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The integers do not have property (T)

Statement

Assume the Axiom of Choice (The Axiom of Choice), and give the additive group $Z$ (The integers as equivalence classes of pairs of naturals) the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). For every compact subset $Q \subseteq Z$ (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) and every $ε > 0$ , there is a complex number $z$ with $∣ z ∣ = 1$ and $z \neq 1$ such that the character $χ_{z} (n) = z^{n}$ defines a strongly continuous unitary representation $π_{z} (n) u = z^{n} u$ on the standard complex Hilbert space $C$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space) with no nonzero invariant vector, while its unit vector $1$ is $(Q, ε)$ -invariant (Almost invariant vectors for a unitary representation): $∣ z^{n} - 1 ∣ < ε (n \in Q) .$ Consequently, no compact subset of $Z$ is a Kazhdan set, and $Z$ does not have property (T) (Kazhdan's property (T)) by Property (T) is equivalent to the existence of a compact Kazhdan pair. No compact Kazhdan pair exists (Kazhdan pairs, Kazhdan sets and Kazhdan constants).

Facts & Assumptions

Given: AC, the additive group $Z$ with discrete topology, a compact subset $Q \subseteq Z$ , and a real $ε > 0$ .

[F1]

A compact subset of a discrete space is finite: the singleton sets form an open cover in the subspace topology, and compactness gives a finite subcover (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[F2]

Give $C$ the pairing $⟨ u, v ⟩ = u \overline{v}$ . Field arithmetic and conjugation make it linear in the first variable and conjugate symmetric, and $⟨ u, u ⟩ = ∣ u ∣^{2}$ is nonnegative and vanishes exactly at $u = 0$ ; hence this is a complex inner product with induced norm $∣ u ∣$ . The complex plane is complete for that norm, so it is a complex Hilbert space (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , Real and complex inner-product spaces and their induced length, The induced length is a norm, Real and imaginary parts, complex conjugation, and modulus, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts, Hilbert space, Linear map between vector spaces over the same field).

[F3]

For $z \neq 0$ , integer powers agree with the powers in the multiplicative group $C^{\times}$ ; $z^{n + m} = z^{n} z^{m}$ , $z^{- n} = (z^{n})^{- 1}$ , and the complex modulus is multiplicative and subadditive. If $∣ z ∣ = 1$ , then $∣ z^{n} ∣ = 1$ for every integer $n$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ , $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , Integer powers in the complex field, Group and abelian group, Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , Exponent laws in a group: $g^{m + n} = g^{m} g^{n}$ and $(g^{m})^{n} = g^{m n}$ for all $m, n \in Z$ , and $(g h)^{n} = g^{n} h^{n}$ when $g$ and $h$ commute, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F4]

The discrete topology makes every map from $Z$ continuous; the addition map is continuous because $Z \times Z$ is discrete in the product topology, and negation is continuous for the same reason. Thus $Z$ is a topological group and every scalar character on it is continuous. The integer operations form an additive group by the commutative ring theorem (The integers as equivalence classes of pairs of naturals, Arithmetic on the integers, The integers form a commutative ring, Group and abelian group); the topology assertions use The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, 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, and Topological group: multiplication and inversion are continuous.

[F5]

The order-preserving embeddings $Z ↪ Q ↪ R$ turn the finite set of integer magnitudes from [F1] into a finite linearly ordered subset of $R$ , so it has a maximum $M$ . Also $M \geq 0$ and $1 > 0$ , hence $M + 1 > 0$ . These facts use the ordered-field structure of $R$ , the integer order and absolute value, and the finite set convention (The integers embed in the rationals, The unique embedding of ℚ into an ordered field, The reals form a totally ordered field, Order on the integers, The integers form a totally ordered ring, Basic properties of the absolute value, Finite, countably infinite, countable, uncountable).

[F6]

Under AC, property (T) implies the existence of a compact Kazhdan pair; this is the forward implication in the property-(T)/Kazhdan-pair equivalence (The Axiom of Choice, Kazhdan's property (T), Kazhdan pairs, Kazhdan sets and Kazhdan constants, Property (T) is equivalent to the existence of a compact Kazhdan pair).

Proof

Proof technique: choose a nontrivial scalar of modulus one explicitly and bound its integer powers on the finite compact test set.

1.1F4

The additive group $Z$ with the discrete topology is a topological group: every subset of $Z$ and of $Z \times Z$ is open, so addition and negation are continuous by [F4].

1.2F1F3F5algebra

The compact set $Q$ is finite by [F1]. Let $M : = \max ({0} \cup {∣ n ∣ : n \in Q})$ in $R$ , using the order-preserving embeddings in [F5]. This maximum exists because the displayed set is finite and linearly ordered; it also gives $M = 0$ when $Q = \emptyset$ . Set $t : = ε / (4 (M + 1)) > 0$ and $z : = ((1 - t^{2}) + 2 i t) / (1 + t^{2})$ . Then $∣ z ∣^{2} = ((1 - t^{2})^{2} + 4 t^{2}) / (1 + t^{2})^{2} = 1$ , so $∣ z ∣ = 1$ by nonnegativity of the modulus; the imaginary part $2 t / (1 + t^{2})$ is nonzero, so $z \neq 1$ . Moreover, $∣ z - 1 ∣^{2} = (4 t^{4} + 4 t^{2}) / (1 + t^{2})^{2} = 4 t^{2} / (1 + t^{2}) \leq 4 t^{2}$ , and hence $∣ z - 1 ∣ \leq 2 t = ε / (2 (M + 1))$ .

2.1F3step 1.2algebra

For every integer $n$ , the geometric-sum identity and [F3] give $∣ z^{n} - 1 ∣ \leq ∣ n ∣ ∣ z - 1 ∣$ : for $n = k \geq 1$ factor $z^{k} - 1 = (z - 1) (1 + z + \dots + z^{k - 1})$ and use $∣ z^{j} ∣ = 1$ ; for $n = - k < 0$ , $∣ z^{- k} - 1 ∣ = ∣ z^{- k} (1 - z^{k}) ∣ = ∣ 1 - z^{k} ∣$ ; for $n = 0$ both sides are zero. Therefore, for every $n \in Q$ , $∣ z^{n} - 1 ∣ \leq ∣ n ∣ ε / (2 (M + 1)) \leq M ε / (2 (M + 1)) < ε$ . The last strict inequality also holds when $M = 0$ .

2.2F2F3F4step 1.1

Since $∣ z ∣ = 1$ , the map $χ_{z} (n) = z^{n}$ is a homomorphism from the additive group $Z$ into the unit scalars by [F3], and it is continuous because its domain is discrete by [F4]. On $H = C$ with the inner product from [F2], define $π_{z} (n) u : = χ_{z} (n) u$ . Each $π_{z} (n)$ is complex linear by the field laws and preserves the norm since $∣ z^{n} u ∣ = ∣ z^{n} ∣ ∣ u ∣ = ∣ u ∣$ ; its inverse is $π_{z} (- n)$ . The homomorphism law for $χ_{z}$ gives the representation law, and its orbit maps are continuous by [F4], so $π_{z}$ is a strongly continuous unitary representation.

3.1step 1.2step 2.1step 2.2F2

The vector $1 \in C$ has norm one by [F2] and satisfies $∥ π_{z} (n) 1 - 1 ∥ = ∣ z^{n} - 1 ∣ < ε$ for each $n \in Q$ by step 2.1. If $u \in C$ were invariant under $π_{z}$ , invariance under $1 \in Z$ would give $z u = u$ ; since $z \neq 1$ and $C$ is a field, this forces $u = 0$ . Thus $1$ is a $(Q, ε)$ -invariant unit vector in a representation with no nonzero invariant vector.

4.1F1F6step 3.1∎

Since $Q$ and $ε > 0$ were arbitrary, the witness in step 3.1 shows that no compact $Q$ and positive tolerance form a Kazhdan pair. Hence $Z$ has no compact Kazhdan set. By [F6], this rules out property (T). The character construction itself uses no Choice; AC is used only in this final implication.

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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)
Terence Tao, 254B, Notes 2: Cayley graphs and Kazhdan's property (T) (standard reference, not scraped)