Spectral gaps of simplicial complexes without large missing faces
aa r X i v : . [ m a t h . C O ] J un Spectral gaps of simplicial complexeswithout large missing faces
Alan Lew ∗ Abstract
Let X be a simplicial complex on n vertices without missing facesof dimension larger than d . Let L j denote the j -Laplacian acting onreal j -cochains of X and let µ j ( X ) denote its minimal eigenvalue. Westudy the connection between the spectral gaps µ k ( X ) for k ≥ d and µ d − ( X ). In particular, we establish the following vanishing result: If µ d − ( X ) > (1 − (cid:0) k +1 d (cid:1) − ) n , then ˜ H j ( X ; R ) = 0 for all d − ≤ j ≤ k . Asan application we prove a fractional extension of a Hall-type theorem ofHolmsen, Mart´ınez-Sandoval and Montejano for general position setsin matroids. Let X be a simplicial complex on vertex set V . A simplex σ ⊂ V is calleda missing face of X if σ / ∈ X but for any τ ( σ , τ ∈ X . The set of missingfaces M X of the complex X completely determines X : X = { τ ⊂ V : σ τ for all σ ∈ M X } . Let h ( X ) = max { dim( σ ) : σ ∈ M X } .For k ≥ − C k ( X ) be the space of real valued k -cochains of thecomplex X and let d k : C k ( X ) → C k +1 ( X ) be the coboundary operator.For k ≥ k -dimensional Laplacian of X is defined by L k ( X ) = d k − d ∗ k − + d ∗ k d k .L k is a positive semidefinite operator from C k ( X ) to itself. The k -th spectralgap of X , denoted by µ k ( X ), is the smallest eigenvalue of L k .Let G = ( V, E ) be a graph on n vertices. Its clique complex (or flagcomplex) X ( G ) is the simplicial complex on vertex set V whose simplicesare the cliques of G . Note that clique complexes are exactly the complexeswith h ( X ) = 1. Indeed, the missing faces of X ( G ) are the edges of thecomplement of G . Aharoni, Berger and Meshulam [2] prove the followingresult: ∗ Department of Mathematics, Technion, Haifa 32000, Israel. e-mail:[email protected] . Supported by ISF grant no. 326/16. heorem 1.1 (Aharoni, Berger, Meshulam [2]) . Let G = ( V, E ) be a graph,where | V | = n , and let X = X ( G ) be its clique complex. Then for k ≥ kµ k ( X ) ≥ ( k + 1) µ k − ( X ) − n. Our main result is a generalization of Theorem 1.1 to complexes withoutlarge missing faces.
Theorem 1.2.
Let X be a simplicial complex with h ( X ) = d on vertex set V , where | V | = n . Then for k ≥ d ( k − d + 1) µ k ( X ) ≥ ( k + 1) µ k − ( X ) − dn. Our proof combines the approach of [2] with additional new ideas. Bothresults can be thought of as global variants of Garland’s method, which inits original form relates the spectral gaps of a complex with the spectralgaps of the links of its faces; See [8, 14]. As a consequence of Theorem 1.2we obtain
Theorem 1.3.
Let X be a simplicial complex with h ( X ) = d , on vertex set V , where | V | = n . If µ d − ( X ) > − (cid:18) k + 1 d (cid:19) − ! n, then ˜ H j ( X ; R ) = 0 for all d − ≤ j ≤ k . Remark.
In the case d = 1 it is shown in [2] that the condition in The-orem 1.3 is the best possible: Let G be the complete r -partite graph on n = ℓr vertices, with all sides of size ℓ . Then µ ( X ( G )) = r − r n , but˜ H r − ( X ( G ); R ) = 0.For d = 2 we have found such extremal examples only for a few cases:1. Let X be the simplicial complex whose vertices V are the points of theaffine plane over F , and whose missing faces are the lines of the affineplane. On the one hand, one can check that µ ( X ) = 6 = | V | . On theother hand, ˜ H ( X ; R ) = R = 0 (computer checked).2. Let X be the simplicial complex whose vertices V are the points of theprojective space of dimension 3 over F , and whose simplices are the setsof points containing at most two points from each line (so the missingfaces are the subsets of size 3 of the lines in the projective space). Onecan show that µ ( X ) = 36 = | V | . On the other hand, ˜ H ( X ; R ) = 0(computer checked). 2e next give some applications of Theorem 1.2 to connectivity boundsand Hall-type theorems for general simplicial complexes.Let η ( X ) = conn R ( X ) + 2, whereconn R ( X ) = min { i : ˜ H i ( X ; R ) = 0 } − homological connectivity of X over R .A subset of vertices S ⊂ V in a graph G = ( V, E ) is called a totallydominating set if for all v ∈ V there is some u ∈ S such that vu ∈ E . The total domination number of G , denoted by ˜ γ ( G ), is the minimal size of atotally dominating set. Let I ( G ) be the independence complex of the graph,i.e. the simplicial complex whose faces are all the independent sets σ ⊂ V .The total domination number gives a lower bound on the connectivity of I ( G ) (see [13, Theorem 1.2]): η ( I ( G )) ≥ ˜ γ ( G ) / . (1.1)(For additional lower bounds on η ( I ( G )) in terms of other domination pa-rameters, see e.g. [3, 13]).The inequality (1.1) had been generalized to general simplicial com-plexes: Let X be a complex on vertex set V . We say that a subset S ⊂ V is totally dominating if for every v ∈ V there is some σ ⊂ S such that σ ∈ X but vσ / ∈ X . The total domination number of X , denoted ˜ γ ( X ),is the minimal size of a totally dominating set in X . For a graph G wehave ˜ γ ( G ) = ˜ γ ( I ( G )) (the totally dominating sets of I ( G ) are the same asthe totally dominating sets of G ). In [1] it is shown that for any simplicialcomplex X , η ( X ) ≥ ˜ γ ( X ) / G ), has been introduced in[2] as follows. A vector representation of the graph G is an assignment P : V → R ℓ such that P ( v ) · P ( w ) ≥ v and w are adjacent in G ,and P ( v ) · P ( w ) ≥ α ∈ R V is called dominating for P if P v ∈ V α ( v ) P ( v ) · P ( w ) ≥ w ∈ V . The value of P is | P | = min { X v ∈ V α ( v ) : α is dominating for P } . Let Γ( G ) be the supremum of | P | over all vector representations of G . It iseasy to check that Γ( G ) ≤ ˜ γ ( G ) (see Proposition 1.5). In [2] the followingwas proved: Theorem 1.4 (Aharoni, Berger, Meshulam [2]) . η ( I ( G )) ≥ Γ( G ) . With a view towards generalizing Theorem 1.4 to an arbitrary simplicialcomplex X , we define a new domination parameter Γ( X ).3or k ∈ N let M X ( k ) be the set of missing faces of X of dimension k . Let J X = { i ∈ N : M X ( i ) = ∅ } be the set of dimensions of simplices in M X .Define S ( X ) = ∪ i ∈ J X (cid:0) Vi − (cid:1) .Let σ ∈ S ( X ) and fix ℓ = ℓ ( σ ) ∈ N . A vector representation of X with respect to σ is an assignment P σ : V → R ℓ such that the inner product P σ ( v ) · P σ ( w ) ≥ vwσ ∈ M X ( | σ | +1), and P σ ( v ) · P σ ( w ) ≥ P σ with the matrix P σ ∈ R | V |× ℓ whose rows arethe vectors P σ ( v ), for v ∈ V . We call the collection P = { P σ : σ ∈ S ( X ) } a vector representation of X .For each σ ∈ S ( X ), let α σ ∈ R V be a non-negative vector. The set { α σ : σ ∈ S ( X ) } is called dominating for P if X σ ∈ S ( X ) α σ P σ P Tσ ≥ (where ∈ R V is the all 1 vector). The value of P is | P | = min X σ ∈ S ( X ) α σ · : { α σ } σ ∈ S ( X ) is dominating for P . Let Γ( X ) be the supremum of | P | over all vector representations P of X . Remarks.
1. If X = I ( G ) for a graph G , then Γ( X ) coincides with the parameter Γ( G )defined in [2].2. In the case when all the missing faces are of the same size, we can boundΓ( X ) by the total domination number ˜ γ ( X ): Proposition 1.5.
Let X be a simplicial complex with all its missing facesof dimension equal to d . Then Γ( X ) ≤ (cid:18) ˜ γ ( X ) d (cid:19) . Our main application of Theorem 1.2 is the following extension of The-orem 1.4.
Theorem 1.6. X r ∈ J X r (cid:18) η ( X ) r (cid:19) ≥ Γ( X ) . Let V , . . . , V m be a partition of the vertex set V . We say that a subset σ ⊂ V is colorful if | σ ∩ V i | = 1 for all i ∈ { , , . . . , m } . Theorem 1.6gives rise to the following Hall-type condition for the existence of colorfulsimplices: 4 heorem 1.7. If for every ∅ 6 = I ⊂ { , , . . . , m } Γ( X [ ·∪ i ∈ I V i ]) > X r ∈ J X [ ·∪ i ∈ IVi ] r (cid:18) | I | − r (cid:19) , then X has a colorful simplex. Next we show an application of Theorem 1.7. Let M be a matroid onvertex set V with rank function ρ . Assume ρ ( V ) = d + 1. We identify M with the simplicial complex of its independent sets. For S ⊂ V , define its closure by cl( S ) = { v ∈ V : ρ ( S ) = ρ ( S ∪ { v } ) } . A subset F ⊂ V is a flat of M if F = cl( F ), i.e. ρ ( F ∪ { v } ) > ρ ( F ) for all v / ∈ F .We say that a subset S ⊂ V is in general position with respect to M iffor any 1 ≤ k ≤ d every flat of M of rank k contains at most k points of S . This is equivalent to requiring that any S ′ ⊂ S with | S ′ | ≤ d + 1 is anindependent set in M .For S ⊂ V denote by ϕ M ( S ) the maximal size of a subset of S in generalposition.Let V , . . . , V m be a partition of V . The following Hall-type theorem isproved in [10]. Theorem 1.8 (Holmsen, Mart´ınez–Sandoval, Montejano [10] ) . If for every ∅ 6 = I ⊂ { , , . . . , m } ϕ M ( ·∪ i ∈ I V i ) > ( | I | − if | I | ≤ d + 1 ,d (cid:0) | I |− d (cid:1) if | I | ≥ d + 2 , then V has a colorful subset in general position. Let S ⊂ V . A weight function f : S → R ≥ is in fractional generalposition with respect to M if for any 1 ≤ k ≤ d and for any flat F of M ofrank k and σ ⊂ F ∩ S of size k − X v ∈ S, cl( vσ )= F f ( v ) ≤ d. Denote by ϕ ∗ M ( S ) the maximum of P v ∈ S f ( v ) over all functions f : S → R ≥ in fractional general position. Let f be the characteristic function of aset S ′ ⊂ S in general position. Let F be a flat of M of rank k for 1 ≤ k ≤ d and σ ⊂ F ∩ S of size k −
1. Then X v ∈ S, cl( vσ )= F f ( v ) = (cid:12)(cid:12) { v ∈ S ′ : cl( vσ ) = F } (cid:12)(cid:12) ≤ | S ′ ∩ F | ≤ k ≤ d, so f is in fractional general position. Therefore ϕ ∗ M ( S ) ≥ ϕ M ( S ) . (1.2)Here we prove the following: 5 heorem 1.9. If for every ∅ 6 = I ⊂ { , , . . . , m } ϕ ∗ M ( ·∪ i ∈ I V i ) > d d X r =1 r (cid:18) | I | − r (cid:19) , then V contains a colorful subset in general position. In particular, we obtain a strengthening of Theorem 1.8:
Theorem 1.10.
If for every ∅ 6 = I ⊂ { , , . . . , m } ϕ M ( ·∪ i ∈ I V i ) > ( | I | − if | I | ≤ d + 1 ,d P dr =1 r (cid:0) | I |− r (cid:1) if | I | ≥ d + 2 , then V contains a colorful subset in general position. The paper is organized as follows. In Section 2 we review some basicfacts concerning simplicial cohomology and high dimensional Laplacians.We also introduce some notation and results about complexes without largemissing faces that we will need later. In Section 3 we prove our main result,Theorem 1.2, and its corollary Theorem 1.3. Section 4 deals with the vectordomination parameter Γ( X ) of the complex X . In it we prove Proposition1.5, Theorem 1.6 and Theorem 1.7. In Section 5 we apply the results ofthe previous section in order to prove Theorems 1.9 and 1.10, which providesufficient conditions for the existence of colorful sets in general position ina matroid. Let X be a finite simplicial complex on the vertex set V . We denote the setof k -dimensional simplices in X by X ( k ). For each σ ∈ X ( k ) we choose anorder of its vertices v , . . . , v k , which induces an orientation on σ .For σ ∈ X ( k ), let lk( X, σ ) = { τ ∈ X : τ ∪ σ ∈ X, τ ∩ σ = ∅ } be the link of σ in X , and deg X ( σ ) = | { η ∈ X ( k + 1) : σ ⊂ η } | be the degree of σ in X .For U ⊂ V , let X [ U ] = { σ ∈ X : σ ⊂ U } be the subcomplex of X inducedby U .For two ordered simplices σ ∈ X , τ ∈ lk( X, σ ), denote by [ σ, τ ], or simplyby στ , their ordered union. Similarly, for v ∈ V denote by vσ the orderedunion of { v } and σ .For σ ∈ X , and τ ⊂ σ , both given an order on their vertices, we define( σ : τ ) to be the sign of the permutation on the vertices of σ which mapsthe ordered simplex σ to the ordered simplex [ σ \ τ, τ ] (where the order onthe vertices of σ \ τ is the one induced by the order on σ ).6 simplicial k -cochain is a real valued skew-symmetric function on allordered k -simplices. That is, φ is a k -cochain if for any two k -simplices σ, ˜ σ in X that are equal as sets, it satisfies φ (˜ σ ) = (˜ σ : σ ) φ ( σ ).For k ≥ C k ( X ) denote the space of k -cochains on X . For k = − C − ( X ) = R .We will use the following lemma implicitly in future calculations. Lemma 2.1.
Let τ, η ∈ X ( k ) and φ ∈ C k ( X ) . Let σ, θ ∈ X be orderedsimplices such that τ, η ⊂ σ and θ ⊂ τ ∩ η , and let ˜ σ , ˜ τ , ˜ η , ˜ θ be equal as setsto σ , τ , η and θ respectively. Then1. ( σ : τ ) = ( σ : ˜ τ ) · (˜ τ : τ ) , and if | σ \ τ | = 1 then ( σ : τ ) = ( σ : ˜ σ ) · (˜ σ : τ ) . φ ( τ ) = φ (˜ τ ) . ( σ : τ ) φ ( τ ) = ( σ : ˜ τ ) φ (˜ τ ) , and if | τ \ θ | = 1 then ( τ : θ ) φ ( τ ) = (˜ τ : θ ) φ (˜ τ ) .
4. If | σ \ τ | = 1 and | σ \ η | = 1 then ( σ : τ )( σ : η ) φ ( τ ) φ ( η ) = (˜ σ : ˜ τ )(˜ σ : ˜ η ) φ (˜ τ ) φ (˜ η ) .
5. If | τ \ θ | = 1 and | η \ θ | = 1 then ( τ : θ )( η : θ ) φ ( τ ) φ ( η ) = (˜ τ : ˜ θ )(˜ η : ˜ θ ) φ (˜ τ ) φ (˜ η ) . Proof.
1. Let π be the permutation on the vertices of σ that maps σ to [ σ \ ˜ τ , ˜ τ ] =[ σ \ τ, ˜ τ ], and let π be the permutation on the vertices of τ that maps˜ τ to τ . Extend π to a permutation ˜ π on the vertices of σ , which maps[ σ \ τ, ˜ τ ] to [ σ \ τ, τ ]. It satisfies sign( π ) = sign(˜ π ). Define π = ˜ π ◦ π . π maps σ to [ σ \ τ, τ ], therefore( σ : τ ) = sign( π ) = sign(˜ π ) · sign( π )= sign( π ) · sign( π ) = (˜ τ : τ ) · ( σ : ˜ τ ) . Assume now that | σ \ τ | = 1 and let { v } = σ \ τ . Let π be the permutationon the vertices of σ that maps σ to ˜ σ , and π be the permutation whichmaps ˜ σ to [˜ σ \ τ, τ ] = vτ = [ σ \ τ, τ ]. Then the permutation π ′ = π ◦ π maps σ to [ σ \ τ, τ ], therefore( σ : τ ) = sign( π ′ ) = sign( π ) · sign( π ) = ( σ : ˜ σ ) · (˜ σ : τ ) .
2. Since φ is a cochain, we have φ ( τ ) = ( τ : ˜ τ ) φ (˜ τ ) = φ (˜ τ ) .
7. By the first part of this lemma( σ : τ ) φ ( τ ) = ( σ : ˜ τ )(˜ τ : τ ) φ ( τ ) , and since φ is a cochain( σ : ˜ τ )(˜ τ : τ ) φ ( τ ) = ( σ : ˜ τ ) φ (˜ τ ) . The second equality is similar: By the first part of the lemma( τ : θ ) φ ( τ ) = ( τ : ˜ τ )(˜ τ : θ ) φ ( τ ) , and since φ is a cochain( τ : ˜ τ )(˜ τ : θ ) φ ( τ ) = ( τ : ˜ τ )(˜ τ : θ )( τ : ˜ τ ) φ (˜ τ ) = (˜ τ : θ ) φ (˜ τ ) .
4. By part 3 of this lemma we have( σ : τ )( σ : η ) φ ( τ ) φ ( η ) = ( σ : ˜ τ )( σ : ˜ η ) φ (˜ τ ) φ (˜ η ) . Then by part 1( σ : ˜ τ )( σ : ˜ η ) φ (˜ τ ) φ (˜ η ) = ( σ : ˜ σ )(˜ σ : ˜ τ )( σ : ˜ σ )(˜ σ : ˜ η ) φ (˜ τ ) φ (˜ η ) =(˜ σ : ˜ τ )(˜ σ : ˜ η ) φ (˜ τ ) φ (˜ η ) .
5. The proof is similar to the proof of part 4.For k ≥ d k : C k ( X ) → C k +1 ( X ) be thelinear operator defined by d k φ ( σ ) = k +1 X i =0 ( − i φ ( σ i ) , where for an ordered ( k +1)-simplex σ = [ v , . . . , v k +1 ], σ i is the ordered sim-plex obtained by removing the vertex v i , that is σ i = [ v , . . . , ˆ v i , . . . , v k +1 ].Equivalently, we can write d k φ ( σ ) = X τ ∈ σ ( k ) ( σ : τ ) φ ( τ ) , where σ ( k ) ⊂ X ( k ) is the set of all k -dimensional faces of σ , each given somefixed order on its vertices.For k = − d − : C − ( X ) = R → C ( X ) by d − a ( v ) = a , forevery a ∈ R , v ∈ V .Let ˜ H k ( X ; R ) = Ker( d k ) / Im( d k − ) be the k -th reduced cohomologygroup of X with real coefficients. 8 .2 Higher Laplacians For each k ≥ − C k ( X ) by h φ, ψ i = X σ ∈ X ( k ) φ ( σ ) ψ ( σ ) . This induces a norm on C k ( X ): k φ k = X σ ∈ X ( k ) φ ( σ ) / . Let d ∗ k : C k +1 ( X ) → C k ( X ) be the adjoint of d k with respect to thisinner product. We can write d ∗ k φ explicitly: d ∗ k φ ( τ ) = X v ∈ lk( X,τ ) φ ( vτ ) . For k ≥ k -Laplacian of X by L − k ( X ) = d k − d ∗ k − andthe upper k -Laplacian of X by L + k ( X ) = d ∗ k d k . The reduced k -Laplacianof X is the positive semidefinite operator on C k ( X ) given by L k ( X ) = L − k ( X ) + L + k ( X ).Let k ≥ σ ∈ X ( k ). We define the k -cochain 1 σ by1 σ ( τ ) = ( ( σ : τ ) if σ = τ (as sets) , { σ } σ ∈ X ( k ) forms a basis of the space C k ( X ), which we will call thestandard basis.For a linear operator T : C k ( X ) → C k ( X ), let [ T ] be the matrix repre-sentation of T with respect to the standard basis. We denote by [ T ] σ,τ thematrix element of [ T ] at index (1 σ , τ ).One can write explicitly the matrix representation of the Laplacian op-erators in the standard basis (see e.g. [6, 9]): Claim 2.2.
For k ≥ (cid:2) L − k (cid:3) σ,τ = k + 1 if σ = τ, ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k, otherwise, (cid:2) L + k (cid:3) σ,τ = deg X ( σ ) if σ = τ, − ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k, σ ∪ τ ∈ X ( k + 1) , otherwise, nd [ L k ] σ,τ = k + 1 + deg X ( σ ) if σ = τ, ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k, σ ∪ τ / ∈ X ( k + 1) , otherwise. The following upper bound on the eigenvalues of the Laplacian is implicitin [6]:
Lemma 2.3.
Let X be a simplicial complex on vertex set V , with | V | = n .Let k ≥ and let λ be an eigenvalue of L k ( X ) . Then λ ≤ n. The following discrete version of Hodge’s theorem had been observed byEckmann in [7].
Theorem 2.4 (Simplicial Hodge theorem) . ˜ H k ( X ; R ) ∼ = Ker L k . As a consequence of Hodge theorem we obtain
Corollary 2.5. ˜ H k ( X ; R ) = 0 if and only if µ k ( X ) > . Let X be a complex on vertex set V with h ( X ) = d . Let k ≥ d and θ ∈ (cid:0) Vk +1 (cid:1) .Define T ( θ ) = (cid:26) τ ∈ (cid:18) θd + 1 (cid:19) : τ / ∈ X ( d ) (cid:27) . So T ( θ ) is the set of all d -dimensional simplices in θ that do not belong to X , and θ ∈ X if and only if T ( θ ) = ∅ . LetMis ( θ ) = \ τ ∈ T ( θ ) τ and m ( θ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) \ τ ∈ T ( θ ) τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Since every τ ∈ T ( θ ) has d + 1 vertices it follows that m ( θ ) ≤ d + 1. Anothersimple observation is the following: Lemma 2.6.
Let σ, τ ∈ X ( k ) such that | τ ∩ σ | = k . Then if σ ∪ τ ∈ X ( k +1) , m ( σ ∪ τ ) = 0 , otherwise ≤ m ( σ ∪ τ ) ≤ d + 1 . roof. Denote σ \ τ = { v } and τ \ σ = { w } . If σ ∪ τ ∈ X ( k + 1) then T ( σ ∪ τ ) = ∅ , therefore m ( σ ∪ τ ) = 0. If σ ∪ τ / ∈ X ( k + 1), then every η ∈ T ( σ ∪ τ ) must contain both v and w (otherwise η will be contained in σ or in τ , a contradiction to η / ∈ X ( d )). Therefore m ( σ ∪ τ ) ≥ Lemma 2.7.
Let X be a clique complex with n vertices and let σ ∈ X ( k ) .Then X τ ∈ σ ( k − deg X ( τ ) − k deg X ( σ ) ≤ n. We will need a version of this lemma for complexes without large missingfaces:
Lemma 2.8.
Let X be a simplicial complex on vertex set V with h ( X ) = d .Let k ≥ d and σ ∈ X ( k ) . Then X τ ∈ σ ( k − deg X ( τ ) = k + 1 + ( k + 1) deg X ( σ )+ d +1 X r =2 ( r − · |{ v ∈ V : m ( vσ ) = r }| . Proof. X τ ∈ σ ( k − deg X ( τ ) = X τ ∈ σ ( k − X v ∈ lk( X,τ ) X v ∈ V X τ ∈ σ ( k − ,τ ∈ lk( X,v ) X v ∈ σ X τ ∈ σ ( k − ,τ ∈ lk( X,v ) X v ∈ lk( X,σ ) X τ ∈ σ ( k − ,τ ∈ lk( X,v ) X v ∈ V \ σ,v / ∈ lk( X,σ ) X τ ∈ σ ( k − ,τ ∈ lk( X,v ) . (2.1)We consider separately the three summands on the right hand side of (2.1):1. For v ∈ σ , there is only one τ ∈ σ ( k −
1) such that τ ∈ lk( X, v ), namely τ = σ \ { v } . Thus the first summand is k + 1.2. For v ∈ lk( X, σ ), any τ ∈ σ ( k −
1) is in lk(
X, v ), therefore the secondsummand is ( k + 1) deg X ( σ ).3. Let v ∈ V \ σ such that v / ∈ lk( X, σ ). Let τ ∈ σ ( k −
1) and let u bethe unique vertex in σ \ τ . If τ ∈ lk( X, v ) then every missing face of X contained in vσ must contain u , so u ∈ Mis ( vσ ). If τ / ∈ lk( X, v ), thenthere is a missing face of X contained in vτ , and therefore it doesn’tcontain the vertex u . Hence, u / ∈ Mis ( vσ ). Since v ∈ Mis ( vσ ), the11umber of τ ∈ σ ( k −
1) such that τ ∈ lk( X, v ) is exactly m ( vσ ) − X v ∈ V \ σ,v / ∈ lk( X,σ ) ( m ( vσ ) −
1) = d +1 X r =2 ( r − | { v ∈ V : m ( vσ ) = r } | . We obtain X τ ∈ σ ( k − deg X ( τ ) = k + 1 + ( k + 1) deg X ( σ )+ d +1 X r =2 ( r − |{ v ∈ V : m ( vσ ) = r }| . In this section we prove Theorems 1.2 and 1.3.Let X be a simplicial complex with h ( X ) = d on vertex set V , where | V | = n , and let k ≥ d . For φ ∈ C k ( X ) and u ∈ V we define φ u ∈ C k − ( X )by φ u ( τ ) = ( φ ( uτ ) if u ∈ lk( X, τ ) , B k : C k ( X ) → C k ( X ) be the linear transformation whose matrix repre-sentation in the standard basis is[ B k ] τ,σ = k deg X ( σ ) − P η ∈ σ ( k − deg X ( η ) if σ = τ, ( m ( σ ∪ τ ) − σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k,σ ∪ τ / ∈ X ( k +1) , R k = ( d − L k − B k , and let λ k be the largest eigenvalue of R k .The proof of Theorem 1.2 depends on the following two ingredients: Proposition 3.1.
Let φ ∈ C k ( X ) . Then ( k − d + 1) h L k φ, φ i = X u ∈ V h L k − φ u , φ u i − h R k φ, φ i . Proposition 3.2. λ k ≤ dn . We postpone the proof of these propositions to the end of this section,and first show how they imply Theorem 1.2.12 roof of Theorem 1.2.
Let 0 = φ ∈ C k ( X ) be an eigenvector of L k witheigenvalue µ k ( X ). By Proposition 3.1 we obtain( k − d + 1) µ k ( X ) k φ k = ( k − d + 1) h L k φ, φ i = X u ∈ V h L k − φ u , φ u i − h R k φ, φ i ≥ µ k − ( X ) X u ∈ V k φ u k − λ k k φ k . But X u ∈ V k φ u k = X u ∈ V X τ ∈ X ( k − φ u ( τ ) = X τ ∈ X ( k − X u ∈ lk( X,τ ) φ ( uτ ) = ( k + 1) X σ ∈ X ( k ) φ ( σ ) = ( k + 1) k φ k . Therefore ( k − d + 1) µ k ( X ) ≥ ( k + 1) µ k − ( X ) − λ k , and by Proposition 3.2( k − d + 1) µ k ( X ) ≥ ( k + 1) µ k − ( X ) − dn. For the proof of Theorem 1.3 we will need the following result, whichwill also be used in Section 4.
Claim 3.3.
For k ≥ d − , µ k ( X ) ≥ (cid:18) k + 1 d (cid:19) µ d − ( X ) − (cid:18)(cid:18) k + 1 d (cid:19) − (cid:19) n. (3.1) If in addition X has complete ( d − -dimensional skeleton, then there isequality in (3.1) for ≤ k ≤ d − .Proof. We argue by induction on k . The case k = d − k ≥ d .By Theorem 1.2 and the induction hypothesis we obtain µ k ( X ) ≥ k + 1 k − d + 1 µ k − ( X ) − dk − d + 1 n ≥ k + 1 k − d + 1 (cid:20)(cid:18) kd (cid:19) µ d − ( X ) − (cid:18)(cid:18) kd (cid:19) − (cid:19) n (cid:21) − dk − d + 1 n = (cid:18) k + 1 d (cid:19) µ d − ( X ) − (cid:18)(cid:18) k + 1 d (cid:19) − (cid:19) n. Now assume that X has complete ( d − k < d −
1. Then we have (cid:0) k +1 d (cid:1) = 0, therefore the inequality in the claim isjust µ k ( X ) ≥ n . But one can see by Claim 2.2 that in this case L k is thescalar matrix with diagonal elements n , thus µ k ( X ) = n .13 roof of Theorem 1.3. Let d − ≤ j ≤ k . We have by Claim 3.3 µ j ( X ) ≥ (cid:18) j + 1 d (cid:19) µ d − ( X ) − (cid:18)(cid:18) j + 1 d (cid:19) − (cid:19) n> (cid:18) j + 1 d (cid:19) · − (cid:18) k + 1 d (cid:19) − ! n − (cid:18)(cid:18) j + 1 d (cid:19) − (cid:19) n ≥ (cid:18) j + 1 d (cid:19) · − (cid:18) j + 1 d (cid:19) − ! n − (cid:18)(cid:18) j + 1 d (cid:19) − (cid:19) n = 0 . Thus, by Corollary 2.5, ˜ H j ( X ; R ) = 0.In order to prove Proposition 3.1 we will need the following claims. Claim 3.4 (see [2, Claim 3.1]) . For φ ∈ C k ( X ) k d k φ k = X σ ∈ X ( k ) deg X ( σ ) φ ( σ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) φ ( vη ) φ ( wη ) . Proof. k d k φ k = X τ ∈ X ( k +1) d k φ ( τ ) = X τ ∈ X ( k +1) X θ ∈ τ ( k ) ( τ : θ ) φ ( θ ) X θ ∈ τ ( k ) ( τ : θ ) φ ( θ ) = X τ ∈ X ( k +1) X σ ∈ τ ( k ) φ ( σ ) + X τ ∈ X ( k +1) X θ ∈ τ ( k ) X θ ∈ τ ( k ) ,θ = θ ( τ : θ )( τ : θ ) φ ( θ ) φ ( θ )= X σ ∈ X ( k ) deg( σ ) φ ( σ ) + X τ ∈ X ( k +1) X θ ∈ τ ( k ) X θ ∈ τ ( k ) ,θ = θ ( τ : θ )( τ : θ ) φ ( θ ) φ ( θ ) . Now look at the map (cid:26) ( η, v, w ) : η ∈ X ( k − ,v,w ∈ V, v = w,vw ∈ lk( X,η ) (cid:27) → n ( τ, θ , θ ) : τ ∈ X ( k +1) ,θ ,θ ∈ τ ( k ) , θ = θ o defined by ( η, v, w ) ( vwη, vη, wη ). For each ( τ, θ , θ ) in the codomain,let η = θ ∩ θ , { v } = θ \ θ and { w } = θ \ θ . ( η, v, w ) is the unique14lement sent to ( τ, θ , θ ). So the map is a bijection, therefore we obtain k d k φ k = X σ ∈ X ( k ) deg X ( σ ) φ ( σ ) + X η ∈ X ( k − X v ∈ V X w ∈ V \{ v } ,vw ∈ lk( X,η ) ( vwη : vη )( vwη : wη ) φ ( vη ) φ ( wη )= X σ ∈ X ( k ) deg X ( σ ) φ ( σ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) φ ( vη ) φ ( wη ) . Claim 3.5.
For φ ∈ C k ( X ) X u ∈ V k d k − φ u k = X σ ∈ X ( k ) X τ ∈ σ ( k − deg X ( τ ) φ ( σ ) − k X τ ∈ X ( k − X vw ∈ lk( X,τ ) φ ( vτ ) φ ( wτ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) u/ ∈ lk( X,vwη ) φ ( vuη ) φ ( wuη ) . Proof.
First we apply Claim 3.4 to φ u ∈ C k − ( X ): k d k − φ u k = X τ ∈ X ( k − deg X ( τ ) φ u ( τ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) φ u ( vη ) φ u ( wη ) . Summing over all vertices we obtain X u ∈ V k d k − φ u k = X u ∈ V X τ ∈ X ( k − deg X ( τ ) φ u ( τ ) − X u ∈ V X η ∈ X ( k − X vw ∈ lk( X,η ) φ u ( vη ) φ u ( wη )= X u ∈ V X τ ∈ X ( k − ∩ lk( X,u ) deg X ( τ ) φ ( uτ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) φ ( vuη ) φ ( wuη )= X σ ∈ X ( k ) X τ ∈ σ ( k − deg X ( τ ) φ ( σ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) φ ( vuη ) φ ( wuη ) . Let η ∈ X ( k − vw ∈ lk( X, η ), and u ∈ lk( X, vη ) ∩ lk( X, wη ). Wesplit into two different cases: u ∈ lk( X, vwη ) or u / ∈ lk( X, vwη ). Assume15 ∈ lk( X, vwη ), and let τ = uη . Then we have vw ∈ lk( X, τ ). This definesa map n ( η, vw, u ) : η ∈ X ( k − ,vw ∈ lk( X,η ) ,u ∈ lk( X,vwη ) o → n ( τ, vw ) : τ ∈ X ( k − ,vw ∈ lk( X,τ ) o . Each pair ( τ, vw ) has a preimage of size k (these are the tuples ( τ \ u, vw, u )for each u ∈ τ ). Therefore we obtain X u ∈ V k d k − φ u k = X σ ∈ X ( k ) X τ ∈ σ ( k − deg X ( τ ) φ ( σ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vwη ) φ ( vuη ) φ ( wuη ) − X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) u/ ∈ lk( X,vwη ) φ ( vuη ) φ ( wuη )= X σ ∈ X ( k ) X τ ∈ σ ( k − deg X ( τ ) φ ( σ ) − k X τ ∈ X ( k − X vw ∈ lk( X,τ ) φ ( vτ ) φ ( wτ ) − X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) u/ ∈ lk( X,vwη ) φ ( vuη ) φ ( wuη ) . Remark. If X is a clique complex and u ∈ lk( X, vη ) ∩ lk( X, wη ) for η ∈ X ( k −
2) and vw ∈ lk( X, η ), then all the 1-dimensional faces of the simplex uvwη belong to X , therefore uvwη ∈ X (i.e. u ∈ lk( X, vwη )). Therefore inthis case the last term of the previous equation vanishes (see [2, Claim 3.2]).
Claim 3.6 (see [2, Claim 3.3]) . For φ ∈ C k ( X ) X u ∈ V (cid:13)(cid:13) d ∗ k − φ u (cid:13)(cid:13) = k (cid:13)(cid:13) d ∗ k − φ (cid:13)(cid:13) . Proof. (cid:13)(cid:13) d ∗ k − φ (cid:13)(cid:13) = X τ ∈ X ( k − d ∗ k − φ ( τ ) = X τ ∈ X ( k − X v ∈ lk( X,τ ) φ ( vτ ) . X u ∈ V (cid:13)(cid:13) d ∗ k − φ u (cid:13)(cid:13) = X u ∈ V X η ∈ X ( k − X v ∈ lk( X,η ) φ u ( vη ) = X η ∈ X ( k − X u ∈ lk( X,η ) X v ∈ lk( X,uη ) φ ( uvη ) = k X τ ∈ X ( k − X v ∈ lk( X,τ ) φ ( vτ ) = k (cid:13)(cid:13) d ∗ k − φ (cid:13)(cid:13) . Let A k : C k ( X ) → C k ( X ) be the linear transformation whose matrixrepresentation in the standard basis is[ A k ] σ,τ = ( ( m ( σ ∪ τ ) − · ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k,σ ∪ τ / ∈ X ( k +1) , Claim 3.7.
For φ ∈ C k ( X ) h A k φ, φ i = 2 X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) ,u/ ∈ lk( X,vwη ) φ ( vuη ) φ ( wuη ) . Proof. h A k φ, φ i = X τ ∈ X ( k ) X σ ∈ X ( k ) , | σ ∩ τ | = k,σ ∪ τ / ∈ X ( k +1) ( m ( σ ∪ τ ) − σ : σ ∩ τ ) · ( τ : σ ∩ τ ) φ ( τ ) φ ( σ )= X θ ∈ X ( k − X v ∈ lk( X,θ ) X w ∈ lk( X,θ ) ,vwθ / ∈ X ( k +1) ( m ( vwθ ) − vθ : θ ) · ( wθ : θ ) φ ( vθ ) φ ( wθ )= X θ ∈ X ( k − X v ∈ lk( X,θ ) X w ∈ lk( X,θ ) ,vwθ / ∈ X ( k +1) ( m ( vwθ ) − φ ( vθ ) φ ( wθ ) . Let < be an order on the vertices of X . Look at the map (cid:26) ( η, u, vw ) : η ∈ X ( k − , vw ∈ lk( X,η ) ,u ∈ lk( X,vη ) ∩ lk( X,wη ) ,u/ ∈ lk( X,vwη ) (cid:27) → (cid:26) ( θ, v, w ) : θ ∈ X ( k − ,v,w ∈ lk( X,θ ) , v 2. So we have h A k φ, φ i = X θ ∈ X ( k − X v ∈ lk( X,θ ) X w ∈ lk( X,θ ) ,vwθ / ∈ X ( k +1) ( m ( vwθ ) − φ ( vθ ) φ ( wθ )= 2 X θ ∈ X ( k − X v ∈ lk( X,θ ) X w ∈ lk( X,θ ) ,v Let φ ∈ C k ( X ). By Claim 3.7 we have h B k φ, φ i = X σ ∈ X ( k ) k deg X ( σ ) − X τ ∈ σ ( k − deg X ( τ ) φ ( σ ) + 2 X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) u/ ∈ lk( X,vwη ) φ ( vuη ) φ ( wuη ) . By Claims 3.4 and 3.5 we obtain k k d k φ k = X u ∈ V k d k − φ u k + X σ ∈ X ( k ) k deg X ( σ ) − X τ ∈ σ ( k − deg X ( τ ) φ ( σ ) + 2 X η ∈ X ( k − X vw ∈ lk( X,η ) X u ∈ lk( X,vη ) ∩ lk( X,wη ) u/ ∈ lk( X,vwη ) φ ( vuη ) φ ( wuη )= X u ∈ V k d k − φ u k + h B k φ, φ i . Then by the previous equation and Claim 3.6, we obtain k h L k φ, φ i = k (cid:10) d ∗ k d k φ + d k − d ∗ k − φ, φ (cid:11) = k k d k φ k + k (cid:13)(cid:13) d ∗ k − φ (cid:13)(cid:13) = X u ∈ V k d k − φ u k + h B k φ, φ i + X u ∈ V (cid:13)(cid:13) d ∗ k − φ u (cid:13)(cid:13) = X u ∈ V h L k − φ u , φ u i + h B k φ, φ i . d − h L k φ, φ i from both sides of the equation we get( k − d + 1) h L k φ, φ i = X u ∈ V h L k − φ u , φ u i − h (( d − L k − B k ) φ, φ i = X u ∈ V h L k − φ u , φ u i − h R k φ, φ i . For the proof of Proposition 3.2 we will need the next result, whichfollows from the definition of B k and Claim 2.2. Claim 3.8. The matrix representation of R k in the standard basis is [ R k ] σ,τ = P η ∈ σ ( k − deg X ( η ) − ( k − d +1) deg X ( σ )+( d − k +1) if σ = τ, ( d + 1 − m ( σ ∪ τ ))( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k,σ ∪ τ / ∈ X ( k +1) , otherwise.Proof of Proposition 3.2. Let K r be the ( k + 1)-dimensional simplicial com-plex on vertex set V , with full k -skeleton, whose ( k + 1)-dimensional facesare the simplices η ∈ (cid:0) Vk +2 (cid:1) such that m ( η ) = r . By Claim 2.2, we have (cid:2) L + k ( K r ) (cid:3) σ,τ = deg K r ( σ ) if σ = τ, − ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k, m ( σ ∪ τ ) = r, |{ v ∈ V : m ( vσ ) = r }| if σ = τ, − ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = k, m ( σ ∪ τ ) = r, M k,r the principal submatrix of (cid:2) L + k ( K r ) (cid:3) obtained by keepingonly the rows and columns corresponding to simplices in X ( k ). M k,r is apositive semidefinite matrix (as a principal submatrix of a positive semidef-inite matrix).Define a new matrix M k = [ R k ] + d X r =2 ( d + 1 − r ) M k,r . For a matrix A , denote by λ max ( A ) the largest eigenvalue of A . Since M k,r is positive semidefinite it follows that λ max ( − M k,r ) ≤ ≤ r ≤ d and therefore λ k = λ max ( R k ) ≤ λ max ( M k )+ d X r =2 ( d +1 − r ) λ max ( − M k,r ) ≤ λ max ( M k ) . (3.3)19y equation (3.2), Lemma 2.6 and Claim 3.8 we see that the matrix M k isdiagonal, and( M k ) σ,σ = X η ∈ σ ( k − deg X ( η ) − ( k − d + 1) deg X ( σ ) + ( d − k + 1)+ d X r =2 ( d + 1 − r ) · |{ v ∈ V : m ( vσ ) = r }| . Let σ ∈ X ( k ). We can writedeg X ( σ ) = | { v ∈ V : v ∈ lk( X, σ ) } | and k + 1 = | { v ∈ V : v ∈ σ } | , and by Lemma 2.8 X η ∈ σ ( k − deg X ( η ) = |{ v ∈ V : v ∈ σ }| + ( k + 1) · |{ v ∈ V : v ∈ lk( X, σ ) }| + d +1 X r =2 ( r − · |{ v ∈ V : m ( vσ ) = r }| . Hence,( M k ) σ,σ = d · |{ v ∈ V : v ∈ σ }| + d · |{ v ∈ V : v ∈ lk( X, σ ) }| + d +1 X r =2 d · |{ v ∈ V : m ( vσ ) = r }| ≤ d · | V | = dn. Therefore λ max ( M k ) ≤ dn , so by inequality (3.3): λ k ≤ dn . In this section we study the vector domination number Γ( X ) of a simplicialcomplex X , leading up to the proof of Theorem 1.6 that provides an upperbound on Γ( X ) in terms of the homological connectivity of X . First weprove Proposition 1.5, relating Γ( X ) to the total domination number ˜ γ ( X ). Proof of Proposition 1.5. Let S be a totally dominating set in X . Let σ ∈ S ( X ) = (cid:0) Vd − (cid:1) . Let f σ be the characteristic vector of S \ σ . Define α σ = d f σ if σ ⊂ S , and α σ = 0 otherwise. Then for every vector representation P of X and every w ∈ V we have X σ ∈ S ( X ) X v ∈ V α σ ( v ) P σ ( v ) · P σ ( w ) = X σ ∈ ( Sd − ) X v ∈ S \ σ d P σ ( v ) · P σ ( w ) . is totally dominating, therefore there is some τ ⊂ S such that τ ∈ X but wτ / ∈ X . Since all the missing faces are of dimension d we must have | τ | ≥ d , and by taking a subset if necessary we may assume | τ | = d . Forevery σ ∈ (cid:0) τd − (cid:1) , let u be the unique vertex in τ \ σ . Then wuσ = wτ is amissing face of X , thus P σ ( u ) · P σ ( w ) ≥ 1. Hence X σ ∈ ( Sd − ) X v ∈ S \ σ d P σ ( v ) · P σ ( w ) ≥ X σ ∈ ( τd − ) 1 d = d · d = 1 . So P σ ∈ S ( X ) α σ P σ P Tσ ≥ 1, therefore { α σ } σ ∈ S ( X ) is dominating for P . So wehave | P | ≤ X σ ∈ S ( X ) X v ∈ V α σ ( v ) = X σ ∈ ( Sd − ) X v ∈ S \ σ d = (cid:18) | S | d − (cid:19) | S | − d + 1 d = (cid:18) | S | d (cid:19) . Therefore Γ( X ) ≤ (cid:0) ˜ γ ( X ) d (cid:1) .Let X be a simplicial complex. For each i ∈ J X , let X i be the complexwhose missing faces are M X ( i ). Note that X i has full ( i − X = ∩ i ∈ J X X i .We want to bound the spectral gaps of X by the spectral gaps of thecomplexes X i . We will need the following lemma: Lemma 4.1. Let A , . . . , A m be simplicial complexes on vertex set V , where | V | = n . Then µ k ( ∩ mi =1 A i ) ≥ m X i =1 µ k ( A i ) − ( m − n. Proof. We argue by induction on m . For m = 1 the statement is trivial.Assume m = 2. For any complex C on vertex set V containing A ∩ A ,denote by ˜ L k ( C ) the principal submatrix of [ L k ( C )] obtained by keepingonly the rows and columns corresponding to simplices of A ∩ A .Let λ min ( ˜ L k ( C )) and λ max ( ˜ L k ( C )) be respectively the minimal and max-imal eigenvalues of ˜ L k ( C ).We have λ min ( ˜ L k ( C )) ≥ µ k ( C ) and λ max ( ˜ L k ( C )) ≤ λ max ( L k ( C )) ≤ n (by Lemma 2.3).It is easy to check by Claim 2.2 that[ L k ( A ∩ A )] = ˜ L k ( A ) + ˜ L k ( A ) − ˜ L k ( A ∪ A ) . Therefore µ k ( A ∩ A ) ≥ λ min ( ˜ L k ( A )) + λ min ( ˜ L k ( A )) − λ max ( ˜ L k ( A ∪ A )) ≥ µ k ( A ) + µ k ( A ) − n. m > m = 2 and the induction hypothesis µ k ( ∩ mi =1 A i ) ≥ µ k ( A ) + µ k ( ∩ mi =2 A i ) − n ≥ µ k ( A ) + m X i =2 µ k ( A i ) − ( m − n ! − n = m X i =1 µ k ( A i ) − ( m − n. For i ∈ J X let Y i be the i -dimensional complex on vertex set V withfull ( i − i -dimensional faces are the sets in M X ( i ). Denote the maximal eigenvalue of L + i − ( Y i ) by λ i max ( X ). Claim 4.2. For all i ∈ J X µ i − ( X i ) = n − λ i max ( X ) . Proof. By Claim 2.2 we have (cid:2) L + i − ( Y i ) (cid:3) σ,τ = deg Y i ( σ ) if σ = τ, − ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = i − ,σ ∪ τ ∈ Y i ( i ) , n − i − deg X i ( σ ) if σ = τ, − ( σ : σ ∩ τ ) · ( τ : σ ∩ τ ) if | σ ∩ τ | = i − ,σ ∪ τ / ∈ X i ( i ) , ( n − [ L i − ( X i )] σ,τ if σ = τ, − [ L i − ( X i )] σ,τ otherwise.Therefore L + i − ( Y i ) = nI − L i − ( X i ). So every eigenvector of L i − ( X i )with eigenvalue λ is an eigenvector of of L + i − ( Y i ) with eigenvalue n − λ . Inparticular, n − µ i − ( X i ) is the largest eigenvalue of L + i − ( Y i ). Claim 4.3. For k ≥ , µ k ( X ) ≥ n − X i ∈ J X (cid:18) k + 1 i (cid:19) λ i max ( X ) . Proof. By Lemma 4.1 we obtain µ k ( X ) = µ k ( ∩ i ∈ J X X i ) ≥ X i ∈ J X µ k ( X i ) − ( | J X | − n. Applying Claim 3.3 to each of the complexes X i (note that h ( X i ) = i and X i has full ( i − µ k ( X ) ≥ X i ∈ J X (cid:20)(cid:18) k + 1 i (cid:19) µ i − ( X i ) − (cid:18)(cid:18) k + 1 i (cid:19) − (cid:19) n (cid:21) − ( | J X | − n. µ k ( X ) ≥ X i ∈ J X (cid:20)(cid:18) k + 1 i (cid:19) ( n − λ i max ( X )) − (cid:18)(cid:18) k + 1 i (cid:19) − (cid:19) n (cid:21) − ( | J X | − n = n − X i ∈ J X (cid:18) k + 1 i (cid:19) λ i max ( X ) . Claim 4.4. X i ∈ J X (cid:18) η ( X ) i (cid:19) λ i max ( X ) ≥ n. Proof. Let k be the integer such that X i ∈ J X (cid:18) k − i (cid:19) λ i max ( X ) < n ≤ X i ∈ J X (cid:18) ki (cid:19) λ i max ( X ) . Let j ≤ k − 2. By Claim 4.3, µ j ( X ) ≥ n − X i ∈ J X (cid:18) j + 1 i (cid:19) λ i max ( X ) > , therefore by Corollary 2.5 we have ˜ H j ( X ; R ) = 0. So η ( X ) ≥ k , thus X i ∈ J X (cid:18) η ( X ) i (cid:19) λ i max ( X ) ≥ X i ∈ J X (cid:18) ki (cid:19) λ i max ( X ) ≥ n. Claim 4.5. Let i ∈ J X . Then for φ ∈ C i − ( Y i ) , (cid:10) L + i − ( Y i ) φ, φ (cid:11) ≤ X σ ∈ ( Vi − ) X vw ∈ lk( Y i ,σ ) ( φ ( vσ ) − φ ( wσ )) . Proof. X σ ∈ Y i ( i − X vw ∈ lk( Y i ,σ ) ( φ ( vσ ) − φ ( wσ )) = X σ ∈ Y i ( i − X v ∈ lk( Y i ,σ ) deg Y i ( vσ ) φ ( vσ ) − X σ ∈ Y i ( i − X vw ∈ lk( Y i ,σ ) φ ( vσ ) φ ( wσ )= i · X η ∈ Y i ( i − deg Y i ( η ) φ ( η ) − X σ ∈ Y i ( i − X vw ∈ lk( Y i ,σ ) φ ( vσ ) φ ( wσ ) . 23y Claim 3.4 we have (cid:10) L + i − ( Y i ) φ, φ (cid:11) = k d i − φ k = X η ∈ Y i ( i − deg Y i ( η ) φ ( η ) − X σ ∈ Y i ( i − X vw ∈ lk( Y i ,σ ) φ ( vσ ) φ ( wσ ) . Hence (cid:10) L + i − ( Y i ) φ, φ (cid:11) = X σ ∈ Y i ( i − X vw ∈ lk( Y i ,σ ) ( φ ( vσ ) − φ ( wσ )) − ( i − · X η ∈ Y i ( i − deg Y i ( η ) φ ( η ) ≤ X σ ∈ Y i ( i − X vw ∈ lk( Y i ,σ ) ( φ ( vσ ) − φ ( wσ )) .Y i has full ( i − Y i ( i − 2) = (cid:0) Vi − (cid:1) . Thus (cid:10) L + i − ( Y i ) φ, φ (cid:11) ≤ X σ ∈ ( Vi − ) X vw ∈ lk( Y i ,σ ) ( φ ( vσ ) − φ ( wσ )) Claim 4.6. Let P be a vector representation of X . Then for all i ∈ J X λ i max ( X ) ≤ i · max σ ∈ ( Vi − ) , v ∈ V P σ ( v ) · X w ∈ V P σ ( w ) ! . Proof. Let φ ∈ C i − ( Y i ). For σ ∈ Y i ( i − 2) = (cid:0) Vi − (cid:1) and v, w ∈ V \ σ , v = w ,we have, by the definition of P , P σ ( v ) · P σ ( w ) ≥ vw ∈ lk( Y i , σ ), and P σ ( v ) · P σ ( w ) ≥ X σ ∈ ( Vi − ) X vw ∈ lk( Y i ,σ ) ( φ ( vσ ) − φ ( wσ )) ≤ X σ ∈ ( Vi − ) X v,w ∈ V \ σ ( φ ( vσ ) − φ ( wσ )) P σ ( v ) · P σ ( w )= X σ ∈ ( Vi − ) X v ∈ V \ σ φ ( vσ ) P σ ( v ) · X w ∈ V \ σ P σ ( w ) − X σ ∈ ( Vi − ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X v ∈ V \ σ φ ( vσ ) P σ ( v ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ X σ ∈ ( Vi − ) X v ∈ V \ σ φ ( vσ ) P σ ( v ) · X w ∈ V \ σ P σ ( w ) ≤ X σ ∈ ( Vi − ) X v ∈ V \ σ φ ( vσ ) · max σ ∈ ( Vi − ) , v ∈ V \ σ P σ ( v ) · X w ∈ V \ σ P σ ( w ) . (4.1)24ince Y i has full ( i − X σ ∈ ( Vi − ) X v ∈ V \ σ φ ( vσ ) = X σ ∈ Y i ( i − X v ∈ lk( Y i ,σ ) φ ( vσ ) = i X η ∈ Y i ( i − φ ( η ) = i k φ k . (4.2)Combining (4.1),(4.2) and Claim 4.5 we obtain (cid:10) L + i − ( Y i ) φ, φ (cid:11) ≤ X σ ∈ ( Vi − ) X vw ∈ lk( Y i ,σ ) ( φ ( vσ ) − φ ( wσ )) ≤ i k φ k · max σ ∈ ( Vi − ) , v ∈ V \ σ P σ ( v ) · X w ∈ V \ σ P σ ( w ) ≤ i k φ k · max σ ∈ ( Vi − ) , v ∈ V P σ ( v ) · X w ∈ V P σ ( w ) . Thus λ i max ( X ) = max = φ ∈ C i − ( Y i ) (cid:10) L + i − φ, φ (cid:11) k φ k ≤ i · max σ ∈ ( Vi − ) ,v ∈ V P σ ( v ) · X w ∈ V P σ ( w ) ! . Lemma 4.7. Let P be a vector representation of X . Then | P | = max { α · : α ≥ , αP σ P Tσ ≤ ∀ σ ∈ S ( X ) } . Proof. Let σ , . . . , σ m be all the sets in S ( X ). For each i ∈ { , , . . . , m } let A i = P σ i P Tσ i ∈ R | V |×| V | . Note that A i = A Ti . Define the matrix A = ( A | A | · · · | A m ) T ∈ R ( m | V | ) ×| V | . Let x ∈ R m | V | . Write x = ( α σ | α σ | · · · | α σ m ), where α σ i ∈ R | V | for each i ∈ { , , . . . , m } . We have xA = m X i =1 α σ i A i = X σ ∈ S ( X ) α σ P σ P Tσ , therefore | P | = min { X α ∈ S ( X ) α σ · : α σ ≥ ∀ σ ∈ S ( X ) , X σ ∈ S ( X ) α σ P σ P Tσ ≥ } = min { x · : x ≥ , xA ≥ } . 25y linear programming duality | P | = max { y · : y ≥ , yA T ≤ } . But yA T = ( yA | yA | · · · | yA m ), so yA T ≤ if and only if yA i ≤ for all i ∈ { , , . . . , m } . Therefore | P | = max { y · : y ≥ , yP σ P Tσ ≤ ∀ σ ∈ S ( X ) } . Let Z + denote the positive integers, and Q + the positive rationals. Let a ∈ Z V + and V a = { ( v, i ) : v ∈ V, ≤ i ≤ a ( v ) } . Define the projection π : V a → V by π (( v, i )) = v , and let X a = π − ( X ) = { σ ⊂ V a : π ( σ ) ∈ X } . The missing faces of X a are the sets σ ⊂ V a such that | π ( σ ) | = | σ | and π ( σ )is a missing face of X . π induces an homotopy equivalence between X a and X (see [11, Lemma2.6]), therefore η ( X a ) = η ( X ). Proof of Theorem 1.6. Let P = { P σ } σ ∈ S ( X ) be a vector representation of X . Let α ∈ Q V + such that αP σ P Tσ ≤ for all σ ∈ S ( X ). Write α = a/k where k ∈ Z + and a ∈ Z V + . Denote N = | V a | = P v ∈ V a ( v ). For σ ∈ S ( X a )and ( v, j ) ∈ V a define Q σ (( v, j )) = ( P π ( σ ) ( v ) if | π ( σ ) | = | σ | , Q = { Q σ : σ ∈ S ( X a ) } is a vector representation of X a : Let σ ∈ S ( X a ) ofsize r − 1, and let ˜ v = ( v, i ) , ˜ u = ( u, j ) ∈ V a such that ˜ u ˜ vσ ∈ M X a ( r ). Then π (˜ u ˜ vσ ) = uvπ ( σ ) ∈ M X ( r ). In particular | π ( σ ) | = | σ | , therefore, since P isa representation of X , Q σ (˜ v ) · Q σ (˜ u ) = P π ( σ ) ( v ) · P π ( σ ) ( u ) ≥ . Let r ∈ J X . By Claim 4.6 λ r max ( X a ) ≤ r · max σ ∈ ( Var − ) , ( v,j ) ∈ V a Q σ (( v, j )) · X ( w,k ) ∈ V a Q σ (( w, k )) = r · max τ ∈ ( Vr − ) ,v ∈ V P τ ( v ) · X w ∈ V a ( w ) P τ ( w ) ! ≤ r · k. 26y Claim 4.4 we obtain X r ∈ J Xa (cid:18) η ( X a ) r (cid:19) r · k ≥ X r ∈ J Xa (cid:18) η ( X a ) r (cid:19) λ r max ( X a ) ≥ N. Therefore α · = 1 k X v ∈ V a ( v ) = Nk ≤ X r ∈ J Xa r (cid:18) η ( X a ) r (cid:19) = X r ∈ J X r (cid:18) η ( X ) r (cid:19) . Thus by Lemma 4.7 | P | = max { α · : α ≥ , αP σ P Tσ ≤ ∀ σ ∈ S ( X ) } = sup { α · : α ∈ Q V + , αP σ P Tσ ≤ ∀ σ ∈ S ( X ) } ≤ X r ∈ J X r (cid:18) η ( X ) r (cid:19) , therefore Γ( X ) ≤ P r ∈ J X r (cid:0) η ( X ) r (cid:1) .For the proof of Theorem 1.7 we need the following Hall-type conditionfor the existence of colorful simplices, which appears in [3, 12], and moreexplicitly in [13]: Proposition 4.8. Let Z be a simplicial complex on vertex set W = ·∪ mi =1 W i .If for all ∅ 6 = I ⊂ { , , . . . m } η ( Z [ ·∪ i ∈ I W i ]) ≥ | I | then Z contains a colorful simplex.Proof of Theorem 1.7. Let ∅ 6 = I ⊂ { , , . . . , m } . By Theorem 1 . X r ∈ J X [ ·∪ i ∈ IVi ] r (cid:18) η ( X [ ·∪ i ∈ I V i ])) r (cid:19) ≥ Γ( X [ ∪ i ∈ I V i ]) > X r ∈ J X [ ·∪ i ∈ IVi ] r (cid:18) | I | − r (cid:19) , therefore η ( X [ ·∪ i ∈ I V i ])) > | I | − . Thus by Proposition 4 . X has a colorful simplex. Let M be a matroid of rank d + 1 on vertex set V . Let ˜ M be the simplicialcomplex on vertex set V whose simplices are the subsets S ⊂ V in generalposition with respect to M . The missing faces of ˜ M are the dependent sets S ⊂ V with | S | ≤ d + 1 such that any | S | − S are independentin M . 27 laim 5.1. For U ⊂ V , ϕ ∗ M ( U ) ≤ d · Γ( ˜ M [ U ]) . Proof. We construct a vector representation of the complex ˜ M [ U ]. Let 1 ≤ r ≤ d and let F r be the set of flats of M of rank r .Let σ ∈ S ( ˜ M [ U ]) with | σ | = r − 1, and let v ∈ U . Define P σ ( v ) ∈ R F r by P σ ( v )( F ) = ( vσ ) = F, v, w ∈ U , if vwσ is a missing face of ˜ M [ U ] of dimension r then vwσ liesin a flat of rank r , which is spanned by any r points in vwσ . In particularcl( vσ ) = cl( wσ ) ∈ F r , therefore P σ ( v ) · P σ ( w ) = 1 . Hence P is a vector representation of ˜ M [ U ].Let f : U → R ≥ be a function in fractional general position with P v ∈ U f ( v ) = ϕ ∗ M ( U ). Define α ∈ R U by α ( v ) = f ( v ) /d .Let w ∈ U , and let F = cl( wσ ). If F / ∈ F r then P σ ( w ) = 0, therefore P v ∈ U α ( v ) P σ ( v ) · P σ ( w ) = 0 ≤ 1. If F ∈ F r then X v ∈ U α ( v ) P σ ( v ) · P σ ( w ) = X v ∈ U, cl( vσ )= F α ( v ) = 1 d X v ∈ U, cl( vσ )= F f ( v ) ≤ . So αP σ P Tσ ≤ for each σ ∈ S ( ˜ M [ U ]), therefore by Lemma 4.7Γ( ˜ M [ U ]) ≥ | P | ≥ α · = ϕ ∗ M ( U ) d . Proof of Theorem 1.9. Let ∅ 6 = I ⊂ { , , . . . , m } . By Claim 5.1Γ( ˜ M [ ·∪ i ∈ I V i ]) ≥ ϕ ∗ M ( ·∪ i ∈ I V i ) d > d X r =1 r (cid:18) | I | − r (cid:19) . Thus by Theorem 1.7 there is a colorful simplex of ˜ M , i.e. a colorful subsetof V in general position. Proof of Theorem 1.10. Let ∅ 6 = I ⊂ { , , . . . , m } . Assume | I | ≤ d + 1.The d -dimensional skeleton of ˜ M [ ·∪ i ∈ I V i ] is M [ ·∪ i ∈ I V i ], therefore for all 0 ≤ k ≤ d − H k (cid:16) ˜ M [ ·∪ i ∈ I V i ]; R (cid:17) = ˜ H k ( M [ ·∪ i ∈ I V i ]; R ) . is a matroid, therefore ˜ H k ( M [ ·∪ i ∈ I V i ]; R ) = 0 for 0 ≤ k ≤ ρ ( ·∪ i ∈ I V i ) − η ( ˜ M [ ·∪ i ∈ I V i ]) ≥ ρ ( ·∪ i ∈ I V i ). But ρ ( ·∪ i ∈ I V i ) = min { d + 1 , ϕ M ( ·∪ i ∈ I V i ) } , so if ϕ M ( ·∪ i ∈ I V i ) > | I | − 1, then η ( ˜ M [ ·∪ i ∈ I V i ]) > | I | − | I | ≥ d + 2. If ϕ M ( ·∪ i ∈ I V i ) > d P dr =1 r (cid:0) | I |− r (cid:1) , then, byinequality (1.2), ϕ ∗ M ( ·∪ i ∈ I V i ) > d P dr =1 r (cid:0) | I |− r (cid:1) , and therefore by Theorem1.6 and Claim 5.1 d X r =1 r (cid:18) η ( ˜ M [ ·∪ i ∈ I V i ]) r (cid:19) ≥ Γ( ˜ M [ ·∪ i ∈ I V i ]) ≥ ϕ ∗ M ( ·∪ i ∈ I V i ) d > d X r =1 r (cid:18) | I | − r (cid:19) , so η ( ˜ M [ ·∪ i ∈ I V i ]) > | I | − 1. Therefore by Proposition 4.8 there is a colorfulsubset of V in general position. 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