f-vectors implying vertex decomposability

We prove that if a pure simplicial complex of dimension d with n facets has the least possible number of (d-1)-dimensional faces among all complexes with n faces of dimension d, then it is vertex decomposable. This answers a question of J. Herzog and T. Hibi. In fact we prove a generalization of their theorem using combinatorial methods.


Introduction
We call a simplicial complex pure if all its facets are of the same dimension.Definition 1.A pure simplicial complex ∆ of dimension d and n facets is called extremal if it has the least possible number of (d − 1)-dimensional faces among all complexes with n faces of dimension d.
In particular, for d = 0 all zero dimensional complexes are extremal, since all of them have exactly one (−1)-dimensional face, namely the empty set.
In this paper we generalize and prove by only combinatorial means the following theorem of Herzog and Hibi form 1999.
Their proof is algebraic and uses results from [1] and [6].In fact they asked for a combinatorial proof.We give it by proving that an extremal simplicial complex is vertex decomposable.It is well-known that vertex decomposable complexes are Cohen-Macaulay.Our proof goes along the lines of the proof of Kruskal-Katona inequality.We start with a presentation of some necessary preliminaries.
Definition 2. A pure simplicial complex ∆ is vertex decomposable if one of the following holds: (1) ∆ is empty, (2) ∆ is a single vertex, (3) for some vertex x both lk ∆ {x} and ∆\x are pure and vertex decomposable.
Definition 3.For a simplicial complex ∆ on the set of vertices {1, . . ., n} and a given field K, the Stanley-Reisner ring (the face ring) is K[∆] := K[x 1 , . . ., x n ]/I, where I is generated by all square-free monomials x i1 • • • x i l for which {i 1 , . . ., i l } is not a face in ∆.
When we say that a simplicial complex is Cohen-Macaulay we always mean that its Stanley-Reisner ring has this property.
The following is a folklore result (we refer the reader to e.g.[2]).
Theorem 2. For a simplicial complex ∆ the following implications hold: We recall a combinatorial description of Cohen-Macaulay complexes.
be the face ring of ∆.Then the following conditions are equivalent: ( For classical techniques of counting homologies we refer the reader to [7], [14].For entertaining ones we advise Section 3.2 of [12].

Kruskal-Katona theorem. One of the most natural questions concerning simplicial complexes is:
What is the minimum number of (k − 1)-element faces in simplicial complex with n faces of size k?This question was answered independently by Kruskal [11] and Katona [10] in 1960's.For a positive integer k, they enlisted all k-element subsets of integers in the following order, called the squashed order : Let S k (n) be the set of first n sets in this list.For a given set U of k-element sets, denote by ∆U the set of all (k − 1)-element sets which are contained in some member of U. The Kruskal-Katona theorem reads as follows.
Theorem 4. For a positive integers n, k and a set U of n sets of size k we have This result was further generalized by Clements and Lindström in [3].Daykin [4,5] gave two simple proofs, and later Hilton [9] gave another one.For an algebraic proof we refer the reader to [1].We will work mainly with Hilton's idea.
Note that the cardinality of ∆S k (n) may be easily determined.For a given k, each positive integer n can be uniquely expressed as with 1 ≤ t ≤ a t and a t < • • • < a k .We have As a consequence of Kruskal-Katona theorem we get:

The main result
For a better understanding of the assumption that ∆ is extremal we will use Hilton's idea from his proof [9] of the Kruskal-Katona theorem.First we define sets similar to S k (n).Let S i k (n) denote the first n sets of k-element subsets of integers in the squashed order (A < B if max(A \ B) < max(B \ A)) which do not contain i.We also denote by {i}(∪)U the set {{i} ∪ A : A ∈ U}.
Let U be a n-element set of k-element sets, let V = A∈U A be an underlying set, and let v be its cardinality.For i ∈ V , let B i = {A ∈ U : i / ∈ A}, C i = {A \ {i} : i ∈ A ∈ U}, and let b i , c i be the respective cardinalities.Note that c i = 0. We want to find an index i such that Proof.We are going to count the sum of cardinalities of both sets when i runs over all elements of V .Then since at left hand side each A ∈ U gives k distinct sets in its boundary, and it is counted once for every i / ∈ A. Some sets in boundaries of sets from B i can be the same, but their number is at most (v − 1) − (k − 1) = v − k.On the other hand each A ∈ U is counted k times at the right side.Hence we can find a desired i, or the above bounds are tight.In the latter case, when A ∈ ∆B i , all v − k possibilities of completing it to a k-element set has to be in U.This means that U consists of all possible k-element subsets of V because from any set in U we can delete any element and insert any other.Lemma 2. If ∆ is an extremal simplicial complex of positive dimension, then there exists a vertex x such that both lk ∆ {x} and ∆ \ x are extremal.
Proof.Let ∆ be of dimension d − 1 > 0 and let U be the set of all d-element sets in ∆.If U consists of all possible d-element subsets of a given v-element set, then the assertion of the lemma is clearly true (we can take any vertex).Otherwise, due to Lemma 1, there exists an i ∈ V such that Since ∆B i and {i}(∪)∆C i are disjoint, it follows that for some e, there are now two possibilities: (1) ) , which contradicts the assumption that ∆ is extremal, since a complex generated by sets ) , and equality holds if and only if . The complex ∆ is extremal, so equality holds, and we get that C i ⊂ ∆B i and Finally, we are ready to prove the generalization of Theorem 1.
Theorem 5.An extremal simplicial complex is vertex decomposable.
Proof.The proof goes by an induction on d the dimension of ∆ and secondly on the number of facets.If d = 0 then ∆ consists of points and by the definition it is vertex decomposable.When d > 0, then by Lemma 2 there exists a vertex x, such that both complexes lk ∆ {x} and ∆ \ x are extremal.The first is of lower dimension; and the second either has the same dimension as ∆ but fewer facets, or it has smaller dimension.By the inductive hypothesis, both lk ∆ {x} and ∆ \ x are vertex decomposable, and as a consequence ∆ also is.The above result is best possible in the following sense.Let ∆ be a pure simplicial complex of dimension d > 0 with f -vector (f 0 , . . ., f d ), and with f d−1 = δ d (f d )+c, c ∈ N. Due to Corollary 1 the meaning of Theorem 5 is that if c = 0, then ∆ is vertex decomposable.But even for c = 1 complex ∆ does not have to be Cohen-Macaulay, which by Theorem 2 is a weaker property then vertex decomposability.We show the following.
Example 1.We have that δ d (2) = 2d+1.Let ∆ be a pure simplicial complex of dimension d with the set of facets U consisting of two disjoint ones.Then |∆U| = 2d + 2, and H0 (lk ∆ ∅; K) = 1, so due to Theorem 3 complex ∆ is not Cohen-Macaulay over any field K, and as a consequence it is also not vertex decomposable.
are extremal, where [A] means the simplicial complex generated by the set of faces A. Observe that lk ∆ {i} = [C i ] and∆ \ i = [B i ].The first equality is obvious, while the second is not as clear.If σ = {v 1 , . . ., v k } is a face in ∆ \ i then it is a subface of some facet F = {v 1 , . . ., v d }.If i does not belong to F then F ∈ [B i ] and so σ does.Otherwise, F \ {i} ∈ C i ⊂ ∆B i , so F \ {i} ∪ {j} ∈ B i for some j.Hence σ ∈ [B i ].Now i = x gives the assertion.