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A well-known application of Kazhdan's Property (T) is the construction of expander graphs. Background on this is discussed, for example, in this post on Terry Tao's blog. Essentially, Cayley graphs of finite quotients of property (T) groups can give us families of expanders (see Exercise 14 of Tao's blog post). The construction seems to critically use the finite quotients to obtain the unitary representations required to employ the definition of property (T). It would be very nice to have an answer to the following:

Question: Is any of the behavior of expander graphs reflected in the (infinite) Cayley graph of a property (T) group with respect to a finite, symmetric generating set?

Please provide a reference to anything in the literature that sheds some light on this.

A somewhat broader (related) question that may be helpful:

What are some qualitative properties of the Cayley graph of a property (T) group?

For example, does the Cayley graph of a property (T) group exhibit any sort of (local) concentration of measure phenomena using the word metric w.r.t. a finite generating set?

What are some useful intuitions for the Cayley graph of a property (T) group? (Here I'm wondering if there is anything akin to the image of "thin triangles" for hyperbolic groups.)

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This is just a suggestion, not an answer and I do not know if it was implemented. Given a finite graph $C$ and its subgraph $A$ one can define the relative Cheeger constant $h(C,A)$. Use it to define relative expander property for sequences $(C_n,A_n)$. Then, you can try to show that if a f.g. group $G$ has property $T$ then its Cayley graph $C$ admits an exhaustion by finite graphs $C_n$ so that $(C_n, \partial C_n)$ is a relative expander. The idea here is that $C_n$'s could be fundamental domains for finite-index subgroups $G_n\subset G$ and $C_n$ resembles the graph $C/G_n$ except... –  Misha Jun 1 '12 at 18:59
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@Agol: which result from Żuk's paper are you referring to? –  Michal Kotowski Jun 1 '12 at 19:12
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@Ian and @Jon: On the second thought, maybe the spectral link condition (used for establishing property T long before Zuk's paper) could be useful. Maybe there is a local-to-global principle which would say that if you have a 2-dimensional simplicial complex $C$ (admitting a cocompact group action) whose links satisfy spectral condition then, say, large metric balls $C_n$ in $C$, satisfy relative expander property as in my comment or some yet-to-be-defined expander-like property. –  Misha Jun 2 '12 at 14:02
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@Jon: You have it now. –  Misha Jun 2 '12 at 14:35
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@everybody: Exercise: show that if two groups with given finite generating subsets have isomorphic (unlabeled) Cayley graphs and one has Property T, then the other also has Property T. –  Yves Cornulier Jun 3 '12 at 15:15
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2 Answers

If Kazhdan's property (T) is reflected in the structure of the Cayley graph, then not in a very geometric way.

Steve Gersten (that is what I read in the book by B. Bekka, P. de la Harpe and A. Valette) was the first who found that Kazhdan's property (T) is not invariant under quasi-isometry. The reason is not complicated. If a central extension $$1 \to \mathbb Z \to \Gamma \to \Lambda $$ is obtained from a bounded cocycle $c \colon \Lambda \times \Lambda \to \mathbb Z$, then $\Gamma$ is quasi-isometric to $\Lambda \times \mathbb Z$. This situation arises for $\Lambda$ a cocompact lattice in a simple real Lie group with infinite fundamental group; such as $SU(2,2)$. Then, $\Gamma$ is the inverse image of $\Lambda$ in the universal covering. In this situation, one actually obtains (for suitable generating sets) a bi-Lipschitz equivalence of Cayley graphs.

Now, $\Lambda \times \mathbb Z$ does not have Kazhdan's property (T) since it surjects onto $\mathbb Z$, but $\Gamma$ has Kazhdan's property (T) inheriting it from the universal cover of $SU(2,2)$.

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@Andreas: it depends on what you call "geometric". Here you mean it by "QI-invariant". But it does not mean that features of Property T groups are "not very geometric". Also, for a graph, being a tree is not a QI-invariant property... but would you say that being a tree is "not very geometric"? –  Yves Cornulier Jun 3 '12 at 17:24
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@Andreas: This example came as a result of a conversation between Gersten and Ramachandran, I think, in 1995, when Mohan was visiting Utah. There is no known geometric characterization of property T; the fact that T is not QI invariant does not mean that such characterization does not exist. For instance, being CAT(0) is not QI invariant but, clearly, geometric. –  Misha Jun 3 '12 at 17:27
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Thanks for the comments. Sure, geometric is not a technical term. However, some people say a property is geometric if it is QI-invariant; which make some sense. And actually I meant what I wrote, Kazhdan's property is a property of the whole unitary dual and not so much of the left-regular representation alone. The left-regular representation is clearly captured by the geometry of the Cayley graph (vanishing of l2-Betti numbers is QI-invariant). However, how the unitary representation theory is encoded in the Cayley graph is less clear. –  Andreas Thom Jun 3 '12 at 18:03
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This is not an answer but a collection of observations, some of which, are, hopefully, interesting.

First, recall that Cheeger constant $h(X)$ for a complete open Riemannian manifold $X$ is $$ \inf_D \frac{|\partial D|}{|D|} $$ where infimum is taken overall bounded domains $D\subset X$ with smooth boundary and vertical bars denote volume (of appropriate dimension). In the context of infinite connected graphs, the same definition applies. Then $h(X)=0$ iff $X$ is amenable. Cheeger constant is also defined for closed manifolds (finite graphs) where one looks at the infimum of the quantity $$ \frac{|S|}{|M_S|} $$ where $S$ is a cooriented hypersurface in $M$ and $M_S$ the "positive side" of $S$ in $M$, and the infimum is taken over all cooriented hypersurfaces $S$. A sequence of closed connected manifolds $M_i$ (of "uniformly bounded geometry", i.e., constant dimension, with fixed bounds on curvature and a lower bound of the injectivity radius) is an "expander" if $h(M_i)$ is uniformly bounded below, away from zero, while diameters of $M_i$ diverge to infinity. Same definition applies to graphs where bounds on geometry are replaced by uniform bounds on valence. In other words, in the context of finite graphs or sequences of closed manifolds of bounded geometry, being an "expander" is essentially the negation of amenability. The question raised by Jon is if one can define an expander in the context of a single infinite graph or, I would say, a single open Riemannian manifold $X$ of bounded geometry. Defining an expander as before, by saying $X$ is nonamenable, is too weak. The usual link between expanders and open Riemannian manifolds/infinite graphs is via taking quotients: If $\Gamma_i$ be a infinite sequence of finite-index subgroups of a fixed discrete cocompact subgroup $\Gamma\subset Isom(X)$, then the sequence of quotients $M_i:=X/\Gamma_i$, is an expander. One then could say that $X$ itself an expander if it admit a sequence of group actions $\Gamma_i$ satisfying the above conditions. Note that the quotient manifolds $M_i$ "approximate" $X$, provided that intersection of the subgroups $\Gamma_i$ is trivial.

The question is how to eliminate groups in this definition, since some infinite f.g. groups with property T could be simple or at least, contain no proper finite-index subgroups.

I will now proceed in the category of Riemannian manifolds, but the discussion is equivalent to the one in the case of graphs. Let $Q_i\subset X$ be fundamental domains for the groups $\Gamma_i$. Let $S\subset M_i$ be a cooriented compact hypersurface, transversal to the projection of the boundary of $Q_i$, and let $T\subset Q=Q_i$ be preimage of $S$ in $Q$. Note that $T$ is not properly embedded in $Q$. For each $T\subset Q$ we have the ratio $$ \rho_{T,i}=\frac{|T|}{|Q_T|}, $$ where $Q_T$ is "positive side" of $T$ in $Q$. Then, $$ \inf_{T,i} \rho_{T,i} $$ is still bounded away from zero, provided that $\Gamma$ has property T. Here infimum is taken over all $T_i\subset Q_i$, so that $T_i$ is the preimage of a closed surface in $M_i$. (Yves exercise amounts to saying that $\Gamma$ having property T depends only on $X$ and not on the particular group acting cocompactly on $X$.)

One then can attempt to promote this to a definition: We say that a sequence of domains $Q_i\subset X$ is "anti-Folner" if the infimum of ratios $\frac{|T|}{|Q_{i,T}|}$ defined as above (using arbitrary cooriented hypersurfaces $T\subset Q_i$) is positive and so that $diam(Q_i)\to \infty$. One then would say that $X$ is an expander itself if it admits an anti-Folner sequence of connected bounded domains $Q_i\subset X$.

One can give a "spectral" interpretation of the anti-Folner property. Namely, for each $Q$ define
$$ \mu(Q):=\inf \frac{ \int_Q |\nabla f|^2}{\int_Q |f|} $$ where infimum is taken over all smooth functions on $Q$. Then as in the proof of Cheeger's theorem, one gets: If $\inf \mu(Q_i)>0$ then the sequence $Q_i$ is anti-Folner.

Note that amenability of $X$ is equivalent to $\lambda(X)=\inf \lambda(D)=0$, where $\lambda(D)$ is the lowest positive eigenvalue of a bounded domain $D\subset X$ and $$ \lambda(D)= \inf \frac{ \int_Q |\nabla f|^2}{\int_Q |f|}, $$ where infimum is taken over all smooth compactly supported functions on $D$. So the difference between $\lambda$ and $\mu$ is that in the definition of $\mu$ we allow functions in bounded domains in $X$ that do not have compact support.

What's missing here is a proof that $\mu$ is bounded away from zero in some interesting examples, since, in the context of sequences of discrete groups $\Gamma_i$, I was using not all but only some hypersurfaces in $Q_i$ (the ones which came from compact hypersurfaces in $M_i$).

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Maybe one can characterize property $\tau$ like this, but not property $(T)$. –  Andreas Thom Jun 5 '12 at 12:42
    
@Andreas: Yes, of course, property T would require something else. –  Misha Jun 5 '12 at 14:39
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