aa r X i v : . [ m a t h . F A ] S e p Geometric property (T) for non-discrete spaces
Jeroen Winkel
Abstract
Geometric property (T) was defined by Willett and Yu, first for sequences of graphs and laterfor more general discrete spaces. Increasing sequences of graphs with geometric property (T) areexpanders, and they are examples of coarse spaces for which the maximal coarse Baum-Connesassembly map fails to be surjective. Here, we give a broader definition of bounded geometryfor coarse spaces, which includes non-discrete spaces. We define a generalisation of geometricproperty (T) for this class of spaces and show that it is a coarse invariant. Additionally, wecharacterise it in terms of spectral properties of Laplacians. We investigate geometric property(T) for manifolds and warped systems.
In their paper [7], Willett and Yu introduce geometric property (T) for sequences of graphs, in orderto provide examples of coarse spaces for which the maximal coarse Baum-Connes assembly mapfails to be surjective. The concept is studied further by the same authors in [8]. There, geometricproperty (T) is defined for coarse spaces that are discrete with bounded geometry, monogenic, andcountable. Such a space has geometric property (T) if each unitary representation of the translationalgebra C u [ X ] (we call it C cs [ X ], see Definition 5.3) that has almost invariant vectors, has a non-zero invariant vector. For increasing sequences of finite graphs, this is a strictly stronger propertythan being an expander. In fact, in this case it is equivalent to the graph Laplacian having spectralgap in the maximal completion of C u [ X ] (see [8, Proposition 5.2]). A sequence of finite graphs withgeometric property (T) can never have large girth (see [7, Corollary 7.5]).The aim of this article is to generalise geometric property (T) to spaces satisfying a broadernotion of bounded geometry , see Definition 6.7. This broader notion of bounded geometry is ex-plained in Section 3. Spaces with bounded geometry include sequences of graphs with boundeddegree, manifolds whose Ricci curvature and injectivity radius are uniformly bounded from below,and warped systems coming from an action of a finitely generated group on a compact manifold.We do not assume monogenicity of the spaces. However, we do assume that the spaces are coveredby a countable number of “bounded” sets.In [8], translation operators are used extensively to study geometric property (T) for discretespaces. For non-discrete spaces, these operators do not behave as nicely. Therefore we introducethe concept of “operators in blocks”, which are operators with a convenient decomposition (seeDefinition 5.5).Any coarse space with bounded geometry can be equipped with a measure µ that is uniformlybounded and for which a gordo set exists (see Definition 3.6 and Proposition 3.7). We will firstlydefine geometric property (T) for a coarse space X equipped with such a measure µ . However,Theorem 8.6 will show that it is in fact independent of the chosen measure µ .1e will consider the Hilbert space L ( X, µ ). An operator T in B ( L ( X, µ )) has controlledsupport if the support of a function is not changed much by T (see Definition 5.1). Let C cs [ X ] ⊆ B ( L ( X, µ )) be the algebra of operators with controlled support. A representation of this algebra isa unital ∗ -homomorphism ρ : C cs [ X ] → B ( H ) for some Hilbert space H . For such a representation,the subspace of constant vectors H c is defined (see Definition 6.6). The coarse space X has geometricproperty (T) if every unit vector v ∈ H ⊥ c is “moved enough” by a suitable operator in C cs [ X ] (seeDefinition 6.7).An important notion in coarse geometry is that of coarse invariance (see Section 2). In [8], itwas shown that geometric property (T) is a coarse invariant (see [8, Theorem 4.1]). In this paper weshow that our generalisation of geometric property (T) is still a coarse invariant. This essentiallyensures that we have the “right” definition. Theorem 8.6.
Suppose ( X, µ ) and ( X ′ , µ ′ ) are coarsely equivalent spaces with bounded geometry,equipped with uniformly bounded measures for which gordo sets exist. Then ( X, µ ) has property (T)if and only if ( X ′ , µ ′ ) has property (T). In particular, whether X has property (T) is independentof the chosen uniformly bounded measure µ for which a gordo set exists. Analogous to [8], we establish a complete characterisation of geometric property (T) in terms ofnon-amenability for connected coarse spaces of bounded geometry. In fact, for unbounded connectedspaces, geometric property (T) is the complete converse of amenability for coarse spaces. Thisgeneralizes [8, Corollary 6.1]. The notion of amenability for coarse spaces with bounded geometryis introduced in Lemma 9.1.
Theorem 9.4.
Let X be a connected coarse space and let µ be a uniformly bounded measure forwhich a gordo set exists. If X is bounded, then it is amenable and has geometric property (T). If X is unbounded, it has geometric property (T) if and only if it is not amenable. Towards the end of this article, we investigate geometric property (T) for Riemannian manifolds.Let M be a Riemannian manifold or a countable disjoint union of Riemannian manifolds with thesame dimension. The Riemannian metric defines a coarse structure on M . If the Ricci curvatureand the injectivity radius of M are uniformly bounded from below, then M has bounded geometry(see Section 10). In this case the manifold Laplacian ∆ M is an element of the algebra C ∗ max ( X ),and the manifold M has geometric property (T) if and only if ∆ M has spectral gap in this algebra. Theorem 10.4.
Let ( M, g ) be a Riemannian manifold of bounded geometry equipped with the coarsestructure coming from the geodesic metric. Then M has geometric property (T) if and only if theLaplacian ∆ M has spectral gap in C ∗ max ( M ) . In the last section we discuss warped systems. A group Γ acting by homeomorphisms on aRiemannian manifold M gives rise to a coarse spaces called a warped system , see Section 11. InSection 11 we will show that warped systems have bounded geometry if the manifold M is compact.One might expect that the warped system has geometric property (T) if and only if the group hasproperty (T) and the action is ergodic. This remains an open question, though some partial resultsare given. A coarse space is a space with a large-scale geometric structure. They were first described in [3]and further developed in [4]. Let us first recall the definition from [4, Definition 2.3].2 efinition 2.1.
Let X be a set. A coarse structure on X consists of a collection E of subsets of X × X , whose elements are called the controlled sets , such that:(i) The diagonal { ( x, x ) | x ∈ X } is an element of E .(ii) For every E ∈ E and F ⊆ E , we have F ∈ E .(iii) For every E ∈ E , the inverse E − = { ( x, y ) ∈ X × X | ( y, x ) ∈ E } is an element of E .(iv) For every E, F ∈ E , we have E ∪ F ∈ E .(v) For every E, F ∈ E , the composition E ◦ F = { ( x, z ) ∈ X × X | there is y ∈ X such that ( x, y ) ∈ E and ( y, z ) ∈ F } is an element of E .A set X equipped with a coarse structure is called a coarse space . Note that parts ( iii ) and ( v )give E a groupoid structure. Example 2.2.
Every metric space (
X, d ) can be given a coarse structure: let E = { E ⊆ X × X | sup ( x,y ) ∈ E d ( x, y ) < ∞} . For any
R >
0, we denote E R = { ( x, y ) ∈ X × X | d ( x, y ) ≤ R } . Then the controlled sets areexactly the subsets of X × X that are contained in some E R . Definition 2.3.
Let X and Y be coarse spaces. A map f : X → Y is a coarse equivalence if thereis a map g : Y → X such that for any controlled set E ⊆ X × X the set ( f × f )( E ) is controlled,for any controlled set F ⊆ Y × Y the set ( g × g )( F ) is controlled, the set { ( x, gf ( x )) | x ∈ X } iscontrolled and the set { ( y, f g ( y )) | y ∈ Y } is controlled. The spaces X and Y are called coarselyequivalent if such an f exists.This definition is easily seen to be equivalent to Definition 2.21 in [4].An important example of the above is when Y is coarsely dense in X . Definition 2.4.
Let X be a coarse space and Y ⊆ X . The subset Y is called coarsely dense in X if there is a controlled set E ⊆ X × X such that for all x ∈ X there is y ∈ Y with ( x, y ) ∈ E .If Y is coarsely dense in X , it is easy to show that X is coarsely equivalent to Y with the inducedcoarse structure.In this paper we will use the following properties and notation: Definition 2.5.
Let X be a coarse space and E ⊆ X × X a controlled set.(i) The controlled set E is called symmetric if E − = E .(ii) For any x ∈ X , we write E x = { y ∈ X | ( y, x ) ∈ E } .(iii) For any U ⊆ X , we write E U = S x ∈ U E x .(iv) A subset U ⊆ X is called bounded if U × U is a controlled set.3v) A subset U ⊆ X is called E -bounded if U × U ⊆ E .(vi) We write E ◦ n for the n -fold composition E ◦ E ◦ · · · ◦ E .(vii) The set E generates the coarse structure if for any controlled set F , there is an integer n suchthat F ⊆ E ◦ n .For later use we give some basic properties using these definitions. Lemma 2.6.
Let X be a coarse space.(i) Each controlled set is contained in a symmetric controlled set.(ii) A subset U ⊆ X is bounded if and only if it is E -bounded for some controlled set E .(iii) If E is a controlled set and x ∈ X then E x is E ◦ E − -bounded (though it is not necessarily E -bounded).(iv) If U is an E -bounded set, it is contained in E x for any x ∈ U .Proof. These follow directly from the definitions.Most of the time, we will only consider symmetric controlled sets.
Consider a coarse space X . In [8] it is defined that X is a discrete space of bounded geometry iffor every controlled set E ⊆ X × X , there is an integer N such that E x ≤ N for all x ∈ X . Wewill define geometric property (T) for a wider class of spaces. We still need a concept of boundedgeometry, which we define below. Definition 3.1.
Let X be a coarse space. A controlled set F ⊆ X × X is called covering if it issymmetric and for every controlled E there is an N such that each E x can be covered by at most N sets that are F -bounded. Lemma 3.2.
Let X be a coarse space and F ⊆ X × X a symmetric controlled set. The followingare equivalent:(i) The controlled set F is covering.(ii) For each controlled set E there is a constant N such that each E -bounded set U can be coveredby at most N sets that are F -bounded.Proof. Suppose F is covering and let E be a controlled set. Let N be such that each E x can becovered by at most N sets that are F -bounded. Let U be a non-empty E -bounded set. For any x ∈ U we have U ⊆ E x , so U is also covered by at most N sets that are F -bounded.Conversely, suppose that ( ii ) is true and let E be a controlled set. There is a constant N suchthat each E ◦ E − -bounded set can be covered by at most N sets that are F -bounded. Then each E x can also be covered by at most N sets that are F -bounded. Definition 3.3.
A space X has bounded geometry if there is a controlled covering set F ⊆ X × X .4 xample 3.4. Consider a metric space X with the corresponding coarse structure. If there existsa controlled covering set, we can enlarge it to E R for some R >
0. Therefore, X has boundedgeometry if and only if E R is covering for some R > E R will be covering for any R > Example 3.5.
A discrete space of bounded geometry certainly has bounded geometry in thisdefinition, as we can take F to simply be the diagonal.A different way to characterise bounded geometry is by the existence of a measure on X satisfyingcertain properties. Definition 3.6.
Let µ be a measure on the coarse space ( X, E ).(i) The measure is called E -uniformly bounded or simply uniformly bounded if for each controlledset E there is a constant C > E -bounded measurable set U has measure atmost C .(ii) A symmetric controlled set E is called µ -gordo or simply gordo if it is measurable and µ ( E x )is bounded away from zero independently of x , for all x ∈ X . Proposition 3.7.
Let X be a coarse space. The following are equivalent:(i) The space X has bounded geometry.(ii) There is a uniformly bounded measure µ on X and a gordo set E for µ .(iii) There is a coarsely dense subspace Y ⊆ X that is a discrete space of bounded geometry.Proof. Suppose X has bounded geometry and F ⊆ X × X is a controlled covering set. By Zorn’sLemma there is a maximal subset Y ⊆ X satisfying ( y, y ′ ) F for all y = y ′ ∈ Y . By maximalitywe know that for each x ∈ X there exists y ∈ Y such that ( x, y ) ∈ F , so Y is coarsely dense in X . Toshow that Y is a discrete space of bounded geometry, consider a controlled set E ⊆ X × X . Thereexists N such that every E y can be covered by F -bounded sets U , . . . , U N . Since each F -boundedset can contain at most 1 element of Y , we see that E y ∩ Y ) ≤ N , so Y has bounded geometry,proving ( i ) = ⇒ ( iii ).Suppose Y ⊆ X is a coarsely dense discrete space of bounded geometry and let µ be the countingmeasure of Y . Any controlled set E on X restricts to a controlled set on Y . Since Y is a discretespace of bounded geometry any E -bounded set can contain a bounded number of points of Y . Thisshows that µ is uniformly bounded. Now let E ⊆ X × X be a symmetric controlled set such thatfor all x ∈ X there is y ∈ Y with ( x, y ) ∈ E . Then each E x has measure at least 1, so E is gordo,showing ( iii ) = ⇒ ( ii ).Finally, let µ be a uniformly bounded measure on X and let F be a gordo set for µ . We willshow that F ◦ is a covering set. There is ε > µ ( F x ) ≥ ε for all x . Let E be a symmetriccontrolled set. Since µ is uniformly bounded, there is a constant C such that µ ( V ) ≤ C for all F ◦ E ◦ F -bounded V . Now let U ⊆ X be E -bounded. Then F U is F ◦ E ◦ F -bounded, so µ ( F U ) ≤ C .Let Y ⊆ U be maximal such that the F y are pairwise disjoint when the y range over Y . Since the F y are all contained in F U and have measure at least ε we have Y ≤ N , with N = Cε . Since Y is maximal we know that for all x ∈ U there is a y ∈ Y with F x ∩ F y = ∅ , so ( x, y ) ∈ F ◦ F . Thisimplies that the sets ( F ◦ F ) y , y ∈ Y cover U . These sets are F ◦ -bounded. By Lemma 3.2 we seethat F ◦ is covering, showing ( ii ) = ⇒ ( i ). 5 Blocking collections
From now on, let X be a coarse space of bounded geometry that can be covered by a countablenumber of bounded sets. Let µ be a uniformly bounded measure on X for which a gordo set exists.This measure exists by Proposition 3.7; in applications, there is often already a measure given thatsatisfies the condition above. For example, if X is a metric space, µ satisfies the conditions if thereis a radius R > R -balls are bounded from below in measure, and for all radii R > R -balls are bounded from above in measure.The definition of geometric property (T) uses the measure µ , but we will show later that it doesnot depend on it (see Theorem 8.6).We will start by defining the notion of a blocking collection of subsets of X and proving someelementary results about them. Definition 4.1.
Let E be a controlled set and let ε >
0. A collection ( A i ) of measurable subsetsof X is called ( E, ε ) -blocking if S i A i × A i ⊆ E , the A i are pairwise disjoint and µ ( A i ) ≥ ε for all i .A collection ( A i ) of measurable subsets of X is called blocking if it is ( E, ε )-blocking for some
E, ε .The conditions on the measure µ ensure that there are sufficiently large blocking collections: Lemma 4.2.
There is a blocking collection ( A i ) with union X .Proof. Let E ⊆ X × X be a gordo set for µ . By Zorn’s Lemma, there is a maximal subset I ⊆ X such that all the E i are pairwise disjoint when i ranges over I . Then for all x ∈ X there exists an i ∈ I such that E x ∩ E i = ∅ , so ( x, i ) ∈ E ◦ E . So the ( E ◦ E ) i cover X , while the E i are pairwisedisjoint. Hence, we can find measurable A i with E i ⊆ A i ⊆ ( E ◦ E ) i such that X = F i A i . Toconstruct the A i explicitly, choose a well-ordering on I and define A i = E i ∪ ( E ◦ E ) i \ [ i ′ = i E i ′ ∪ [ i ′
Let E ⊆ X × X . We say that E is a controlled set in blocks if E is of the form F i A i × A i , where the ( A i ) form a blocking collection.Note that F i A i × A i is always a controlled set if the ( A i ) form a blocking collection. Lemma 4.4.
Let E ⊆ X × X be a controlled set. Then E is contained in a finite union of controlledsets in blocks.Proof. Assume without loss of generality that E is symmetric. By Lemma 4.2, there is a blockingcollection ( A i ) with union X . Let F = F i A i × A i which is a controlled set. Define a graph structureon the index set I by connecting i and j if E A i ∩ E A j = ∅ .If i and j are connected by an edge, then ( E ◦ E ) A i ∩ A j = ∅ , and since A j is F -bounded itfollows that A j ⊆ ( F ◦ E ◦ E ) A i . Since µ is uniformly bounded and the A i are F -bounded, we knowthat the measure of ( F ◦ E ◦ E ) A i is bounded from above, say by M , where M does not dependon i . Moreover, the measure of A j is bounded from below uniformly in j , say by ε >
0. Since the6 j are also pairwise disjoint, the degree of any vertex i in the graph can be at most Mε . Then wecan colour the graph in N colours, for some integer N , such that no two adjacent vertices have thesame colour. Now for 1 ≤ k ≤ N define C k = { E A i | i is coloured with the k -th colour } . Then because of the way we coloured the graph, the sets in C k are disjoint. Moreover they are E ◦ F ◦ E -bounded and bounded from below in measure, so C k is a blocking collection for each k .For 1 ≤ k ≤ N we now define E k = G { E A i × E A i | i is coloured with the k -th colour } . These are controlled sets in blocks. For all ( x, y ) ∈ E , there is an i with y ∈ A i . Let k be the colourof i . Then ( x, y ) ∈ E A i × E A i ⊆ E k . So E is contained in F k E k . Analogous to the discrete setting, we now define the algebra of operators with controlled support.We will use the Hilbert space L X = L ( X, µ ). Definition 5.1.
Let E ⊆ X × X be a measurable controlled set and T ∈ B ( L X ). We say that T is supported on E if for all measurable U ⊆ X , all ξ ∈ L X supported on U and all η ∈ L X wehave h ξ, T η i = h ξ, T η | E U i . We say that T has controlled support if it is supported on some measurable controlled set.We can give some equivalent definitions for this: Lemma 5.2.
Let E be a measurable symmetric controlled set and T ∈ B ( L X ) . The following areequivalent:(i) The operator T is supported on E .(ii) For all measurable U ⊆ X , all ξ ∈ L X and all η ∈ L X supported on U we have h ξ, T η i = h ξ | E U , T η i . (iii) The operator T ∗ is supported on E .(iv) For all measurable U ⊆ X and all η ∈ L X supported on U the function T η is supported on E U .Proof. To prove ( i ) = ⇒ ( ii ), we have h ξ, T η i = h ξ | E U , T η i + h ξ | X \ E U , T η i = h ξ | E U , T η i + h ξ | X \ E U , T η | E X \ EU i = h ξ | E U , T η i . The other direction is similar, and the equivalences ( ii ) ⇐⇒ ( iii ) and ( ii ) ⇐⇒ ( iv ) are clear.7learly linear combinations of operators with controlled support again have controlled support,as does a product of operators with controlled support. So the operators with controlled supportform a unital ∗ -algebra. Definition 5.3.
Denote by C cs [ X ] the subalgebra { T ∈ B ( L X ) | T has controlled support } of B ( L X ). We will denote it C cs [ X, µ ] when the measure is not clear from the context.
Remark . In the case that X is a discrete space of bounded geometry, the algebra C cs [ X ] is thesame as the algebra C u [ X ] as defined in [8, Definition 3.1]. Definition 5.5.
Let T ∈ C cs [ X ], let E be a controlled set and let ε >
0. We say that T isan operator in ( E, ε ) -blocks if it is supported on F i A i × A i where ( A i ) is some ( E, ε )-blockingcollection. We say that T is an operator in blocks if it is supported on some controlled set in blocks.For a measurable subset U ⊆ X , denote by π U : L X → L U the restriction map and by i U : L U → L X extension by 0. An operator T that is supported on F i A i × A i can be writtenas T = P i T i , where T i = π A i T i A i : L A i → L A i , using Lemma 5.2. This decomposition makes itvery convenient to do computations with operators in blocks. Moreover, every operator in C cs [ X ]can be written as a finite sum of operators in blocks: Lemma 5.6.
Let T ∈ C cs [ X ] . We can write T as a finite sum T + . . . + T N where each T i ∈ C cs [ X ] is an operator in blocks.Proof. Let T be supported on the measurable and symmetric controlled set E . Let ( A i ) be ablocking collection as in the proof of Lemma 4.4 and consider the same colouring on I as in thatproof. Let T k = X i with colour k π A i T. Then T k is controlled on ⊔ i with colour k E A i × E A i , so it is an operator in blocks. Moreover we have T = T + . . . + T N .We define some spaces of functions on X : Definition 5.7. (i) Let L b X denote the L -functions on X with bounded support.(ii) Let LL X denote the measurable functions on X that are locally L : for each ϕ ∈ LL X andeach bounded measurable U , we have R U | ϕ | < ∞ . Lemma 5.8.
The dual of L b X can be identified with LL X .Proof. For any ξ ∈ L b X and η ∈ LL X , the inner product h η, ξ i is finite, so η defines a function in( L b X ) ∗ .Conversely, let ϕ ∈ ( L b X ) ∗ . For any bounded measurable U , the function ϕ can be pulled backalong the inclusion L U ֒ → L b X to an element of ( L U ) ∗ , which may be viewed as a function η U ∈ L U . If U ⊆ V are bounded and measurable, then we find that η U = η V | U . Therefore, the η U may be glued together to a function η ∈ LL X .If T ∈ B ( L X ) has controlled support, it sends L b X to L b X , and so does the dual T ∗ . Takingthe dual of T ∗ : L b ( X ) → L b ( X ) gives a map LL X → LL X which we denote by T again. Now8efine the linear map Φ : C cs [ X ] → LL X by Φ( T ) = T ( X ) ∈ LL X . Explicitly, if T is supportedon a controlled set E then we have Φ( T )( x ) = ( T E x )( x ) . For all
T, S ∈ C cs [ X ] we have Φ( ST ) = S Φ( T ). Lemma 5.9.
Let T ∈ C cs [ X ] such that Φ( T ) is bounded. Then we can write T = T + . . . + T N where the T k are operators in blocks and Φ( T k ) is bounded.Proof. Construct T k as in the proof of Lemma 5.6. Then the Φ( T k ) are bounded. Definition 6.1.
A representation of C cs [ X ] consists of a Hilbert space H and a unital ∗ -homomorphism ρ : C cs [ X ] → B ( H ). If no confusion arises we will omit the map ρ from the notation.The standard representation of C cs [ X ] is the inclusion C cs [ X ] ֒ → B ( L X ). Note that we do notask the map ρ to be continuous with respect to the norm on C cs [ X ], inherited from B ( L X ). Definition 6.2.
Let
T, S ∈ C cs [ X ].(i) We write T ≥ S if this inequality holds in B ( L X ).(ii) We write T ≥ max S if for every representation ( ρ, H ) of C cs [ X ], we have the inequality ρ ( T ) ≥ ρ ( S ).(iii) Let k T k L be the norm of T in B ( L X ).(iv) Define k T k max = sup (cid:13)(cid:13) ρ ( T ) (cid:13)(cid:13) , where the supremum is over all representations ( ρ, H ). We willsee in Corollary 6.4 that this is always finite, and hence it also defines a norm on C cs [ X ]. Lemma 6.3.
Let T ∈ C cs [ X ] be an operator in blocks.(i) If T ≥ , then T ≥ max .(ii) We have k T k L = k T k max .Proof. (i) Suppose T is supported on F i A i × A i where ( A i ) is a blocking collection. Write T = P i T i with T i : L A i → L A i . Then T = P i T i is again an operator in blocks. Nowfor any representation ( ρ, H ) we have ρ ( T ) = ρ ( T ) ≥
0, so we have T ≥ max k T k L ≤ k T k max . Note that T ∗ T is again an operator in blocks. We have 0 ≤ T ∗ T ≤ k T k L , and by part (i), it follows that 0 ≤ max T ∗ T ≤ max k T k L . So for any representa-tion ( ρ, H ), we have 0 ≤ ρ ( T ∗ T ) ≤ k T k L . So (cid:13)(cid:13) ρ ( T ) (cid:13)(cid:13) = (cid:13)(cid:13) ρ ( T ) ∗ ρ ( T ) (cid:13)(cid:13) = (cid:13)(cid:13) ρ ( T ∗ T ) (cid:13)(cid:13) ≤ k T k L ,and hence k T k L = k T k max .If T is an operator in blocks we will simply write k T k for both the norms k T k L and k T k max . Corollary 6.4.
For any T ∈ C cs [ X ] , the maximal norm k T k max is finite. roof. This follows directly from Lemma 6.3 and Lemma 5.6.Therefore k·k max defines a norm and we can define the maximal C ∗ -algebra: Definition 6.5.
The maximal C ∗ -algebra C ∗ max ( X ) is defined as the completion of C cs [ X ] withrespect to the maximal norm k·k max .For a discrete space X of bounded geometry the C ∗ -algebra C ∗ max ( X ) is the same as the one de-fined in [8] (there it is called C ∗ u, max ( X )). For f ∈ L ∞ X denote by M f ∈ B ( L X ) the multiplicationoperator by f . Similar to [8], we define the subspace of the constant vectors. Definition 6.6.
Let H be a non-degenerate representation of C cs [ X ], and let v ∈ H . We call v a constant vector if T v = M Φ( T ) v for all operators T ∈ C cs [ X ] for which Φ( T ) is bounded. Here Φ isas defined at the end of Section 5. The constant vectors form a closed subspace H c of H .Since we can replace T by T − M Φ( T ) , it is enough to ask that T v = 0 for all T with Φ( T ) = 0.Now we are ready to define geometric property (T). Definition 6.7.
Let X be a coarse space with bounded geometry and let µ be a uniformly boundedmeasure for which a gordo set exists. We say that ( X, µ ) satisfies geometric property (T) if thereare a controlled set E and constants ε, γ > ρ, H ) and every unit vector v ∈ H ⊥ c there is an operator T in ( E, ε )-blocks with Φ( T ) boundedand (cid:13)(cid:13)(cid:13) ( T − M Φ( T ) ) v (cid:13)(cid:13)(cid:13) ≥ γ k T k .In the definition, it is necessary to ask that Φ( T ) is a bounded function, because otherwise M Φ( T ) would not be defined. By considering the operator T − M Φ( T ) , we see that we can equivalently askthere to be an operator T in ( E, ε )-blocks with Φ( T ) = 0 and k T v k ≥ γ k T k .If X is a discrete space of bounded geometry with a generating controlled set, and µ is thecounting measure, we retrieve the definition of [8] (cf. Proposition 3.1 in that paper): Lemma 6.8.
Let X be a discrete space of bounded geometry for which there is a generating con-trolled set, and let µ be the counting measure. The following are equivalent:(i) The space ( X, µ ) satisfies geometric property (T).(ii) There exists a controlled generating set E and a constant γ > such that for any representation H and unit vector v ∈ H ⊥ c , there is an operator T ∈ C cs [ X ] with support in E such that (cid:13)(cid:13)(cid:13) ( T − M Φ( T ) ) v (cid:13)(cid:13)(cid:13) > γ sup x,z | T xz | , where T xz denotes the value at ( x, z ) of the matrix corresponding to the operator T .Proof. We certainly have k T k L ≥ sup x,z | T xz | . With this, the direction ( i ) = ⇒ ( ii ) is trivial, sincewe may enlarge the controlled set E to make it generating.Now suppose ( ii ) is true, take E, γ satisfying the condition, and let ε = 1. Without loss ofgenerality, we may assume E to be symmetric. There is an integer N such that E x ≤ N foreach x ∈ X . Let H be a representation and v ∈ H ⊥ c a unit vector. We know there is T ∈ C cs [ X ]supported on E such that (cid:13)(cid:13) ( T − Φ( T )) v (cid:13)(cid:13) > γ sup x,z | T xz | . For each pair ( x, z ) ∈ E there are fewerthan 4 N other pairs ( x ′ , z ′ ) ∈ E with { x, z } ∩ { x ′ , z ′ } 6 = ∅ . By a graph colouring argument we may10rite T = T + . . . + T N as a sum of operators controlled on E with for each k : if ( x, z ) ∈ E and( x ′ , z ′ ) ∈ E are different pairs where T k is supported, then { x, z } ∩ { x ′ , z ′ } = ∅ . Moreover we getsup x,z | T xz | = sup k k T k k L . We then have (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X k ( T k − M Φ( T k ) ) v (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13) ( T − M Φ( T ) ) v (cid:13)(cid:13)(cid:13) ≥ γ sup x,z | T xz | = γ sup k k T k k L , hence there is a k with (cid:13)(cid:13)(cid:13) ( T k − M Φ( T k ) ) v (cid:13)(cid:13)(cid:13) ≥ γ N k T k k L , proving ( i ). In this section we will give a different characterisation of geometric property (T) using spectralproperties of Laplacians.
Definition 7.1.
Let E be a measurable symmetric controlled set. Define the Laplacian ∆ E ∈ B ( L X ) as ∆ E ( ξ )( x ) = Z E x ( ξ ( x ) − ξ ( y )) dµ ( y ) . The Laplacian corresponding to E is obviously supported on E , in particular we have ∆ E ∈ C cs [ X ]. The Laplacians are positive operators, as is shown by the following lemma. Lemma 7.2.
For any measurable symmetric controlled set E we have ≤ max ∆ E ≤ max M , where M = sup x µ ( E x ) .Proof. A straightforward computation shows that for all measurable symmetric controlled sets F ,we have h ∆ F ξ, ξ i = Z X ∆ F ξ ( x ) ξ ( x ) dµ ( x )= Z X Z F x ξ ( x ) − ξ ( y ) ξ ( x ) dµ ( y ) dµ ( x )= 12 Z X Z F x (cid:16) ξ ( x ) − ξ ( y ) ξ ( x ) + ξ ( y ) − ξ ( x ) ξ ( y ) (cid:17) dµ ( y ) dµ ( x )= 12 Z F | ξ ( x ) − ξ ( y ) | d ( µ × µ )( x, y ) ≥ , so ∆ F ≥
0. Since | ξ ( x ) − ξ ( y ) | ≤ | ξ ( x ) | + 2 | ξ ( y ) | we also have h ∆ F ξ, ξ i ≤ Z F ( | ξ ( x ) | + | ξ ( y ) | ) d ( µ × µ )( x, y ) = 2 Z F | ξ ( x ) | d ( µ × µ )( x, y )by symmetry. This is equal to 2 R F | ξ ( x ) | µ ( F x ) dµ ( x ), which is at most 2 k ξ k sup x µ ( F x ), so wehave (cid:13)(cid:13) ∆ F (cid:13)(cid:13) L ≤ x µ ( F x ). Moreover, if F , F are disjoint measurable controlled sets, we seedirectly that ∆ F ∪ F = ∆ F + ∆ F . 11et E be any measurable symmetric controlled set. By Lemma 4.4, there are controlled sets inblocks E , . . . , E N whose union contains E . Let E ′ , . . . , E ′ N be pairwise disjoint measurable setssuch that E ′ k ⊆ E k for each k and E = F k E ′ k . For each k , the operator ∆ E ′ k is an operator inblocks, so by Lemma 6.3 we have 0 ≤ max ∆ E ′ k ≤ max x µ (( E ′ k ) x ). Taking the sum from k = 1to N , we get 0 ≤ max ∆ E ≤ max x µ ( E x ). Lemma 7.3.
Let E be a controlled set in blocks and let T ∈ C cs [ X ] be a positive operator supportedon E such that Φ( T ) = 0 . Then T ≤ max k T k inf x ( µ ( E x )) ∆ E . Proof.
Write E = F i A i × A i and T = P i T i with T i : L A i → L A i . It is enough to prove theinequality T i ≤ k T i k µ ( A i ) ∆ E i for each i . The space L A i can be written as the direct sum of the constantfunctions and the functions with zero integral. For any ξ ∈ L A i write ξ = ξ c + ξ d where ξ c isconstant and R ξ d = 0. Then T i ξ = T i ξ d and ∆ E i ξ = µ ( A i ) ξ d . So we have h T i ξ, ξ i ≤ k T i k ·k ξ d k = h k T i k µ ( A i ) ∆ E i ξ, ξ i showing that T i ≤ k T i k µ ( A i ) ∆ E i .Since Φ(∆ E ) = 0, we have ∆ E v = 0 for all v ∈ H c , where H is a representation of C cs [ X ].Moreover, we have the following lemma, which shows the converse. Lemma 7.4.
Let ( ρ, H ) be a representation of C cs [ X ] . Then H c = ∩ E ker( ρ (∆ E )) where the intersection ranges over all the measurable controlled sets E .Proof. Let v ∈ ∩ E ker( ρ (∆ E )) and let T ∈ C cs [ X ] such that Φ( T ) is bounded, we will show that T v = M Φ( T ) v , showing that v ∈ H c . By Lemma 5.9, we can assume that T is an operator in blocks.We can also assume that Φ( T ) = 0 by considering the operator T − Φ( T ). By Lemma 7.3, there isa measurable controlled set E and a constant c such that T ∗ T ≤ max c ∆ E , since T ∗ T is a positiveoperator in blocks. Then ∆ E v = 0 and we get k T v k = h T ∗ T v, v i ≤ c h ∆ E v, v i = 0, so T v = 0.
Definition 7.5.
Let T ∈ C ∗ max ( X ) be a positive operator. We say that T has spectral gap if thereis a constant γ > σ max ( T ) ⊆ { } ∪ [ γ, ∞ ), where σ max denotes the spectrum of T inthe maximal C ∗ -algebra.Now we can prove the generalisation of [8, Proposition 5.2]. Proposition 7.6.
Let X be a coarse space of bounded geometry and let µ be a uniformly boundedmeasure for which a gordo set exists. Then ( X, µ ) has geometric property (T) if and only if there ex-ists a measurable controlled set E such that for each representation ( ρ, H ) we have H c = ker( ρ (∆ E )) and ∆ E has spectral gap.Proof. Suppose X has property (T). Let E be a controlled set and ε, γ > H and every unit vector v ∈ H ⊥ c , there is an operator T in ( E, ε )-blockssuch that Φ( T ) = 0 and k T v k ≥ γ k T k . We will show that for each representation ( ρ, H ), we have H c = ker( ρ (∆ E )) and ∆ E has spectral gap. 12et v ∈ H ⊥ c be a unit vector. Let T be an operator in ( E, ε )-blocks with Φ( T ) = 0 and k T v k ≥ γ k T k . Let T be supported on the blocks ( A i ) and let E ′ = F i A i × A i ⊆ E . By Lemma7.3, we have T ∗ T ≤ k T k inf i µ ( A i ) ∆ E ′ ≤ k T k ε ∆ E . Now γ k T k ≤ k T v k = h T ∗ T v, v i ≤ k T k ε h ∆ E v, v i ≤ k T k ε (cid:13)(cid:13)(cid:13) ∆ E v (cid:13)(cid:13)(cid:13) , so (cid:13)(cid:13) ∆ E v (cid:13)(cid:13) ≥ γ ε . This shows that ker( ρ (∆ E )) ∩ H ⊥ c = { } , so ker( ρ (∆ E )) = H c . Moreover, itshows that σ ( ρ (∆ E )) ⊆ { } ∪ [ γ ε, ∞ ). Since this holds for every representation ( ρ, H ), we get σ max (∆ E ) ⊆ { } ∪ [ γ ε, ∞ ), hence ∆ E has spectral gap.Now suppose that E is a measurable controlled set such that for each representation ( ρ, H ) wehave H c = ker( ρ (∆ E )) and ∆ E has spectral gap, say σ max (∆ E ) ⊆ { }∪ [ γ, ∞ ). By Lemma 4.4 thereare controlled sets in blocks E , . . . , E N whose union contains E . Let ε = min k inf x µ (( E k ) x ) > k ranges from 1 to N and x ranges over X . Now let ( ρ, H ) be a representation and let v ∈ H ⊥ c be a unit vector. Then v ∈ ker( ρ (∆ E )) ⊥ so h ∆ E v, v i ≥ γ . We also have ∆ E ≤ ∆ E + . . . + ∆ E N so there is 1 ≤ k ≤ N such that h ∆ E k v, v i ≥ γN . Now E k is an ( E, ε )-operator in blocks and (cid:13)(cid:13) ∆ E k v (cid:13)(cid:13) ≥ γN , so we see that X has geometric property (T).In [8, Proposition 5.2] it is actually shown that H c = ker( ρ (∆ E )) holds for all generatingcontrolled sets E and that if X has geometric property (T), then ∆ E has spectral gap for allgenerating controlled sets E . In our case, we have a similar result, for which we first need anotherlemma and its corollary. Lemma 7.7.
Let E be a gordo set and F a symmetric measurable controlled set satisfying F ◦ E = E ◦ F . Let f ∈ L ∞ ( X ) be the function f ( x ) = µ (( E ◦ F ) x ) . Then there is a constant δ such that δ ∆ F ◦ F ≤ max (2 M f − ∆ E ◦ F ) M f ∆ E ◦ F . Proof.
For every ξ ∈ L X and x ∈ X , we have(2 M f − ∆ E ◦ F ) ξ ( x ) = f ( x ) ξ ( x ) + Z ( E ◦ F ) x ξ ( y ) dµ ( y ) . Then we have(2 M f − ∆ E ◦ F ) M f ∆ E ◦ F ξ ( x )=∆ E ◦ F ξ ( x ) + Z ( E ◦ F ) x f ( y ) ∆ E ◦ F ξ ( y ) dµ ( y )= f ( x ) ξ ( x ) − Z ( E ◦ F ) x ξ ( y ) dµ ( y ) + Z ( E ◦ F ) x ξ ( y ) dµ ( y ) − Z ( E ◦ F ) x f ( y ) Z ( E ◦ F ) y ξ ( z ) dµ ( z ) dµ ( y )= f ( x ) ξ ( x ) − Z ( E ◦ F ) ◦ x α ( x, z ) ξ ( z ) dµ ( z ) , where α ( x, z ) = Z ( E ◦ F ) x ∩ ( E ◦ F ) z f ( y ) dµ ( y ) . z ∈ ( F ◦ F ) x . There is y ∈ X with ( x, y ) ∈ F and ( y, z ) ∈ F . It follows that E y ⊆ ( E ◦ F ) x ∩ ( E ◦ F ) z , so α ( x, z ) ≥ R E y f ( y ′ ) dµ ( y ′ ). Since E is gordo and f is bounded from above, there is aconstant δ such that α ( x, z ) ≥ δ for all z ∈ ( F ◦ F ) x . Now define β ( x, z ) = ( α ( x, z ) − δ if ( x, z ) ∈ F ◦ F,α ( x, z ) otherwise.Then β is symmetric, β ( x, z ) ≥ x, z ) ∈ X × X and(2 M f − ∆ E ◦ F ) M f ∆ E ◦ F = δ ∆ F ◦ F + T, where T ∈ C cs [ X ] is the operator defined by T ξ ( x ) = Z ( E ◦ F ) ◦ x β ( x, y )( ξ ( x ) − ξ ( y )) dµ ( y ) . It remains to show that T ≥ max
0. This is proved similarly to Lemma 7.2. Let E , . . . , E N becontrolled sets in blocks whose union contains ( E ◦ F ) ◦ . Let E ′ , . . . , E ′ N be measurable subsetsof E , . . . , E N respectively, that are disjoint from each other and whose union is ( E ◦ F ) ◦ . For1 ≤ k ≤ N define T k ∈ C cs [ X ] by T k ξ ( x ) = Z ( E ′ k ) x β ( x, y )( ξ ( x ) − ξ ( y )) dµ ( y ) . An easy computation shows that h T k ξ, ξ i = 12 Z E ′ k β ( x, y )( ξ ( x ) − ξ ( y )) d ( µ × µ )( x, y ) , so T k ≥
0. Since T k is an operator in blocks, we get T k ≥ max T = P Nk =1 T k ≥ max Corollary 7.8.
Let E be a gordo set and F a symmetric measurable controlled set satisfying F ◦ E = E ◦ F and E ⊆ F . Then for every representation ( ρ, H ) , we have ker( ρ (∆ F ◦ F )) = ker( ρ (∆ E ◦ F )) .Moreover, ∆ F ◦ F has spectral gap if and only if ∆ E ◦ F has spectral gap.Proof. Since E ⊆ F we have E ◦ F ⊆ F ◦ F , and therefore ∆ E ◦ F ≤ max ∆ F ◦ F . It follows thatker( ρ (∆ F ◦ F )) ⊆ ker( ρ (∆ E ◦ F )) and that if ∆ E ◦ F has spectral gap, then ∆ F ◦ F also has spectral gap.The other direction follows from Lemma 7.7. Proposition 7.9.
Let X be a coarse space of bounded geometry and let µ be a uniformly boundedmeasure. Let E be a gordo set that generates the coarse structure on X . Let E ′ = E ◦ . Then forevery representation ( ρ, H ) of C cs [ X ] , we have ker( ρ (∆ E ′ )) = H c , and X has geometric property(T) if and only if ∆ E ′ has spectral gap.Proof. We apply Corollary 7.8 with F = E ◦ n for n ≥
2. By induction it follows that for all n ≥ ρ (∆ E ◦ n )) = ker( ρ (∆ E ′ )) and that ∆ E ◦ n has spectral gap if and only if ∆ E ′ has spectralgap.If X has geometric property (T), then there is a controlled set G such that ker( ρ (∆ G )) = H c for every representation ( ρ, H ) and ∆ G has spectral gap. Since E is generating, there is some n such that G ⊆ E ◦ n . Then ∆ G ≤ max ∆ E ◦ n . So ker( ρ (∆ E ′ )) = ker( ρ (∆ E ◦ n )) = H c for everyrepresentation ( ρ, H ) and ∆ E ◦ n and ∆ E ′ have spectral gap.14e will now give a third characterisation of geometric property (T) based more directly on theC*-algebra C ∗ max ( X ). Note that ker(Φ) is a left ideal in C cs [ X ]. Let I c ( X ) = C ∗ max ( X ) ker(Φ) bethe left ideal in C ∗ max ( X ) generated by ker(Φ). Proposition 7.10.
Let X be a coarse space of bounded geometry and let µ be a uniformly boundedmeasure for which a gordo set exists. Then ( X, µ ) has geometric property (T) if and only if thereis a projection p ∈ C ∗ max ( X ) generating I c ( X ) .Proof. First suppose that (
X, µ ) has geometric property (T). Let ( ρ, H ) be a faithful unital repre-sentation of C ∗ max ( X ). By Proposition 7.6 there is a controlled set E such that ker( ρ (∆ E )) = H c and ∆ E has spectral gap. By functional calculus there is a projection p ∈ I c ( X ) such that pv = 0for all v ∈ H c and pv = v for all v ∈ H ⊥ c . Now for any t ∈ I c ( X ) we have tv = 0 for all v ∈ H c ,hence t = tp . So p generates I c ( X ).Conversely, suppose p ∈ C ∗ max ( X ) is a projection that generates I c ( X ). Write p = tS with t ∈ C ∗ max ( X ) and S ∈ ker(Φ) ⊆ C cs [ X ]. We can choose T ′ ∈ C cs [ X ] with (cid:13)(cid:13) t − T ′ (cid:13)(cid:13) ≤ k S k . Let T = T ′ S , then T ∈ C cs [ X ], and Φ( T ) = 0, and k T − p k ≤ (cid:13)(cid:13) t − T ′ (cid:13)(cid:13) ·k S k ≤ . Write T = T + . . . + T N as a sum of operators in blocks and choose a controlled set E and a constant ε > E, ε )-blocks. Now let ( ρ, H ) be any representation. For any v ∈ ker( p ) we have sv = 0for all s ∈ I c ( X ), hence v ∈ H c . So ker( p ) = H c . Now let v ∈ H ⊥ c be a unit vector. Then pv = v so k T v k ≥ . Then there is 1 ≤ k ≤ N such that k T k v k ≥ N , showing that ( X, µ ) has geometricproperty (T).
In this section, we will prove that geometric property (T), as defined in Definition 6.7, does notdepend on the measure µ chosen, and moreover, that it is a coarse invariant. Our method is inspiredby the one employed in [8]. Let X be a coarse space with bounded geometry, and let µ be a uniformlybounded measure for which a gordo set exists. We will start by constructing a discrete space Y with bounded geometry that is coarsely dense in X , with an appropriate measure ν . We will show acorrespondence between representations of C cs [ Y ] and some of the representations of C cs [ X ]. Thiswill allow us to see C ∗ max ( Y ) as a subalgebra of C ∗ max ( X ). Then we can apply the criterion forgeometric property (T) given in Proposition 7.10 to conclude that ( X, µ ) has geometric property(T) if and only if (
Y, ν ) has geometric property (T). After this we will show that geometric property(T) does not depend on the measure as long as the space is discrete. Finally, we will conclude that if X is coarse equivalent to another space X ′ , with a uniformly bounded measure µ ′ for which a gordoset exists, then ( X, µ ) has geometric property (T) if and only if ( X ′ , µ ′ ) has geometric property(T).By Lemma 4.2, there is a blocking collection with union X . Write X = F y ∈ Y U y , where ( U y ) isa blocking collection and y is a point in U y (the notation U y is not to be confused with the notation E x introduced in Definition 2.5(ii)). Note that Y is coarsely dense in X and that µ ( U y ) is boundedfrom above and below, uniformly in y . Let ν be the measure on Y defined by ν ( { y } ) = µ ( U y ). Wewill first show that ( X, µ ) has property (T) if and only if (
Y, ν ) has property (T).
Lemma 8.1.
Let X be a coarse space with bounded geometry and let µ be a uniformly boundedmeasure on X for which a gordo set exists. Let Y be a coarsely dense discrete space of boundedgeometry with a measure ν . Suppose X = F y U y is a disjoint union of measurable sets with y ∈ U y nd ν ( y ) = µ ( U y ) , such that F y U y × U y is a controlled set. Then ( X, µ ) has property (T) if andonly if ( Y, ν ) has property (T). We need some preparation before we can prove Lemma 8.1. For x ∈ X , let y ( x ) ∈ Y be theunique point such that x ∈ U y . Define the linear maps a : L X → L ( Y, ν ) and b : L ( Y, ν ) → L X by aξ ( y ) = ν ( { y } ) R U y ξ and bη ( x ) = η ( y ( x )). Note that these are adjoint to each other and that ab is the identity on L ( Y, ν ). Let A = ba . This is a projection in C cs [ X ]. The maps α : A C cs [ X ] A → C cs [ Y ] , α ( T ) = aT b and β : C cs [ Y ] → A C cs [ X ] A, β ( S ) = bSa are ∗ -isomorphisms and inverseto each other. Now if H X is a representation of C cs [ X ], then H Y = A H X is a representation of A C cs [ x ] A and via β also a representation of C cs [ Y ]. Conversely, every representation of C cs [ Y ] givesrise to a representation of C cs [ X ]. Lemma 8.2.
Let ( ρ Y , H Y ) be a representation of C cs [ Y ] . Then there is a representation ( ρ X , H X ) such that A H X = H Y . Here H X is a completion of a quotient of the algebraic tensor product C cs [ X ] ⊙ H Y that we will denote by C cs [ X ] ⊗ H Y .Proof. First consider the algebraic tensor product C cs [ X ] ⊙ H Y . We equip this with the sesquilinearform defined on simple tensors by h T ⊙ v, T ′ ⊙ v ′ i = h α ( A ( T ′ ) ∗ T A ) v, v ′ i and extended to ( C cs [ X ] ⊙ H Y ) by sesquilinearity. It is easy to see that this is well defined anddefines a conjugate symmetric sesquilinear form. We will now show that it is positive semi-definite.For any operator T ∈ C cs [ X ] there is an integer M , such that for every y ∈ Y there are at most M different y ′ ∈ Y such that π U y T i U y ′ = 0 or π U y ′ T i U y = 0. By a graph colouring argument wemay write T = T + . . . + T N , such that for every 1 ≤ k ≤ N , and every y ∈ Y , there is at mostone y ′ such that π U y T k i U y ′ = 0 or π U y ′ T k i U y = 0.It follows that each element of C cs [ X ] ⊙H Y is of the form P Nk =1 T k ⊙ v k with T k ∈ C cs [ X ] , v k ∈ H Y and the T k satisfy that for every y there is at most one y ′ such that π U y T i U y ′ = 0 or π U y ′ T i U y = 0.Let this y ′ be f k ( y ) if it exists, otherwise put f k ( y ) = y . Then each f k is an involutive function(i.e. f k ◦ f k = id Y ) and T k sends L U y to L U f k ( y ) . Define Q = T . . . T N . . . . . . ∈ B ( L X ) ⊗ M N +1 ( C ) = B ( L ( X × { , , . . . , N } )) . For y ∈ Y define V y = U y ×{ }⊔ (cid:16)F Nk =1 U f k ( y ) × { k } (cid:17) ⊆ X ×{ , , . . . , N } . Then X ×{ , , . . . , N } = F y V y and Q sends each L V y to L V y . Let A ⊗ I = A . . . . . . A ∈ B ( L ( X × { , . . . , N } )) . Then also ( A ⊗ I ) Q ∗ Q ( A ⊗ I ) sends each L V y to L V y . It is also a positive element so we candefine the square root R = (cid:0) ( A ⊗ I ) Q ∗ Q ( A ⊗ I ) (cid:1) ∈ ( A ⊗ I ) B ( L ( X × { , , . . . , N } ))( A ⊗ I ). It16ends each L V y to L V y . Denote the entries of R by R kl ∈ AB ( L X ) A . Then R kl sends L U y to L U f k ( f l ( y )) . So R kl has controlled support, and we have R kl ∈ A C cs [ X ] A . We have R =( A ⊗ I ) Q ∗ Q ( A ⊗ I ) and looking at the entry at ( k, l ) gives P Ns =1 R ks R sl = AT ∗ k T l A and R ∗ kl = R lk for all 1 ≤ k, l ≤ N .Now we can compute *X k T k ⊙ v k , X k T k ⊙ v k + = X k,l h α ( AT ∗ k T l A ) v l , v k i = X k,l,s h α ( R ks R sl ) v l , v k i = X k,l,s h α ( R sl ) v l , α ( R sk ) v k i = X s *X k α ( R sk ) v k , X k α ( R sk ) v k + ≥ . So the sesquilinear form is semi-positive definite. Define H X to be the Hilbert space obtainedby dividing C cs [ X ] ⊙ H Y by the kernel of the semi-norm k v k = h v, v i and completing with respectto the resulting norm. Now we will define the representation ρ X : C cs [ X ] → B ( H X ).Let T ∈ C cs [ X ]. This acts on C cs [ X ] ⊙ H Y by T · ( S ⊙ v ) = T S ⊙ v and extending by linearity. Toshow that this defines a bounded map on H X we need to show that there is a constant C dependingon T such that k T w k ≤ C · k w k for all w ∈ C cs [ X ] ⊙ H Y . For this we can assume that T is anoperator in blocks, by Lemma 5.6. Then S = ( k T k − T ∗ T ) is an element in C cs [ X ] and we havefor all w ∈ C cs [ X ] ⊙ H Y : k T k ·k w k −k T w k = h ( k T k − T ∗ T ) w, w i = h Sw, Sw i ≥ . So k T w k ≤ k T k ·k w k , showing that the action of T on C cs [ X ] ⊙ H Y defines a bounded linear map ρ X ( T ) ∈ B ( H X ). It is now easy to check that this defines a representation of C cs [ X ] on H X .Finally, note that we have an injection H Y ֒ → C cs [ X ] ⊙ H Y sending v to 1 ⊙ v . This respects thesesquilinear form, to it induces an injection H Y ֒ → H X . For a simple tensor T ⊙ v ∈ C cs [ X ] ⊙ H Y it is easy to check that A ( T ⊙ v ) = AT ⊙ v = 1 ⊙ α ( AT A ) v , hence A ( C cs [ X ] ⊙ H Y ) = H Y and A H X = H Y . Lemma 8.3.
The map β : C cs [ Y ] ∼ −→ A C cs [ X ] A extends to an isometry β : C ∗ max ( Y ) ∼ −→ AC ∗ max ( X ) A .Proof. Let S ∈ C cs [ Y ] and let ( ρ Y , H Y ) be a representation such that k S k max = (cid:13)(cid:13) ρ ( S ) (cid:13)(cid:13) . Now H X = C cs [ X ] ⊗ H Y is a representation of C cs [ X ] by Lemma 8.2. Moreover, the restriction of ρ X ( β ( S )) to H Y = A H X equals ρ Y ( S ), so (cid:13)(cid:13) β ( S ) (cid:13)(cid:13) max ≥ (cid:13)(cid:13) ρ X ( β ( S )) (cid:13)(cid:13) ≥ (cid:13)(cid:13) ρ Y ( S ) (cid:13)(cid:13) max = k S k max .Conversely, let ( ρ X , H X ) be a representation of C cs [ X ] such that (cid:13)(cid:13) β ( S ) (cid:13)(cid:13) max = (cid:13)(cid:13) ρ X ( β ( S )) (cid:13)(cid:13) . Wehave H X = A H X + (1 − A ) H X where H Y = A H X is a representation of C cs [ Y ]. The restriction of ρ X ( β ( S )) to H Y equals ρ Y ( S ) and the restriction of ρ X ( β ( S )) to (1 − A ) H X equals zero, because β ( S ) ∈ A C cs [ X ] A . So k S k max ≥ (cid:13)(cid:13) ρ Y ( S ) (cid:13)(cid:13) = (cid:13)(cid:13) ρ X ( β ( S )) (cid:13)(cid:13) = (cid:13)(cid:13) β ( S ) (cid:13)(cid:13) max .Hence β : C cs [ Y ] → A C cs [ X ] A is an isometry, and it extends to an isometric embedding β : C ∗ max ( Y ) ֒ → C ∗ max ( X ). Its image is the closure of A C cs [ X ] A , this is AC ∗ max ( X ) A .17 emma 8.4. We have C ∗ max ( X ) A ker(Φ X ) A = I c ( X ) A and β ( I c ( Y )) = AI c ( X ) A .Proof. Recall that I c ( X ) = C ∗ max ( X ) ker(Φ X ), so clearly C ∗ max A ker(Φ X ) A ⊆ I c ( X ) A . To show theconverse, it is enough to show that ker(Φ X ) A ⊆ C cs [ X ] A ker(Φ X ) A . Let T ∈ ker(Φ X ) A . For every y, y ′ ∈ Y the map π U y T i U y ′ : L U y ′ → L U y is the same when composed with i U y ′ Ai U y ′ : L U y ′ → L U y ′ , so it has rank at most 1. There is an integer N , such that for every y ∈ Y there are atmost N different y ′ ∈ Y such that π U y T i U y ′ = 0, because T has controlled support. Thereforeim( T ) ∩ L U y has rank at most N . Write im( T ) ∩ L U y = f y C ⊕ · · · ⊕ f Ny C where the f ky arepairwise perpendicular elements of L X (some of them may be zero). For all 1 ≤ k ≤ N and y ∈ Y let B ky : L U y → L U y be a rank-one isometry sending f ky C to U y C (or let B ky = 0 if f ky = 0).Then B ∗ ky AB ky = B ∗ ky B ky is the projection on f ky C . Then P Nk =1 B ∗ ky AB ky is the projection onim( T ) ∩ L U y . Now define B k = P y B ky . This is a bounded operator with controlled support sowe have B k ∈ C cs [ X ]. Moreover T = P y P Nk =1 B ∗ ky AB ky = P Nk =1 B ∗ k AB k T ∈ C cs [ X ] A ker(Φ X ) A .For the second claim note that for any S ∈ C cs [ Y ] we have Φ Y ( S ) = 0 if and only if Φ X ( β ( S )) =0. Thus β ( I c ( Y )) = β ( C ∗ max ( Y ) ker(Φ Y )) = AC ∗ max ( X ) A ker(Φ X ) A = AI c ( X ) A .Now we can prove Lemma 8.1. Proof of Lemma 8.1.
We will use the characterisation of geometric property (T) given in Proposi-tion 7.10.Suppose that (
X, µ ) has geometric property (T). Let p ∈ C ∗ max ( X ) be a projection generatingthe left ideal I c ( X ). Since 1 − A ∈ I c ( X ), it is a multiple of p , and from this it follows that ApA is again a projection. Then we see that
ApA ∈ AI c ( X ) A and for all t ∈ AI c ( X ) A we have tApA = tA = t , so ApA generates AI c ( X ) A as a left AC ∗ max ( X ) A -module. So, by Lemma 8.4 weknow that I c ( Y ) is generated by a projection, hence ( Y, ν ) has geometric property (T).Conversely, suppose that (
Y, ν ) has geometric property (T). Let q ∈ C ∗ max ( Y ) be a projectiongenerating the left ideal I c ( Y ). By Lemma 8.4 we have β ( q ) ∈ AI c ( X ) A . In particular, β ( q )(1 − A ) =0, and from this it follows that β ( q ) + 1 − A is again a projection. We have C ∗ max ( X )( β ( q ) + 1 − A )= C ∗ max ( X ) β ( q ) + C ∗ max ( X )(1 − A )= C ∗ max ( X ) AI c ( X ) A + C ∗ max ( X )(1 − A )= I c ( X ) A + C ∗ max ( X )(1 − A )= I c ( X )by Lemma 8.4, so ( X, µ ) has geometric property (T).Now we show that geometric property (T) is independent of the measure for discrete spaces ofbounded geometry.
Lemma 8.5.
Let Y be a discrete space of bounded geometry and let ν be a measure on Y such thatall points are measurable and have measure uniformly bounded from above and below. Denote thecounting measure on Y by κ . Then ( Y, ν ) has property (T) if and only if ( Y, κ ) has property (T).Proof. Define N ∈ L ∞ ( Y ) by N ( y ) = ν ( y ) . We will consider this as a multiplication operator in C cs [ Y ]. Consider the linear map a : L ( Y, ν ) → L ( Y, κ ) given by a ( η ) = N η . This is a unitary map18nd it induces an isomorphism C cs [ Y, κ ] → C cs [ Y, ν ] sending T to a ∗ T a . For any y ∈ Y and largeenough V ⊆ Y , we haveΦ ν ( a ∗ T a )( y ) = a ∗ T a ( V )( y ) = N − T N ( V )( y ) = Φ κ ( N − T N )( y ) , so Φ ν ( a ∗ T a ) = Φ κ ( N − T N ).We now identify the algebras C cs [ Y, κ ] ∼ = C cs [ Y, ν ] ∼ = C cs [ Y ], just remembering that Φ ν ( T ) =Φ κ ( N − T N ). We get C ∗ max ( Y, ν ) = C ∗ max ( Y, κ ). Moreover we have ker(Φ ν ) = N ker(Φ κ ) N − andtherefore I c ( Y, ν ) =
N I c ( Y, κ ) N − . We see that I c ( Y, ν ) is generated by a projection if and only if I c ( Y, κ ) is generated by a projection. By Proposition 7.10 we are done.Finally we prove that property (T) does not depend on the chosen measure and is also a coarseinvariant.
Theorem 8.6.
Suppose ( X, µ ) and ( X ′ , µ ′ ) are coarsely equivalent spaces with bounded geometry,equipped with uniformly bounded measures for which gordo sets exist. Then ( X, µ ) has property (T)if and only if ( X ′ , µ ′ ) has property (T). In particular, whether X has property (T) is independentof the chosen uniformly bounded measure µ for which a gordo set exists.Proof. Construct Y and Y ′ as before. Then Y, X, X ′ , Y ′ are all coarsely equivalent. Since Y and Y ′ are coarsely equivalent, by a well-known structural result (see e.g. [8, Lemma 4.1]) there are coarsedense Z ⊆ Y and Z ′ ⊆ Y ′ and a bijective coarse equivalence f : Z → Z ′ . Taking the countingmeasure on Z and Z ′ , it is easy to see that f induces an isomorphism between the algebras C cs [ Z ]and C cs [ Z ′ ] commuting with Φ Z and Φ Z ′ , so that Z has property (T) if and only if Z ′ has property(T).We can write Y as a bounded disjoint union S z ∈ Z V z such that z ∈ V z . Define the measure λ on Z by λ ( z ) = ν ( V z ). Similarly define a measure λ ′ on Z ′ . Now by Lemmas 8.1 and 8.5 the followingare equivalent:– The space ( X, µ ) has property (T).– The space (
Y, ν ) has property (T).– The space (
Z, λ ) has property (T).– The space Z has property (T) with the counting measure.– The space Z ′ has property (T) with the counting measure– The space ( Z ′ , λ ′ ) has property (T).– The space ( Y ′ , ν ′ ) has property (T).– The space ( X ′ , µ ′ ) has property (T). 19 Amenability
In this section, we will consider which connected coarse spaces X have geometric property (T).Recall that a coarse space X is connected if { ( x, y ) } is a controlled set for all x, y ∈ X . As in [8], itturns out that an unbounded connected coarse space X has geometric property (T) precisely if itdoes not satisfy a version of amenability. First, we give two equivalent definitions of amenability forconnected coarse spaces. It is a generalisation of amenability for discrete metric spaces of boundedgeometry, as defined e.g. in [8]. Proposition 9.1.
Let X be a connected coarse space and µ a uniformly bounded measure for whicha gordo set exists. The following are equivalent:(i) There is a positive unital linear map ϕ : L ∞ X → C such that if ( A i ) is a blocking collectionand f ∈ L ∞ X satisfies R A i f = 0 for all i , and f is zero outside of the union F i A i , then ϕ ( f ) = 0 .(ii) For each controlled measurable set E ⊆ X × X and each ε > there is a non-empty boundedmeasurable U ⊆ X such that µ ( E U ) ≤ (1 + ε ) µ ( U ) .In this case, we say that X is amenable .Proof. First assume ( ii ). Define the directed setΛ = { ( E, ε ) | E ⊆ X × X gordo , ε > } with ( E, ε ) ≥ ( E ′ , ε ′ ) if E ⊇ E ′ and ε ≤ ε ′ . For each λ = ( E, ε ) ∈ Λ, choose a measurable set U λ ⊆ X such that µ ( E U ) ≤ (1 + ε ) µ ( U ). Define ϕ λ ∈ ( L ∞ X ) ∗ by ϕ λ ( f ) = µ ( U λ ) R U λ f . Then the ϕ λ form a net in the unit ball of ( L ∞ X ) ∗ . By the Banach-Alaoglu theorem there is a limit point ϕ .We will now show that ϕ satisfies the condition. So let ( A i ) be a blocking collection and f ∈ L ∞ X such that R A i f = 0 for all i and f is zero outside F i A i . Let E be the gordo set ⊔ i A i × A i ∪ { ( x, x ) | x ∈ X } . Let λ = ( E ′ , ε ) with E ′ ⊇ E . Since E U λ is a union of some of the A i and points outside of F i A i , we have R E Uλ f = 0. Then ϕ λ ( f ) = µ ( U λ ) R U λ f = − µ ( U λ ) R E Uλ \ U λ f and | ϕ λ ( f ) | ≤ µ ( E Uλ \ U λ ) µ ( U λ ) ·k f k ∞ ≤ ε k f k ∞ . This shows that ϕ λ ( f ) → λ → ∞ , hence ϕ ( f ) = 0.Conversely, assume ( i ). Let ϕ : L ∞ X → C be a positive unital linear map satisfying the con-dition. Let E be a gordo set and assume for contradiction that there is ε > U ⊆ X we have µ ( E U ) > (1 + ε ) µ ( U ). Let ( A i ) be a blocking collection withunion X , such that each A i has measure at least ε ′ >
0. We may assume without loss of generalitythat E = { A i × A j | ( A i × A j ) ∩ E = ∅} . Because of bounded geometry, we can use a colouringargument to find involutions σ , . . . , σ N on the index set I such that E A i = A σ ( i ) ∪ · · · ∪ A σ N ( i ) forall i ∈ I . For 1 ≤ k ≤ N let E k = F i A i × A σ k ( i ) . Note that ∆ E k can be viewed as an operator on L and also as an operator on L ∞ X . For any f ∈ L ∞ X we have R A i ∪ A σk ( i ) ∆ E k f = 0 for all i , soby the condition on ϕ we have ϕ (∆ E k f ) = 0. Equivalently, (∆ E k ) ∗ ϕ = 0.Let P ( X ) ⊆ L X denote the positive measurable functions on X with integral 1. We can view L X as a subset of ( L ∞ X ) ∗ using the standard pairing between L X and L ∞ X . By the Goldstinetheorem, ϕ is in the weak closure of P ( X ). Note that (∆ E k ) ∗ is a weakly continuous function on( L ∞ X ) ∗ and on L X it is the same as ∆ E k . Therefore, 0 is in the weak closure of the convex set { (∆ E k ψ ) | ψ ∈ P ( X ) } ⊆ N M k =1 L ( X ) . ψ ∈ P ( X ) such that all ∆ E k ψ are small in the L -norm. In particular,we can choose ψ ∈ P ( X ) such that P i M i ≤ η , where M i = sup k R A i | ∆ E k ψ | and we will choosethe constant η > δ = ε <
1. For n ∈ Z define U n = F { A i | R A i ψ ≥ δ n µ ( A i ) } . Since R X ψ = 1,the U n have bounded measure, they consist of finitely many of the A i and they are bounded.Moreover, U n = ∅ for n small enough. By assumption, we have µ ( E U n ) ≥ (1 + ε ) µ ( U n ) for each n . For each i , define n i as the minimal number such that A i ⊆ E U ni and N i as the maximalnumber such that A i U N i (these may be ∞ ). Then R A i ψ < δ N i µ ( A i ), and there is k such that R A σk ( i ) ψ ≥ δ n i µ ( A σ k ( i ) ). Now M i ≥ Z A i | ∆ E k ψ | = Z A i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) µ ( A σ k ( i ) ) ψ − Z A σk ( i ) ψ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ ε ′ ( δ n i − δ N i ) . Define D n = E U n \ U n +1 = G { A i | n i ≤ n ≤ N i − } . Then X n δ n µ ( D n ) = X i µ ( A i ) N i − X n = n i δ n = 11 − δ X i µ ( A i )( δ n i − δ N i ) ≤ − δ ) ε ′ X i M i ≤ η (1 − δ ) ε ′ . For all n we have (1 + ε ) µ ( U n ) ≤ µ ( E U n ) ≤ µ ( U n +1 ) + µ ( D n ) . Multiplying this inequality by δ n on both sides and adding it for all n we get(1 + ε ) X n δ n µ ( U n ) ≤ δ X n δ n µ ( U n ) + X n δ n µ ( D n ) , so (1 + ε − δ ) X n δ n µ ( U n ) ≤ η (1 − δ ) ε ′ and X n δ n µ ( U n ) ≤ η (1 − δ ) εε ′ . On the other hand, we have1 = Z ψ = X n Z U n +1 \ U n ψ ≤ X n δ n ( µ ( U n +1 ) − µ ( U n )) = ( 1 δ − X n δ n µ ( U n ) . Combining these inequalities gives 1 ≤ ηδεε ′ . Picking η = δεε ′ gives the desired contradiction.The following proposition is a generalisation of [8, Proposition 6.1]. It gives several characteri-sations of amenability of a coarse space X in terms of its algebra. Proposition 9.2.
Let X be a connected coarse space and µ a uniformly bounded measure for whicha gordo set exists. The following are equivalent: i) The space X is amenable.(ii) There is a net ξ λ of unit vectors in L X satisfying (cid:13)(cid:13)(cid:13) T ξ λ − M Φ( T ) ξ λ (cid:13)(cid:13)(cid:13) → for all T ∈ C cs [ X ] for which Φ( T ) is bounded.(iii) For all T ∈ C cs [ X ] with Φ( T ) = 0 , we have ∈ σ ( T ) .(iv) For all gordo sets E , we have ∈ σ max (∆ E ) .(v) There is a unital representation H of C cs [ X ] that has a non-zero constant vector.(vi) There is a positive unital linear map ϕ : L ∞ X → C satisfying ϕ ( T f ) = ϕ (Φ( T ∗ ) · f ) for all T ∈ C cs [ X ] for which ϕ ( T ∗ ) is bounded.Proof. First, assume that X is amenable. We will use criterion ( ii ) from Proposition 9.1. As in theproof of Proposition 9.1, define the directed setΛ = { ( E, ε ) | E ⊆ X × X gordo , ε > } with ( E, ε ) ≥ ( E ′ , ε ′ ) if E ⊇ E ′ and ε ≤ ε ′ . For each λ = ( E, ε ) ∈ Λ, choose a bounded measurablesubset U λ ⊆ X such that µ (( E ◦ E ) U λ ) ≤ (1 + ε ) µ ( U λ ). Define ξ λ = µ ( U λ ) − U λ . This is a unitvector in L X .Let T ∈ C cs [ X ] such that Φ( T ) is bounded. We will show that (cid:13)(cid:13)(cid:13) T ξ λ − M Φ( T ) ξ λ (cid:13)(cid:13)(cid:13) →
0. Wecan assume without loss of generality that Φ( T ) = 0. Let T be supported on E , let E ′ ⊇ E and ε >
0. Then for λ = ( E ′ , ε ), the function ξ λ is constant on E X \ U λ , so T ξ λ is supported on E U λ . Since T ( E ◦ E ) Uλ is zero on E U λ , we know that T ξ λ is the restriction to E U λ of − T η λ , where η λ = µ ( U λ ) − ( E ◦ E ) Uλ \ U λ . Now k T ξ λ k ≤ k T η λ k ≤ k T k ·k η λ k = k T k µ ( U λ ) − µ (( E ◦ E ) U λ \ U λ ) ≤ k T k · ε , so indeed T ξ λ → ii ) = ⇒ ( iii ) = ⇒ ( iv ) are trivial.Suppose ( iv ) is true. Let Λ be the set { ( E, ε ) | E ⊆ X × X measurable controlled , ε > } . We equip this with the partial order (
E, ε ) ≥ ( E ′ , ε ′ ) if E ⊇ E ′ and ε ≤ ε ′ . Then Λ is a directedset. Let U be an ultrafilter on P (Λ) containing { λ ′ ≥ λ } for each λ ∈ Λ. By assumption, forevery λ = ( E, ε ) ∈ Λ there is a representation H λ of C cs [ X ] and a unit vector v λ ∈ H λ satisfying (cid:13)(cid:13) ∆ E v λ (cid:13)(cid:13) ≤ ε . Note that for λ ′ ≥ λ we also have (cid:13)(cid:13) ∆ E v λ ′ (cid:13)(cid:13) ≤ ε . Now let H = Q λ ∈ Λ H λ / U be theultraproduct of the H λ and let v = lim U v λ ∈ H . Then v is a unit vector and for each measurablecontrolled set E we have ∆ E v = lim U ∆ E v λ = 0. By Lemma 7.4, v is a constant vector, showing( iv ) = ⇒ ( v ).For ( v ) = ⇒ ( vi ), let v ∈ H be a constant vector of norm 1 and define ϕ : L ∞ X → C by ϕ ( f ) = h M f v, v i . Then ϕ is a positive unital linear map and for all f ∈ L ∞ X and T ∈ C cs [ X ] suchthat Φ( T ∗ ) is bounded, we have ϕ ( T f ) = h M Φ( T M f ) v, v i = h T M f v, v i = h M f v, M Φ( T ∗ ) v i = h M Φ( T ∗ ) · f v, v i = ϕ (Φ( T ∗ ) · f ) . vi ) = ⇒ ( i ), let ϕ : L ∞ X → C be a positive unital linear map satisfying ϕ ( T f ) = ϕ (Φ( T ∗ ) · f ) for all T ∈ C cs [ X ] such that Φ( T ∗ ) is bounded. Let ( A i ) be a blocking collection and E = F i A i × A i . Let f ∈ L ∞ X be a function such that R A i f = 0 for all i , and f is zero outside of F i A i . Then ∆ E f = f and ϕ (∆ E ) = 0, so ϕ ( f ) = ϕ (∆ E f ) = 0. Therefore, X is amenable (usingpart ( i ) of Proposition 9.1). Remark . If E ′ is as in proposition 7.9, the properties in Theorem 9.2 are also equivalent to0 ∈ σ max (∆ E ′ ).Now we can give the generalisation of [8, Corollary 6.1]. Theorem 9.4.
Let X be a connected coarse space and let µ be a uniformly bounded measure forwhich a gordo set exists. If X is bounded, then it is amenable and has geometric property (T). If X is unbounded, it has geometric property (T) if and only if it is not amenable.Proof. If X is bounded, we have already shown that it has geometric property (T), and it isamenable because X ∈ L X is a constant vector in the standard representation.Suppose that X is unbounded and not amenable. Then there is a gordo set E such that0 σ max (∆ E ). Then for all representations ( ρ, H ) of C cs [ X ] we have ker( ρ (∆ E )) = H c = ∅ and ∆ E has spectral gap. By proposition 7.6 the space X has geometric property (T).Suppose that X is unbounded, has geometric property (T) and is amenable. By Proposition 7.6there is a measurable controlled set E such that for all representations ( ρ, H ) of C cs [ X ], we have H c = ker( ρ (∆ E )), and ∆ E has spectral gap. Since X is amenable, we have 0 ∈ σ (∆ E ), but then 0is an isolated point in the spectrum of ∆ E ∈ B ( L X ). So there is a non-zero ξ ∈ ker(∆ E ), which isthen a constant vector. Since ξ ∈ L X and X is unbounded, ξ can not be constant as a function.So there are measurable sets A, B ⊆ X of positive measure such that the convex hulls of ξ ( A ) and ξ ( B ) are disjoint. If we intersect A with all of the members of a blocking collection, one of theintersections must have positive measure, since blocking collections are countable. Therefore, wecan assume that A is bounded, and similarly, that B is bounded. Since X is connected, A ∪ B isalso bounded. Let T ∈ C cs [ X ] be the operator defined by T η ( x ) = ( µ ( B ) R B η − µ ( A ) R A η if x ∈ A T ) = 0 but T ξ = 0, hence ξ is not a constant vector and we have a contradiction. Corollary 9.5.
Amenability for connected coarse spaces of bounded geometry does not depend onthe measure chosen and is a coarse invariant.Proof.
This follows from Theorem 8.6 and Theorem 9.4. It is also straightforward to show it directly,using criterion ( ii ) of Proposition 9.1.
10 Manifolds
Let (
M, g ) be an n -dimensional complete Riemannian manifold, not necessarily connected. Considerthe geodesic distance d and volume µ . The metric d defines a coarse structure on M , where E ⊆ M × M is controlled if sup ( x,y ) ∈ E d ( x, y ) < ∞ . For any R > E R be the controlled set23 ( x, y ) ∈ M × M | d ( x, y ) ≤ R } . Let B ( x, R ) = ( E R ) x denote the ball of radius R around a point x ∈ M . In [2], it is defined that M has bounded geometry if the Ricci curvature and the injectivityradius are uniformly bounded from below. In this section, we assume that this is the case.By Bishop’s inequality (see [1, Theorem 3.101.i]) there is a constant B such that µ ( B ( x, R )) ≤ BR n for every x ∈ M and R >
0. Hence, the measure µ is uniformly bounded. Moreover, byGunther’s inequality (see [1, Theorem 3.101.ii]), there is a constant A such that µ ( B ( x, R )) ≥ AR n for all x ∈ M and small enough R >
0. In particular, the controlled set E R is gordo. So themanifold M has bounded geometry as a coarse space.Let ∆ M be the (positive) Laplacian operator on M and let ∆ H = 1 − exp( − ∆ M ). Since thespectrum of ∆ M is contained in [0 , ∞ ], the spectrum of ∆ H is contained in [0 , H is a bounded operator. We can write∆ H f ( x ) = f ( x ) − Z M p ( x, y ) f ( y ) dµ ( y )where p ( x, y ) is the heat kernel on M . We need some estimates on this function. Lemma 10.1. (i) There are constants c, R > such that p ( x, y ) ≥ c whenever d ( x, y ) ≤ R .(ii) The manifold is stochastically complete : for every x ∈ M we have R M p ( x, y ) dµ ( y ) = 1 .(iii) For every ε > there is an R such that R B ( x,R ) p ( x, y ) dµ ( y ) ≥ − ε for all x ∈ M .Proof. (i) This follows from [2, Equation 7.43].(ii) This follows from [2, Theorem 3.4] and the bound on µ ( B ( x, R )) given above.(iii) By [2, Equation 5.19] there are some constants C, D > p ( x, y ) ≤ C exp − d ( x, y ) D ! for all x, y ∈ M . Now for any x ∈ M and R > Z M \ B ( x,R ) p ( x, y ) dµ ( y ) ≤ C Z M \ B ( x,R ) exp − d ( x, y ) D ! dµ ( y ) ≤ C Z ∞ R rD exp − r D ! V r ( x ) dr ≤ BCD Z ∞ R r n +1 exp − r D ! dr. As this last integral converges, we know that for every ε >
R > R M \ B ( x,R ) p ( x, y ) dµ ( y ) ≤ ε . Then by part (ii) we get R B x,R p ( x, y ) dµ ( y ) ≥ − ε for all x ∈ M . Lemma 10.2.
We have ∆ H ∈ C ∗ max ( M ) . roof. For any
R > HR ∈ C cs [ M ] be the operator defined by∆ HR f ( x ) = Z B ( x,R ) p ( x, y )( f ( x ) − f ( y )) dµ ( y ) . Then for R ′ > R we have(∆ HR ′ − ∆ HR ) f ( x ) = Z B ( x,R ′ ) \ B ( x,R ) p ( x, y )( f ( x ) − f ( y )) dµ ( y ) . For large enough R we have R M \ B ( x,R ) p ( x, y ) dµ ( y ) ≤ ε . Then we have h (∆ HR ′ − ∆ HR ) ξ, ξ i = 12 Z { ( x,y ) ∈ M | R
We have the following estimates on ∆ H :(i) There are constants c, R > such that ∆ H ≥ max c ∆ R .(ii) For every ε > there are constants d, R > such that ∆ H ≤ max ε + d ∆ R .Proof. (i) Take the same constants c, R as in Lemma 10.1 part (i). Then(∆ H − c ∆ R ) f ( x ) = Z M α ( x, y )( f ( x ) − f ( y )) dµ ( y ) , where α ( x, y ) = p ( x, y ) if d ( x, y ) > R and α ( x, y ) = p ( x, y ) − c if d ( x, y ) ≤ R . Since α ≥ H ≥ max c ∆ R .(ii) Let ∆ HR be as in the proof of Lemma 10.2. There is a large enough R such that ∆ H ≤ max ε + ∆ HR . The function p ( x, y ) is bounded from above by some constant d , and then we get∆ HR ≤ max d ∆ R .For any representation H of C ∗ max ( M ), we can consider the unbounded symmetric operator∆ M = − log(1 − ∆ H ) on H . We say that ∆ M has spectral gap if there is a constant γ > { } ∪ [ γ, ∞ ) for each representation H . Theorem 10.4.
Let ( M, g ) be a Riemannian manifold of bounded geometry equipped with the coarsestructure coming from the geodesic metric. Then M has geometric property (T) if and only if theLaplacian ∆ M has spectral gap in C ∗ max ( M ) . roof. For any
R >
0, the controlled set E R generates the coarse structure on M and is gordo.So we can apply Proposition 7.9 to see that M has geometric property (T) if and only if ∆ R hasspectral gap. Clearly, ∆ H has spectral gap if and only if ∆ M has spectral gap. For the remainderof the proof, we use the estimates of Lemma 10.3.Suppose that M has geometric property (T). By Lemma 10.1 part (i), there are constants c, R > H ≥ c ∆ R . Now ∆ R has spectral gap, and so does ∆ H , and also ∆ M .Suppose that ∆ H has spectral gap. There is ε > σ max (∆ H ) ⊆ { } ∪ [2 ε, ∞ ). ByLemma 10.1 part (ii) there are b, R > H ≤ ε + b ∆ R . Then ∆ R has spectral gap, so M has geometric property (T).
11 Warped systems
Let Γ be a finitely generated group with finite generating set S . Let ( M, g ) be a compact Riemannianmanifold and let α : G y M be an action by diffeomorphisms. Let d be the distance function on M .Let C M = M ×{ , , . . . } be the discrete cone. This is equipped with the Riemannian metric g C thatis t · g on M t = M ×{ t } . Let d C be the corresponding distance function, given by d C ( x, y ) = t · d ( x, y )when x, y ∈ M t = M ×{ t } and d C ( x, y ) = ∞ if x, y are in different M t . Now define the warped system Warp(Γ y M ) as the space C M equipped with the largest metric δ Γ for which δ Γ ( x, y ) ≤ d C ( x, y )and δ Γ ( x, s · x ) ≤ s ∈ Γ and x, y ∈ C M . We will consider Warp(Γ y M ) as a coarse space.In [6], it is shown that that X = Warp(Γ y M ) does not depend on the generating set of Γ asa coarse space. It is also coarsely equivalent to the subset M × { , , . . . } of the warped cone asintroduced by Roe in [5] (see [6, Lemma 6.5]).Let µ be the measure on X defined through the Riemannian metric g C . Since the manifold M iscompact, the Ricci curvature and injectivity radius are uniformly bounded from below. It followsas in the previous section that µ is uniformly bounded and E R = { ( x, y ) ∈ X × X | d C ( x, y ) ≤ R } is gordo for any R >
0. In particular, X has bounded geometry.Let ∆ Γ = P s ∈ S − s ∈ C [Γ]. This can also be viewed as an element in C cs [ X ]. In [6], it isshown that a coarsely dense subset of X with discrete bounded geometry, is a family of expanders ifand only if there is a constant γ > h ∆ Γ ξ, ξ i ≥ γ k ξ k for all ξ ∈ L M with R M ξdµ = 0.We want to consider when these graphs have geometric property (T), or equivalently by Theorem8.6, when X has geometric property (T). Definition 11.1.
Let Γ be a group and (
M, µ ) a measure space. Let α : Γ y M be a measurableaction. The action is called ergodic if the only measurable Γ-invariant subsets of M are the emptyset and M itself.If Γ has property (T) and the action on M is ergodic, we might expect the warped system X to have geometric property (T). At the moment, this is an open question. Question 11.2.
Suppose Γ has property (T) and the action α : Γ y M is ergodic. Does X havegeometric property (T)? Since Γ has property (T), we know that ∆ Γ has spectral gap in C ∗ max (Γ). The map C [Γ] → C cs [ X ]extends to C ∗ max (Γ) → C ∗ max ( X ). So ∆ Γ also has spectral gap in C ∗ max ( X ). If ker( ρ (∆)) = H c forany representation ( ρ, H ) of C cs [ X ], then we can conclude that X has property (T) by Proposition7.6. We can prove the following partial result. 26 emma 11.3. Let H be a representation, and v ∈ H such that ∆ Γ v = 0 . For any f ∈ L ∞ X suchthat R X t f dµ = 0 for all t ∈ N , we have h M f v, v i = 0 .Proof. The composition Γ α −→ C cs [ X ] ρ −→ B ( H ) gives an action of Γ on H , still denoted by α . For any s ∈ S , we have (cid:13)(cid:13) (1 − α s ) v (cid:13)(cid:13) = h (1 − α ∗ s )(1 − α s ) v, v i = h (2 − α s − α ∗ s ) v, v i . Since 2 − α s − α ∗ s ≤ ∆ Γ it follows that (1 − α s ) v = 0. So v is Γ-invariant.Now define ϕ ∈ ( L ∞ X ) ∗ by ϕ ( f ) = h M f v, v i . Note that ϕ is a mean on L ∞ X . Moreover, ϕ isΓ-invariant: for any g ∈ Γ , f ∈ L ∞ X we have ϕ ( α g f ) = h M α g f v, v i = h α g M f α ∗ g v i = h M f v, v i = ϕ ( f ) . Let P ( X ) = { ψ ∈ L X | ψ ≥ , R X ψdµ = 1 } . By the Goldstine theorem, ϕ is in the weak closureof P ( X ). Let f ∈ L ∞ X such that R X t f dµ = 0 for all t ∈ N . Let ε >
0. Then 0 is in the weakclosure of the convex set ((1 − α s ) ψ ) ∈ X s ∈ S L X | ψ ∈ P ( X ) , | ψ ( f ) − ϕ ( f ) | < ε . Since norm-closed convex sets are also weakly closed, 0 is also in the norm closure of the set above.Hence there is ψ ∈ P ( X ) with | ψ ( f ) − ϕ ( f ) | < ε and (cid:13)(cid:13) (1 − α s ) ψ (cid:13)(cid:13) < ε for all s ∈ S . Now let ξ = ψ ∈ L X . This is a unit vector satisfying (cid:13)(cid:13) (1 − α s ) ξ (cid:13)(cid:13) = Z X ( ξ ( s · x ) − ξ ( x )) dµ ( x ) ≤ Z X | ξ ( s · x ) − ξ ( x ) | = (cid:13)(cid:13) (1 − α s ) ψ (cid:13)(cid:13) < ε for all s ∈ S . Then k ∆ Γ ξ k ≤ ε | S | . Let ξ c ∈ L X be the locally constant function defined by ξ c ( x ) = µ ( X t ) R X t ξ for x ∈ X t . There is γ > σ (∆ Γ ) ⊆ { } ∪ [ γ, ∞ ). Now ξ − ξ c is perpendicular to the locally constant functions, and the locally constant functions are the onlyΓ-invariant functions in L X , since the action of Γ is ergodic. From (cid:13)(cid:13) ∆ Γ ( ξ − ξ c ) (cid:13)(cid:13) ≤ ε | S | it nowfollows that (cid:13)(cid:13) ( ξ − ξ c ) (cid:13)(cid:13) ≤ γ − ε | S | . Now | ϕ ( f ) | ≤ | ψ ( f ) | + ε = (cid:12)(cid:12)(cid:12)(cid:12)Z X ξ ( x ) f ( x ) dµ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) + ε = (cid:12)(cid:12)(cid:12)(cid:12)Z X ( ξ ( x ) − ξ c ( x )) f ( x ) dµ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) + 2 (cid:12)(cid:12)(cid:12)(cid:12)Z X ( ξ ( x ) − ξ c ( x )) ξ c ( x ) f ( x ) dµ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) + ε ≤ γ − ε | S | ·k f k ∞ + γ − ε | S | ·k f k ∞ + ε. Since this holds for any ε >
0, we conclude that ϕ ( f ) = 0. Remark . In the above, we only used that σ (∆ Γ ) ⊆ { }∪ [ γ, ∞ ) instead of the stronger condition σ max (∆ Γ ) ⊆ { } ∪ [ γ, ∞ ). Remark . From the lemma above it is not hard to show that if v ∈ ker( ρ (∆ Γ )) ∩ H ⊥ c , then infact v ∈ ( ρ ( C ∗ max ( X )) H c ) ⊥ .The converse is also an open question. 27 uestion 11.6. Suppose the action α is free and X has geometric property (T). Is it necessarythat Γ has property (T) and that the action is ergodic? We have some partial results on this question. Before we state them, we give some preliminaryresults.
Lemma 11.7.
For
R > and g ∈ Γ define the controlled sets E R = { ( x, y ) ∈ X | δ Γ ( x, y ) ≤ R } , E g = { ( g · x, x ) | x ∈ X } and F R = { ( x, y ) ∈ X | d C ( x, y ) ≤ R } . For any R > there is R ′ > such that E R ⊆ S g ∈ S ⌊ R ⌋ F R ′ ◦ E g .Proof. For any s ∈ S the map M → M, x → s · x is Lipschitz. So there is a constant L > d C ( s · x, s · y ) ≤ L · d C ( x, y ). Let R ′ = L R R . For any ( y, x ) ∈ E R , there are x = x, x , . . . , x n ∈ X and y , y , . . . , y n = y ∈ X and s , s , . . . , s n ∈ S such that s k y k − = x k for k = 1 , , . . . , n and n + d C ( x , y ) + . . . + d C ( x n , y n ) ≤ R. It follows by induction that d C ( s k · · · s · s x , y k ) ≤ L k d C ( x , y ) + . . . + Ld C ( x k − , y k − ) + d C ( x k , y k ) . In particular d C ( g · x, y ) ≤ L n d C ( x , y ) + . . . + Ld C ( x n − , y n − ) + d C ( x n , y n ) ≤ L R R = R ′ for g = s n · · · s · s . Now ( g · x, x ) ∈ E g and ( y, g · x ) ∈ F R ′ so ( y, x ) ∈ F R ′ ◦ E g . Since g ∈ S ⌊ R ⌋ this shows that E R ⊆ [ g ∈ S ⌊ R ⌋ F R ′ ◦ E g . Note that I = L t ∈ N B ( L X t ) is a two-sided ideal in C cs [ X ]. Lemma 11.8.
Suppose that the action α is free. Then C cs [ X ] /I is naturally isomorphic to thealgebraic cross product C cs [ C M ] /I ⋊ Γ .Proof. The natural map C cs [ C M ] ⋊ Γ → C cs [ X ] descends to C cs [ C M ] /I ⋊ Γ → C cs [ X ] /I .To show that it is surjective, let T ∈ C cs [ X ]. Then T is supported on E R for some integer R > R ′ > E R ⊆ S g ∈ S R F R ′ E g . We can use this decompositionto write T as a finite sum T = P g A g α g , with A g ∈ C cs [ C M ] showing that the map is indeedsurjective.For injectivity, suppose that P g ∈ B A g α g ∈ I , with B ⊂ Γ finite and A g ∈ C cs [ C M ]. Then wewill show that A g ∈ I for all g ∈ B . We may as well assume that P g ∈ B A g α g = 0. Since the actionis free and by homeomorphisms, for any g ∈ Γ we have min x ∈ M d ( g · x, x ) >
0. There is some
R > A g are supported on F R . Now let T = R min g = h ∈ B min x ∈ M d ( gh − · x,x ) . For any g ∈ B we know that A g α g is supported on F R E g . We also have A g α g = − P h ∈ B \{ g } A h α h , so A g α g issupported on F R E g ∩ S h ∈ B \{ g } F R E h . Suppose that ( x, y ) ∈ F R E g ∩ F R E h . Then d C ( x, g · y ) ≤ R and d C ( x, h · y ) ≤ R , so d C ( g · y, h · y ) ≤ R . For some t we have x, y ∈ X t and then d C ( g · y, h · y ) ≥ t · min z ∈ M d ( gh − · z, z ) ≥ RtT . So t ≤ T . Hence we have F R E g ∩ F R E h ⊆ S t ≤ T X t × X t , and F R E g ∩ S h ∈ B \{ g } F R E h ⊆ S t ≤ T X t × X t . So A g α g is supported on S t ≤ T X t × X t , and we concludethat A g ∈ L t ∈ N B ( L X t ). 28 emma 11.9. Suppose the action α is free. Then the natural map C ∗ max (Γ) → C ∗ max ( X ) is aninjection.Proof. The proof is similar to the proof of [8, Lemma 7.1]. Let I denote the closure of I in C ∗ max ( X ).The previous lemma gives C cs [ X ] /I ∼ = C cs [ C M ] /I ⋊ Γ, hence C ∗ max ( X ) /I ∼ = C ∗ max ( C M ) /I ⋊ max Γ.Now it suffices to prove that the composition C ∗ max (Γ) → C ∗ max ( X ) → C ∗ max ( X ) /I ∼ = C ∗ max ( C M ) /I ⋊ max Γ is injective.For any t ∈ N define the map ϕ t : C cs [ C M ] → C by ϕ t ( T ) = µ ( X t ) h X t , T X t i . This ex-tends to a positive unital map ϕ t : C ∗ max ( C M ) → C . Let ϕ be a limit point of these maps.Then ϕ : C ∗ max ( C M ) /I → C is a ucp map. Since maximal crossed products are functorial forucp maps, this gives a map C ∗ max ( C M ) /I ⋊ max Γ → C ∗ max (Γ). Now the composition C ∗ max (Γ) → C ∗ max ( C M ) /I ⋊ max Γ → C ∗ max (Γ) is the identity, so the first map is an injection.As a result, we get that for free actions, ∆ Γ has spectral gap in C ∗ max ( M ) if and only if Γ hasproperty (T). So if Γ does not have property (T), for any ε >
0, there is some representation ( ρ, H )and a unit vector v ∈ H ⊥ c with k ∆ Γ v k < ε . Suppose that k ∆ R v k < < R <
1. Thenone can show that k ∆ R v k → R →
0. Let u : L X → L X be given by u ( ξ )( x, t ) = ξ ( x, t ).Precomposing the representation ρ by the map τ : C cs [ X ] → C cs [ X ] , τ ( T ) = uT u ∗ multiple timesgives vectors v = v, v , . . . such that k ∆ Γ v n k < ε and k ∆ R v n k < ε . In this case it follows that X does not have geometric property (T). A similar strategy would work for non-ergodic actions. Themain obstacle is, that there may be vectors v for which k ∆ R v k = 1 for all 0 < R < Acknowledgement
I would like to thank my advisor Tim de Laat for his support and suggestions. I also thank for RufusWillett for helpful suggestions. The author is supported by the Deutsche Forschungsgemeinschaftunder Germany’s Excellence Strategy - EXC 2044 - 390685587, Mathematics M¨unster: Dynamics -Geometry - Structure.
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Address : Einsteinstrasse 62, 48149 M¨unster