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SL2(R) does not have property (T)

Statement

Assume the Axiom of Choice (The Axiom of Choice). The group ${S L}_{2} (R)$ does not have property (T) (Kazhdan's property (T)). Explicitly, for every compact $Q \subseteq {S L}_{2} (R)$ and every $ε > 0$ there is $ν \in (0, 1)$ such that the spherical complementary-series representation $I_{0, ν}$ (The normalized principal series I(epsilon, nu), Unitarity of the complementary series) has no nonzero invariant vector but has a $(Q, ε)$ -invariant unit vector (Almost invariant vectors for a unitary representation), namely its $K$ -fixed vector $f_{0}$ for $ν$ sufficiently close to $1$ . Consequently no pair $(Q, ε)$ with $Q$ compact is a Kazhdan pair (Kazhdan pairs, Kazhdan sets and Kazhdan constants).

Facts & Assumptions

Given: AC, $G = {S L}_{2} (R)$ with its standard topology, a compact set $Q \subseteq G$ , and $ε > 0$ .

[F1]

For $0 < ν < 1$ , the completion of $I_{0, ν}$ in the normalized complementary-series form is an irreducible strongly continuous unitary representation, and $f_{0} = 1$ has norm one (Unitarity of the complementary series).

[F2]

The spherical coefficient $φ_{ν} (g) = ⟨ Π_{ν} (g) f_{0}, f_{0} ⟩$ converges to $1$ uniformly on each compact subset of $G$ as $ν ↑ 1$ (The spherical complementary series converge to the trivial representation).

[F3]

For a unitary representation and a unit vector $ξ$ , $∥ π (g) ξ - ξ ∥^{2} = 2 (1 - Re ⟨ π (g) ξ, ξ ⟩)$ ; this is the expansion of the squared Hilbert norm and uses $∥ π (g) ξ ∥ = 1$ .

[F4]

A pair $(Q, ε)$ is Kazhdan if every strongly continuous unitary representation with a $(Q, ε)$ -invariant unit vector has a nonzero invariant vector (Kazhdan pairs, Kazhdan sets and Kazhdan constants).

[F5]

A strongly continuous unitary representation has almost invariant vectors exactly when every compact test set and every positive tolerance admit a near-invariant unit vector (Almost invariant vectors for a unitary representation).

[F6]

Property (T) requires every strongly continuous unitary representation with almost invariant vectors to have a nonzero invariant vector (Kazhdan's property (T)).

[F7]

The Hilbert direct sum carries the componentwise unitary action, and a vector is invariant exactly when each coordinate is invariant (Hilbert direct sums of unitary representations).

[F8]

In the smooth spherical compact picture, $f_{2 j} (k_{θ}) = e^{2 i j θ}$ has right $K$ -character $k_{ϕ} \mapsto e^{2 i j ϕ}$ for every $j \in Z$ . In the complementary completion at $0 < ν < 1$ , its squared norm is $B_{ν} (f_{2 j}, f_{2 j}) = a_{2 j} (ν) > 0$ , so each of these distinct smooth $K$ -lines survives as a nonzero line. This transfers the characters, not the ordinary $L^{2} (K)$ norm, to the positive weighted completion (The normalized principal series I(epsilon, nu), K-type decomposition of the SL2(R) principal series, Unitarity of the complementary series).

[A1]

AC is assumed in the normalized principal-series, complementary-series, and Hilbert direct-sum interfaces (The Axiom of Choice).

Proof

technique · use compact-uniform convergence of the normalized spherical coefficient, then form one direct-sum representation with almost invariant vectors and no invariant vector

1.1F1F2F3F5choosealgebra

If $Q = \emptyset$ , take $ν = 1 / 2$ and $f_{0}$ ; the invariance condition is vacuous. Otherwise [F2] lets us choose $0 < ν < 1$ sufficiently close to $1$ that $\sup_{g \in Q} ∣ φ_{ν} (g) - 1 ∣ < ε^{2} / 2$ . By [F1], $f_{0}$ is a unit vector in $I_{0, ν}$ , and [F3] gives $∥ Π_{ν} (g) f_{0} - f_{0} ∥^{2} = 2 (1 - Re φ_{ν} (g)) \leq 2 ∣ 1 - φ_{ν} (g) ∣ < ε^{2}$ for every $g \in Q$ . Thus the promised fixed-parameter vector is $(Q, ε)$ -invariant.

2.1F1F4F8step 1.1algebra

The invariant subspace of each $I_{0, ν}$ is closed and $G$ -invariant. By irreducibility in [F1], it is either zero or the whole representation; the latter would make every $K$ -action the identity, contrary to the nonzero even-weight lines in [F8]: for example $f_{2}$ has positive norm and $Π_{ν} (k_{π / 2}) f_{2} = - f_{2}$ . Hence every fixed $I_{0, ν}$ has no nonzero invariant vector. Together with step 1.1 and [F4], this shows that no compact $(Q, ε)$ is a Kazhdan pair.

3.1F1F2F3F5F6F7A1step 2.1construct∎

Set $ν_{j} = 1 - 1 / (j + 2)$ for $j \geq 1$ and let $Π = {⨁^{^}}_{j \geq 1} I_{0, ν_{j}}$ . By [F7] this is a strongly continuous unitary representation. For each compact $Q$ and $ε > 0$ , [F2] and [F3] show that the unit vector $f_{0}$ from [F1] in a sufficiently late summand has displacement less than $ε$ on $Q$ ; therefore $Π$ has almost invariant unit vectors by [F5]. An invariant vector in $Π$ would have every coordinate invariant, so step 2.1 forces it to be zero. By [F6], this single representation witnesses that $G$ fails property (T).

Remarks

No fixed $I_{0, ν}$ is asserted to have almost invariant vectors or to weakly contain the trivial representation; the property-(T) failure is witnessed by the direct sum of a cofinal parameter sequence.

Depends on

Used by

The spherical complementary series destroys property (T) for SL2(R) Counterexample

Dependency tree · two levels

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Sources

Bachir Bekka, Pierre de la Harpe and Alain Valette, Kazhdan's Property (T) (Cambridge University Press 2008; author-hosted complete text) (standard reference, not scraped)
Pavel Etingof, Representations of Lie Groups (MIT 18.757 lecture notes, Fall 2023) (standard reference, not scraped)
Emmanuel Breuillard, PCMI Lecture Notes on Property (T), Expander Graphs and Approximate Groups (standard reference, not scraped)