Decay to zero of matrix coefficients at Adjoint infinity

by Scot Adams

I. Introduction

The main theorems below are Theorem 9 and Theorem 11. As far as I know, Theorem 9 represents a slight improvement over what currently appears in the literature, and gives a fairly easy proof of Theorem 11 which is due to R. Howe and C. Moore [4]. In the semisimple case, the Howe-Moore result follows from [6] or [7].

The proofs appearing here are relatively elementary and some readers will recognize the influence of R. Ellis and M. Nerurkar [2]. What may be less evident is the connection to N. Kowalsky’s work [5] on dynamics on Lorentz manifolds: Lemma 1 below is a unitary analogue of the fact that an expansive Adjoint action can result in much of the Lie algebra being lightlike. In Hilbert space, the situation is even nicer: an isotropic vector must equal zero, so we get that the Lie algebra of the stabilizer contains all “Kowalsky” vectors for the Adjoint sequence, and not just a codimension one subspace. Moreover, much of the rest of proof of Theorem 9 also uses ideas that were originally developed to describe effectively the collection of simply connected Lie groups admitting an orbit nonproper action on a connected Lorentz manifold. Thus the debt to Kowalsky is significant.

In this note, we assume some familiarity with Lie theory and basic unitary representation theory. The exposition should otherwise be self-contained.

Please send comments, suggestions or questions to adams@math.umn.edu.

Throughout, “Lie group” means “C^{∞} real Lie group”, “Lie algebra” means “real Lie algebra” and “Hilbert space” means “complex Hilbert space”. Lie groups are denoted by capital Roman letters and, for any Lie group, its Lie algebra is denoted, without comment, by the corresponding small letter in the fraktur font. The sesquilinear form on a Hilbert space is denoted ⟨ · , · ⟩. Convergence in the weak topology on a Hilbert space is denoted with ⇀, so v_{i} ⇀ v in a Hilbert space H means: for all w ∈ H, ⟨v_{i}, w⟩ → ⟨v, w⟩. The group of bounded operators on H is denoted B(H). The unitary group of a Hilbert space H is denoted U(H). The weak-operator and strong-operator topologies agree on U(H), and U(H) is given this topology. Unitary representations are always assumed continuous in the weak-operator topology (or, equivalently, in the strong-operator topology).

I do not plan to publish this note.

II. Preliminary dynamical results

Let G be a Lie group acting by a unitary representation on a Hilbert space H. Let w ∈ H and let 𝔪 be the Lie algebra of Stab_{G}(w).

Lemma 1: Let U_{i} → 0 in 𝔤. Let g_{i} be a sequence in G. Assume (Ad g_{i})U_{i} → X in 𝔤. Let v ∈ H and assume that g_{i}v ⇀ w in H. Then X ∈ 𝔪.

Proof: Fix t ∈ ℝ and let h := exp(tX). We wish to show that hw = w.

As exp(tU_{i}) → 1_{G} in G, we get (exp(tU_{i}))v → v in H. So, for all x ∈ H, since {g_{i}^{−1}x} is norm-bounded in H, Cauchy-Schwartz yields: ⟨[(exp(tU_{i}))v] − v, g_{i}^{−1}x⟩ → 0; then

⟨g_{i}(exp(tU_{i}))v, x⟩ − ⟨g_{i}v, x⟩ → 0.

Moreover, as g_{i}v ⇀ w, we see, for all x ∈ H, that

⟨g_{i}v, x⟩ − ⟨w, x⟩ → 0.

Adding the last two displayed limits, for all x ∈ H, we have ⟨g_{i}(exp(tU_{i}))v, x⟩−⟨w, x⟩ → 0. That is, g_{i}(exp(tU_{i}))v ⇀ w in H.

For all i, let X_{i} := (Ad g_{i})U_{i} and let h_{i} := exp(tX_{i}). Then, for all i, we have g_{i}(exp(tU_{i}))g_{i}^{−1} = h_{i}, so h_{i}g_{i} = g_{i}(exp(tU_{i})). Then h_{i}g_{i}v = g_{i}(exp(tU_{i}))v ⇀ w in H.

We have X_{i} → X in 𝔤, and so h_{i} → h in G. For all x ∈ H, since h_{i}^{−1}x → h^{−1}x and since {g_{i}v} is norm-bounded in H, Cauchy-Schwartz gives: ⟨g_{i}v, h_{i}^{−1}x − h^{−1}x⟩ → 0; then

⟨h_{i}g_{i}v, x⟩ − ⟨hg_{i}v, x⟩ → 0.

Since g_{i}v ⇀ w, it follows, for all x ∈ H, that ⟨g_{i}v, h^{−1}x⟩ − ⟨w, h^{−1}x⟩ → 0; then

⟨hg_{i}v, x⟩ − ⟨hw, x⟩ → 0.

Adding the last two displayed limits, for all x ∈ H, we have ⟨h_{i}g_{i}v, x⟩ − ⟨hw, x⟩ → 0. That is, h_{i}g_{i}v ⇀ hw. So, recalling that h_{i}g_{i}v ⇀ w, we get hw = w. QED

Corollary 2: For all T ∈ 𝔪, we have (𝔠_{𝔤}(T)) ∩ ((ad T)𝔤) ⊆ 𝔪.

Proof: Let X ∈ (𝔠_{𝔤}(T)) ∩ ((ad T)𝔤). We wish to show that X ∈ 𝔪.

We have (ad T)X = 0 and X ∈ (ad T)𝔤. Choose S ∈ 𝔤 such that (ad T)S = X. Let r_{i} be a sequence of nonzero real numbers such that r_{i} → +∞. For all i, let g_{i} := exp(r_{i}T) and let U_{i} := S/r_{i}. Then U_{i} → 0 in 𝔤. Since T ∈ 𝔪, it follows, for all i, that g_{i}w = w.

Since (ad T)S = X and (ad T)X = 0, it follows, for all i, that (Ad g_{i})S = S + r_{i}X. Then (Ad g_{i})U_{i} = (S/r_{i})+X → X in 𝔤, so, by Lemma 1 (with v := w), we are done. QED

III. More preliminaries

Remark 3: Let 𝔤 be a Lie algebra. Let U_{i} → 0 in 𝔤 and let X ∈ 𝔤. Let g_{i} be a sequence in G. Assume that (Ad g_{i})U_{i} → X in 𝔤. Then ad X : 𝔤 → 𝔤 is nilpotent.

Proof: Let F_{i} ⊆ ℂ be the set of eigenvalues of ad U_{i} : 𝔤 → 𝔤. Then, as U_{i} → 0, it follows that F_{i} → {0} in the topological space of finite subsets of ℂ. For all i, in End(𝔤), we have

ad((Ad g_{i})U_{i})  =  (Ad g_{i}) (ad U_{i}) (Ad g_{i})^{−1},

so F_{i} is also the set of eigenvalues of ad((Ad g_{i})U_{i}) : 𝔤 → 𝔤. Passing to the limit, {0} is the set of eigenvalues of ad X : 𝔤 → 𝔤. QED

Remark 4: Let V be a finite-dimensional real vector space.

(i) Let T_{i} be a sequence in GL(V). Assume that {T_{i}} is not precompact in GL(V). Then either {T_{i}} or {T_{i}^{−1}} is not precompact in End(V).

(ii) Let E_{i} be a sequence in End(V). Assume that {E_{i}} is not precompact in End(V). Then there exists v ∈ V such that E_{i}(v) is not precompact in V.

Proof: Proof of (i): For all i, let U_{i} := T_{i} ⊕ T_{i}^{−1} ∈ GL(V ⊕ V), so U_{i} is defined by U_{i}(v, w) = (T_{i}v, T_{i}^{−1}w). Then {U_{i}} is not precompact in GL(V ⊕ V). For all i, we have det(U_{i}) = 1, so {U_{i}} is not precompact in SL(V ⊕ V). So, since SL(V ⊕ V) is closed in End(V ⊕ V), we see that {U_{i}} is not precompact in End(V ⊕ V). Then, as {U_{i}} = {T_{i} ⊕ T_{i}^{−1}}, the result follows. End of proof of (i).

Proof of (ii): Let d := dim(V). Let v_{1}, …, v_{d} be a basis for V. Let ψ_{1}, …, ψ_{d} : V → ℝ be a basis for the dual space of V. For all i, let M_{i} := [ψ_{k}(E_{i}(v_{j}))]_{jk} be the matrix of E_{i} with respect to the two bases. Then {M_{i}}_{i} is not precompact in ℝ^{d×d}. Choose j, k ∈ {1, …, d} such that {ψ_{k}(E_{i}(v_{j}))}_{i} is not precompact in ℝ. Then {E_{i}(v_{j})}_{i} is not precompact in V. Let v := v_{j}. End of proof of (ii). QED

Let 𝔤 be a Lie algebra and let X, Y, T ∈ 𝔤. We will say that (X, Y, T) is a standard triple for 𝔤 if [T, X] = 2X, [T, Y] = −2Y and [X, Y] = T ≠ 0.

Let G be a Lie group and let X, Y, T ∈ 𝔤. For all u ∈ ℝ, let

n(u) := exp(uX),  \overline{n}(u) := exp(uY),  a(u) := exp(uT).

We will say that (X, Y, T) is a basic triple for G if, for all τ ∈ ℝ, for all δ ∈ ℝ\{0},

[n (\frac{e^{τ} − 1}{δ})] · [\overline{n}(δ)] · [n (\frac{e^{−τ} − 1}{δ})] = [\overline{n}(e^{−τ} δ)] · [a(τ)].

Remark 5: Let G be a Lie group. Then any standard triple for 𝔤 is a basic triple for G.

Proof: Let (X, Y, T) be a standard triple for 𝔤. We wish to show that (X, Y, T) is a basic triple for G. Let S := SL_{2}(ℝ) and let

X_{0} = [ \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ] ,  Y_{0} = [ \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ] ,  T_{0} = [ \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ] .

Then, for all u ∈ ℝ, we have

exp(uX_{0}) = [ \begin{matrix}1 & u \\ 0 & 1 \end{matrix} ] ,  exp(uY_{0}) = [ \begin{matrix}1 & 0 \\ u & 1 \end{matrix} ] ,  exp(uT_{0}) = [ \begin{matrix}e^{u} & 0 \\ 0 & e^{−u} \end{matrix} ] ,

so straightforward matrix computations prove that (X_{0}, Y_{0}, T_{0}) is a standard triple for 𝔰 and a basic triple for S. Let S̃ be the universal cover of S and let p : S̃ → S be a covering homomorphism. Then dp : 𝔰̃ → 𝔰 is an isomorphism of Lie algebras. Let X̃_{0} := (dp)^{−1}(X_{0}), let Ỹ_{0} := (dp)^{−1}(Y_{0}) and let T̃_{0} := (dp)^{−1}(T_{0}). Then (X̃_{0}, Ỹ_{0}, T̃_{0}) is a standard triple for 𝔰̃ and, by uniquness of liftings across covering maps, (X̃_{0}, Ỹ_{0}, T̃_{0}) is a basic triple for S̃.

Since (X̃_{0}, Ỹ_{0}, T̃_{0}) is a standard triple for 𝔰̃ and (X, Y, T) is a standard triple for 𝔤, there is a homomorphism of Lie algebras ϕ : 𝔰̃ → 𝔤 such that ϕ(X̃_{0}) = X, ϕ(Ỹ_{0}) = Y and ϕ(T̃_{0}) = T. By the Monodromy Theorem (see Theorem 2.7.5, on p. 71 of [8]), there is a Lie group homomorphism f : S̃ → G such that df = ϕ. Then, by naturality of the exponential map, we see that (X, Y, T) is a basic triple for G. QED

Proposition 6: Let S be a connected simple Lie group acting on a Hilbert space H by a unitary representation. Let w ∈ H. Let 𝔪 be the Lie algebra of Stab_{S}(w). Assume ∃X ∈ 𝔪\{0} such that ad X : 𝔫 → 𝔫 is nilpotent. Then S fixes w.

Proof: By Jacobson-Morozov (see, e.g., Theorem 7.4 of Chapter IX on p. 432 of [3]), choose Y, T ∈ 𝔰 such that (X, Y, T) is a standard triple for 𝔰. Then T ≠ 0. For all u ∈ ℝ, let

n(u) := exp(uX),  \overline{n}(u) := exp(uY),  a(u) := exp(uT).

Let A := {a(u)}_{u∈ℝ} be the image of a : ℝ → S.

Claim α: w is A-invariant. Proof of Claim α: Let N := {n(u)}_{u∈ℝ} be the image of n : ℝ → S. Since X ∈ 𝔪, we see that w is N-invariant.

Define f : S → ℝ by f(s) = Re(⟨sw, w⟩). For all s ∈ S, and all p, q ∈ N, we have ⟨psqw, w⟩ = ⟨s(qw), p^{−1}w⟩ = ⟨sw, w⟩, so f(psq) = f(g). That is, f is bi-invariant under N.

Let a_{0} ∈ A. We wish to show that a_{0}w = w. We have f(1_{S}) = ∥w∥^{2} and

2(f(a_{0})) = −∥a_{0}w − w∥^{2} + ∥a_{0}w∥^{2} + ∥w∥^{2} = −∥a_{0}w − w∥^{2} + 2∥w∥^{2},

so 2(f(a_{0})) − 2(f(1_{S})) = −∥a_{0}w − w∥^{2}. It therefore suffices to prove f(a_{0}) = f(1_{S}).

Fix τ ∈ ℝ such that a_{0} = a(τ). Let δ_{i} be a sequence of nonzero real numbers such that δ_{i} → 0 in ℝ. For all i, let

n_{i} := n (\frac{e^{τ} − 1}{δ_{i}}) ,  \overline{n}_{i} := \overline{n}(δ_{i}),  n′_{i} := n (\frac{e^{−τ} − 1}{δ_{i}}) .

By Remark 5, since (X, Y, T) is a standard triple for 𝔰, (X, Y, T) is a basic triple for S. Then, for all i, we have n_{i}\overline{n}_{i}n′_{i} = [\overline{n}(e^{−τ} δ_{i})] · a_{0}, so, since δ_{i} → 0 in ℝ, we get n_{i}\overline{n}_{i}n′_{i} → a_{0} in S. Moreover, \overline{n}_{i} = \overline{n}(δ_{i}) → \overline{n}(0) = 1_{S}. By bi-invariance of f under N, for all i, we have f(n_{i}\overline{n}_{i}n′_{i}) = f(\overline{n}_{i}), so, passing to the limit, f(a_{0}) = f(1_{S}). End of proof of Claim α.

For all λ ∈ ℝ, let 𝔰_{λ} := {A ∈ 𝔰 | (ad T)A = λA}.

Claim β: For all λ ∈ ℝ\{0}, we have 𝔰_{λ} ⊆ 𝔪. Proof of Claim β: Fix λ ∈ ℝ\{0} and A ∈ 𝔰_{λ}. We wish to show that A ∈ 𝔪.

Fix a sequence r_{i} in ℝ such that r_{i}λ → +∞. For all i, let g_{i} := a(r_{i}) = exp(r_{i}T); then, as A ∈ 𝔰_{λ}, we get (ad T)A = λA, so (Ad g_{i})A = e^{r_{i}λ}A, so (Ad g_{i})(e^{−r_{i}λ}A) = A. Moreover, by Claim α, we have g_{i}w = w. Then, by Lemma 1 (with v := w, with U_{i} replaced by e^{−r_{i}λ}A and with X replaced by A), we are done. End of proof of Claim β.

Let 𝔰_{∗} := (ad T)𝔰 and let 𝔰′ be the Lie subalgebra of 𝔰 generated by the set 𝔰_{∗}. By, e.g., Corollary 4.11.7 on p. 130 of [1], ad T : 𝔰 → 𝔰 is real diagonalizable. Then 𝔰_{∗} = ∑_{λ∈ℝ\{0}} 𝔰_{λ}

and 𝔰 = 𝔰_{0} + 𝔰_{∗} and, from Claim β, we get 𝔰_{∗} ⊆ 𝔪. Then 𝔰′ ⊆ 𝔪.

The Jacobi identity implies that 𝔠_{𝔰}(T) normalizes the set (ad T)𝔰, i.e., that 𝔰_{0} normalizes 𝔰_{∗}. Then 𝔰_{0} normalizes 𝔰′. Moreover, 𝔰_{∗} ⊆ 𝔰′, so 𝔰_{∗} normalizes 𝔰′. Then 𝔰_{0} + 𝔰_{∗} normalizes 𝔰′, i.e., 𝔰 normalizes 𝔰′. That is, 𝔰′ is an ideal in 𝔰. Since T ≠ 0, we have ad_{𝔰}(T) ≠ 0, so 𝔰_{∗} ≠ {0}, so 𝔰′ ≠ {0}. Then, by simplicity of 𝔰, it follows that 𝔰′ = 𝔰. Then 𝔰 = 𝔰′ ⊆ 𝔪. That is, S fixes w. QED

The proof of Claim α above is an algebraic argument derived from the geometry of the proof of Theorem 2.4.2 on p. 29 of [9].

The following is Lemma 4.15.7 on p. 144 of [1]:

Lemma 7: Let 𝔤 be a Lie algebra with no simple direct summand and let 𝔫 be the nilradical of 𝔤. Then 𝔠_{𝔤}(𝔫) = 𝔷(𝔫).

Proof: Let 𝔰 be a Levi factor of 𝔤 and let 𝔯 be the solvable radical of 𝔤. Then, because 𝔠_{𝔤}(𝔫) is an ideal in 𝔤, it follows (by, e.g., Lemma 4.10.19 on p. 119 of [1]) that

𝔠_{𝔤}(𝔫)  =  [(𝔠_{𝔤}(𝔫)) ∩ 𝔰]  +  [(𝔠_{𝔤}(𝔫)) ∩ 𝔯],

i.e., that 𝔠_{𝔤}(𝔫) = (𝔠_{𝔰}(𝔫)) + (𝔠_{𝔯}(𝔫)).

By (ii) of Theorem 3.8.3 on p. 206 of [8], we have [𝔤, 𝔯] ⊆ 𝔫. It follows that [𝔰, 𝔯] ⊆ 𝔫, i.e., that ad : 𝔰 → 𝔤𝔩(𝔯/𝔫) is zero. On the other hand, as 𝔤 has no simple direct summand, it follows that ad : 𝔰 → 𝔤𝔩(𝔯) is faithful. Therefore, by complete reducibility of ad : 𝔰 → 𝔤𝔩(𝔯), ad : 𝔰 → 𝔤𝔩(𝔫) is faithful. That is, 𝔠_{𝔰}(𝔫) = {0}. Then 𝔠_{𝔤}(𝔫) = (𝔠_{𝔰}(𝔫)) + (𝔠_{𝔯}(𝔫)) = 𝔠_{𝔯}(𝔫).

We have [𝔠_{𝔯}(𝔫), 𝔠_{𝔯}(𝔫)] ⊆ [𝔤, 𝔯] ⊆ 𝔫 and [𝔠_{𝔯}(𝔫), 𝔫] = {0}. Then 𝔠_{𝔯}(𝔫) is a nilpotent ideal of 𝔤, so 𝔠_{𝔯}(𝔫) ⊆ 𝔫. Then 𝔠_{𝔤}(𝔫) = 𝔠_{𝔯}(𝔫) ⊆ 𝔫, so 𝔠_{𝔤}(𝔫) = (𝔠_{𝔤}(𝔫)) ∩ 𝔫 = 𝔷(𝔫). QED

Lemma 8: Let 𝔤 be a Lie algebra with no simple direct summand. Let 𝔫 be the nilradical of 𝔤. Let 𝔪 be a vector subspace of 𝔤. Assume ∀T ∈ 𝔪, (𝔠_{𝔤}(T)) ∩ ((ad T)𝔤) ⊆ 𝔪. Assume ∃X ∈ 𝔪\{0} such that ad X : 𝔫 → 𝔫 is nilpotent. Then 𝔪 ∩ (𝔷(𝔫)) is a nonzero ideal of 𝔤.

Proof: As 𝔷(𝔫) is an Abelian ideal of 𝔤, for all T ∈ 𝔷(𝔫), we get (ad T)𝔤 ⊆ (𝔠_{𝔤}(T)) ∩ (𝔷(𝔫)). Then, for all T ∈ 𝔪 ∩ (𝔷(𝔫)), we have (ad T)𝔤 ⊆ (𝔠_{𝔤}(T)) ∩ ((ad T)𝔤) ∩ (𝔷(𝔫)) ⊆ 𝔪 ∩ (𝔷(𝔫)). That is, 𝔪 ∩ (𝔷(𝔫)) is an ideal of 𝔤. It remains to show that 𝔪 ∩ (𝔷(𝔫)) ≠ {0}.

Define the descending central series of 𝔫 by 𝔫^{(1)} := 𝔫 and 𝔫^{(i+1)} := [𝔫, 𝔫^{(i)}]. Fix an integer l ≥ 1 such that 𝔫^{(l)} = {0}. Say a sequence P_{0}, P_{1}, P_{2}, … in 𝔤 is good if (1) P_{0} ∈ 𝔪\{0};

(2) ad P_{0} : 𝔫 → 𝔫 is nilpotent; and

(3) for all integers i ≥ 0, we have P_{i+1} ∈ (𝔠_{𝔫}(P_{i})) ∩ ((ad P_{i})𝔫). By assumption, X, 0, 0, 0, … is good. For any good sequence P_{i}, by induction, for all integers i ≥ 1, we have P_{i} ∈ 𝔫^{(i)}; in particular, P_{l} = 0.

Fix a good sequence P_{i} such that k := max{i | P_{i} ≠ 0} is as large as possible. Then (𝔠_{𝔫}(P_{k})) ∩ ((ad P_{k})𝔫) = {0}. If k = 0, then, by (2), ad_{𝔫}(P_{k}) is nilpotent. If k ≥ 1, then P_{k} ∈ 𝔫^{(k)} ⊆ 𝔫, and so, again, ad_{𝔫}(P_{k}) is nilpotent.

Then (𝔠_{𝔫}(P_{k})) ∩ ((ad P_{k})𝔫) is intersection of the kernel and image of the nilpotent map ad_{𝔫}(P_{k}). However, (𝔠_{𝔫}(P_{k})) ∩ ((ad P_{k})𝔫) = {0}, while the intersection of the kernel and image of a nonzero nilpotent endomorphism is never {0}. Then ad_{𝔫}(P_{k}) = 0, i.e., P_{k} ∈ 𝔠_{𝔤}(𝔫). So, as 𝔤 has no simple direct summand, by Lemma 7, we get P_{k} ∈ 𝔷(𝔫).

For all integers i ≥ 0, we have P_{i+1} ∈ (𝔠_{𝔫}(P_{i})) ∩ ((ad P_{i})𝔫) ⊆ (𝔠_{𝔤}(P_{i})) ∩ ((ad P_{i})𝔤). So, by induction, for all integers i ≥ 0, we have P_{i} ∈ 𝔪. Then 0 ≠ P_{k} ∈ 𝔪 ∩ (𝔷(𝔫)). QED

IV. Decay to zero at Adjoint infinity for connected Lie groups

Theorem 9: Let G be a connected Lie group acting on a Hilbert space H by a unitary representation π : G → U(H). Assume that (∗) no nonzero vector of H is fixed by a nontrivial normal connected subgroup of G. Let g_{i} be a sequence in G and assume that Ad_{𝔤}(g_{i}) leaves compact subsets of GL(𝔤). Then π(g_{i}) → 0 in the weak-operator topology on B(H).

Notes:

1. Condition (∗) is satisfied if G is simple and H admits no G-invariant vectors.

2. Condition (∗) is satisfied if π is faithful and irreducible. (The set of vectors fixed by a normal subgroup is a G-invariant subspace. By irreducibility, if such a subspace were nonzero, it would equal H. Then, by faithfulness, the subgroup would be trivial.)

3. I am not sure whether, in Theorem 9, we need that G is connected.

4. One may summarize Theorem 9 as asserting: If a unitary representation of a connected Lie group satisfies (∗), then its matrix coefficients decay to zero at “Ad-infinity”.

Proof of Theorem 9: Let B(H) have the weak-operator topology. Assume, for a contradiction, that π(g_{i}) ↛ 0 in B(H).

Passing to a subsequence, assume that {π(g_{i})} is bounded away from 0 in B(H). For all i, π(g_{i}^{−1}) = (π(g_{i}))^{∗}, so, because L ↦ L^{∗} : B(H) → B(H) is continuous, we see that {π(g_{i}^{−1})} is bounded away from 0 in B(H), as well. By (i) of Remark 4, either {Ad_{𝔤}(g_{i})} or {Ad_{𝔤}(g_{i}^{−1})} is not precompact in End(𝔤). By, if necessary, replacing each g_{i} with g_{i}^{−1}, assume that {Ad_{𝔤}(g_{i})} is not precompact in End(𝔤). Passing to a subsequence, assume that Ad_{𝔤}(g_{i}) leaves compact subsets of End(𝔤). Fix v ∈ H such that g_{i}v ⇀̸ 0 in H. Passing to a subsequence, assume that g_{i}v ⇀ w ≠ 0 in H.

Choose normal connected Lie subgroups G′ and G′′ of G such that 𝔤 = 𝔤′ + 𝔤′′, such that 𝔤′ ∩ 𝔤′′ = {0}, such that 𝔤′ is semisimple or trivial and such that 𝔤′′ has no simple direct summand. It follows either that {Ad_{𝔤′}(g_{i})} is not precompact in End(𝔤′) or that {Ad_{𝔤′′}(g_{i})} is not precompact in End(𝔤′′).

Case A: {Ad_{𝔤′}(g_{i})} is not precompact in End(𝔤′). Choose a simple normal connected Lie subgroup S of G′ such that {Ad_{𝔰}(g_{i})} is not precompact in End(𝔰). By (ii) of Remark 4, choose U ∈ 𝔰 such that {(Ad g_{i})U} is not precompact in 𝔰.

Passing to a subsequence, choose t_{i} → 0 in ℝ such that t_{i}(Ad g_{i})U → X ≠ 0 in 𝔰. By Remark 3 (with 𝔤 replaced by 𝔰 and U_{i} replaced by t_{i}U), ad X : 𝔰 → 𝔰 is nilpotent. Let 𝔪 be the Lie algebra of Stab_{S}(w). By Lemma 1 (with G replaced by S and U_{i} replaced by t_{i}U), we see that X ∈ 𝔪. Then, by Proposition 6, S fixes w. However, S is normal in G′, and therefore in G, contradicting (∗). End of Case A.

Case B: {Ad_{𝔤′′}(g_{i})} is not precompact in End(𝔤′′). By (ii) of Remark 4, choose U ∈ 𝔤′′ such that {(Ad g_{i})U} is not precompact in 𝔤′′.

Passing to a subsequence, choose t_{i} → 0 in ℝ such that t_{i}(Ad g_{i})U → X ≠ 0 in 𝔤′′. By Remark 3 (with 𝔤 replaced by 𝔤′′ and U_{i} replaced by t_{i}U), ad X : 𝔤′′ → 𝔤′′ is nilpotent. Let 𝔪 be the Lie algebra of Stab_{G′′}(w). By Lemma 1 (with G replaced by G′′ and U_{i} replaced by t_{i}U), we get X ∈ 𝔪. By Corollary 2 (with G replaced by G′′), for all T ∈ 𝔪, we have (𝔠_{𝔤′′}(T)) ∩ ((ad T)𝔤′′) ⊆ 𝔪. Let 𝔫 be the nilradical of 𝔤′′. Then ad X : 𝔫 → 𝔫 is nilpotent. Then, by Lemma 8 (with 𝔤 replaced by 𝔤′′), 𝔪∩(𝔷(𝔫)) is a nonzero ideal of 𝔤′′, and therefore of 𝔤. Then the connected Lie subgroup of G corresponding to 𝔪 ∩ (𝔷(𝔫)) is a nontrivial normal connected subgroup of G fixing w, contradicting (∗). End of Case B. QED

Thanks to D. Witte Morris for help in developing the following example:

Example 10: In the statement of Theorem 9, (∗) cannot be replaced by (∗′) no nonzero vector of H is fixed by a noncompact normal connected subgroup of G even under the assumption that π is faithful.

Proof: Let H be the 3-dimensional Heisenberg group, and let Z denote the center of H. Let ϕ : H → ℝ^{2} be a surjective homomorphism whose kernel is Z, and let D be a nontrivial discrete subgroup of Z. Let G := H/D and let p : H → G be the canonical homomorphism. Let ψ : G → ℝ^{2} be defined by ψ(p(g)) = ϕ(g).

Let H′ be a Hilbert space and let ρ : ℝ^{2} → U(H′) be a unitary representation of ℝ^{2} with no nonzero invariant vectors such that not all matrix coefficients decay to zero at infinity. Let π′ := ρ ∘ ψ : G → U(H′). Let H′′ be another Hilbert space and fix a faithful unitary representation π′′ : G → U(H′′).

Let H := H′ ⊕ H′′. Then π := π′ ⊕ π′′ : G → U(H) is faithful and satisfies (∗′), but does not enjoy the property that all matrix coefficients decay to zero at Ad-infinity. QED

V. Decay to zero at projective infinity for algebraic groups

Theorem 11: Let G be the connected real points of a linear algebraic ℝ-group. Let π : G → U(H) be an irreducible unitary representation on a Hilbert space H. Then π(g) → 0 in the weak-operator topology on B(H), as g leaves compact subsets of G modulo the projective kernel of π.

Proof: Let K := ker(π). Let \overline{G} := G/K. Let p : G → \overline{G} be the canonical homomorphism. For all g ∈ G, let \overline{g} := p(g). Define \overline{π} : \overline{G} → U(H) by \overline{π}(\overline{g}) = π(g).

The Adjoint representation Ad : G → GL(𝔤) is algebraic and 𝔨 is an invariant subspace, so Ad : G → GL(𝔤/𝔨) is algebraic and therefore has closed image. So, since \overline{𝔤} = 𝔤/𝔨 and since Ad\overline{𝔤}(\overline{G}) = Ad\overline{𝔤}(G), we see that Ad\overline{𝔤}(\overline{G}) is closed in GL(\overline{𝔤}). Then Ad : \overline{G} → GL(\overline{𝔤}) factors to a proper, injective Lie group homomorphism F : \overline{G}/(Z(\overline{G})) → GL(\overline{𝔤}).

Assume that g_{i} leaves compact subsets of G modulo the projective kernel of π. We wish to show that π(g_{i}) → 0 in the weak-operator topology on U(H).

The sequence \overline{g_{i}} leaves compact subsets of \overline{G} modulo the projective kernel of \overline{π}. By Schur’s Lemma, since \overline{π} is faithful and irreducible, its projective kernel is Z(\overline{G}). So, since F is proper, Ad \overline{g_{i}} leaves compact subsets of GL(\overline{𝔤}). Then, by Theorem 9, \overline{π}(\overline{g_{i}}) → 0 in the weak-operator topology on U(H). For all i, \overline{π}(\overline{g_{i}}) = π(g_{i}), so we are done. QED

The proof of Theorem 11 also works for any connected Lie group that is locally isomorphic to a real linear algebraic group.

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