Betti Numbers of Complexes
with Highly Connected Links
Amir Abu-Fraiha
Department of Mathematics, Technion, Haifa
32000, Israel. e-mail: [email protected]Β .
ββ
Roy Meshulam
Department of Mathematics, Technion, Haifa
32000, Israel. e-mail: [email protected]Β . Supported by
ISF and GIF grants.
Abstract
Let Ξnβ1(k)β denote the k-dimensional skeleton of the (nβ1)-simplex Ξnβ1β and consider a complex
Ξnβ1(kβ1)ββXβΞnβ1(k)β. Let \twelvebbK be a field and let 0β€β<k. It is shown that if H~kβββ2β(lk(X,Ο);\twelvebbK)=0 for all β-dimensional faces Ο of X then
[TABLE]
with equality iff lk(X,Ο) is a (kβββ1)-hypertree for all β-dimensional simplices Ο of Ξnβ1β.
Examples based on sum complexes show that the bound is asymptotically tight for all fixed k,β as nββ.
1 Introduction
Let X be a simplicial complex on the vertex set V. Numerous problems in topological combinatorics ask for estimates on some global invariants of X, e.g. its connectivity or Betti numbers, given that X satisfies certain local properties. One remarkable local to global result of this nature is Garlandβs theorem [3].
We first recall some definitions. The induced subcomplex of X on Vβ²βV is
X[Vβ²]={ΟβX:ΟβVβ²}.
Denote the star, link and costar of a simplex ΟβX by
[TABLE]
Let X(j) denote the j-th skeleton of X and let X(j) be the family of j-dimensional simplices of X. Denote fjβ(X)=β£X(j)β£.
Assume that X is a pure k-dimensional complex, and define a positive weight function on its simplices by
[TABLE]
For ΟβX let cΟβ be the induced weight function on lk(X,Ο) given by cΟβ(Ξ±)=c(ΟβͺΞ±).
For β1β€jβ€k let Cj(X;\twelvebbR) denote the space of real valued j-cochains of X and let djβ:Cj(X;\twelvebbR)βCj+1(X;\twelvebbR) denote the j-th coboundary map
of X. Let djββ:Cj+1(X;\twelvebbR)βCj(X;\twelvebbR) be the adjoint of djβ with respect to the weight function c.
Let Ljβ=djβ1βdjβ1ββ+djββdjβ:Cj(X;\twelvebbR)βCj(X;\twelvebbR) be the j-th Laplacian of X and let ΞΌjβ(X) denote its minimal eigenvalue.
Theorem 1** (Garland [3]).**
Let β1β€β<kβ1. If
ΞΌkβββ2β(lk(X,Ο))>kβ+1β for all ΟβX(β), then H~kβ1β(X;\twelvebbR)=0.
Garlandβs Theorem and its variants have applications in a wide range of areas including representation theory, geometric group theory, hypergraph matching and random complexes
(see e.g. [3, 2, 1, 4]). Here we study the following question that naturally arises in connection with Theorem 1: What can be said concerning H~kβ1β(X) if, instead of ΞΌkβββ2β(lk(X,Ο))>kβ+1β, it is only assumed that H~kβββ2β(lk(X,Ο))=0 for all ΟβX(β)?
Β
Let \twelvebbK be a fixed field. Let H~jβ(X)=H~jβ(X;\twelvebbK) and Ξ²~βjβ(X)=dim\twelvebbKβH~jβ(X;\twelvebbK) be the reduced homology groups and reduced Betti numbers of X over \twelvebbK.
For 0β€β<k and β1β€j let
[TABLE]
Let
[TABLE]
and
[TABLE]
Let Ξ(V) denote the simplex on the vertex set V. Let [n]={0,β¦,nβ1} and let Ξnβ1β=Ξ([n]) be the (nβ1)-simplex on [n].
Theorem 2**.**
If Ξnβ1(kβ1)ββXβΞnβ1β and 0β€β<k then
[TABLE]
Theorem 2 implies the following
Corollary 3**.**
Suppose Ξnβ1(kβ1)ββXβΞnβ1β satisfies Ξ»β,kβββ2β(X)=0. Then
[TABLE]
The next three results concern some aspects of the equality cases in Corollary 3. A complex Ξ(V)(rβ1)βYβΞ(V)(r) is an
r-hypertree over \twelvebbK on the vertex set V (abbreviated r-hypertree for a fixed field \twelvebbK) if
H~ββ(Y;\twelvebbK)=0. It is easy to check that Y is an r-hypertree iff frβ(Y)=(rβ£Vβ£β1β) and either H~rβ1β(Y;\twelvebbK)=0 or H~rβ(Y;\twelvebbK)=0.
See Kalaiβs paper [5] for further discussion, including a Cayley type formula for the weighted enumeration of rational hypertrees.
Theorem 4**.**
Let 0β€β<k and suppose Ξnβ1(kβ1)ββXβΞnβ1(k)β satisfies
Ξ»β,kβββ2β(X)=0.
Then the following three conditions are equivalent.
- (a)
Ξ²~βkβ1β(X)=Bn,k,ββ.**
2. (b)
Ξ²~βkβ(X)=0* and fkβ(X)=Fn,k,ββ.*
3. (c)
lk(X,Ο)* is a (kβββ1)-hypertree on [n]βΟ for all ΟβΞnβ1β(β).*
The next result asserts that the bound (2) is asymptotically tight for fixed
k,β and nββ.
Theorem 5**.**
Let 0β€β<k be fixed. Then for any prime number n>k there exists a complex
Ξnβ1(kβ1)ββXn,k,βββΞnβ1(k)β such that Ξ»β,kβββ2β(Xn,k,ββ)=0 and
[TABLE]
Finally, we give examples that show the optimality of (2) for β=0 and kβ€3.
Theorem 6**.**
Let 1β€kβ€3. Then for infinitely many nβs there exist complexes
Ξnβ1(kβ1)ββJn,kββΞnβ1(k)β such that Ξ²~βkβ2β(lk(Jn,kβ,v))=0 for all vβΞnβ1β(0) and
[TABLE]
The paper is organized as follows: In Section 2 we prove a monotonicity result (Proposition 7) that directly implies Theorem 2. The characterization of equality cases (Theorem 4) is established in Section 3.
In Section 4 we recall the notion of sum complexes and prove an upper bound on the Betti number of their links (Proposition 11). This result is the main ingredient in the proof of Theorem 5 given in Section 5. In Section 6 we describe the constructions that yield Theorem 6. We conclude in Section 7 with some remarks and open problems.
2 The Upper Bound
The main ingredient in the proof of Theorem 2 is the following monotonicity result.
Proposition 7**.**
Let 0β€β<k. If XβΞnβ1(k)β and ΟβX(k) then
[TABLE]
Proof: First note that
[TABLE]
Let ΟβX(β). If ΟβΟ then
[TABLE]
On the other hand, if Οξ βΟ then lk(X,Ο)=lk(XβΟ,Ο).
Summing (4) over all ΟβX(β) we obtain
[TABLE]
Consider two cases:
(i) Ξ²~βkβ1β(XβΟ)=Ξ²~βkβ1β(X)+1. Then by (5)
[TABLE]
Thus (3) holds.
(ii) Ξ²~βkβ1β(XβΟ)=Ξ²~βkβ1β(X). To establish (3) it suffices to show that Ξ»β,kβββ2β(XβΟ)=Ξ»β,kβββ2β(X), or equivalently
that if ΟβΟ(β) then
[TABLE]
Consider the decompositions
[TABLE]
and
[TABLE]
Then
[TABLE]
and
[TABLE]
Note that H~kβ2β(βΟβY)β
H~kβββ2β(Y) for any Y.
Hence by Mayer-Vietoris we obtain a commutative diagram
[TABLE]
where the rows are exact and the iβββs are induced by inclusion maps. Clearly (i1β)ββ and (i4β)ββ are the identity maps. As the removal of the k-dimensional simplex Ο does not effect the (kβ2)-homology, it follows that (i5β)ββ is an isomorphism. The assumption Ξ²~βkβ1β(XβΟ)=Ξ²~βkβ1β(X) implies that
(i2β)ββ is an isomorphism. It follows by the 5-lemma that (i3β)ββ is an isomorphism as well, and thus (6) holds.
This completes the proof of (3).
β‘
Proof of Theorem 2:
First note that if ΟβΞnβ1β(β) and β<k then lk(Ξnβ1(kβ1)β,Ο)β
Ξnβββ2(kβββ2)β and that Ξ²~βjβ(Ξm(j)β)=(j+1mβ).
Secondly, as both H~kβ1β(X) and H~kβββ2β(lk(X,Ο)) for ΟβΞnβ1β(β) depend only on the k-dimensional skeleton of X, we may assume that
XβΞnβ1(k)β. By repeatedly removing k-simplicies from X and using (3) it follows that
[TABLE]
β‘
Theorem 2 can be also formulated as the following upper bound on Ξ²~βkβ(X).
Theorem 8**.**
Let XβΞnβ1(k)β. Then for any β1β€β<k
[TABLE]
Proof:
As both sides of (7) do not depend on the (kβ1)-skeleton of X, we may assume that XβΞnβ1(kβ1)β.
The exact sequence for the pair (X,Ξnβ1(kβ1)β)
[TABLE]
implies that
[TABLE]
Similarly, for each ΟβΞnβ1β(β)
[TABLE]
Summing (9) over all ΟβΞnβ1β(β) we obtain
[TABLE]
Combining (8), (1) and (10) it follows that
[TABLE]
β‘
3 Characterizations of Equality
In this section we prove Theorem 4. Denote the support of a k-chain z=βΟβY(k)βaΟβΟ of a complex Y by supp(z)={ΟβY(k):aΟβξ =0}.
We shall need the following observation.
Claim 9**.**
Let YβΞnβ1(k)β and let 0ξ =zβH~kβ(Y). If Οβsupp(z) and ΟβΟ(β)
then Ξ²~βkβββ1β(lk(Y,Ο))>0.
Proof: The assumptions imply that Ξ²~βkβ(cost(Y,Ο))<Ξ²~βkβ(Y).
Using, as in the proof of Proposition 7, the exact sequence
[TABLE]
it follows that
[TABLE]
β‘
Proof of Theorem 4:
(a) β (b): We first show that (a) implies Ξ²~βkβ(X)=0. Otherwise choose an inclusion-wise minimal
Ξnβ1(kβ1)ββXβΞnβ1(k)β such that both
Ξ»β,kβββ2β(X)=0 and Ξ²~βkβ1β(X)=Bn,k,ββ, but Ξ²~βkβ(X)>0.
Let 0ξ =zβH~kβ(X) and let Οβsupp(z). Then
Ξ²~βkβ(XβΟ)=Ξ²~βkβ(X)β1 and hence by (8)
[TABLE]
Proposition 7 thus implies
[TABLE]
Therefore
[TABLE]
Combining (11), (12) and the minimality of X, it follows that
Ξ²~βkβ(XβΟ)=0. Using (8) for the complex XβΟ it follows that
[TABLE]
Let ΟβΞnβ1β(β). Using (8) for lk(X,Ο) we obtain
[TABLE]
Summing (13) over all ΟβΞnβ1β(β) it follows that
[TABLE]
Choose a k-simplex Οξ =Οβ²βsupp(z) and an β-simplex ΟβΟβ²(β)βΟ(β). Then on one hand
lk(XβΟ,Ο)=lk(X,Ο), hence by Claim 9
[TABLE]
On the other hand it follows from (14) that Ξ²~βkβββ1β(lk(XβΟ,Ο))=0, a contradiction.
Thus Ξ²~βkβ(X)=0 and hence
[TABLE]
(b) β (c): Suppose fkβ(X)=Fn,k,ββ. As Ξ»β,kβββ2β(X)=0, it follows that
Ξ²~βkβββ2β(lk(X,Ο))=0 and hence fkβββ1β(lk(X,Ο)β₯(kβββ1nβββ2β) for all ΟβΞnβ1β(β).
Therefore
[TABLE]
Hence fkβββ1β(lk(X,Ο))=(kβββ1nβββ2β) and therefore lk(X,Ο) is a (kβββ1)-hypertree for all ΟβΞnβ1β(β).
(c) β (a): Assume that lk(X,Ο) is a (kβββ1)-hypertree for all ΟβΞnβ1β(β). Then, as in (15), it follows that
fkβ(X)=Fn,k,ββ. Furthermore, by (7)
[TABLE]
Hence
[TABLE]
β‘
4 Links of Sum Complexes
Let n be a prime and let A be a subset of the cyclic group V=\twelvebbZnβ.
Identify the vertex set of Ξnβ1β with the elements of \twelvebbZnβ. For sβ€nβ2 define the sum complex YA,s+1ββΞnβ1(s)β
by
[TABLE]
The homology groups H~ββ(YA,s+1β;\twelvebbK) were determined in [7, 8].
When A is a cyclic interval in \twelvebbZnβ, the Betti numbers Ξ²~βββ(YA,s+1β;\twelvebbK) do not depend on \twelvebbK and take the following simple form.
Theorem 10** ([7, 8]).**
Let n be a prime and let A={t,t+1,β¦,t+r} be an interval of size r+1 in \twelvebbZnβ. Then for any field \twelvebbK
[TABLE]
Let ββ€kβ2 and let ck,ββ=(kβββ1)!(β+1)(kββ)β.
Proposition 11**.**
Let B={0,β¦,kβββ1}. Then for any ΟβΞnβ1β(β)
[TABLE]
Proof: Let y=βxβΟβx and let C={bβy:bβB}. Then
[TABLE]
Applying Theorem 10 with A=C, r=s=kβββ1 and iβ{sβ1,s}, it follows that
[TABLE]
Hence, the exact sequence
[TABLE]
implies
[TABLE]
For a,cβ\twelvebbZnβ let
[TABLE]
Note that any Ξ·βΞnβ1β(kβββ2) is contained in at most one ΟβFa,cβ.
Therefore
[TABLE]
Combining (16) and (17) it follows that
[TABLE]
β‘
5 The Lower Bound
Proof of Theorem 5: For the case β=kβ1 see the remark in Section 7.
Assume that ββ€kβ2.
Let n be a prime and let B={0,β¦,kβββ1}.
For any ΟβΞnβ1β(β)
choose a set of (kβββ1)-dimensional simplices
SΟββlk(Ξnβ1β,Ο)(kβββ1)
of size
β£SΟββ£=Ξ²~βkβββ2β(lk(YB,k+1β,Ο))
such that
[TABLE]
Proposition 11 implies that
[TABLE]
Let
[TABLE]
Then for all ΟβΞnβ1β(β)
[TABLE]
Applying Theorem 10 with A=B, r=kβββ1, s=k and i=sβ1, it follows that
[TABLE]
Hence
[TABLE]
β‘
6 Constructions of Jn,kβ for kβ€3
In this section we describe the constructions that establish Theorem 6.
Β
(i) Let k=1 and n be even. Let Jn,1β be a perfect matching on the vertex set [n].
Then Ξ²~ββ1β(lk(Jn,1β,v))=0 for all vβ[n] and Ξ²~β0β(Jn,1β)=2nββ1=Bn,1,0β.
Β
(ii) Let k=2 and n=3t+2. Let Jn,2β be the 2-dimensional complex on the vertex set \twelvebbZnβ
(see Figure 1(a)) given by
[TABLE]
Proposition 12**.**
Jn,2β* satisfies Ξ²~β0β(lk(Jn,2β,v))=0 for all vβ\twelvebbZnβ
and Ξ²~β1β(Jn,2β)=31β(2nβ2β)=Bn,2,0β.*
Proof:
By Theorem 4 it suffices to show that for all iβ\twelvebbZnβ the graph lk(Jn,2β,i) is a tree on the vertex set \twelvebbZnββ{i}.
By homogeneity it suffices to consider lk(Jn,2β,0). It follows from the definition of Jn,2β that
[TABLE]
where
[TABLE]
Thus, lk(Jn,2β,0) is the tree on \twelvebbZnββ{0} depicted in Figure 1(b).
β‘
(iii) Let k=3 and 4β€n be even. Let Jn,3β be the 3-dimensional complex
on the vertex set \twelvebbZnβ (see Figure 2) given by
[TABLE]
Proposition 13**.**
Jn,3β* satisfies Ξ²~β1β(lk(Jn,3β,v))=0 for all vβ\twelvebbZnβ
and Ξ²~β2β(Jn,3β)=41β(3nβ2β)=Bn,3,0β.*
Proof: As in the proof of Proposition 12, it suffices to show that lk(Jn,3β,0) is a 2-hypertree on the vertex set \twelvebbZnββ{0}. We claim that lk(Jn,3β,0) is in fact collapsible. Partition the 2-simplices of lk(Jn,3β,0) into 3 disjoint families
(see Figure 3):
[TABLE]
where
[TABLE]
If 0<i<j<n/2 then the edge {i,j} is contained in the unique 2-simplex {i,j,i+n/2}βB0β.
If 0<i<n/2<j<i+n/2 then the edge {j,i+n/2} is contained in the unique 2-simplex {i,j,i+n/2}βC0β.
Collapsing all these edges and the corresponding 2-simplices, the resulting complex consists of all simplices in A0β and their faces.
This complex is a cone on the vertex n/2 and is therefore collapsible.
β‘
7 Concluding Remarks
We have shown that if Ξnβ1(kβ1)ββXβΞnβ1(k)β satisfies
Ξ²~βkβββ2β(lk(X,Ο);\twelvebbK)=0 for all ΟβΞnβ1β(β), then Ξ²~βkβ1β(X;\twelvebbK)β€Bn,k,ββ.
Furthermore, this bound is asymptotically tight for fixed k,β and nββ, and exact for (k,β)=(1,0),(2,0),(3,0) and infinitely many nβs.
We suggest the following
Conjecture 14**.**
For any fixed 0β€β<k there exists a constant n0β(k,β) such that if nβ₯n0β(k,β) and if Bn,k,ββ is an integer, then
there exists a complex Ξnβ1(kβ1)ββXβΞnβ1(k)β such that Ξ»β,kβββ2β(X)=0 and Ξ²~βkβ1β(X)=Bn,k,ββ.
Remarks:
-
Let β=kβ1. It follows from Theorem 4 that Ξnβ1(kβ1)ββXβΞnβ1(k)β satisfies both
Ξ»β,kβββ2β(X)=Ξ»kβ1,β1β(X)=0 and Ξ²~βkβ1β(X)=Bn,k,kβ1β=(k+1nββ1)(kβ1nβ1β) iff each (kβ1)-simplex ΟβΞnβ1β(kβ1)
is contained in a unique k-simplex in X(k), i.e. iff S=X(k)β(k+1[n]β) is a Steiner system of type S(k,k+1,n). Thus the case β=kβ1 of Conjecture 14
follows from Keevash groundbreaking work [6] on the existence of designs.
-
It would be interesting and useful for various applications to interpolate between Corollary 2 and Garlandβs Theorem, in particular to obtain sharp upper bounds on the rational Betti number Ξ²~βkβ1β(X;\twelvebbQ) in terms of ΞΌ~βkβββ2β(X)=minΟβX(β)βΞΌkβββ2β(lk(X,Ο))
when 0<ΞΌ~βkβββ2β(X)β€kβ+1β.
-
The characterization given in Theorem 4 and the examples in Section 6 suggest some natural questions concerning hypertrees, e.g. for which (kβ1)-hypertrees
Ξnβ2(kβ2)ββTβΞnβ2(kβ1)β there exists a complex Ξnβ1(kβ1)ββXβΞnβ1(k)β such that
lk(X,v)β
T for all vertices vβX(0) ?