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Property (T) passes to Hausdorff quotients

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $G$ be a locally compact Hausdorff topological group (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Topological group: multiplication and inversion are continuous) with Kazhdan's property (T) (Kazhdan's property (T)), and let $N ⊴ G$ be a closed normal subgroup (Normal subgroup: invariance under conjugation). Then the Hausdorff quotient topological group $G / N$ with the quotient topology (The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection) has property (T).

Facts & Assumptions

Given: AC; a locally compact Hausdorff topological group $G$ with property (T); a closed normal subgroup $N$ ; the algebraic quotient $G / N$ with its quotient topology and quotient map $q : G \to G / N$ .

[F1]

Property (T) says that every strongly continuous unitary representation with almost invariant unit vectors has a nonzero invariant vector. Almost invariance tests every compact subset of the group and every positive tolerance. (Kazhdan's property (T), Almost invariant vectors for a unitary representation, Strongly continuous unitary representations, invariant linear subspaces and intertwiners)

[F2]

The quotient group has coset product $(g N) (h N) = g h N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ; the canonical map $q (g) = g N$ is a continuous surjection and is a quotient map for the quotient topology. (Normal subgroup: invariance under conjugation, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection, For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ , A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps)

[F3]

Left and right translations and inversion in $G$ are homeomorphisms, so translating an open subset of $G$ by a fixed element preserves openness. (Topological group: multiplication and inversion are continuous, Left and right translations and inversion in a topological group are homeomorphisms, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological)

[F4]

In the product topology, basic open sets in $G \times G$ are rectangles; a map into a product is continuous when its coordinate maps are continuous. An open continuous surjection is a quotient map, and a map out of a quotient map is continuous exactly when its composite is continuous. (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps, For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map)

[F5]

The continuous image of a compact subset is compact. (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism)

[F6]

A space is Hausdorff when every two distinct points have disjoint open neighborhoods. (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not)

[F7]

AC means every set-indexed family of nonempty sets has a choice function; the proof below makes no such selection. (The Axiom of Choice)

Proof

technique · Pull a representation of $G/N$ back along the quotient map. Compact subsets of $G$ have compact images, so almost invariance pulls back, and surjectivity transfers invariant vectors back to the quotient

1.1F2F3

The quotient map $q : G \to G / N$ is continuous and surjective by [F2]. If $U \subseteq G$ is open, then $q^{- 1} (q (U)) = U N = ⋃_{n \in N} U n$ ; every $U n$ is open by [F3], so the quotient topology makes $q (U)$ open. Thus $q$ is open.

2.1F2F3F4F6step 1.1

Let $g N \neq h N$ . Then $g^{- 1} h \notin N$ , so closedness of $N$ gives an open set $W$ containing $g^{- 1} h$ and disjoint from $N$ . The map $(u, v) \mapsto u^{- 1} v$ is continuous by the topological-group operations [F3] and the product topology [F4]; hence there are open neighborhoods $U ∋ g$ and $V ∋ h$ with $U^{- 1} V \subseteq W$ . Their images $q (U)$ and $q (V)$ are open by step 1.1. They are disjoint, since a common coset would give $u \in U$ , $v \in V$ with $u^{- 1} v \in N \cap W$ . Therefore $G / N$ is Hausdorff.

2.2F2F3F4step 1.1

The map $q \times q : G \times G \to (G / N) \times (G / N)$ is continuous by [F4]. It is open: each basic rectangle $U \times V$ maps to the open rectangle $q (U) \times q (V)$ by step 1.1, and every open set is a union of basic rectangles. It is surjective since $q$ is. Hence $q \times q$ is a quotient map by [F4]. The quotient multiplication $m_{G / N}$ satisfies $m_{G / N} \circ (q \times q) = q \circ m_{G}$ , and quotient inversion $i_{G / N}$ satisfies $i_{G / N} \circ q = q \circ i_{G}$ . The right sides are continuous, so the quotient universal property [F4] makes both quotient operations continuous. Thus $G / N$ is a topological group.

3.1F1F2F5step 2.2

Let $σ$ be any strongly continuous unitary representation of $G / N$ with almost invariant vectors, and put $π : = σ \circ q$ . This is a strongly continuous unitary representation of $G$ , since each orbit map is the composite of the continuous orbit map for $σ$ with $q$ . Given a compact $K \subseteq G$ and $ε > 0$ , [F5] makes $q (K)$ compact; almost invariance of $σ$ supplies a unit vector $ξ$ with $∥ σ (y) ξ - ξ ∥ < ε$ for every $y \in q (K)$ . Thus $∥ π (g) ξ - ξ ∥ < ε$ for every $g \in K$ . This proves that $π$ has almost invariant vectors.

4.1F1F2F7step 3.1∎

Since $G$ has property (T), [F1] gives a nonzero vector $ξ$ invariant under $π (G)$ . Surjectivity of $q$ gives $π (G) = σ (G / N)$ , so $ξ$ is invariant under $σ (G / N)$ . As $σ$ was arbitrary, $G / N$ has property (T). The Axiom of Choice is included as in the dispatched statement but is not used in this proof.

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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)
Emmanuel Breuillard, PCMI Lecture Notes on Property (T), Expander Graphs and Approximate Groups (standard reference, not scraped)