Topology of complements of skeletons

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 Cohen-Macaulay 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
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, Bayer [2018] 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 higherdimensional faces instead of the low-dimensional ones.We define the kth 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.
Section 2 of this paper sets out the basic definitions we will use.In Section 3, we examine some initial facts about co-skeletons of arbitrary complexes, and their connections with various forms of duality (see Proposition 3.3 and Corollary 3.5).In Section 4, we construct a long exact sequence relating the kth and (k −1)th co-skeletons and the links of k-dimensional faces (Theorem 4.1).In Section 5, 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 (Theorems 5.3,5.10 and 5.11).Finally, Section 6 examines cubical complexes, particularly CAT(0) cubical complexes: we examine the "crossing complex" defined by Rowlands [2020], and show that a CAT(0) cubical complex has one of the properties discussed in Section 5 if and only if its crossing complex shares that property (Theorem 6.10).Figure 1: An example of a polytopal complex

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 R N with the following properties: • If σ is in X and τ is a face of σ, then τ is in X, and • If σ and σ ′ are polytopes in X, then σ ∩ σ ′ is a face of each (possibly the empty face).
See Figure 1 for a small example.In this paper, we will only consider finite polytopal complexes.
If every polytope in X is a simplex, then we say that ∆ := 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 ≥ 0), then := X is a cubical complex.
An abstract simplicial complex ∆ 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 σ ∈ ∆ and τ ⊆ σ implies τ ∈ ∆.The dimension of a face σ is dim σ :− #σ − 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 σ is a polytope, |σ| will denote its relative interior.If S is a collection of faces in a polytopal complex, then their geometric realisation |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 σ is in S, we do not include the boundary of σ in the geometric realisation |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 |S| as a topological space.
Suppose Y is a subset of a polytopal complex X.We write X \ Y to denote the difference of sets, so X \ 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 "\" to denote deletion of sets or topological spaces.Note that |X \ Y | = |X| \ |Y |, but the space |X − Y | is not in general the same; see Figure 2, for an example.However, these spaces are sometimes related by the following lemma: Lemma 2.1 ( [Munkres, 1984b, Lemma 70.1]).If ∆ is a simplicial complex and Λ is an induced subcomplex (in other words, every face of ∆ whose vertices are contained in the vertex set of Λ is a face of Λ), then |∆ − Λ| and |∆| \ |Λ| are homotopy equivalent.
Given a face σ of X, the (open) star of σ is the set of faces The geometric realisation of a star is always contractible, if σ = ∅.The link of σ is the set We will sometimes simply write "star σ" and "link σ" if the space X is clear from context.Note that if ∆ 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 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 bary(X) which has one vertex v σ for each non-empty face σ of X, and a set {v σ1 , . . ., v σm } forms a face of bary(X) whenever {σ 1 , . . ., σ m } is a chain in the poset of faces of X ordered by inclusion (that is, σ 1 ⊂ • • • ⊂ σ m , up to reordering).If each vertex v σ is positioned at the barycentre of the polytope σ, then the geometric realisation of bary(X) is exactly the geometric realisation of X, as a topological space.See Figure 3 for an example.
Given a polytopal complex X, the set of k-dimensional faces of X will be denoted The central definition in this paper is the kth co-skeleton of X, which is the complement of the the k-skeleton: We will mostly be interested in the topological properties of |Skel c k X| = |X| \ |Skel k X|.Note that the coskeleton is not in general a polytopal complex, except in two special cases: Skel c −1 X is X itself (modulo the empty face, which makes no difference to the topology), and if X is d-dimensional, Skel c d X is the polytopal complex with no faces.

Topology
We will use the notation A ≃ 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 Hatcher [2002]; Munkres [1984b].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 Z. Reduced homology is denoted by H i (A).We use the convention that H −1 (∅) = 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.
Theorem 2.2 (Mayer-Vietoris, [Hatcher, 2002, p. 149], [Munkres, 1984b, Theorem 33.1]).If A and B are open subsets of a topological space, then there is a long exact sequence: If A ∩ B = ∅, then we may replace these unreduced homology groups with reduced homology throughout.
Theorem 2.3 (Nerve theorem, [Hatcher, 2002, Corollay 4G.3], [Björner, 1995, Theorem 10.7]).Suppose U = {U 1 , . . ., U n } is a family of open sets whose union is a paracompact space (e.g.any subspace of R n ), or a family of closed sets whose union is a triangulable space.Suppose further that for every index set I ⊆ {1, . . ., n}, the intersection i∈I U i is either contractible or empty.Construct a simplicial complex N (U) (called the "nerve" of U) where the vertex set is {1, . . ., n} and the set I forms a face whenever i∈I U i is non-empty.Then

First results about co-skeletons
Let us begin to study the co-skeleton Skel c k X.This first lemma tells us that while Skel c k X is not itself a polytopal complex, it is homotopy equivalent to one.See Figure 4.
Proof.By definition, Now, bary(Skel k X) is an induced subcomplex of the simplicial complex bary(X): if v σ1 , . . ., v σm are vertices of bary(Skel k X) that form a face of bary(X), then the faces σ 1 , . . ., σ m form a chain in the face poset of X, so they still form a chain in the face poset of Skel k X.Therefore, we may invoke Lemma 2.1: Proof.The vertices of bary(X) − bary(Skel k X) correspond to faces σ of X of dimensions between k + 1 and d; therefore, the largest possible face of bary(X) − bary(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.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 |bary(X)− bary(Skel k X)| is exactly the (d − k − 1)-skeleton of the dual cell structure used in some proofs of Poincaré duality -for example, Munkres [1984b, §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.
Theorem 3.4 (Alexander duality, [Munkres, 1984b, §71, particularly Theorem 71.1 and Exercise 4]).Suppose X is a polytopal complex where dim ∆ = d and |X| is homeomorphic to a sphere, and suppose Y is a proper, non-empty subcomplex of X.Then Taking Y = Skel k X gives us this corollary: Corollary 3.5.If X is a d-dimensional polytopal complex where |X| is homeomorphic to a sphere, then for k = 0, . . ., d − 1, For a first example of an application for co-skeletons, let us consider neighbourly simplicial spheres.Suppose ∆ is a simplicial sphere (that is, a simplicial complex where |∆| is homeomorphic to a sphere) with n vertices.Then ∆ 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 ⌊d/2⌋-neighbourly.A major reason why neighbourly spheres have been studied is their connection with the Upper Bound Conjecture (see e.g.Alon and Kalai [1985]).
The following lemma will let us characterise skeletons of simplices by their homology.(It follows without much difficulty from Björner and Kalai [1988, Theorem 3.1] and Kalai [2002, Equation (3.6)], but we also present a self-contained proof.)Lemma 3.6.Suppose ∆ is a simplicial complex with n vertices, n ≥ 1, and dim ∆ = k < n.Then with equality if and only if ∆ is isomorphic to the k-skeleton of a simplex with n vertices.If the coefficient ring is a field instead of Z, the same holds with rank replaced by dimension.
Proof.For simplicity of notation, assume the coefficient ring is 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 ∆ as {1, . . ., n}, and define the following two open sets: Then A ∪ B is |∆|, and A ∩ B is |star n| \ {n} = |link n|.The Mayer-Vietoris theorem (Theorem 2.2) then gives us the following long exact sequence: Since |star n| is contractible, H i (|star n|) is zero.Now, consider the following segment of the exact sequence: By Lemma 2.1, |∆| \ {n} is homotopy equivalent to |∆ − n|, which is a simplicial complex with n − 1 vertices and dimension less than or equal to k, so by induction, rank H k (|∆| \ {n}) ≤ n−2 k+1 .(Note that if the dimension of ∆ − n is less than k, its degree k homology must be zero.)Similarly, 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, rank To obtain equality for rank H k (|∆|), we need equality to hold for both rank H k (|∆−n|) and rank H k−1 (|link n|), so by induction, both ∆ − n and s-link n need to be skeletons of simplices of the appropriate sizes.Therefore, ∆ 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 ∆ 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 Corollary 3.5, we obtain the following characterisations of neighbourly spheres in terms of their skeletons and their co-skeletons: Corollary 3.7.Suppose ∆ is a d-dimensional simplicial sphere with n vertices.Then for t = 1, . . ., d, the following are equivalent: Proof.By definition, ∆ is t-neighbourly if and only if Skel t−1 ∆ is isomorphic to the (t − 1)-skeleton of a simplex on n vertices.Lemma 3.6 says that this occurs if and only if rank . And Corollary 3.5 says that

The key long exact sequence
We now reach the main tool of this paper, which we will make much use of in the upcoming sections.
Theorem 4.1.If X is a d-dimensional polytopal complex and 0 ≤ k ≤ d, then there is the following long exact sequence: Proof.We will begin with the case k = d.The link of any d-dimensional face is empty, as is Skel c d X.The (d − 1)th co-skeleton 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: 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: See Figures 5 and 6 First, we claim that the union in A is in fact a disjoint union.If two stars, say star bary(X) v σ and star bary(X) v τ , were to intersect, they would have a face of bary(X) in common, and by the definition of an open star, this face would need to contain both v σ and v τ .But faces in bary(X) correspond to chains in the face poset of X; therefore, no face in bary(X) can contain both v σ and v τ , since σ and τ 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 ∩ B. Since star bary(X) v σ does not intersect any faces of X of dimension less than k, and the only k-dimensional face it intersects is σ, we conclude that Therefore, A ∪ B consists of the relative interiors of all faces of dimension k and greater; that is, We can now apply the Mayer-Vietoris theorem, with unreduced homology.This gives us the following long exact sequence: Since unreduced homology of a disconnected space can be decomposed as a direct sum over components, we can simplify this long exact sequence: 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 σ ∈ X k , there are two cases to consider: either σ is not a facet, or it is. If 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 (|star bary(X) v σ | \ |σ|) to a generator of H 0 (|star bary(X) v σ |) ∼ = 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 (|star bary(X) v σ | \ |σ|) becomes reduced, and since |star bary(X) v σ | is contractible, its homology is zero outside degree 0, so we can omit the H i (|star bary(X) v σ |) term.This holds for every k-face σ that is not a facet.
What if σ is a facet?In this case, |star bary(X) v σ | = |σ|, so |star bary(X) v σ | \ |σ| 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: 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 The result is the following: • The H i (|star bary(X) v σ |) term is removed (since it is zero when i = 0), • No net change occurs to H 0 (|Skel c k−1 X|), and • The H −1 (|star bary(X) v σ |\|σ|) 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 ) 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: There is one last thing to do to this long exact sequence: simplifying H i |star bary(X) v σ | \ |σ| .To do this, let us call on the Mayer-Vietoris theorem again.Assume that σ is not a facet, so |star bary(X) v σ | \ |σ| is non-empty.Define two new spaces:  sequence with reduced homology for A ′ and B ′ is the following: But stars of non-empty faces are contractible, so this becomes: ) for all non-facets σ.Finally, note that if σ 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
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.
Remark 5.1.Cohen-Macaulayness is also a topological condition: X is Cohen-Macaulay if and only if H i (|X|) and H i (|X|, |X| \ {p}) are both 0 for all i < dim X and all points p ∈ |X|.This was proved by Munkres [Munkres, 1984a, Corollary 3.4] for simplicial complexes, but a similar proof applies to polytopal complexes as well, with the following modified version of Munkres [1984a, Lemma 3.3]: Lemma 5.2.If X is a polytopal complex, σ a face of X and p a point in the relative interior of σ, then Proof.We saw in the proof of Theorem 4.1 that where "s-link" denotes the simplicial link.Since bary(∂σ) is an induced subcomplex of the simplicial complex s-link bary(X) v σ , Lemma 2.1 says that It is not hard to check that s-link bary(X) v σ = s-link bary(X) v σ − bary(∂σ) * bary(∂σ), where * denotes simplicial join [Munkres, 1984b, §62], so Munkres [1984b, Theorem 62.5] implies that Finally, since bary(X) is a simplicial complex, we can apply Munkres [1984a, Lemma 3.3] to conclude that and note that although v σ is typically defined to be the barycentre of σ, it can be chosen to be any point in the relative interior of σ.
The following theorem allows us to determine whether a complex is Cohen-Macaulay by considering its co-skeletons.Note that this theorem combined with Corollary 3.2 means that the homology of |Skel c k X| is entirely concentrated in degree d − k − 1.
Proof of Theorem 5.3.To begin with, assume X is Cohen-Macaulay.We will prove that H i (|Skel c k X|) = 0 for i < d − k − 1 by induction on k.
For the base case, when k = −1, we have 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 Theorem 4.1: k X|) must be 0 too, and we are done with this direction of the proof.Now, for the reverse direction, assume that H i (|Skel c k X|) = 0 for all i < d − k − 1 and all k = −1, . . ., d.Consider this part of the long exact sequence for k ≥ 0: This is true for all faces σ of dimension k ≥ 0, which leaves the case σ = ∅.In this case, Therefore, X is Cohen-Macaulay.
Remark 5.4.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 kth co-skeletons of X for the d + 2 values of k between −1 and d, or equivalently (by Lemma 3.1) the simplicial complexes bary(X) − bary(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 bary(X) − bary(Skel k X) is at least on the order of 2 d .
Remark 5.5.We saw in Corollary 3.7 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, Figure 8 shows two complexes, one Cohen-Macaulay and one not, whose k-skeletons have isomorphic ith homology groups for all k and i.
Theorem 5.3 says that if X is Cohen-Macaulay, almost all of the long exact sequence from Theorem 4.1 is 0. The remaining non-zero terms are captured in the following corollary: Corollary 5.6.If X is a Cohen-Macaulay polytopal complex and k = 0, . . ., d, we have the following short exact sequence: Proof.Take the following segment of the long exact sequence from Theorem 4.1: The first term, We can combine these short exact sequences into a long exact sequence, using the following homological algebra fact: Lemma 5.7.Suppose we have two exact sequences: Then the horizontal sequence in the following diagram is exact: Corollary 5.8.If X is Cohen-Macaulay, we have the following long exact sequence: Proof.Use Lemma 5.7 to stitch together the short exact sequences from Corollary 5.6.The start of the sequence looks like this: And the end like this: Note that this exact sequence is reminiscent of the "partition complex", a chain complex defined by Adiprasito and Yashfe [2021, Definition 25].
Remark 5.9.If the definition of a Cohen-Macaulay complex is modified to allow H i (|link X σ|) to be non-zero when σ = ∅, 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 Theorem 5.3: a complex X is Buchsbaum if and only if the map H i (|Skel c k X|) → H i (|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, . . ., d.This follows fairly easily from the long exact sequence in Theorem 4.1, although we leave the details to the reader.

Leray complexes
A simplicial complex ∆ is r-Leray if every induced subcomplex Λ of ∆ has H i (|Λ|) = 0 for i ≥ r.For example, if ∆ is the nerve of a family of convex open subsets of R r , then it follows from the nerve theorem (Theorem 2.3) that ∆ is r-Leray (see Kalai and Meshulam [2006]).
Equivalently, ∆ is r-Leray if the homology H i (|link ∆ σ|) is 0 for every face σ ∈ ∆ (including the empty face) and every i ≥ r [Kalai and Meshulam, 2006, 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 H i (|link X σ|) = 0 for all faces σ (including σ = ∅) and all i ≥ 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.
Theorem 5.10.A polytopal complex X is r-Leray if and only if H i (|Skel c k X|) = 0 for all i ≥ r and all k = −1, . . ., d.
Proof.For the forward direction, assume X is r-Leray and let i ≥ r.
which is 0, since X is r-Leray.
For k ≥ 0, consider the following subsequence of the long exact sequence from Theorem 4.1: Since this is true for all k = 0, . . ., d, we get a series of injections: But we saw above that H i (|Skel c −1 X|) = 0. Therefore, H i (|Skel c k X|) injects into 0, so it must itself be 0, for all k = 0, . . ., d.
For the reverse direction, assume that H i (|Skel c k X|) = 0 for all k = −1, . . ., d and all i ≥ r.When k ≥ 0, consider this part of the long exact sequence: By our assumptions, the two outer terms here are both 0, hence the middle term is 0 as well, so H i (|link σ|) = 0 for all k-faces σ.This works for all k = 0, . . ., d, which only leaves k = −1: the only (−1)-dimensional face is ∅, and −1 X|) which we assumed to be 0.
Therefore, H i (|link X σ|) = 0 for all faces σ and all i ≥ r, so X is r-Leray.

Stacked balls
A homology sphere is a polytopal complex X of dimension d where for every face σ including σ = ∅, A homology manifold is defined similarly but only considering σ = ∅.A homology ball of dimension d is a complex X where: • H i (|X|) = 0 for all i, • for every non-empty face σ,

and
• the set of faces σ with H d−dim σ−1 (|link σ|) = 0 forms a subcomplex of X that is a (d − 1)-dimensional homology sphere.
If H d−dim σ−1 (|link σ|) is 0, σ is called a boundary face, and if this homology is R, σ 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.Kalai [1987]), or "capped" for cubical complexes (e.g.Blind and Blind [1998]).Simplicial s-stacked balls are well studied due to their connection to the Lower Bound Conjecture -see e.g.Murai and Nevo [2013].
Just like Cohen-Macaulay complexes, we can characterise s-stacked balls using their co-skeletons.
Theorem 5.11.Suppose X is a homology ball with dimension d.Then X is s-stacked if and only if Proof.By definition, a k-face σ is a boundary face if and only if We could apply this fact directly to the long exact sequence from Theorem 4.1; however, since X is a homology ball, it is Cohen-Macaulay, so we can take a shortcut by using the short exact sequence from Corollary 5.6: The inner term of this short exact sequence is 0 for k ≤ 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.
Remark 5.12.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 Figure 9 have homeomorphic kth co-skeletons for all k, but only one of the spheres is 1-stacked as a simplicial complex.
(a) A 1-stacked sphere, the boundary of the ball made from three stacked 3-simplices (b) A sphere that is not 1-stacked, the boundary of an octahedron

CAT(0) cubical complexes
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 R r , then the ith hyperplane of that cube is the subspace where x i = 1 2 -see Figure 10(b).A hyperplane in a cubical complex is a maximally connected cubical complex obtained by glueing together hyperplanes of its component cubes where they meet along faces -see Figure 11.(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.Lemma 6.6 -so we will not dwell on this issue.) 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 to be a hyperplane of a hyperplane of . . . of .We will say the original hyperplanes are the "1st iterated hyperplanes", the hyperplanes of the hyperplanes are the "2nd iterated hyperplanes", and so on.See Figures 10(a) to 10(d).As these figures show, an iterated hyperplane can sometimes be obtained in more than one way, when considered as a subspace of -for example, each line segment in Figure 10(c) is a hyperplane of two different squares in Figure 10(b).
The following proposition reveals the connection between co-skeletons and iterated hyperplanes.
Proposition 6.1.Suppose is a cubical complex.Then |bary( ) − bary(Skel k )| is equal to the union of the (k + 1)th iterated hyperplanes of , considered as subspaces of .Proof of Proposition 6.1.We will show that the intersections of |bary( ) − bary(Skel k )| and of the union of the (k + 1)th iterated hyperplanes with each cube of are the same.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 1 2 ; therefore, the union of all (k + 1)th iterated hyperplanes is the space Now, consider |bary( ) − bary(Skel k )|.The intersection of this space with some cube of , identified with For each face σ of [0, 1] r , we have a vertex v σ of bary([0, 1] r ) at the barycenter of σ; the coordinates of v σ will all be 0, 1 or 1 2 , and the number of coordinates equal to 1 2 is dim σ.The remaining vertices of bary([0, 1] r ) after deleting bary(Skel k [0, 1] r ), therefore, are precisely those which have at least k + 1 coordinates equal to 1 2 .
For the rest of this proof, we will show that B = Z.First, we claim that B ⊆ Z.For each face {v σ1 , . . ., v σm } of B, the set {σ 1 , . . ., σ m } is a chain, so it has a minimal element -without loss of generality, say this is σ 1 .By construction, the dimension of σ 1 is at least k + 1, so there is some index set I of at least k + 1 elements so that (v σ1 ) i = 1 2 for all i ∈ I. Since {σ 1 , . . ., σ m } is a chain, every v σj for j = 1, . . ., m must also satisfy (v σj ) i = 1 2 for all i ∈ I. Therefore, the entire set {v σ1 , . . ., v σm } lies in the affine subspace {x ∈ R r : x i = 1 2 for all i ∈ I}, so the convex hull of this set is contained in Z. Thus B ⊆ 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 = 1 2 for all i ∈ I.The set of all points of [0, 1] r satisfying x i = 1 2 for i ∈ I is an iterated hyperplane of [0, 1] r , and its barycentric subdivision is the subcomplex of bary([0, 1] r ) − bary(Skel k [0, 1] r ) generated by vertices satisfying (v σ ) i = 1 2 for i ∈ I. Therefore, z is in B, so Z ⊆ B. In summary, the space |bary( ) − bary(Skel k )| and the union of the (k + 1)th iterated hyperplanes agree within each cube of , 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 Bridson and Haefliger [1999, Theorem 5.4 & Theorem 5.18]; see also Gromov [1987]).However, finite CAT(0) cubical complexes are also interesting from a purely combinatorial and topological perspective [Ardila et al., 2012;Hagen, 2014;Sageev, 1995;Roller, 2016;Rowlands, 2020].Figure 13: A CAT(0) cubical complex and its crossing complex One particular construction associated to a CAT(0) cubical complex is its crossing complex, introduced in Rowlands [2020]: given a CAT(0) cubical complex , its crossing complex, here denoted Cross( ), is a simplicial complex whose vertices are in bijection with the hyperplanes of , 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 Theorem 2.3.) We record here some basic facts about CAT(0) cubical complexes and their crossing complexes.
Lemma 6.4 ([Rowlands, 2020, Proposition 6.4, Corollary 6.5]).There is a bijection between the facets of and the facets of Cross( ).If a facet of has dimension k, the corresponding facet of Cross( ) has dimension k − 1.As a consequence, dim Cross( ) = dim − 1, and is pure if and only if Cross is pure.Lemma 6.5 ( [Sageev, 1995, Theorem 4.11]).Every hyperplane (and thus every iterated hyperplane) in a CAT(0) cubical complex is also CAT(0).Ardila et al. [2012, 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 , . . ., H m .Lemma 6.6 ([Rowlands, 2020, Proposition 5.7]).In this embedding of into [0, 1] m , the hyperplane H i of is the intersection of with the ith hyperplane of [0, 1] m .
The resulting embedding of a hyperplane H i into the ith 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 are themselves each an intersection of with some set of hyperplanes of [0, 1] m .Therefore, if S is a subset of {1, . . ., m} of size (k + 1), we define Either this is a (k + 1)th iterated hyperplane of , or it is empty, and all (k + 1)th iterated hyperplanes of are of this form.By definition of Cross( ), H S is an iterated hyperplane if and only if S is a face of Cross( ); therefore, the (k + 1)th iterated hyperplanes of correspond to k-faces of Cross( ), for k = 0, . . ., d − 1.We can now begin to prove some new results about CAT(0) cubical complexes.The following proposition generalises Rowlands [2020, Theorem 6.3], which was the case of k = 0. | is homotopy equivalent to the union of H.Each H σ is a closed set in , and by Proposition 6.1 their union is a triangulable space.Any intersection of spaces in H is itself an iterated hyperplane if it is non-empty, which is CAT(0) by Lemma 6.5 and thus contractible by Lemma 6.2.Therefore, we may apply the nerve theorem to H, and conclude that |Skel But what is N (H)?The vertices are in bijection with k-faces σ of ∆.A of vertices, say {σ 1 , . . ., σ m }, forms a face of N (H) whenever the intersection H σ1 ∩ • • • ∩ H σm is non-empty.This intersection is the set Now, let us turn our attention to Skel c k−1 ∆, the set of faces of ∆ of dimension at least k.For each k-face σ of ∆, let S σ denote |star ∆ σ|, and define S := {S σ : σ ∈ ∆ k }.Every face of ∆ of dimension at least k is contained in star ∆ σ for some k-face σ, and no face of dimension less than k is, so the union of S is |Skel c k−1 ∆|.Each S σ is open in ∆, and any non-empty intersection of stars is itself a star and thus contractible.Therefore, we may apply the nerve theorem to S as well.
Let us examine N (S).The vertices are in bijection with k-faces σ of ∆, and a set {σ 1 , . . ., σ m } forms a face of N (S) whenever star ∆ σ 1 ∩ • • • ∩ star ∆ σ m is non-empty.This happens if and only if there exists a face of ∆ that contains all of σ 1 , . . ., σ m , which happens whenever σ But this means that N (S) = N (H)!Therefore, 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 σ of the cube [0, 1] r to vectors χ σ in {0, 1, * } r , where the ith coordinate of χ σ is 0 or 1 if all points x in σ have x i = 0 or x i = 1 respectively, and χ σ i is * otherwise.If σ and τ are two faces, then σ ⊆ τ if and only if χ σ i χ τ i for all i, where 0 * and 1 * are the only non-equality relations on 0, 1 and * .
Suppose ∆ is an abstract simplicial complex with vertex set {1, . . ., n}.We define the cubical cone over ∆, denoted Cone(∆), to be the subcomplex of [0, 1] n consisting of all faces σ such that the set i : χ σ i ∈ {1, * } is a face of ∆ (including the empty face).For example, see Figure 14.The reason for the name "cubical cone" is the following lemma, analogous to properties of a simplicial cone.Lemma 6.8.The cubical cone over ∆ 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, . . ., 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 ) ∆.
Proof.First, let us check that the cubical cone does satisfy these properties.If σ is a face of Cone(∆), we can replace every 1 in χ σ with * to obtain the vector of a face of Cone(∆) that contains σ; therefore, the vectors of facets of Cone(∆) consist only of 0 and * , so every facet contains v 0 = (0, . . ., 0).Conversely, the faces that properly contain v 0 are precisely the faces σ = v 0 such that χ σ consists of 0 and * , and these faces are in bijection with non-empty faces of ∆, so the link of v 0 is indeed ∆.Now, suppose 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, . . ., 0).As before, the link of v 0 is then the set of faces of whose vectors consist only of 0 and * ; since we assume this set is isomorphic to ∆, we may again use the symmetry of the cube to permute the coordinates and assume that this set is equal to link Cone(∆) v 0 .Since contains these faces of the cube, it must also contain every sub-face of these faces, so contains every face of Cone(∆).Moreover, it cannot contain any other faces: any such face must be contained in some facet of , but that facet cannot be one that we have already accounted for in link v 0 , so the facet would not contain v 0 .Therefore, ∼ = Cone(∆).
While every flag simplicial complex is the crossing complex of some CAT(0) cubical complex (see Hagen [2014, Proposition 2.19] and Rowlands [2020, 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 Cross( ) is a connected manifold, the only possibility is = Cone(∆).Corollary 6.9.Let be a d-dimensional CAT(0) cubical complex and ∆ its crossing complex.If ∆ is a connected homology manifold, then is the cubical cone over ∆.Every vertex link in a CAT(0) cubical complex is an induced subcomplex of the crossing complex [Rowlands, 2020, 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, link v 0 = ∆.Lemma 6.4 says that there is a bijection between the facets of the cubical complex and the facets of its crossing complex.In this case, since link v 0 = ∆, we also have a bijection between the facets of the crossing complex ∆ and facets of that contain v 0 .Therefore, every facet of must contain v 0 .Hence is a cubical cone.

Figure 2 :
Figure 2: A comparison of the different types of deletion
for an example.Both A and B are open in |X|: A is a union of open sets, and B is the complement of a closed set.

Figure 5 :
Figure5: The sets A, B, A ∩ B and A ∪ B in the proof of Theorem 4.1, when k = 0 and X is the polytopal complex in Figure1 and the union A ′ ∪ B ′ is simply |star X σ|.Therefore, the Mayer-Vietoris long exact Three faces of X |star bary(X) v σ1 | |star bary(X) v σ2 | |star bary(X) v σ3 | (b) |star bary(X) vσ| for σ = σ1, σ2, σ3

Figure 7 :
Figure 7: The spaces |star bary(X) v σ | and |star X σ| in the proof of Theorem 4.1, for various choices of σ Theorem 5.3.A d-dimensional polytopal complex X is Cohen-Macaulay if and only if it is pure and H i (|Skel c k X|) = 0 for all i < d − k − 1 and all k = −1, . . ., d.

Figure 8 :
Figure 8: Two simplicial complexes whose skeletons have isomorphic homology groups

Figure 9 :
Figure 9: Two spheres with homeomorphic co-skeletons Figure 11: A cubical complex and its hyperplanes

Proof.|
Examine this section of the long exact sequence from Theorem 4.1, taking k = 0 and using coefficients in Zis simply | |, which is contractible since is CAT(0), so the first and last terms of this sequence are 0. By Proposition 6.7,H d−1 (|Skel c 0 |) is isomorphic to H d−1 (|Skel c −1 ∆|) = H d−1 (|∆|), and since ∆ is a connected homology manifold of dimension d − 1, this homology group is Z/2.Therefore,v∈ 0 H d−1 (|link v|) ∼ = Z/2.Thus there must be one vertex v 0 for which H d−1 (|link v 0 |) = Z/2, and all other vertices have H d−1 (|link v|) = 0.