On Topological Minors in Random Simplicial Complexes^{†}^{†}thanks: Research supported by the Swiss National Science Foundation (SNF Projects 200021125309 and 200020138230)
For random graphs, the containment problem considers the probability that a binomial random graph contains a given graph as a substructure. When asking for the graph as a topological minor, i.e. for a copy of a subdivision of the given graph, it is wellknown that the (sharp) threshold is at . We consider a natural analogue of this question for higherdimensional random complexes , first studied by Cohen, Costa, Farber and Kappeler for .
Improving previous results, we show that is the (coarse) threshold for containing a subdivision of any fixed complete complex. For higher dimensions , we get that is an upper bound for the threshold probability of containing a subdivision of a fixed dimensional complex.
1 Introduction
A basic problem in graph theory is to determine whether a given graph , which may be thought of as “large”, contains a fixed graph as a substructure. The most straightforward form of containment is that contains a copy of as a subgraph. Another important variant is that contains some subdivision of as a subgraph; in this case, one also says that contains as a topological minor.
For random graphs, the containment problem considers the probability that a binomial random graph contains a copy of a given graph . For subgraph containment, it is wellknown [8] that this probability has a (coarse) threshold of , where is the density of the densest subgraph of . The (sharp) threshold for containment of any complete graph of fixed size as a topological minor is by a wellknown result of Ajtai, Komlós and Szemerédi^{3}^{3}3The lower bound on the threshold follows from [12, Theorem 5e]. [1].
Here, a sequence is called a threshold for an increasing graph property if the probability tends to if and to if . A threshold is sharp if for any
Any monotone graph property has a threshold [9]. If a monotone graph property does not have a sharp threshold, it has a coarse threshold. (For more detailed results on thresholds of graph properties, see, e.g., [13, 14] and [20, Chapter 1] for an overview.)
A random model for simplicial complexes of arbitrary fixed dimension which generalizes the random graph model was introduced by Linial and Meshulam [25] and has since then been studied extensively, see, e.g., [3, 4, 11, 23, 27]. Subgraph containment admits a direct generalization to higher dimensions: We can ask whether a given simplicial complex contains a fixed complex as a subcomplex. The proof methods for random graphs extend directly to random complexes of higher dimension, and the threshold probability for to contain a fixed complex as a subcomplex is given by the density (in terms of number of faces versus number of vertices) of the densest subcomplex of , see [4, 11].
As a natural higherdimensional analogue of topological graph minors, we say that a complex has a fixed complex as a topological minor if contains some subdivision of . Cohen, Costa, Farber and Kappeler [11] show that for any and , the random complex asymptotically almost surely (a.a.s.), i.e., with probability tending to as , contains any fixed as a topological minor. Their method also extends to random complexes of higher dimension with .
We improve this upper bound on the threshold probability for containing a fixed dimensional topological minor to :
Theorem 1.
For every and there is a constant such that with a.a.s. contains the complete complex on vertices as a topological minor.
Here, the complete complex is the skeleton of the simplex and consists of all sets of cardinality on vertices. Thus, every finite complex is contained in for large enough .
The result in [11] is proven by reduction to the subcomplex containment problem, by showing that a given complex can be subdivided to decrease its triangle density. For Theorem 1 we use a different approach, based on an idea going back at least to Brown, Erdős and Sós [10] and used also in [4]: For complexes our proof is based on studying common links of pairs of vertices, which form random graphs of the type and then uses results on the phase transition in random graphs. This approach also extends to complexes of dimension . For a (possibly more approachable) sketch of the proof for the case we refer the reader to the extended abstract [16] of this paper.
For complexes we also show a corresponding lower bound and thus establish that the threshold for containing a subdivision of any fixed complete complex is at :
Theorem 2.
There is a constant such that for any the random complex with a.a.s. does not have as a topological minor.
The somewhat technical proof of Theorem 2 is based on bounds on the number of triangulations of a fixed surface.
Before we give the proofs of these results, we present a simpler application of the basic idea behind the proof of Theorem 1. We are grateful to Matt Kahle for asking the second author whether our method could be applied in this way.
We consider the threshold for vanishing of the fundamental group of . In [4], Babson, Hoffman and Kahle show a lower bound of order for any as well as an an upper bound of for any function with as ; as mentioned above, the proof of the upper bound is based on studying the random graphs that appear as common links of pairs of vertices, which are connected for in this regime. Using a slightly more involved argument based on these random graphs, we show that the logarithmic factor in the upper bound is unnecessary:
Theorem 3.
There is a constant such that the random complex with a.a.s. has a trivial fundamental group: .
2 Preliminaries
Abstract Simplicial Complexes.
A (finite abstract) simplicial complex is a finite set system that is closed under taking subsets, i.e., implies . The sets are called faces of . The dimension of a face is . The dimension of is the maximal dimension of any face. A dimensional simplicial complex will also be called a complex.
We denote the set of dimensional faces by . The skeleton of is the simplicial complex . A vertex of is a dimensional face , the singleton set will be identified with its element . The set of vertices , also denoted by , is called the vertex set of . Corresponding to the notation for graphs, we write complexes as with and .
The random complex has vertex set , a complete skeleton, i.e., for , and every is added to independently with probability , which may be constant or, more generally, a function depending on . The complete complex has vertex set and for all . The link of a face in is the complex .
Geometric Simplicial Complexes.
There is a more geometric way to define simplicial complexes: A geometric simplex is the convex hull of a set of affinely independent points, the vertices of , in some Euclidean space . The convex hull of a subset of the vertices of is a face of . A geometric simplicial complex is then a finite collection of geometric simplices in satisfying two conditions: If is in and is a face of , then is also in . Furthermore, the intersection of any two simplices in is a common face of both, or empty.
A geometric simplicial complex defines a topological space, its polyhedron, the union of all its simplices: . It carries the subspace topology inherited from the ambient Euclidean space . We call a triangulation of . Any geometric simplicial complex gives rise to an abstract complex in a straightforward way: A set of vertices forms an (abstract) simplex in iff it is the vertex set of a geometric simplex in . The geometric complex is then called a geometric realization of , or of any abstract complex isomorphic to it. Here, two simplicial complexes are isomorphic if there is a facepreserving bijection between their vertex sets.
Any abstract complex has a geometric realization, e.g., as a subcomplex of a simplex of sufficiently high dimension. We denote by the polyhedron of any geometric realization of . This is welldefined because the polyhedra of (geometric realizations of) two isomorphic complexes are homeomorphic (see, e.g., [26]). Also the abstract complex is called a triangulation of .
Subcomplexes and Subdivisions.
A subcomplex of (or ) is a subset () that is itself a simplicial complex. A subdivision of a geometric simplicial complex is a complex with such that every simplex of is contained in some simplex of . For an abstract complex , a complex is a subdivision of if there exist geometric realizations and of and such that is a subdivision of . A subdivision of an abstract complex can be seen as a complex that is obtained by replacing the edges of with internallydisjoint paths and the triangles of with internallydisjoint triangulated disks such that for every triangle the subdivision of the triangle agrees with the subdivisions of its edges. See Figure 1 for an illustration.
We say that a complex has a fixed complex as a topological minor if contains a copy of some subdivision of .
3 On the Threshold for Vanishing Fundamental Group
Before we come to the proof of Theorem 1, the upper bound for the threshold for topological minor containment, we present a simpler application of its basic idea and prove Theorem 3.
The fundamental group of a topological space at a point is the group of homotopy classes of loops, i.e., continuous maps with the same starting and ending point . The fundamental group of a pathconnected space is independent, up to isomorphism, of the choice of basepoint and is then denoted by . (See, e.g., [17] for more details.)
For a simplicial complex the fundamental group depends only on the skeleton of , and it suffices to consider loops on the skeleton (“edge paths”). (See, e.g., [29].)
To prove Theorem 3, we will show that in with (for sufficiently large) a.a.s. every cycle consisting of three edges can be filled with a subdivision of a disk. This implies that every cycle (and, because the skeleton is the complete graph, also every other cycle) is homotopically trivial. As the property of having a trivial fundamental group is monotone (preserved under adding simplices), it is enough to consider the case .
Consider three vertices . To fill the cycle spanned by , and , we find a fourth vertex and fill the three cycles , and .
For a pair of vertices , we consider the graph . We call a vertex “good” for the pair if there is a vertex with such that and are contained in the same connected component of . Note that for a good vertex , any path connecting and in together with the triangle forms a subdivision of a disk filling the cycle . Hence, finding a vertex that is good for , and gives a filling of the cycle . See Figure 2 for an illustration. (Note that we do not have to (and possibly will not) find disjoint fillings of the three cycles , and as depicted in the figure.)
We denote the set of vertices that are good for by . In we can ensure that for any pair of vertices this set of good vertices is large with high probability:
Lemma 4.
There exist constants such that in with
for any .
Our previous observations will allow us to reduce the problem to studying the graphs , which are random graphs of type . We will employ the following classical result by Erdős and Rényi [12] on the phase transition of the random graph . For a simple proof see [24], in particular Section 3.1 describing how to achieve the bound on the probability.
Theorem 5 ([12, 24]).
For every there is such that with probability at least the random graph has a connected component of size at least .
We will furthermore use the following classical result on the concentration of binomially distributed random variables:
Theorem 6 (Chernoff’s inequality, see, e.g., [19, Theorem 1]).
Let be a binomially distributed random variable with parameters and . Then for any :
Proof of Lemma 4.
Fix . Denote by the largest connected component of the graph . If there are several components of maximum size, let be the one containing the smallest vertex. Note that if there is a vertex such that , then every vertex in is good for , i.e., . Hence, it suffices to consider the following event:
For a fixed set with the number of vertices with is binomially distributed with parameters and . Thus, its expectation is and by Chernoff’s inequality, Theorem 6:
We now condition on being :
Recall that is the largest component of the graph , which is a random graph of type . As , we can now employ Theorem 5: If we choose large enough, there is such that has size greater than with probability . Then we have:
∎
Now, we know that with high probability the set of good vertices is large for any pair and use this to show that we can fill every cycle.
Proposition 7.
There is such that in with a.a.s. every cycle consisting of three edges can be filled with a subdivision of a disk.
Proof.
We observed above that finding a vertex that is good for , and gives a filling of the cycle . It hence suffices to show that the probability
converges to . For fixed we can employ Lemma 4:
Using a union bound, we see that
∎
4 The Upper Bound for Topological Minor Containment
We now address Theorem 1. For fixed , we aim to find a copy of a subdivision of in . We would be allowed to subdivide faces of of any dimension, but there will be no need for this: we find vertices and take all faces of dimension at most spanned by these vertices to form the skeleton of our subdivision of . We then show that the spheres (boundaries of simplices) spanned by the faces between any of them can be filled with disjoint triangulated simplices.
Basic SetUp.
We will only consider complexes with vertex set and complete skeleton. For notational convenience, we assume without loss of generality that is divisible by . Fix a partition of the vertex set into two sets and , each of size . We will choose the vertices of from , whereas the internal vertices for fillings will come from . To ensure disjointness of the fillings of different spheres, we partition into sets , , each of size , and choose the internal vertices of the filling for each from .
For a face , denote by the graph which has vertex set and edge set . Denote by the largest connected component of . If there are several components of maximum size, let be the one containing the smallest vertex.
The Main Idea.
The basic idea of the proof is the following lemma, based on an idea going back at least to Brown, Erdős and Sós [10] that is also used in [4].
Lemma 8.
Let be a complex with vertex set and complete skeleton. Suppose there is a set , with a bijection satisfying the following property: For every face of there is a vertex such that for there are vertices with and . Then contains a subdivision of .
Proof.
For and , , , as above there exists a path in between and which together with the faces for and the face , creates a dimensional disk filling the sphere (boundary of a simplex) created by the faces , . By choosing a distinct for each we ensure disjoint fillings. See Figure 3 for an illustration for the case . ∎
Random Complexes.
We now proceed to show that for a suitable constant and the random complex a.a.s. satisfies the conditions of Lemma 8. We first give a criterion for complexes satisfying these conditions and then show that this criterion is satisfied a.a.s. by a random complex.
For fixed , and , call connected to if for some and let . Consider two families of complexes with vertex set and complete skeleton:

.

For : .
Lemma 9.
Let be a complex with vertex set and complete skeleton. If there is a such that for all , and , then there exists a set satisfying the conditions of Lemma 8.
Proof.
We need to show the existence of a set , with a bijection such that for every there is a vertex such that for :

There is with .

The vertex is connected to .
As for all and , the first condition holds for any choice of ,, and . So we only need to deal with the second condition. We consider tupels with and all pairwise distinct and let . The function is then determined by . We show that for a tupel chosen uniformly at random we have
Thus, there is a tuple that also satisfies the second condition. For fixed and :
By a union bound, we hence have
∎
Now, we finally turn to random complexes . As the property of containing a given topological minor is monotone (preserved under adding simplices), it is enough to consider the case .
Lemma 10.
For every and there is a constant such that for the random complex asymptotically almost surely satisfies the conditions of Lemma 9.
Proof.
Let and . Let . We show that there is and such that for a.a.s. satisfies the conditions of Lemma 9, i.e.,
Fix , and . The probability of the events and depends on the size of , the largest connected component of the graph , which is a random graph of type .
As , for large enough the graph fails to have a giant component of size linear in with exponentially small probability: By Theorem 5 for any there are and such that
As we will later need that , we choose . Then and we choose with .
For denote by the conditional probability when conditioning on . Then is at least
where the sum runs over all with .
As and depend on different kinds of faces and the presences of faces are decided independently, we have
We consider the two terms seperately:
:
Here we consider . The number of vertices with is a binomially distributed variable with parameters and . Hence, its expectation is and by Chernoff’s inequality, Theorem 6:
:
Call connected to if for some . Then we need to consider
For fixed the probability not to be connected to is . For each the decisions over the faces deciding whether is connected to are taken independently. Hence, also the number of vertices that are connected to is a binomially distributed variable with parameters and . As we chose , we get by Chernoff’s inequality, Theorem 6, for large enough :
Notice that the probabilities and do not depend on . Hence we can use and get by the choice of and :
Applying a union bound, we get for some :
∎
5 The Lower Bound for Topological Minor Containment
We now turn to the proof of Theorem 2 on random complexes . Our goal is to show the existence of a constant such that for the probability to find a subdivision of converges to zero. The proof bases on the following simple observation: If a complex contains a subdivision of with , it also contains a subdivision of a triangulation of , the orientable surface of genus ^{4}^{4}4The smallest possible triangulation of has vertices, see [18].. We then use that the number of triangulations of any surface with a fixed number of vertices is known to be at most simply exponential in .
Bounds on the number of triangulations of a fixed closed surface can be drawn from the theory of enumeration of maps on surfaces which has its beginning in Tutte’s famous results on the number of rooted maps on the sphere [30, 31, 32]. As the terminology in these references differs a lot from ours and as furthermore the classes of objects that are counted are not exactly the same, we first explain in detail the enumeration result we will use. We rely on [5, 6, 7, 15].
Maps on Surfaces.
Let be a connected compact manifold without boundary. A map on is a graph together with an embedding of into such that each connected component of is simply connected, i.e., each face is a disk. Graphs are unlabeled, finite and connected, loops and multiple edges are allowed.
A map is rooted if an edge, a direction along the edge and a side of the edge are distinguished. An edge is called double if its image belongs to the boundary of only one face. Any other, single, edge belongs to two faces. The valency of a face is the number of single edges in its boundary plus twice the number of double edges. A triangular map is a map such that each face has valency three.
Two maps and are considered equivalent if there is a homeomorphism and a graph isomorphism such that .
Triangular Maps vs. Triangulations.
Let be a triangular map such that the graph is simple, i.e., does not have loops or multiple edges. Then every face of has a boundary consisting of exactly three edges. Define a complex by letting
Equivalent maps yield isomorphic complexes:
Lemma 11.
Let be a triangular map such that the graph is simple and let be equivalent to . Then and are isomorphic.
Proof.
Since and are equivalent, there is a homeomorphism and a graph isomorphism such that . We show that is also an isomorphism between and . As is a graph isomorphism, all we need to show is that preserves faces. Let , so are the vertices of a face of . This face is mapped to some disk in by . As , this disk is a face of with and as boundary vertices. Hence, . The same argument shows that any face of is mapped to a face of . ∎
For a complex such that is homeomorphic to a surface , define a triangular map , where is the restriction of a homeomorphism to the skeleton of . The following lemma shows that is welldefined and that isomorphic complexes give rise to equivalent maps:
Lemma 12.
Let be a complex such that is homeomorphic to a surface and let be isomorphic to . Let furthermore and be homeomorphisms and define and to be the restrictions of and to the skeleta of and , respectively. Then and are equivalent.
Proof.
Let be an isomorphism between and . Then the affine extension is a homeomorphism (see, e.g., [26, Proposition 1.5.4]). So is also homeomorphic to . Choosing and , we get . ∎
The Number of Triangulations.
In [15] Gao gives an asymptotic enumeration result for rooted triangular maps on any closed surface.
Theorem 13 (Gao, [15]).
Let denote the number of vertex rooted triangular maps on the orientable surface of genus .^{5}^{5}5Gao also considers nonorientable surfaces, in which we are not interested here. There is a constant , independent of , such that for ,
We are interested in the number of vertex triangulations of the orientable surface of genus , i.e., the number of complexes such that and is homeomorphic to . By the considerations above this is the number of triangular maps on with a simple underlying graph. As Gao’s result also allows loops and multiple edges and makes a distinction between equivalent maps that are rooted in a different way, we get and hence:
Corollary 14.
Let be the number of triangulations of , the orientable surface of genus , with vertices. There is a constant , independent of , such that .
Proof of the Lower Bound on the Threshold.
Now that we have established (Corollary 14) that the number of triangulations of any fixed surface with a fixed number of vertices is at most simply exponential in , we can turn to the proof of Theorem 2.
Proof of Theorem 2.
Fix , and let be a triangulation of , the orientable surface of genus , with vertices. As is a subcomplex of , we have:
We show that for sufficiently small the latter probability tends to . Ignoring that we only consider subdivisions of , we get:
where the second sum is over the set of all triangulations of that have vertices. Denote by the number of such triangulations and choose as in Corollary 14 such that is at most . Let for some .
By Euler’s formula, every triangulation of the oriented surface of genus satisfies