,0) – (1,0,0) – (1,0,1) – (0,0,1) – (0,1,1) – (0,1,0) – (0,0,0) – (0,0,1) (1,0,1) – (1,1,1) – (0,1,1); in <span>\*00, 0\*0, 01\*, \*11, 1\*1, 10\*, 00\*, 0\*1, \*01, \*0\*, 0\*\*, \*\*1, \*\*\*</span> at (v) ;

First, consider the union of the $(k+1)$th hyperplanes. In each cube, identified with $[0,1]^r$, a single $(k+1)$th iterated hyperplane is the result of setting $(k+1)$ of the coordinates to be $\frac{1}{2}$; therefore, the union of all $(k+1)$th iterated hyperplanes is the space $$Z \coloneqq \left\{ x \in \abs*{[0,1]^r} : \text{at least $k+1$ coordinates of $x$ are $\tfrac{1}{2}$} \right\}.$$

Now, consider $\abs{\operatorname{bary}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}) - \operatorname{bary}(\operatorname{Skel}_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})}$. The intersection of this space with some cube of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, identified with $\abs{[0,1]^r}$, is $$B \coloneqq \abs*{\operatorname{bary}([0,1]^r) - \operatorname{bary}(\operatorname{Skel}_k [0,1]^r)}.$$ For each face $\sigma$ of $[0,1]^r$, we have a vertex $v_\sigma$ of $\operatorname{bary}([0,1]^r)$ at the barycenter of $\sigma$; the coordinates of $v_\sigma$ will all be $0$, $1$ or $\frac{1}{2}$, and the number of coordinates equal to $\frac{1}{2}$ is $\dim \sigma$. The remaining vertices of $\operatorname{bary}([0,1]^r)$ after deleting $\operatorname{bary}(\operatorname{Skel}_k [0,1]^r)$, therefore, are precisely those which have at least $k+1$ coordinates equal to $\frac{1}{2}$.

For the rest of this proof, we will show that $B = Z$. First, we claim that $B \subseteq Z$. For each face $\{ v_{\sigma_1}, \dotsc, v_{\sigma_m} \}$ of $B$, the set $\{\sigma_1, \dotsc, \sigma_m\}$ is a chain, so it has a minimal element — without loss of generality, say this is $\sigma_1$. By construction, the dimension of $\sigma_1$ is at least $k+1$, so there is some index set $I$ of at least $k+1$ elements so that $(v_{\sigma_1})_i = \frac{1}{2}$ for all $i \in I$. Since $\{\sigma_1, \dotsc, \sigma_m\}$ is a chain, every $v_{\sigma_j}$ for $j = 1, \dotsc, m$ must also satisfy $(v_{\sigma_j})_i = \frac{1}{2}$ for all $i \in I$. Therefore, the entire set $\{ v_{\sigma_1}, \dotsc, v_{\sigma_m} \}$ lies in the affine subspace $\{ x \in \mathbb R^r : \text{$x_i = \frac{1}{2}$ for all $i \in I$} \}$, so the convex hull of this set is contained in $Z$. Thus $B \subseteq Z$.

Conversely, let $z$ be a point in $Z$. By definition, there is an index set $I$ of size $k+1$ so that $z_i = \frac{1}{2}$ for all $i \in I$. The set of all points of $[0,1]^r$ satisfying $x_i = \frac{1}{2}$ for $i \in I$ is an iterated hyperplane of $[0,1]^r$, and its barycentric subdivision is the subcomplex of $\operatorname{bary}([0,1]^r) - \operatorname{bary}(\operatorname{Skel}_k [0,1]^r)$ generated by vertices satisfying $(v_\sigma)_i = \frac{1}{2}$ for $i \in I$. Therefore, $z$ is in $B$, so $Z \subseteq B$.

In summary, the space $\abs{\operatorname{bary}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}) - \operatorname{bary}(\operatorname{Skel}_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})}$ and the union of the $(k+1)$th iterated hyperplanes agree within each cube of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, so they must be the same subspace.

There is a class of cubical complexes where hyperplanes and co-skeletons work particularly well. In general, the link of any vertex in a cubical complex is isomorphic as a poset to the non-empty faces of a simplicial complex. A cubical complex is *CAT(0)* if:

-   it is simply connected, and

-   the link of every vertex is isomorphic to a *flag* simplicial complex.

A simplicial complex is flag if the vertex set of every clique in its $1$-skeleton is the vertex set of a face.

CAT(0) cubical complexes were first studied from the perspective of metric spaces: equivalently, a cubical complex is CAT(0) if and only if it is simply connected and has non-positive curvature at every point (see @book:Bridson-Haefliger [Theorem 5.4 & Theorem 5.18]; see also @art:Gromov). However, finite CAT(0) cubical complexes are also interesting from a purely combinatorial and topological perspective [@art:Ardila-geodesics; @art:Hagen-hyperbolicity; @art:Sageev; @art:Roller; @art:Rowlands-CAT0].

One particular construction associated to a CAT(0) cubical complex is its *crossing complex*, introduced in @art:Rowlands-CAT0: given a CAT(0) cubical complex ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, its crossing complex, here denoted $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$, is a simplicial complex whose vertices are in bijection with the hyperplanes of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, and a set of vertices forms a face if the intersection of the corresponding hyperplanes is non-empty. (In other words, the crossing complex is the nerve of the hyperplanes, as defined in \[thm:nerve\].)

<span>0.4</span>

(0,0,0) – (-1,0,0) – (-1,0,1) – (0,0,1) – cycle; (-1,0,0) – (-1,1,0) – (-1,1,1) – (-1,0,1) – cycle; (-1,0,1) – (0,0,1) – (0,1,1) – (-1,1,1) – cycle; (0,0) rectangle (1,1); (0,1) rectangle (1,2);

(-0.5,0,0)–(-0.5,0,1)–(-0.5,1,1) node \[label=<span>\[label distance=-5pt\]above left:$H_1$</span>\] (-1,0.5,0) node \[label=<span>\[label distance=-5pt\]below left:$H_3$</span>\] –(-1,0.5,1)–(0,0.5,1) (-1,1,0.5) node \[label=<span>\[label distance=-2pt\]left:$H_2$</span>\] –(-1,0,0.5)–(0,0,0.5); (0,0.5)–(1,0.5) (0,1.5)–(1,1.5) node \[label=<span>\[label distance=-2pt\]right:$H_5$</span>\] (0.5,0)–(0.5,2) node \[label=<span>\[label distance=-2pt\]above:$H_4$</span>\] ;

<span>0.4</span>

\[scale=1.5\]

(v1) at (0.1,0.7) ; (v2) at (0.7,0.2) ; (v3) at (0,0) ; (v4) at (1.5,0.1) ; (v5) at (1.6,0.9) ;

(v1)–(v2)–(v3)–(v1); (v2)–(v4)–(v5);

at (v1) ; at (v2) ; at (v3) ; at (v4) ; at (v5) ;

We record here some basic facts about CAT(0) cubical complexes and their crossing complexes.

\[<span>[@book:Bridson-Haefliger Corollary 1.5], [@art:Gromov Section 4]</span>\] \[thm:CAT(0)-contractible\] Any CAT(0) cubical complex is contractible (not just simply connected).

\[<span>[@art:Hagen-hyperbolicity Lemma 2.15]</span>\] \[thm:crossing-complex-flag\] $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$ is always a flag simplicial complex. In particular, it is the clique complex of the "crossing graph" defined by Hagen @art:Hagen-hyperbolicity [Definition 2.16].

\[<span>[@art:Rowlands-CAT0 Proposition 6.4, Corollary 6.5]</span>\] \[thm:CAT(0)-facets-bijection\] There is a bijection between the facets of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ and the facets of $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$. If a facet of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ has dimension $k$, the corresponding facet of $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$ has dimension $k-1$. As a consequence, $\dim \operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}) = \dim {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}- 1$, and ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is pure if and only if $\operatorname{Cross}{{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is pure.

\[<span>[@art:Sageev Theorem 4.11]</span>\] \[thm:CAT(0)-hyperplanes-CAT(0)\] Every hyperplane (and thus every iterated hyperplane) in a CAT(0) cubical complex is also CAT(0).

@art:Ardila-geodesics [p. 9] described a way to embed a CAT(0) cubical complex with $m$ hyperplanes into the cube $[0,1]^m$. This embedding is essentially canonical, up to the choice of a "root vertex" and a choice of labelling for the hyperplanes as $H_1, \dotsc, H_m$.

\[<span>[@art:Rowlands-CAT0 Proposition 5.7]</span>\] \[thm:CAT(0)-hyperplanes-from-cube-embedding\] In this embedding of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ into $[0,1]^m$, the hyperplane $H_i$ of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is the intersection of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ with the $i$th hyperplane of $[0,1]^m$.

The resulting embedding of a hyperplane $H_i$ into the $i$th hyperplane of $[0,1]^m$ agrees with Ardila et al.'s canonical embedding of $H_i$ as a CAT(0) cubical complex into a cube; therefore, the iterated hyperplanes of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ are themselves each an intersection of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ with some set of hyperplanes of $[0,1]^m$. Therefore, if $S$ is a subset of $\{1, \dotsc, m\}$ of size $(k+1)$, we define $$\begin{aligned}
H_S & \coloneqq {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}\cap \bigcap_{i \in S} (\text{$i$th hyperplane of $[0,1]^m$}) \\
& = \bigcap_{i \in S} \big( {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}\cap (\text{$i$th hyperplane of $[0,1]^m$}) \big) = \bigcap_{i \in S} H_i. $$ Either this is a $(k+1)$th iterated hyperplane of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, or it is empty, and all $(k+1)$th iterated hyperplanes of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ are of this form. By definition of $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$, $H_S$ is an iterated hyperplane if and only if $S$ is a face of $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$; therefore, the $(k+1)$th iterated hyperplanes of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ correspond to $k$-faces of $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$, for $k = 0, \dotsc, d-1$.

We can now begin to prove some new results about CAT(0) cubical complexes. The following proposition generalises @art:Rowlands-CAT0 [Theorem 6.3], which was the case of $k=0$.

\[thm:CAT(0)-coskeleton-simeq-crossing-coskeleton\] If ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is a $d$-dimensional CAT(0) cubical complex and $\Delta \coloneqq \operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$ is its crossing complex, then $\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}} \simeq \abs{\operatorname{Skel}^c_{k-1} \Delta}$ for all $k = 0, \dotsc, d$.

If $k = d$, then both $\abs{\operatorname{Skel}^c_d {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$ and $\abs{\operatorname{Skel}^c_{d-1} \Delta}$ are empty, since $\dim \Delta = d-1$ by \[thm:CAT(0)-facets-bijection\]. Therefore, for the rest of this proof, assume that $k < d$.

We will apply the nerve theorem (\[thm:nerve\]) to each of $\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$ and $\abs{\operatorname{Skel}^c_{k-1} \Delta}$, to show that they are both homotopy equivalent to the same nerve.

First, consider $\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$, and define the family $\mathcal H \coloneqq \{ H_\sigma : \sigma \in \Delta_k \}$, where $H_\sigma \coloneqq \bigcap_{i \in \sigma} H_i$ as above. By \[thm:cubical-coskeleton-simeq-iterated-hyperplanes\] and the above discussion, $\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$ is homotopy equivalent to the union of $\mathcal H$. Each $H_\sigma$ is a closed set in ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, and by \[thm:cubical-coskeleton-simeq-iterated-hyperplanes\] their union is a triangulable space. Any intersection of spaces in $\mathcal H$ is itself an iterated hyperplane if it is non-empty, which is CAT(0) by \[thm:CAT(0)-hyperplanes-CAT(0)\] and thus contractible by \[thm:CAT(0)-contractible\]. Therefore, we may apply the nerve theorem to $\mathcal H$, and conclude that $\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}} \simeq \abs{N(\mathcal H)}$.

But what is $N(\mathcal H)$? The vertices are in bijection with $k$-faces $\sigma$ of $\Delta$. A set of vertices, say $\{ \sigma_1, \dotsc, \sigma_m \}$, forms a face of $N(\mathcal H)$ whenever the intersection $H_{\sigma_1} \cap \dotsb \cap H_{\sigma_m}$ is non-empty. This intersection is the set $$\begin{aligned}
H_{\sigma_1} \cap \dotsb \cap H_{\sigma_m} & = \bigcap_{i \in \sigma_1} H_i \cap \dotsb \cap \bigcap_{i \in \sigma_m} H_i \\
& = \bigcap_{i \in \sigma_1 \cup \dotsb \cup \sigma_m} H_i \\
& = H_{\sigma_1 \cup \dotsb \cup \sigma_m}.\end{aligned}$$ So $\{ \sigma_1, \dotsc, \sigma_m \}$ is a face of $N(\mathcal H)$ if and only if $\sigma_1 \cup \dotsb \cup \sigma_m$ is a face of $\Delta$.

Now, let us turn our attention to $\operatorname{Skel}^c_{k-1} \Delta$, the set of faces of $\Delta$ of dimension at least $k$. For each $k$-face $\sigma$ of $\Delta$, let $S_\sigma$ denote $\abs{\operatorname{star}_\Delta \sigma}$, and define $\mathcal S \coloneqq \{ S_\sigma : \sigma \in \Delta_k \}$. Every face of $\Delta$ of dimension at least $k$ is contained in $\operatorname{star}_\Delta \sigma$ for some $k$-face $\sigma$, and no face of dimension less than $k$ is, so the union of $\mathcal S$ is $\abs{\operatorname{Skel}^c_{k-1} \Delta}$. Each $S_\sigma$ is open in $\Delta$, and any non-empty intersection of stars is itself a star and thus contractible. Therefore, we may apply the nerve theorem to $\mathcal S$ as well.

Let us examine $N(\mathcal S)$. The vertices are in bijection with $k$-faces $\sigma$ of $\Delta$, and a set $\{ \sigma_1, \dotsc, \sigma_m \}$ forms a face of $N(\mathcal S)$ whenever $\operatorname{star}_\Delta \sigma_1 \cap \dotsb \cap \operatorname{star}_\Delta \sigma_m$ is non-empty. This happens if and only if there exists a face of $\Delta$ that contains all of $\sigma_1, \dotsc, \sigma_m$, which happens whenever $\sigma_1 \cup \dotsb \cup \sigma_m$ is a face of $\Delta$.

But this means that $N(\mathcal S) = N(\mathcal H)$! Therefore, $$\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}} \simeq \abs{N(\mathcal H)} = \abs{N(\mathcal S)} \simeq \abs{\operatorname{Skel}^c_{k-1} \Delta}. \qedhere$$

Next, we will examine some applications of this result, but first we need one more construction.

There is a standard bijection from the non-empty faces $\sigma$ of the cube $[0,1]^r$ to vectors $\chi^\sigma$ in $\{0,1,*\}^r$, where the $i$th coordinate of $\chi^\sigma$ is $0$ or $1$ if all points $x$ in $\sigma$ have $x_i = 0$ or $x_i = 1$ respectively, and $\chi^\sigma_i$ is $*$ otherwise. If $\sigma$ and $\tau$ are two faces, then $\sigma \subseteq \tau$ if and only if $\chi^\sigma_i \preceq \chi^\tau_i$ for all $i$, where $0 \preceq *$ and $1 \preceq *$ are the only non-equality relations on $0$, $1$ and $*$.

Suppose $\Delta$ is an abstract simplicial complex with vertex set $\{1, \dotsc, n\}$. We define the *cubical cone* over $\Delta$, denoted $\operatorname{Cone}(\Delta)$, to be the subcomplex of $[0,1]^n$ consisting of all faces $\sigma$ such that the set $\big\{ i : \chi^\sigma_i \in \{1,*\} \big\}$ is a face of $\Delta$ (including the empty face). For example, see \[fig:cubical-cone\]. The reason for the name "cubical cone" is the following lemma, analogous to properties of a simplicial cone.

The cubical cone over $\Delta$ is the unique subcomplex of a cube (up to isomorphism) with the property that there is a vertex $v_0$ (the "cone point", which is the point $(0,\dotsc, 0)$ in the standard labelling) such that:

-   every facet contains $v_0$, and

-   the link of $v_0$ is (isomorphic as a poset to the non-empty faces of) $\Delta$.

First, let us check that the cubical cone does satisfy these properties. If $\sigma$ is a face of $\operatorname{Cone}(\Delta)$, we can replace every $1$ in $\chi^\sigma$ with $*$ to obtain the vector of a face of $\operatorname{Cone}(\Delta)$ that contains $\sigma$; therefore, the vectors of facets of $\operatorname{Cone}(\Delta)$ consist only of $0$ and $*$, so every facet contains $v_0 = (0,\dotsc,0)$. Conversely, the faces that properly contain $v_0$ are precisely the faces $\sigma \neq v_0$ such that $\chi^\sigma$ consists of $0$ and $*$, and these faces are in bijection with non-empty faces of $\Delta$, so the link of $v_0$ is indeed $\Delta$.

Now, suppose ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is an arbitrary subcomplex of a cube satisfying these properties. By the symmetry of the cube, we may assume that $v_0$ is the vertex $(0,\dotsc,0)$. As before, the link of $v_0$ is then the set of faces of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ whose vectors consist only of $0$ and $*$; since we assume this set is isomorphic to $\Delta$, we may again use the symmetry of the cube to permute the coordinates and assume that this set is equal to $\operatorname{link}_{\operatorname{Cone}(\Delta)} v_0$. Since ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ contains these faces of the cube, it must also contain every sub-face of these faces, so ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ contains every face of $\operatorname{Cone}(\Delta)$. Moreover, it cannot contain any other faces: any such face must be contained in some facet of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$, but that facet cannot be one that we have already accounted for in $\operatorname{link}_{{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}v_0$, so the facet would not contain $v_0$. Therefore, ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}\cong \operatorname{Cone}(\Delta)$.

While every flag simplicial complex is the crossing complex of some CAT(0) cubical complex (see @art:Hagen-hyperbolicity [Proposition 2.19] and @art:Rowlands-CAT0 [Lemma 3.3], where the construction is the cubical cone), in general, there can be many CAT(0) cubical complexes with the same crossing complex. However, we can now prove that if $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$ is a connected manifold, the only possibility is ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}= \operatorname{Cone}(\Delta)$.

<span>0.3</span>

\[scale=1.2\]

/in <span>0.5/1.8, -0.1/0.9, 1.0/1.1, 1.2/0.2, 0.0/0.0</span> (vi) at (,);

(v1)–(v2)–(v3)–cycle; (v3)–(v4)–(v5)–(v2);

in <span>above, left, above right, below right, below left</span> at (vi) ;

<span>0.5</span>

\[3D, scale=1.5\]

//in <span>0/0/0, 0/1/0, 1/0/0, 1/1/1</span> (,,) – +(1,0,0) – +(1,1,0) – +(0,1,0) – cycle; (1,1,0) – (2,1,0) – (2,1,1) – (1,1,1) – cycle; (1,1,0) – (1,2,0) – (1,2,1) – (1,1,1) – cycle;

////in <span>1/1/0/v\_0 = 00000/above, 1/1/1/10000/above, 1/2/0/01000/above left, 2/1/0/00100/above right, 1/0/0/00010/below right, 0/1/0/00001/below left, 1/2/1/11000/above left, 2/1/1/10100/above right, 2/0/0/00110/right, 0/0/0/00011/below, 0/2/0/01001/left, 2/2/1/11100/above</span> at (,,) ;

Let ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ be a $d$-dimensional CAT(0) cubical complex and $\Delta$ its crossing complex. If $\Delta$ is a connected homology manifold, then ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is the cubical cone over $\Delta$.

Examine this section of the long exact sequence from \[thm:LES\], taking $k = 0$ and using coefficients in $\mathbb Z/2$: $$\begin{tikzcd}
{\widetilde}H_d(\abs{\operatorname{Skel}^c_{-1} {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}) \rar & \bigoplus_{v \in {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}_0} {\widetilde}H_{d-1}(\abs{\operatorname{link}v}) \rar & {\widetilde}H_{d-1}(\abs{\operatorname{Skel}^c_0 {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}) \rar & {\widetilde}H_{d-1}(\abs{\operatorname{Skel}^c_{-1} {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}).
\end{tikzcd}$$ Note that $\abs{\operatorname{Skel}^c_{-1} {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$ is simply $\abs{{{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$, which is contractible since ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is CAT(0), so the first and last terms of this sequence are $0$. By \[thm:CAT(0)-coskeleton-simeq-crossing-coskeleton\], ${\widetilde}H_{d-1}(\abs{\operatorname{Skel}^c_0 {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}})$ is isomorphic to ${\widetilde}H_{d-1}(\abs{\operatorname{Skel}^c_{-1} \Delta}) = {\widetilde}H_{d-1}(\abs{\Delta})$, and since $\Delta$ is a connected homology manifold of dimension $d-1$, this homology group is $\mathbb Z/2$. Therefore, $$\bigoplus_{v \in {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}_0} {\widetilde}H_{d-1}(\abs{\operatorname{link}v}) \cong \mathbb Z/2.$$ Thus there must be one vertex $v_0$ for which ${\widetilde}H_{d-1}(\abs{\operatorname{link}v_0}) = \mathbb Z/2$, and all other vertices have ${\widetilde}H_{d-1}(\abs{\operatorname{link}v}) = 0$.

Every vertex link in a CAT(0) cubical complex is an induced subcomplex of the crossing complex [@art:Rowlands-CAT0 Lemma 4.9]. However, the only induced subcomplex of a connected $(d-1)$-manifold with non-zero homology in degree $d-1$ is the entire manifold. Therefore, $\operatorname{link}v_0 = \Delta$.

says that there is a bijection between the facets of the cubical complex and the facets of its crossing complex. In this case, since $\operatorname{link}v_0 = \Delta$, we also have a bijection between the facets of the crossing complex $\Delta$ and facets of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ that contain $v_0$. Therefore, every facet of ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ must contain $v_0$. Hence ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is a cubical cone.

We conclude this paper by combining \[thm:CAT(0)-coskeleton-simeq-crossing-coskeleton\] with \[thm:CM-characterisation,thm:stacked-characterisation,thm:Leray-characterisation\], to compare the combinatorial properties discussed in \[sec:families\] for ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ and $\operatorname{Cross}({{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}})$.

\[thm:CAT(0)-property-iff-crossing\] Suppose ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is a CAT(0) cubical complex and $\Delta$ is its crossing complex. Then:

(a) \[thm:CAT(0)-CM-iff-crossing\] ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is Cohen–Macaulay if and only if $\Delta$ is Cohen–Macaulay.

(b) \[thm:CAT(0)-Leray-iff-crossing\] ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is $r$-Leray if and only if $\Delta$ is $r$-Leray.

(c) \[thm:CAT(0)-stacked-iff-crossing\] If ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ and $\Delta$ are both homology balls, then ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is $s$-stacked if and only if $\Delta$ is $s$-stacked.

Note that the assumption that both ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ and $\Delta$ are balls in \[thm:CAT(0)-stacked-iff-crossing\] is essential: it is possible for one to be a ball and the other not. For example, if ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ consists of $m$ squares glued in a $1 \times m$ rectangle, it is a $1$-stacked ball, but its crossing complex is a star graph with $m$ pendants, which is not a ball for $m > 2$.

\[Proof of \[thm:CAT(0)-property-iff-crossing\]\]

(a) Let $d = \dim {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$. By \[thm:CM-characterisation\], ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is Cohen–Macaulay if and only if it is pure and ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}) = 0$ for all $k = -1, \dotsc, d$ and all $i < d-k-1$. Any CAT(0) cubical complex is always contractible, so ${\widetilde}H_i(\abs{\operatorname{Skel}^c_{-1} {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}})$ is guaranteed to be $0$, hence we only need to consider $k = 0, \dotsc, d$. Now, \[thm:CAT(0)-coskeleton-simeq-crossing-coskeleton\] says that ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}) = {\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} \Delta})$ for $k$ in this range, and \[thm:CAT(0)-facets-bijection\] says that ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is pure if and only if $\Delta$ is pure; therefore, ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is Cohen–Macaulay if and only if $\Delta$ is pure and ${\widetilde}H_i(\abs{\operatorname{Skel}^c_{k-1} \Delta}) = 0$ for all $k = 0, \dotsc, d$ and all $i < d - k - 1 = (d-1) - (k-1) - 1$. But since $\dim \Delta = d-1$, this is exactly the condition given by \[thm:CM-characterisation\] for $\Delta$ to be Cohen–Macaulay.

(b) By \[thm:Leray-characterisation\], ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is $r$-Leray if and only if ${\widetilde}H_i(\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}) = 0$ for all $i \geq r$ and all $k = -1, \dotsc, d$. Because ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is CAT(0), $\abs{\operatorname{Skel}^c_{-1} {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}$ is guaranteed to be contractible, so we only need to check $k = 0, \dotsc, d$. By \[thm:CAT(0)-coskeleton-simeq-crossing-coskeleton\], this condition is equivalent to ${\widetilde}H_i(\abs{\operatorname{Skel}^c_j \Delta})$ being $0$ for all $i \geq r$ and all $j = -1, \dotsc, d-1$, where $j = k-1$. And this precisely means that $\Delta$ is $r$-Leray.

(c) By \[thm:stacked-characterisation\], ${{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$ is $s$-stacked if and only if ${\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_k {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}}) = 0$ for all $k \leq d-s-1$, where $d = \dim {{\mathord{{\tikz[baseline=0.5pt]{\fill [even odd rule] (0,0) [rounded corners=0.15pt] rectangle (0.68em,0.68em) (0.04em,0.07em) [sharp corners] rectangle (0.68em-0.09em,0.68em-0.04em); \useasboundingbox (-0.08em,0) rectangle (0.68em+0.08em,0.68em);}}}}}$. tells us this is equivalent to ${\widetilde}H_{d-k-1}(\abs{\operatorname{Skel}^c_{k-1} \Delta})$ being $0$ for $k \leq d-s-1$. Equivalently, letting $j = k-1$, we have ${\widetilde}H_{(d-1)-j-1}(\abs{\operatorname{Skel}^c_j \Delta}) = 0$ for $j \leq (d-1)-s-1$. Since $\dim \Delta = d-1$, this is equivalent to $\Delta$ being $s$-stacked.

{'source': 'arXiv_out/2209'}