Property (T)

Today I would like to introduce a property of groups which I am interested in because it implies a lot of rigidity for the actions of the group on spaces. I'm really interested in the case of discrete groups, so in what follows the reader would lose very little by just taking every group to be discrete. This is a huge topic and we will only be able to say a few introductory words today. The standard reference is this book.

Definition Let $\mathbb{H}$ be a complex Hilbert space, $G$ be a topological group, and $\pi: G \to  U(\mathbb{H})$ be a continuous unitary representation of $G$. 

• Given a subset $S \subseteq G$ and a number $\epsilon > 0$, a unit vector $v$ in $\mathbb{H}$ is $(S, \epsilon)$-invariant if  $sup_{g\in S}||\pi(g)v -v|| \leq \epsilon ||v|| $.
•  The representation $(\pi, \mathbb{H})$ has almost invariant vectors if it has $(K, \epsilon)$–invariant vectors for every compact subset $K$ of $G$ and every $\epsilon>0$. We also write this as $1_G \prec \pi$.
• The representation $(\pi, \mathbb{H})$ has invariant vectors if there exists a unit vector $v$ in $\mathbb{H}$ such that $\pi(g)v =v$ for all $g \in G$.

$G$ is then said to have Property (T) if it satisfies either of the following equivalent conditions:

1.  Every unitary representation of $G$ that has an $(\epsilon, K)$-invariant unit vector for any $\epsilon > 0$ and any compact subset $K$, has a non-zero invariant vector.
2. There exists an $\epsilon > 0$ and a compact subset $K$ of $G$ such that every unitary representation of $G$ that has an $(\epsilon, K)$-invariant unit vector, has a nonzero invariant vector.

This says that if you get too close to fixing something you in fact have to fix something. To show the converse, let $G$ be a locally compact group satisfying (1), assume by contradiction that for every $K$ and $\epsilon$ there is a unitary representation that has a $(K, \epsilon)$-invariant unit vector and does not have an invariant vector. Look at the direct sum of all such representation and take Hilbert space completion. This representation will fail to satisfy (2). There are many other equivalent definitions which the reader can find online or in the aforementioned book. 

Historically, Kazhdan introduced Property (T) in order to prove that lattices in higher rank Lie groups are finitely generated by showing that they satisfy Property (T). A precise statement is as follows:

Proposition. Let $G$ be a topological group with Property (T). Let $H$ be an open subgroup. Then $G$ is finitely generated over $H$ (that is, there exists a finite subset $S$ such that $G$ is generated by $H \cup S$).

Proof Suppose not. We consider the unitary representation of $G$ on $\oplus_L \ell^2(G/L)$, where $L$ ranges over subgroups of $G$ containing $H$ that are finitely generated over $H$. It almost has invariant vectors, in the strongest sense, that is every compact subset of $G$ fixes a unitary vector (that is, there's one vector working for all $\epsilon$, which is usually not the case). By Property T, there is a nonzero invariant vector $f \in \oplus_L \ell^2(G/L)$. So $f$ has a nonzero coordinate, say $f' \in \ell^2(G/L)$ is nonzero and $G$-invariant. By assumption, $G/L$ is infinite, so there is no nonzero invariant vector in $\ell^2(G/L)$. This is a contradiction. $\blacksquare$

Corollary: If in addition G is locally compact, then it is compactly generated.

Note that we have again used the trick of adding everything together to falsify one of the conditions. This is a common theme in the area. One can use a similar trick to show that an abelian group with Property (T) must be compact. By playing with the induced representation one can further show that Property (T) passes to finite index subgroups and overgroups, so in particular, discrete groups with Property (T) and all their finite index subgroups have finite abelianisation, also called virtual first Betti number 0. Further structural properties include that for $N$ a closed normal subgroup of $H$ such that $H/N$ has (T), if $N$ also has $T$ then so does $H$, i.e. having $(T)$ is preserved under extensions. 

A variation on this theme is the following result:

Theorem: Let $G$ be a locally compact group, and let $H$ be a closed subgroup of $G$ such that $G/H$ has a finite invariant regular Borel measure. Then $G$ has Property (T) iff $H$ has Property (T). In particular, if $\Gamma$ is a lattice in $G$, then $\Gamma$ has Property (T) if and only if $G$ has Property (T).

The main trick is to average a vector over the finite Borel measure to produce an invariant vector, sort of like how in the rep theory of finite groups one can average over the orbit of the group. The details can be found in the aforementioned book.

Since (discrete) groups with Property (T) are finitely generated, one might next ask about finite presentability. Observing that Property (T) passes to quotients, it isn't true that all groups with (T) are finitely presentable, but it turns out that every (T) group has a finitely presented cover, first proved by Shalom. This is a corollary of something much more general. Let $G$ be a group generated by a finite set $S$. Its non-normalised Laplacian is denoted by

\[\Delta=\sum_{x\in S} (1-x)^*(1-x)=\sum_{x\in S} (2-x-x^{-1})\in{\mathbb Z}[\Gamma]\]

Ozawa wrote a paper combining real algebraic geometry with functional analysis to show that $G$ has (T) if and only if the equation 

\[m \Delta^2 = n \Delta + \sum_{i=1}^k l_i \xi_i^*\xi_i\] has a solution in the integral group ring with $k,m,n,l_i\in{\mathbb Z}_{>0}$ and $\xi_i\in{\mathbb Z}[G]$. In particular, this is a finite check, so imposing finitely many relations suffices to make this hold in some cover of $G$. This also means that one can check Property (T) with a computer, and that's exactly what people did to solve the longstanding open problem of showing Aut(F_n) for $n>3$ has (T). It was previously known that it doesn't have (T) for $n=2,3$. No silicon-free proof is known. It remains to be seen whether any mapping class group has (T), but attempts to use this method have failed to produce results.

There are two other main ways of showing that a group has (T). One is to show that the group is boundedly generated. This means that there exists $n$ and $g_1, \dots g_n \in G$ such that for any $g \in G$, there exist integers $m_i$ such that $g = \prod_{i=1}^n g_i^{m_i}$. This and the connection to expander graphs are nicely exposed here. The other is via a sort of spectral criterion, i.e. the spectrum of certain operators have certain properties. We will discuss this in later posts, but the main ones are due to Ershov-Jaikin Zapirain and Zuk.  Also to be discussed at a later point is the property of always having a fixed point when acting on spaces of non-positive curvature, including trees.

To close, we show that having (T) isn't geometric, i.e. isn't preserved under quasi-isometries.

Let $G$ be a connected semisimple Lie group with finite centre, and let $K$ be a maximal compact subgroup of $G$. Recall that $G$ has an Iwasawa decomposition $G = ANK$, where $S = AN$ is a solvable simply connected closed subgroup of $G$, and that there exists a $G-$invariant Riemannian metric on $G/K$, and that, for any such metric, $G/K$ is a Riemannian symmetric space 

Lemma Let $G$ be a connected semisimple Lie group with finite centre, and let $K$ be a maximal compact subgroup of $G$. Let $G$ and $G/K$ be endowed with $G-$invariant Riemannian metrics, and let $K$ be endowed with a Riemannian metric. Then $G$ and $G/K \times K$ are bi-Lipschitz equivalent, where $G/K \times K$ is endowed with the product Riemannian metric.

Proof: Let $G=SK$ be an Iwasawa decomposition of $G$ as above. Let $\psi : G \to S\times K$

 denote the inverse of the product mapping $S \times K → G,(s,k) \mapsto sk$; observe

 that the diffeomorphism $\psi$ is equivariant for the actions of $S$ on $G$ and on $S\times K$ by left translation. We identify $S$ with $G/K$, and denote by $Q_0$ and $Q_1$ the Riemannian metrics on $G$ and $S \times K, respectively. Fix $x \in G$. Since any two scalar products on $\mathbb{R}^n$ are bi-Lipschitz equivalent, there exists a constant $C >0$ such that, for every $X \in T_x(G)$, we have

 \[ \frac{1} {C} ∥X∥_{Q_0} \leq ∥d\psi_x(X)∥_{Q_1} \leq  C∥X∥_{Q_0} \] Set

\[C_x = \sup_{X \in T_x(G) \setminus \{0\}} \left( \frac{\| d\psi_x(X) \|_{Q_1}}{\| X \|_{Q_0}} + \frac{\| X \|_{Q_0}}{\| d\psi_x(X) \|_{Q_1}} \right) < \infty.\]

Since \( Q_0 \) and \( Q_1 \) are \( S \)-invariant, we have

\[C_{sx} = C_x, \quad \text{for all } s \in S.\]

As \( G = SK \), this implies that

\[\sup_{x \in G} C_x = \sup_{x \in K} C_x.\]

Since \( K \) is compact, \( \sup_{x \in K} C_x < \infty \). Hence,

\[\lambda = \sup_{x \in G} C_x < \infty\]

and we have, for all \( x \in G \) and all \( X \in T_x(G) \),

\[\frac{1}{\lambda} \| X \|_{Q_0} \leq \| d\psi_x(X) \|_{Q_1} \leq \lambda \| X \|_{Q_0}.\]

This implies that \( \psi \) is a bi-Lipschitz mapping. \hfill \(\blacksquare\)

Now, let $G$ be a connected simple Lie group with infinite cyclic fundamental group, and let $\tilde{G}$ be the universal covering group of $G$. Let $\Gamma$ be a cocompact lattice in $G$. It is

 easy to see that $\Gamma$ is finitely generated. Let $\tilde{Gamma}$ be the inverse

 image of $\Gamma$ in $\tilde{G}$, which is also a cocompact lattice in $\tilde{G}$. Since $\tilde{\Gamma}$ contains $\pi_1(G) = \mathbb{Z}$, we have a central extension

\[ 1 \to \mathbb{Z} \to \tilde{\Gamma} \to \Gamma \to 1 \]

 Lemma Let $\tilde{\Gamma}$ and $\Gamma$ be the finitely generated groups defined above.

 Then $\tilde{\Gamma}$ and $\Gamma \times \mathbb{Z}$ are quasi-isometric.

Proof Let $K$ be a maximal compact subgroup of $G$. As in the previous lemma, we endow $G,K$ and $G/K$ with appropriate Riemannian metrics. We lift the metrics on $G$ and $K$ to left-invariant Riemannian metrics on $\tilde{G}$ and on $\tilde{K}$.

 On the one hand, $\tilde{\Gamma}$ acts freely by isometries (via left translations) on $\tilde{G}$,

 with compact quotient. Hence, $\tilde{\Gamma}$ is quasi-isometric to the metric space $\tilde{G}$.  On the other hand,  $\Gamma$ acts properly by isometries on $G/K$, with compact quotient. Moreover, since $\pi_1(K) = \pi_1(G) = \mathbb{Z}$, the group $\mathbb{Z}$ acts freely by isometries on $\tilde{K}$, with compact quotient. Hence $\Gamma \times \mathbb{Z}$ is quasi-isometric to $G/K \times \tilde{K}$. By the previous lemma, $G$ and $G/K \times K$ are bi-Lipschitz equivalent. Hence $\tilde{G}$ and $G/K \times \tilde{K}$ are bi-Lipschitz equivalent and therefore quasi-isometric. $\blacksquare$

It suffices now to exhibit a single such Lie group $G$ with Property (T). In the book, such examples are given, and these include simple Lie groups with trivial centre whose Lie algebras are isomorphic to $\mathfrak{so}(2,r), r \geq 3$. In fact, a complete classification of such simple Lie groups in terms of their Lie algebras is available.