# The Flip-Graph of the 4-Dimensional Cube is Connected

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## Abstract

Flip-graph connectedness is established here for the vertex set of the 4-dimensional cube. It is found as a consequence that this vertex set has 92,487,256 triangulations, partitioned into 247,451 symmetry classes.

## Keywords

4-Dimensional hypercube Tesseract Flip-graph connectivity Triangulations Enumeration TOPCOM \(k\)-Regularity## 1 Introduction

It is well known that the vertex set of the \(3\)-dimensional cube has exactly 74 triangulations, all regular, partitioned into 6 symmetry classes [1, 2]. The case of the \(4\)-dimensional cube turns out to be significantly more complicated. The first non-regular triangulation of its vertex set has been found almost two decades ago [1], and the total number of such triangulations was, up to now, unknown. Prior work [5] provided all the regular triangulations and TOPCOM [10] made it possible to conjecture the number of the non-regular ones, already several years ago. This conjecture is shown here to be true. Indeed, the only triangulation enumeration method efficient enough to be tractable in the case of the 4-dimensional cube actually consists in exploring the flip-graph of its vertex set [2]. In this perspective, completely enumerating the triangulations of the vertex set of the 4-dimensional cube is a task conditioned to the connectedness of this graph, which remained an open problem until now [2].

The \(4\)-dimensional cube is identified hereafter with the polytope \([0,1]^4\) and its vertices with the elements of \(\{0,1\}^4\). It is proven in this paper that the flip-graph of \(\{0,1\}^4\) is connected. It is found as a consequence, that the vertex set of the \(4\)-dimensional cube has 92,487,256 triangulations, partitioned into 247,451 symmetry classes. Some of the results in the paper require computer assistance. These computations were mostly done using TOPCOM [10].

The proof consists in finding paths in the flip-graph of \(\{0,1\}^4\) from any triangulation to a corner-cut triangulation [2]. To this end, one needs to perform a sequence of flips that introduces new corner simplices in a triangulation \(T\) of \(\{0,1\}^4\). Consider the \(3\)-dimensional triangulation \(U\) obtained intersecting \(T\) with a hyperplane that cuts the corner of \([0,1]^4\) with apex \(x\). As will be shown below, one can choose \(x\) so that, up to an isometry, \(U\) belongs to a well characterized set of 1,588 triangulations. In addition, conditions will be given such that flips in these triangulations carry over to \(T\) and reduce the star of \(\{x\}\) in \(T\) when they are performed. Hence, the problem is narrowed to checking a simple property regarding flips over the above mentioned 1,588 triangulations. This verification is performed within minutes using a computer, and does not require any arithmetic operations.

The paper is organized as follows. In Sect. 2, formal definitions of the mathematical objects involved in the problem are stated. Section 3 presents preliminary results about the triangulations of \(\{0,1\}^4\). It turns out, in particular, that the corner-cut triangulations of \(\{0,1\}^4\) are regular. In Sect. 4, it is shown how the flips in a triangulation \(T\) of \(\{0,1\}^4\) are related to flips in the triangulation obtained intersecting \(T\) with a hyperplane that cuts a corner of the \(4\)-dimensional cube. Using this, it is proven in Sect. 5 that every triangulation of \(\{0,1\}^4\) can be made regular by performing a sequence of flips, and connectedness follows for the flip-graph of \(\{0,1\}^4\).

## 2 Preliminary Definitions

*point configuration*is a finite subset of \(\mathbb{R }^4\). The elements of a point configuration are referred to as its

*vertices*. Consider a point configuration \(\mathcal{A }\). A simplicial complex on \(\mathcal{A }\) is a set of affinely independent subsets of \(\mathcal{A }\) whose convex hulls collectively form a polyhedral complex. The elements of a simplicial complex \(C\) are called its

*faces*and their vertices are also referred to as the

*vertices of*\(C\). The faces of \(C\) that are maximal for the inclusion are called its

*maximal faces*. The

*domain*of a simplicial complex \(C\) on \(\mathcal A \), denoted by \(\mathrm{dom }(C)\), is the union of the convex hulls of its faces:

*link*of \(t\) in triangulation \(T\) is the following set:

*star*of \(t\) in \(T\), that is, the subset of the faces of \(T\) whose union with \(t\) still belongs to \(T\):

*flippable in T*if there exist two subsets \(\tau \) and \(\lambda \) of \(T\) so that \(\tau \) is a triangulation of \(z\) and every maximal face of \(\tau \) admits \(\lambda \) as its link in \(T\). Assuming, without loss of generality, that \(\tau \) is equal to \(\tau ^-\), the following set is a triangulation of \(\mathcal A \) distinct from \(T\) (see [2, 11]):

*flip*. According to the definitions of \(\tau ^-\) and \(\tau ^+\), triangulations \(T\) and \(\mathfrak F (T, z)\) respectively contain \(z^-\) and \(z^+\). Moreover, all the faces that are removed from \(T\) when \(z\) is flipped in this triangulation admit \(z^-\) as a subset. This property will be useful hereafter.

The interested reader is referred to [2] for further definitions and results on the subject of triangulations and flips.

The flip-graph of \(\mathcal A \), denoted by \(\gamma (\mathcal A )\) in the following, is the graph whose vertices are the triangulations of \(\mathcal A \) and whose edges correspond to flips. The main result in this paper is the connectedness of \(\gamma (\{0,1\}^4)\). On the way to establishing this result, the first step is to prove that the corner-cut triangulations [2] of \(\{0,1\}^4\) are regular. The notion of regularity can be defined using *height functions* that is, real-valued maps on point configurations. A triangulation \(T\) of a point configuration \(\mathcal A \) is regular if there exists a height function \(w:\mathcal A \rightarrow \mathbb R \) so that for all \(t\in {T}\), some affine map \(\xi :\mathbb R ^4\rightarrow \mathbb R \) is smaller than \(w\) on \(\mathcal A \mathord {\setminus }{t}\) and coincides with \(w\) on \(t\). It is well known that the graph \(\rho (\mathcal A )\) induced by regular triangulations in \(\gamma (\mathcal A )\) is connected. Indeed, \(\rho (\mathcal A )\) is isomorphic to the \(1\)-skeleton of the so-called secondary polytope [3, 4]. This connectedness property will be the last argument used in the proof.

## 3 First Properties of \(\{0,1\}^4\) and of Its Triangulations

The \(f\)-vector of \([0,1]^4\) is (1,16,32,24,8,1). Since distinct diagonals of a cube are never found in a same triangulation of this cube, then:

### **Proposition 1**

It \(T\) is a triangulation of \(\{0,1\}^4, \sigma _2(T)\) admits at most \(24\) edges, \(\sigma _3(T)\) admits at most \(8\) edges, and \(\sigma _4(T)\) admits at most \(1\) edge.

Now, consider a point \(x\in \{0,1\}^4\). Further consider the points in \(\{0,1\}^4\) whose squared distance to \(x\) is \(1\). There are exactly four such points that are precisely the vertices adjacent to \(x\) in the \(1\)-skeleton of \([0,1]^4\). The set \(\kappa (x)\) made up of these four points and of point \(x\) is called the *corner simplex* of \(\{0,1\}^4\) with apex \(x\). Observe that if \(y\in \{0,1\}^4\) is a point whose distance to \(x\) is larger than \(1\), then the convex hull of \(\{x,y\}\) and a face of \(\mathrm{conv }(\kappa (x)\mathord {\setminus }\{x\})\) have non-disjoint relative interiors. Therefore, if \(T\) is a triangulation of \(\{0,1\}^4\) that contains \(\kappa (x)\), then \(x\) is an isolated vertex in the three graphs \(\sigma _2(T), \sigma _3(T)\), and \(\sigma _4(T)\). Inversely, if \(T\) is a triangulation of \(\{0,1\}^4\) so that \(x\) is isolated in \(\sigma _2(T), \sigma _3(T)\), and \(\sigma _4(T)\), then the only \(4\)-dimensional face of \(T\) that possibly contains \(x\) is necessarily \(\kappa (x)\). Hence:

### **Proposition 2**

Let \(x\) be a vertex of \(\{0,1\}^4\). A triangulation \(T\) of \(\{0,1\}^4\) contains \(\kappa (x)\) if and only if \(x\) is isolated in \(\sigma _2(T)\), in \(\sigma _3(T)\), and in \(\sigma _4(T)\).

*corner-cut*. It is well known that \(\{0,1\}^4\) admits precisely eight corner-cut triangulations, and that each of the eight diagonals of \([0,1]^4\) is found in exactly one of these triangulations [2]. It turns out that every such triangulation of \(\{0,1\}^4\) is regular:

### **Lemma 1**

All the corner-cut triangulations of \(\{0,1\}^4\) are regular.

### *Proof*

It is proven in Sect. 5 that any triangulation of \(\{0,1\}^4\) is connected in \(\gamma (\{0,1\}^4)\) to a corner-cut triangulation. According to Lemma 1, the connectedness of \(\gamma (\{0,1\}^4)\) will then naturally follow from the connectedness of \(\rho (\{0,1\}^4)\). The proof that a triangulation \(T\) of \(\{0,1\}^4\) can be transformed by flips into a corner-cut triangulation will require a careful study of the way \(T\) decomposes the corners of \([0,1]^4\). In particular, bounds on the degrees of the vertices of graphs \(\sigma _q(T)\) will be needed. These bounds are provided by the two following lemmas.

### **Lemma 2**

For any triangulation \(T\) of \(\{0,1\}^4\), point configurations \(E\) and \(O\) each admit at least four vertices whose degree in \(\sigma _3(T)\) is at most \(1\).

### *Proof*

Consider an element \(s\) of \(\{E,O\}\). Assume that more than four vertices of \(s\) have degree at least \(2\) in \(\sigma _3(T)\). In this case, the sum over \(s\) of the degrees in \(\sigma _3(T)\) is greater than \(8\). According to Proposition 1 though, \(\sigma _3(T)\) contains at most eight edges. Hence, at least one edge of \(\sigma _3(T)\) has its two vertices in \(s\). Since the squared distance between two points of \(s\) is even and since all the edges in \(\sigma _3(T)\) have squared length \(3\), one obtains a contradiction. This proves that at most four vertices of \(s\) have degree at least \(2\) in \(\sigma _3(T)\). Since \(s\) has cardinality \(8\), the desired result follows. \(\square \)

Consider an element \(s\) of \(\{E,O\}\). The bound given in Lemma 2 on the degrees in graph \(\sigma _3(T)\) does not hold for every point of \(s\). Under a condition on the way \(T\) triangulates the corners of \(\{0,1\}^4\), a bound on the degrees in graph \(\sigma _2(T)\) can be obtained that holds for every point of \(s\).

### **Lemma 3**

Let \(T\) be a triangulation of \(\{0,1\}^4\) and \(s\) an element of \(\{E,O\}\). If at least four corner simplices of \(\{0,1\}^4\) with apex in \(s\) are contained in \(T\), then all the vertices of \(s\) have degree at most \(3\) in \(\sigma _2(T)\).

### *Proof*

Assume that at least four corner simplices of \(\{0,1\}^4\) with apex in \(s\) are found in triangulation \(T\). According to Proposition 2, these four apices are isolated in \(\sigma _2(T)\). As \(s\) contains exactly eight points, then at most four vertices of \(s\) are not isolated in \(\sigma _2(T)\). Now recall that the squared distance between a vertex of \(s\) and a vertex of \(\{0,1\}^4\mathord {\setminus }{s}\) is odd. As a consequence, two adjacent points in \(\sigma _2(T)\) are necessarily both in \(s\) or both in \(\{0,1\}^4\mathord {\setminus }{s}\). Since at most four vertices of \(s\) are not isolated in graph \(\sigma _2(T)\), every vertex of \(s\) is adjacent to at most three points in this graph, which completes the proof. \(\square \)

## 4 Flips in the Corners of the \(4\)-Dimensional Cube

*homogeneous contraction*of \(s\) at point \(x\):

### **Lemma 4**

Let \(x\) be a vertex of \(\{0,1\}^4\). For every \(z\in {Z(x)}\), \(z/x\) is a circuit whose Radon partition is \(\{\varepsilon _x^-(z),\varepsilon _x^+(z)\}/x\).

### *Proof*

Let \(z\in {Z(x)}\) be a circuit. Consider a subset \(s\) of \(z\) that contains \(x\) and recall that the homogeneous contraction at point \(x\) induces an bijection from \(\{0,1\}^4\mathord {\setminus }\{x\}\) onto \(\{0,1\}^4/x\). As a consequence, \(s/x\) has exactly one element less than \(s\). Further recall that the affine hull of \(\kappa (x)\mathord {\setminus }\{x\}\) is an hyperplane of \(\mathbb R ^4\) and denote by \(H\) this hyperplane. Since \(x\) belongs to \(s\), the affine hull of \(s/x\) is the intersection of \(H\) with the affine hull of \(s\). Hence the homogeneous contraction at point \(x\) not only decreases by one the cardinality of \(s\) but also the dimension of its affine hull. As a consequence, \(s/x\) is affinely dependent if and only if \(s\) is affinely dependent. Replacing \(s\) by \(z\) in this assertion, one obtains that \(z/x\) is affinely dependent. Now observe that every proper subset of \(z/x\) can be obtained as the homogeneous contraction at point \(x\) of a proper subset of \(z\) that contains \(x\). As \(z\) is a circuit, all its proper subsets are affinely independent. According to the above assertion, this property carries over to the proper subsets of \(z/x\), which proves that \(z/x\) is a circuit.

Respectively denote \(\varepsilon _x^-(z)\) and \(\varepsilon _x^+(z)\) by \(z^-\) and by \(z^+\). Recall that the convex hulls of \(z^-\) and \(z^+\) have non-disjoint relative interiors. As in addition, \(x\) belongs to \(z^-\) and does not belong to \(z^+\), there exists a \(1\)-dimensional affine subspace \(N\) of \(\mathbb R ^4\) that contains \(x\) and that intersects the relative interiors of \(\mathrm{conv }(z^-\mathord {\setminus }\{x\})\) and of \(\mathrm{conv }(z^+)\). Moreover, \(z^-/x\) and \(z^+/x\) are the central projections on \(H\) with respect to point \(x\) of respectively \(z^-\mathord {\setminus }\{x\}\) and \(z^+\). Further recall that point \(x\) is the only vertex of \(\{0,1\}^4\) contained in one of the open half-spaces of \(\mathbb R ^4\) bounded by hyperplane \(H\). In this case, the convex hulls of \(z^-/x\) and \(z^+/x\) both contain the intersection of \(N\) with \(H\), proving that these convex hulls are non-disjoint. Finally, since the homogeneous contraction at point \(x\) induces a bijection from \(\{0,1\}^4\mathord {\setminus }\{x\}\) onto \(\{0,1\}^4/x\), sets \(z^-/x\) and \(z^+/x\) partition \(z/x\). Hence, by definition, the Radon partition of \(z/x\) is \(\{z^-,z^+\}/x\), which completes the proof. \(\square \)

Note that, while this lemma is stated here for \(\{0,1\}^4\), a similar result holds in general for arbitrary point configurations of any dimension.

Let \(x\) be a vertex of \(\{0,1\}^4\) and \(T\) a triangulation of \(\{0,1\}^4\). Consider the set obtained as the homogeneous contraction of \(\mathrm{link }_T(\{x\})\) at point \(x\). This set can be alternatively obtained by intersecting the convex hulls of the faces of \(\mathrm star _T(\{x\})\) with the affine hull \(H\) of \(\kappa (x)\mathord {\setminus }\{x\}\) and by subsequently taking the vertex sets of these intersections. As a consequence, the convex hulls of the elements of \(\mathrm{link }_T(\{x\})/x\) collectively form a polyhedral complex. In addition, every convex combination of \(\{0,1\}^4/x\) belongs to the intersection of \(H\) with the convex hull of some face of \(\mathrm star _T(\{x\})\). Hence \(\mathrm{link }_T(\{x\})/x\) is a triangulation of \(\{0,1\}^4/x\). Note in particular that the homogeneous contraction at point \(x\) not only induces a bijection from \(\{0,1\}^4\mathord {\setminus }\{x\}\) onto \(\{0,1\}^4/x\) but also an isomorphism from \(\mathrm{link }_T(\{x\})\) onto triangulation \(\mathrm{link }_T(\{x\})/x\). This triangulation and its properties are studied in remainder of the section.

Consider a circuit \(z\) in \(Z(x)\). A consequence of Lemma 4 is that, if \(z\) is flippable in \(T\), then \(z/x\) is flippable in \(\mathrm{link }_T(\{x\})/x\). While the converse implication is not true in general, the next theorem states that it becomes so if one further assumes that \(\mathrm{link }_T(\{x\})/x\) contains \(\varepsilon _x^-(z)/x\). Indeed, in this case, \(\varepsilon _x^-(z)\) must be a face of \(T\). Hence, if \(z\) can be flipped in \(T\), all the faces removed by this flip necessarily admit \(\varepsilon _x^-(z)\) as a subset. As \(x\) is a vertex of \(\varepsilon _x^-(z)\), every such face belongs to the star of \(\{x\}\) in \(T\) and can be reconstructed from \(\mathrm{link }_T(\{x\})/x\). It can be proven, as a consequence, that the flippability of \(z/x\) in triangulation \(\mathrm{link }_T(\{x\})/x\) carries over to the flippability of \(z\) in triangulation \(T\). Note that the result fails if \(\varepsilon _x^-(z)\) is replaced by \(\varepsilon _x^+(z)\) in the statement of the theorem, precisely because \(x\) does not belong to \(\varepsilon _x^+(z)\). In this case, flipping \(z\) in \(T\) would remove \(z\mathord {\setminus }\{x\}\). However, this simplex cannot be reconstructed from \(\mathrm{link }_T(\{x\})/x\) because it does not belong to the star of \(\{x\}\) in \(T\)(see, for instance, Sect. 3.6.2. in [2]).

### **Theorem 1**

Let \(x\) be a vertex of \(\{0,1\}^4\), \(T\) a triangulation of \(\{0,1\}^4\), and \(z\in {Z(x)}\) a circuit. If \(z/x\) is flippable in \(\mathrm{link }_T(\{x\})/x\) and if \(\varepsilon _x^-(z)/x\) belongs to \(\mathrm{link }_T(\{x\})/x\), then \(z\) is flippable in \(T\) and \(\varepsilon _x^-(z)\) belongs to \(T\).

### *Proof*

As was the case for Lemma 4, this theorem can be generalized without much more effort to any point configuration of arbitrary dimension. The dimension of \(\mathrm{link }_T(\{x\})/x\) and the number of its vertices have to remain small, though, for this result to provide benefit from a computational point of view.

### **Theorem 2**

Let \(x\) be a vertex of \(\{0,1\}^4\) and \(T\) a triangulation of \(\{0,1\}^4\). If a circuit \(z\in {Z(x)}\) is flippable in \(T\) and if \(\varepsilon _x^-(z)\in {T}\), then \(D_x(\mathfrak F (T,z))\) is a proper subset of \(D_x(T)\) and \(V_x(\mathfrak F (T,z))\) is a subset of \(V_x(T)\).

### *Proof*

Assume that some circuit \(z\in {Z(x)}\) is flippable in \(T\) and that \(\varepsilon _x^-(z)\) belongs to \(T\). Therefore, every face removed from \(T\) when flipping \(z\) in \(T\) admits \(\varepsilon _x^-(z)\) as a subset. Hence, these faces necessarily contain \(x\) and as a consequence, the union \(U\) of their convex hulls is a subset of \(D_x(T)\). As the convex hull of any simplex introduced in \(T\) by the same flip is a subset of \(U\), one therefore obtains that \(D_x(\mathfrak F (T,z))\subset {D_x(T)}\).

Further denote \(t=z\mathord {\setminus }\{x\}\). As \(x\) does not belong to \(t\), then \(\varepsilon _x^-(z)\) is not a subset of \(t\). By definition of the triangulations of a circuit, \(t\) is a face of \(\mathfrak F (T,z)\). In addition, if the convex hull of \(t\) is a subset of \(D_x(\mathfrak F (T,z))\), then \(t\) necessarily belongs to the link of \(\{x\}\) in \(\mathfrak F (T,z)\). In this case, according to the definition of the link, \(t\cup \{x\}\) must be a face of \(\mathfrak F (T,z)\). As \(t\cup \{x\}\) is affinely dependent, this is impossible. One thus obtains an indirect proof that \(\mathrm{conv }(t)\) is not a subset of \(D_x(\mathfrak F (T,z))\). Further observe that \(\mathrm{conv }(t)\subset {U}\). As \(U\) is, in turn, a subset of \(D_x(T)\), this proves that \(D_x(\mathfrak F (T,z))\) is a proper subset of \(D_x(T)\).

Now observe that \(V_x(\mathfrak F (T,z))\) is always a subset of \(V_x(T)\) except if some point is introduced into the link of \(x\) when \(z\) is flipped in \(T\). This occurs only if for some \(y\in \{0,1\}^4\mathord {\setminus }\{x\}\), either \(\varepsilon _x^+(z)=\{y\}\) or \(\varepsilon _x^+(z)=\{x,y\}\). These situations are both impossible, though. Indeed, following its definition, \(\varepsilon _x^+(z)\) does not contain \(x\). In addition, if \(\varepsilon _x^+(z)\) is a singleton then, the unique point it contains must lie in the interior of \([0,1]^4\), which cannot occur because every point in \(\varepsilon _x^+(z)\) is a vertex of \([0,1]^4\). As a consequence, \(V_x(\mathfrak F (T,z))\) is indeed a subset of \(V_x(T)\). \(\square \)

Let \(x\) be a point in \(\{0,1\}^4\). Further consider a set \(S\subset \{0,1\}^4/x\) that admits \(\kappa (x)\mathord {\setminus }\{x\}\) as a subset and denote by \(L_x(S)\) the set of all the triangulations \(U\) of \(S\) so that for every circuit \(z\in {Z(x)}\), either \(z/x\) is not flippable in \(U\), or \(\varepsilon _x^-(z)/x\) does not belong to \(U\). The following result is obtained as a consequence of Theorems 1 and 2:

### **Corollary 1**

Let \(T\) be a triangulation of \(\{0,1\}^4\). For any \(x\in \{0,1\}^4\), there exists a path in \(\gamma (\{0,1\}^4)\) from \(T\) to a triangulation \(T^{\prime }\) so that \(\mathrm{link }_{T^{\prime }}(\{x\})/x\) belongs to \(L_x(V_x(T))\), and for all \(t\in {T}\), if \(x\not \in {t}\) then \(t\in {T^{\prime }}\).

### *Proof*

Let \(x\) be a vertex of \(\{0,1\}^4\). Consider the directed graph \(\Gamma \) whose vertices are the triangulations of \(\{0,1\}^4\) and whose arcs connect a triangulation \(U\) to a triangulation \(V\) whenever there exists a circuit \(z\in {Z(x)}\) flippable in \(U\) so that \(\varepsilon _x^-(z)\in {U}\) and \(V=\mathfrak F (U,z)\). Consider the paths in \(\Gamma \) that start at triangulation \(T\). Among the triangulations at the other end of such paths, let \(T^{\prime }\) be one for which \(D_{x}(T^{\prime })\) is minimal.

If \(\mathrm{link }_{T^{\prime }}(\{x\})/x\) does not belong to \(L_x(V_x(T))\) then, by definition of this set, there exists a circuit \(z\in {Z(x)}\) so that \(z/x\) is flippable in \(\mathrm{link }_{T^{\prime }}(\{x\})/x\) and \(\varepsilon _x^-(z)/x\) is a face of \(\mathrm{link }_{T^{\prime }}(\{x\})/x\). It then follows from Theorem 1 that circuit \(z\) is flippable in \(T^{\prime }\) and that \(\varepsilon _x^-(z)\) belongs to \(T^{\prime }\). Moreover, according to Theorem 2, \(D_x(\mathfrak F (T^{\prime },z))\) is a proper subset of \(D_x(T^{\prime })\), which produces a contradiction. Hence, triangulation \(\mathrm{link }_{T^{\prime }}(\{x\})/x\) belongs to \(L_x(V_x(T))\).

Now, according to the way \(\Gamma \) has been oriented, whenever a flip is performed in a triangulation \(U\) on the path that connects \(T\) to \(T^{\prime }\), this flip only removes faces of \(U\) that contain \(x\). As a consequence, all the faces of \(T\) that do not contain \(x\) are still found in \(T^{\prime }\). \(\square \)

Let \(s\) be an element of \(\{E,O\}\), \(x\) a vertex of \(s\), and \(T\) a triangulation of \(\{0,1\}^4\). Observe that \(\kappa (x)\) is a face of \(T\) if and only if \(\mathrm{link }_{T}(\{x\})/x\) is the unique triangulation of \(\kappa (x)\mathord {\setminus }\{x \}\) (whose unique \(3\)-dimensional simplex is \(\kappa (x)\mathord {\setminus }\{x \}\) itself). It then follows from Corollary 1 that if \(\kappa (x)\not \in {T}\) and \(L_x(V_x(T))\) is a singleton whose element is the unique triangulation of \(\kappa (x)\mathord {\setminus }\{x\}\), then one can use a sequence of flips to increase by one the number of corner simplices of \(\{0,1\}^4\) with apex in \(s\) that are contained in \(T\). The success of this method is conditioned by the content of \(L_x(V_x(T))\), though. The first part of the next section is devoted to the enumeration of \(L_x(S)\) for well chosen subsets \(S\) of \(\{0,1\}^4/x\). It will then be shown that Corollary 1 indeed provides a way to flip any triangulation of \(\{0,1\}^4\) to a corner-cut triangulation.

## 5 The Flip-Graph of \(\{0,1\}^4\) is Connected

### **Lemma 5**

Let \(T\) be a triangulation of \(\{0,1\}^4\) and \(s\in \{E,O\}\). If \(T\) is not corner-cut and every vertex of \(s\) is isolated in \(\sigma _4(T)\), then there exist \(x\in {s}\) and \(q\in \{1,2,3\}\) so that \(\kappa (x)\not \in {T}\) and \(V_x(T)\) is isometric to a subset of \(S_q\).

### *Proof*

Assume that \(T\) is not corner-cut and that every vertex of \(s\) is isolated in \(\sigma _4(T)\). One can check using Fig. 1 that for any point \(x\) in \(\{0,1\}^4\), \(V_x(T)\) is isometric to a subset of \(S_1\) if and only if \(x\) is isolated in \(\sigma _4(T)\) and has degree at most \(1\) in \(\sigma _3(T)\). According to Lemma 2, at least four vertices of \(s\) have degree at most \(1\) in \(\sigma _3(T)\). If in addition one of these vertices is not the apex of a corner simplex of \(\{0,1\}^4\) found in \(T\), then one can find \(x\in {s}\) so that \(\kappa (x)\) does not belong to \(T\) and \(V_x(T)\) is isometric to a subset of \(S_1\), proving that the desired property holds.

Assume that all the points \(x\in {s}\) with degree at most \(1\) in \(\sigma _3(T)\) are so that \(\kappa (x)\in {T}\). According to Lemma 2, triangulation \(T\) then contains at least four corner simplices of \(\{0,1\}^4\) with apex in \(s\). In this case, it follows from Lemma 3 that all the vertices of \(s\) have degree at most \(3\) in \(\sigma _2(T)\). Now observe that if a point \(x\in {s}\) has degree \(1\) or \(2\) in \(\sigma _2(T)\) then \(V_x(T)\) is isometric to a subset of \(S_2\) or to a subset of \(S_3\). Since \(T\) is not corner-cut, the wanted result therefore holds when all the vertices of \(s\) have degree at most \(2\) in \(\sigma _2(T)\).

Assume that some vertex of \(s\) has degree \(3\) in \(\sigma _2(T)\). Using the symmetries of \(\{0,1\}^4\), one can require that this point be \(a\). It can be seen using Fig. 1 that, if \(y\) is one of the three vertices adjacent to \(a\) in \(\sigma _2(T)\), then \(y/a\) is the centroid of an edge of tetrahedron \(\kappa (a)\mathord {\setminus }\{a\}\). Denote the set of these three edges by \(W\). These edges can be placed in three ways: they can have a common vertex, be the edges of a triangle, or form an acyclic path. These three placements will now be discussed one by one. First, if the three elements of \(W\) have a common vertex, one can assume by symmetry that this vertex is \(e\). In this case \(V_a(T)\) is a subset of \(S_2\). Since \(a\) is not isolated in \(\sigma _2(T)\), then \(\kappa (a)\) is not a face of \(T\) and the result follows.

The second placement turns out to be impossible. Indeed if the three elements of \(W\) are the edges of a triangle, then one can require using the symmetries of \(\{0,1\}^4\), that this triangle be \(\{b,c,i\}\). In this case, the vertices adjacent to \(a\) in \(\sigma _2(T)\) are \(d\), \(j\), and \(k\) (see Fig. 1). It follows that the points \(x\in {s}\) so that \(\kappa (x)\) is not a face of \(T\) are precisely \(a\), \(d\), \(j\), and \(k\). These four points belong to the hyperplane of equation \(x\cdot {u_3}=0\). Hence, none of them is a vertex of the \(3\)-dimensional face of \([0,1]^4\) found in the hyperplane of equation \(x\cdot {u_3}=1\). Therefore, the sum of their degrees in \(\sigma _3(T)\) is at most \(7\) and one of these points has degree at most \(1\) in \(\sigma _3(T)\). This produces a contradiction with the assumption that every point of \(s\) with degree at most \(1\) in \(\sigma _3(T)\) is the apex of a corner simplex of \(\{0,1\}^4\) found in \(T\).

Finally, assume that the elements of \(W\) are the edges of an acyclic path. By symmetry, one can require that these three edges be \(\{b,e\}, \{c,i\}\), and \(\{e,i\}\). Hence, the vertices adjacent to \(a\) in \(\sigma _2(T)\) are \(f\), \(k\), and \(m\) (see Fig. 1). As a consequence, the points \(x\in {s}\) so that \(\kappa (x)\) does not belong to \(T\) are precisely \(a\), \(f\), \(k\), and \(m\). Now observe that \(f\) and \(k\) have squared distance \(4\). As a consequence, \(f\) is not adjacent to \(k\) in \(\sigma _2(T)\), and its degree in \(\sigma _2(T)\) is at most \(2\). As already mentioned above, this implies that \(V_f(T)\) is isometric to a subset of \(S_2\) or to a subset of \(S_3\). Since \(\kappa (f)\) is not contained in \(T\), this completes the proof. \(\square \)

### **Proposition 3**

- i.
\(L_a(S_1)\) and \(L_a(S_2)\) are equal to \(\{U_0,U_1^-,U_1^+\}\),

- ii.
\(U_0\) is the unique triangulation found in \(L_a(S_3)\).

Let \(s\) be an element of \(\{E,O\}\). Consider a triangulation \(T\) of \(\{0,1\}^4\) and a point \(x\in {s}\) so that \(\kappa (x)\) does not belong to \(T\). Further assume that for some integer \(q\in \{1,2,3\}\), \(V_x(T)\) is isometric to a subset of \(S_q\). Following Corollary 1 and Proposition 3, \(T\) can be flipped to a triangulation \(T^{\prime }\) that contains all the faces of \(T\) that do not admit \(x\) as a vertex and so that \(\mathrm{link }_{T^{\prime }}(\{x\})/x\) is isometric to either \(U_0\) or \(U_1^-\). Observe that for all \(y\in {s\mathord {\setminus }\{x\}}\), point \(x\) is not a vertex of \(\kappa (y)\). As a consequence, \(T^{\prime }\) contains all the corner simplices of \(\{0,1\}^4\) with apex in \(s\mathord {\setminus }\{x\}\) that are already found in \(T\). If in addition, \(\mathrm{link }_{T^{\prime }}(\{x\})/x\) is isometric to \(U_0\), then \(\kappa (x)\in {T^{\prime }}\) and the sequence of flips that transforms \(T\) into \(T^{\prime }\) increases by one the number of corner simplices of \(\{0,1\}^4\) with apex in \(s\) found in the triangulation. This property fails, though, when \(\mathrm{link }_{T^{\prime }}(\{x\})/x\) is not isometric to \(U_0\) but to \(U_1^-\). A careful study of the triangulations \(T\) of \(\{0,1\}^4\) so that \(\mathrm{link }_T(\{x\})/x\) is isometric to \(U_1^-\) is a first step towards solving this problem. Thanks to the symmetries of \([0,1]^4\), it can be assumed without loss of generality that \(x\) is equal to \(a\) and that \(\mathrm{link }_T(\{x\})/x\) is precisely triangulation \(U_1^-\). The following lemma, that will be invoked in the Proof of Theorem 3, further requires that \(T\) contains \(\kappa (d)\). This additional condition will also turn out not to be restrictive.

### **Lemma 6**

- i.
\(V_a(\mathfrak F (\mathfrak F (T,z_1),z_2))\) is obtained removing \(f/a\) from \(V_a(T)\),

- ii.
For all \(t\in {T}\), if \(\{a,f\}\cap {t}=\emptyset \), then \(t\in \mathfrak F (\mathfrak F (T,z_1),z_2)\).

### *Proof*

This proves that the two tetrahedra in \(\mathrm{link }_T(\{f\})/f\) that admit \(\{b,c,l\}/f\) and \(\{c,g,l\}/f\) as respective subsets and that do not admit \(a/f\) as their last vertex necessarily both contain point \(h/f\). As discussed above, it follows that circuit \(z_1/f\) is flippable in \(\mathrm{link }_T(\{f\})/f\). Now observe that the Radon partition of circuit \(z_1\) is \(\{\{b,g\},\{c,f\}\}\). Further note that \(\varepsilon _f^-(z_1)=\{c,f\}\). Hence, \(\varepsilon _f^-(z_1)/f\) is equal to \(\{c/f\}\). As this singleton is a face of triangulation \(\mathrm{link }_T(\{f\})/f\), it follows from Theorem 1 that circuit \(z_1\) is flippable in \(T\).

Denote \(T^{\prime }=\mathfrak F (T,z_1)\). Observe that \(\mathrm{link }_{T^{\prime }}(\{a\})/a\) is obtained by flipping circuit \(\{b,c,f,g\}/a\) in \(\mathrm{link }_T(\{a\})/a\). One can therefore see in Fig. 2 that circuit \(z_2/a\), that was not flippable in \(\mathrm{link }_T(\{a\})/a\) becomes flippable in \(\mathrm{link }_{T^{\prime }}(\{a\})/a\). Moreover the Radon partition of \(z_2\) is \(\{\{a,f\},\{b,e\}\}\) and \(\varepsilon _a^-(z_2)=\{a,f\}\). Hence, \(\varepsilon _a^-(z_2)/a\) is equal to \(\{f/a\}\). Since \(\mathrm{link }_{T^{\prime }}(\{a\})/a\) contains this singleton, it follows from Theorem 1 that circuit \(z_2\) is flippable in \(T^{\prime }\). Observe that flipping \(z_2\) in \(T^{\prime }\) removes edge \(\{a,f\}\). As \(f/a\) is the only vertex that has been removed from \(V_a(T)\) by the two consecutive flips, the first assertion in the statement of the lemma holds. Further observe that all the faces removed when flipping \(z_1\) in \(T\) and then \(z_2\) in \(T^{\prime }\) contain point \(a\) or point \(f\). As a consequence, the second assertion also holds. \(\square \)

Thanks to Lemma 6, it is possible to avoid being stuck by a triangulation isometric to \(U_1^-\) on the way to introduce a new corner simplex into a triangulation of the \(4\)-dimensional cube:

### **Theorem 3**

Let \(T\) be a triangulation of \(\{0,1\}^4\), \(s\) an element of \(\{E,O\}\), and \(x\) a vertex of \(s\). If \(\mathrm{link }_T(\{x\})/x\) is isometric to \(U_1^-\) or to \(U_1^+\), then there exists a path in \(\gamma (\{0,1\}^4)\) that connects \(T\) to a triangulation \(T^{\prime }\) so that \(\kappa (x)\in {T^{\prime }}\) and for all \(t\in {T}\), if \(t\cap {s}=\emptyset \) then \(t\in {T^{\prime }}\).

### *Proof*

Further observe that \(\{e,l\}\) is an edge of graph \(\sigma _4(T)\) (indeed, as can be seen in Fig. 2, \(\mathrm{link }_T(\{a\})/a\) contains \(\{e,l\}/a\)). It then follows from Proposition 1 that \(e\) and \(l\) are the only two non-isolated vertices of \(\sigma _4(T)\). In particular, \(d\) is not only isolated in \(\sigma _3(T)\), but also in \(\sigma _4(T)\). This proves that \(V_d(T)\) is isometric to a subset of \(S_1\mathord {\setminus }\{l/a\}\). Now observe that \(L_a(S_1\mathord {\setminus }\{l/a\})\) is a subset of \(L_a(S_1)\). Following Proposition 3, the only triangulation of \(S_1\mathord {\setminus }\{l/a\}\) contained in \(L_a(S_1)\) is \(U_0\). As a consequence, this triangulation is the unique element of \(L_a(S_1\mathord {\setminus }\{l/a\})\). Since \(V_d(T)\) is isometric to a subset of \(S_1\mathord {\setminus }\{l/a\}\), one obtains that \(L_d(V_d(T))\) is a singleton whose element is the unique triangulation of \(\kappa (d)\mathord {\setminus }\{d\}\). Hence, according to Corollary 1, \(T\) can be flipped to a triangulation \(T^{\prime \prime }\) so that \(\kappa (d)\in {T^{\prime \prime }}\) and for all \(t\in {T}\), if \(d\not \in {t}\) then \(t\in {T^{\prime \prime }}\).

As \(\{a,d\}\) is not an edge of \(T\) (see the left of Fig. 2), the star of \(\{a\}\) in \(T\) is not affected by the flips that transform \(T\) into \(T^{\prime \prime }\). In particular, \(\mathrm{link }_{T^{\prime \prime }}(\{a\})/a\) is equal to \(U_1^-\). Hence \(T^{\prime \prime }\) satisfies the conditions of Lemma 6. Moreover, it follows from Proposition 3 that \(U_{0}\) is the unique element of \(L_a(V_a(T^{\prime \prime })\mathord {\setminus }\{f/a\})\). Therefore, invoking Lemma 6 and Corollary 1 provides a path in \(\gamma (\{0,1\}^4)\) from triangulation \(T^{\prime \prime }\) to a triangulation \(T^{\prime }\) so that \(\kappa (a)\in {T^{\prime }}\) and for all \(t\in {T^{\prime \prime }}\), if \(\{a,f\}\cap {t}\) is empty then \(t\in {T^{\prime }}\). It follows that all the elements of \(T\) that do not contain \(a\), \(d\), or \(f\) are still found in \(T^{\prime \prime }\). As these three points belong to \(s\), \(T^{\prime }\) has the desired properties. \(\square \)

According to Proposition 3, the next result is an immediate consequence of Corollary 1 and of Theorem 3:

### **corollary 2**

Let \(T\) be a triangulation of \(\{0,1\}^4\), \(s\) an element of \(\{E,O\}\), and \(x\in {s}\) a point. If for some \(q\in \{1,2,3\}\), \(V_x(T)\) is isometric to a subset of \(S_q\), then there exists a path in \(\gamma (\{0,1\}^4)\) that connects \(T\) to a triangulation \(T^{\prime }\) so that \(\kappa (x)\) belongs to \(T^{\prime }\) and for all \(t\in {T}\), if \(t\cap {s}=\emptyset \) then \(t\in {T^{\prime }}\).

### *Proof*

Assume that there exists \(q\in \{1,2,3\}\) so that \(V_x(T)\) is isometric to a subset of \(S_q\). According to Corollary 1, there exists a path in \(\gamma (\{0,1\}^4)\) from \(T\) to a triangulation \(T^{\prime \prime }\) so that \(\mathrm{link }_{T^{\prime \prime }}(\{x\})/x\in {L_x(V_x(T))}\) and for all \(t\in {T}\), if \(x\not \in {t}\) then \(t\in {T^{\prime \prime }}\). It then follows from Proposition 3 that \(\mathrm{link }_{T^{\prime \prime }}(\{x\})/x\) is isometric to either \(U_1^-\) or to \(U_0\). In the latter case, \(x\) is necessarily isolated in \(\sigma _2(T^{\prime \prime })\), in \(\sigma _3(T^{\prime \prime })\), and in \(\sigma _4(T^{\prime \prime })\) and according to Proposition 2, \(\kappa (x)\) belongs to \({T^{\prime \prime }}\). As in addition all the faces of \(T\) that do not contain \(x\) are necessarily found in \(T^{\prime \prime }\), then the wanted properties hold by taking \(T^{\prime }=T^{\prime \prime }\).

Now assume that \(\mathrm{link }_{T^{\prime \prime }}(\{x\})/x\) is isometric to \(U_1^-\). In this case, Theorem 3 provides a path in \(\gamma (\{0,1\}^4)\) connects \(T^{\prime \prime }\) to a triangulation \(T^{\prime }\) so that \(\kappa (x)\) is a face of \(T^{\prime }\) and for all \(t\in {T^{\prime \prime }}\), if \(t\cap {s}=\emptyset \), then \(t\in {T^{\prime }}\). Since \(x\in {s}\), then every face of \(T\) disjoint from \(s\) still belongs to \(T^{\prime }\), which completes the proof. \(\square \)

Combining Lemma 5 and Corollary 2, one finds that every triangulation of \(\{0,1\}^4\) can be transformed into a corner-cut triangulation by a sequence of flips. This leads to the main result of this paper:

### **Theorem 4**

The flip-graph of \(\{0,1\}^4\) is connected.

### *Proof*

Let \(T\) be a triangulation of \(\{0,1\}^4\) and \(s\) an element of \(\{E,O\}\). Consider the triangulations that are connected to \(T\) by a path in \(\gamma (\{0,1\}^4)\) and that contain all the faces of \(T\) disjoint from \(s\). Among these triangulations, let \(T^{\prime \prime }\) be one that contains the largest possible number of corner simplices of \(\{0,1\}^4\) with apex in \(s\).

Since \(\sigma _4(T^{\prime \prime })\) has at most one edge, and since the squared length of such an edge is even, then either all the points of \(s\) are isolated in \(\sigma _4(T^{\prime \prime })\) or exactly two of them are adjacent. If all the points of \(s\) are isolated in \(\sigma _4(T^{\prime \prime })\) then denote \(T^{\prime }=T^{\prime \prime }\) and \(s^{\prime }=s\). Otherwise, let \(s^{\prime }\) be the element of \(\{E,O\}\) distinct from \(s\), and consider the triangulations that are connected to \(T^{\prime \prime }\) by a path in \(\gamma (\{0,1\}^4)\) and that contain all the faces of \(T^{\prime \prime }\) disjoint from \(s^{\prime }\). Among these triangulations, let \(T^{\prime }\) be one that contains the largest possible number of corner simplices of \(\{0,1\}^4\) with apex in \(s^{\prime }\). In this case, the vertices of \(s^{\prime }\) are necessarily isolated in \(\sigma _4(T^{\prime })\). Indeed, \(\sigma _4(T^{\prime \prime })\) contains an edge whose vertices are not in \(s^{\prime }\). Therefore this edge still belongs to \(\sigma _4(T^{\prime })\) and according to Proposition 2, no other edge is found in \(\sigma _4(T^{\prime })\).

According to this construction, there exists a path in \(\gamma (\{0,1\}^4)\) from \(T\) to \(T^{\prime }\), so that all the elements of \(s^{\prime }\) are isolated in \(\sigma _4(T^{\prime })\). Moreover, the number of corner simplices of \(\{0,1\}^4\) with apex in \(s^{\prime }\) found in \(T^{\prime }\) cannot be increased by a sequence of flips without removing from \(T^{\prime }\) a face disjoint from \(s^{\prime }\). Now, assume that \(T^{\prime }\) is not corner-cut. Since all the vertices of \(s^{\prime }\) are isolated in \(\sigma _4(T^{\prime })\), it follows from Lemma 5 that there exists a point \(x\in {s^{\prime }}\) and an integer \(q\in \{1,2,3\}\) so that \(V_x(T^{\prime })\) is isometric to a subset of \(S_q\) and \(\kappa (x)\not \in {T^{\prime }}\). In particular, \(T^{\prime }\) satisfies the conditions of Corollary 2 and one can find a triangulation \(T^{(3)}\) connected to \(T^{\prime }\) by a path in \(\gamma (\{0,1\}^4)\) so that \(T^{(3)}\) contains \(\kappa (x)\) and every face of \(T^{\prime }\) disjoint from \(s^{\prime }\). In particular, for any \(y\in {s^{\prime }}\) so that \(\kappa (y)\in {T^{\prime }}\), \(T^{(3)}\) contains \(\kappa (y)\mathord {\setminus }\{y\}\). As a consequence, \(y\) is isolated in graphs \(\sigma _2(T^{(3)}), \sigma _3(T^{(3)})\), and \(\sigma _4(T^{(3)})\), and it follows from Proposition 2 that \(\kappa (y)\in {T^{(3)}}\). In other words, a sequence of flips was found that increases the number of corner simplices of \(\{0,1\}^4\) with apex in \(s^{\prime }\) found in \(T^{\prime }\). Moreover, this sequence of flips does not remove from \(T^{\prime }\) any face disjoint from \(s^{\prime }\).

This produces a contradiction, proving that \(T^{\prime }\) is corner-cut. According to Lemma 1, a path in \(\gamma (\{0,1\}^4)\) therefore connects triangulation \(T\) to a regular triangulation. The result then follows from the connectedness of the subgraph \(\rho (\{0,1\}^4)\) induced by regular triangulations in \(\gamma (\{0,1\}^4)\). \(\square \)

As mentioned in the introduction, the only algorithm efficient enough to completely enumerate the triangulations of \(\{0,1\}^4\) consists in exploring the flip-graph of \(\{0,1\}^4\), and its validity is based on the connectedness of this graph. Theorem 4 solves this issue. Using TOPCOM, one can therefore perform this enumeration in (much) less than an hour, with the following result:

### **Corollary 3**

The vertex set of the \(4\)-dimensional cube admits 92,487,256 triangulations, partitioned into 247,451 symmetry classes.

## 6 Discussion

The main result of this paper is the connectedness of \(\gamma (\{0,1\}^4)\). Computer assistance was used on the way, in order to obtain Proposition 3. This computer-assisted part consists in checking a property regarding flips over 1,588 triangulations of point configuration \(\{0,1\}^4/a\). This number is small enough to allow for a fastidious, but possible, verification by hand. However, the computer is certainly more reliable in such a task. In addition, the computer-assisted verification can be carried out within a few minutes. A second important result obtained here is the complete enumeration of the triangulations of \(\{0,1\}^4\): 92,487,256 triangulations were found. It was established a few years ago that the number of regular triangulations of \(\{0,1\}^4\) is 87,959,448 [5]. Hence, more than \(95\%\) of the triangulations of \(\{0,1\}^4\) are regular. This does not come as a surprise, since non-regular triangulations of \(\{0,1\}^4\) are difficult to find [1].

These observations lead to a natural question. Call a triangulation of a \(d\)-dimensional point configuration *2-regular* when its faces are projected from the boundary complex of a \((d+2)\)-dimensional polytope. It was found recently that the subgraph induced by \(2\)-regular triangulations in the flip-graph of a point configuration is connected [6], providing new ways to investigate flip-graph connectivity [7, 8]. Since so few triangulations of \(\{0,1\}^4\) are non-regular, it seems natural to conjecture that *all the triangulations of* \(\{0,1\}^4\) *are 2-regular*. Settling this conjecture positively would provide another proof for the connectedness of \(\gamma (\{0,1\}^4)\).

## Supplementary material

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