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Uniformly bounded representations

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

A representation of a group by bounded invertible operators on a Hilbert space is uniformly bounded if the norms of all its operators have a common bound. Every unitary representation is uniformly bounded, and so is every representation of the form \(g\mapsto S\,u(g)\,S^{-1}\) with \(u\) unitary and \(S\) a fixed bounded invertible operator. Day and Dixmier showed in 1950 that for an amenable group nothing else occurs: every uniformly bounded representation is of that form (Theorem 1.2). Dixmier asked whether this property characterizes amenable groups. Ehrenpreis and Mautner found uniformly bounded representations of \(SL_2(\mathbb R)\) that are not similar to unitary ones, and such representations were later found for free groups, hence for every group with a free subgroup of rank two; the survey [Pi] describes this work, and [EM] and [MO] treat further classes of groups. OpenAI proved in September 2026 that the answer is yes for every discrete group [OpenAI-U]:

Theorem. A discrete group is amenable if and only if every uniformly bounded representation of it on a complex Hilbert space is similar to a unitary representation. If the group is not amenable, then for every \(\varepsilon>0\) it has such a representation with all norms at most \(1+\varepsilon\) that is not similar to a unitary one, on a separable Hilbert space if the group is countable.

This course proves the theorem. The present lesson collects the general facts: similarity to a unitary representation as an invariant inner product (Proposition 1.1), the theorem of Day and Dixmier (Theorem 1.2), representations built from operator cocycles (Proposition 2.2), and two ways of passing between a group and its subgroups (Lemmas 3.1 and 3.3). The lesson Contracting averages and sparse assignments turns nonamenability into a combinatorial structure, the lesson Random sign frames and masked operators turns that structure into operators, and the lesson Unitarizability implies amenability proves the theorem.

We use: Theorem 3.1 of Hilbert spaces and compact operators (a bounded sesquilinear form \(B\) is \(B(\xi,\eta)=\langle t\xi,\eta\rangle\) for a unique bounded \(t\), with \(\langle t\xi,\xi\rangle\ge0\) when \(B\) is positive); Proposition 8.5(10) of C*-algebras: continuous functional calculus (positive square roots in a C*-algebra, commuting with everything that commutes with the given element); the inverse mapping theorem, Corollary 5.2 of the Banach space lesson; the Banach–Alaoglu theorem, Theorem 3.1 of the weak topologies lesson; and the definition of amenability by invariant means in Section 1 of Means, Følner sets, and regular representations.

1. Similarity to unitary representations

Throughout, \(G\) is a group, regarded as a discrete group, and Hilbert spaces are complex. For a Hilbert space \(H\), \(GL(H)\) is the group of bounded operators on \(H\) with bounded inverse. A representation \(\pi:G\to GL(H)\) is a homomorphism, and it is uniformly bounded if \[ |\pi|=\sup_{g\in G}\|\pi(g)\|<\infty. \] Then \(\|\pi(g)^{-1}\|=\|\pi(g^{-1})\|\le|\pi|\) as well. The representation is unitarizable if some \(S\in GL(H)\) makes every \(S\pi(g)S^{-1}\) unitary, and the group \(G\) is unitarizable if every uniformly bounded representation of \(G\) on every complex Hilbert space is unitarizable.

An inner product \([\cdot,\cdot]\) on \(H\) is equivalent to \(\langle\cdot,\cdot\rangle\) if \(a\|\xi\|^2\le[\xi,\xi]\le b\|\xi\|^2\) for some constants \(0<a\le b\).

Proposition 1.1. For a uniformly bounded representation \(\pi:G\to GL(H)\) the following are equivalent:

(a) \(\pi\) is unitarizable;

(b) there is a bounded self-adjoint operator \(Q\) with \(a\le Q\le b\) for constants \(0<a\le b\) such that \(\pi(g)^*Q\pi(g)=Q\) for every \(g\);

(c) \(H\) carries an equivalent inner product \([\cdot,\cdot]\) with \([\pi(g)\xi,\pi(g)\eta]=[\xi,\eta]\) for all \(g,\xi,\eta\).

Proof. (a) ⇒ (b). Let \(V(g)=S\pi(g)S^{-1}\) be unitary and put \(Q=S^*S\). Then \(\pi(g)=S^{-1}V(g)S\) and \(\pi(g)^*Q\pi(g)=S^*V(g)^*V(g)S=Q\). Moreover \(\langle Q\xi,\xi\rangle=\|S\xi\|^2\) lies between \(\|S^{-1}\|^{-2}\|\xi\|^2\) and \(\|S\|^2\|\xi\|^2\).

(b) ⇒ (c). Put \([\xi,\eta]=\langle Q\xi,\eta\rangle\).

(c) ⇒ (b). The form \([\cdot,\cdot]\) is bounded: by the Cauchy–Schwarz inequality for it, \(|[\xi,\eta]|\le[\xi,\xi]^{1/2}[\eta,\eta]^{1/2}\le b\|\xi\|\|\eta\|\). Theorem 3.1 of the Hilbert space lesson gives a bounded \(Q\) with \([\xi,\eta]=\langle Q\xi,\eta\rangle\); it is self-adjoint because \(\langle Q\xi,\xi\rangle=[\xi,\xi]\) is real, and \(a\le Q\le b\). Invariance gives \(\langle\pi(g)^*Q\pi(g)\xi,\eta\rangle=[\pi(g)\xi,\pi(g)\eta]\), which is \([\xi,\eta]=\langle Q\xi,\eta\rangle\).

(b) ⇒ (a). First, \(Q\) is invertible: \(a\|\xi\|^2\le\langle Q\xi,\xi\rangle\le\|Q\xi\|\|\xi\|\) gives \(\|Q\xi\|\ge a\|\xi\|\), so \(Q\) is injective with closed range, and the range is dense because its orthogonal complement is \(\ker Q^*=\ker Q=0\). So \(Q\) is a bijection, and \(\|Q^{-1}\|\le1/a\). Let \(S=Q^{1/2}\) be the positive square root (Proposition 8.5(10) of the C*-algebra lesson). It commutes with \(Q\), hence with \(Q^{-1}\), and \(S\cdot SQ^{-1}=SQ^{-1}\cdot S=1\); so \(S\in GL(H)\). Now \(V(g)=S\pi(g)S^{-1}\) is invertible and \[ \begin{aligned} V(g)^*V(g)&=S^{-1}\pi(g)^*S^2\pi(g)S^{-1}\\ &=S^{-1}QS^{-1}=1, \end{aligned} \] so \(V(g)\) is an invertible isometry, that is, a unitary. \(\square\)

Theorem 1.2 (Day, Dixmier). Every amenable group is unitarizable.

Proof. Let \(m\) be a left invariant mean on \(\ell^\infty(G)\): a positive linear functional with \(m(1)=1\) and \(m(L_hf)=m(f)\), where \((L_hf)(x)=f(h^{-1}x)\). A positive functional takes real values on real functions, so \(m(\bar f)=\overline{m(f)}\). Put \(m_{\rm r}(f)=m(x\mapsto f(x^{-1}))\). It is again positive with \(m_{\rm r}(1)=1\), and it is right invariant: the function \(x\mapsto f(x^{-1}h)\) is \(L_h\) applied to \(x\mapsto f(x^{-1})\), so \(m_{\rm r}(x\mapsto f(xh))=m_{\rm r}(f)\).

Let \(\pi:G\to GL(H)\) be uniformly bounded, \(M=|\pi|\). For \(\xi,\eta\in H\) the function \(g\mapsto\langle\pi(g)\xi,\pi(g)\eta\rangle\) is bounded by \(M^2\|\xi\|\|\eta\|\); put \[ [\xi,\eta]=m_{\rm r}\bigl(g\mapsto\langle\pi(g)\xi,\pi(g)\eta\rangle\bigr). \] This is linear in \(\xi\), conjugate-linear in \(\eta\), and \([\eta,\xi]=\overline{[\xi,\eta]}\). Since \(\|\xi\|=\|\pi(g)^{-1}\pi(g)\xi\|\le M\|\pi(g)\xi\|\), the function \(g\mapsto\|\pi(g)\xi\|^2\) takes values in \([M^{-2}\|\xi\|^2,M^2\|\xi\|^2]\), and positivity of \(m_{\rm r}\) gives \(M^{-2}\|\xi\|^2\le[\xi,\xi]\le M^2\|\xi\|^2\). So \([\cdot,\cdot]\) is an equivalent inner product. For \(h\in G\), the function \(g\mapsto\langle\pi(gh)\xi,\pi(gh)\eta\rangle\) is the right translate by \(h\) of the function defining \([\xi,\eta]\), so \([\pi(h)\xi,\pi(h)\eta]=[\xi,\eta]\). Proposition 1.1 shows that \(\pi\) is unitarizable. \(\square\)

2. Operator cocycles

Let \(U:G\to GL(H)\) be a unitary representation. A bounded operator cocycle for \(U\) is a map \(D:G\to B(H)\) with \(\sup_g\|D(g)\|<\infty\) and \[ \begin{aligned} &D(gh)=D(g)\\ &\quad +U(g)D(h)U(g)^{-1}\\ &(g,h\in G). \end{aligned}\tag{2.1} \] It is inner if there is \(B\in B(H)\) with \(D(g)=B-U(g)BU(g)^{-1}\) for all \(g\).

Example 2.1. For every \(T\in B(H)\), the map \(D_T(g)=T-U(g)TU(g)^{-1}\) satisfies (2.1): \[ \begin{aligned} &D_T(g)+U(g)D_T(h)U(g)^{-1}=T\\ &\quad -U(g)TU(g)^{-1}+U(g)TU(g)^{-1}\\ &\quad -U(gh)TU(gh)^{-1}=D_T(gh). \end{aligned} \] It is inner, implemented by \(T\). The construction of the course takes direct sums of cocycles \(D_{T_j}\) on spaces \(H_j\) for which \(\sup_{j,g}\|D_{T_j}(g)\|<\infty\) while \(\|T_j\|\) is unbounded. The direct sum of the \(D_{T_j}\) is a bounded cocycle, and the question is whether some other bounded operator implements it.

Proposition 2.2. Let \(D\) be a bounded operator cocycle for a unitary representation \(U\) on \(H\). On \(H\oplus H\), \[ \pi(g)=\begin{pmatrix}U(g)&D(g)U(g)\\0&U(g)\end{pmatrix} \] defines a uniformly bounded representation with \(\|\pi(g)\|\le1+\|D(g)\|\). If \(\pi\) is unitarizable, then \(D\) is inner.

Proof. Setting \(g=h=e\) in (2.1) gives \(D(e)=0\), so \(\pi(e)=1\). The upper right entry of \(\pi(g)\pi(h)\) is \(U(g)D(h)U(h)+D(g)U(g)U(h)\), which equals \(D(gh)U(gh)\) by (2.1) applied after multiplication by \(U(gh)\) on the right. So \(\pi(g)\pi(h)=\pi(gh)\), and \(\pi(g)^{-1}=\pi(g^{-1})\). The factorization \[ \pi(g)=\begin{pmatrix}1&D(g)\\0&1\end{pmatrix}\begin{pmatrix}U(g)&0\\0&U(g)\end{pmatrix} \] into the identity plus an operator of norm \(\|D(g)\|\) and a unitary gives the norm bound.

Suppose \(S\in GL(H\oplus H)\) makes every \(V(g)=S\pi(g)S^{-1}\) unitary. The closed subspace \(M=H\oplus0\) satisfies \(\pi(g)M\subseteq M\) for every \(g\), hence \(\pi(g)^{-1}M=\pi(g^{-1})M\subseteq M\). So the closed subspace \(SM\) is invariant under every \(V(g)\) and \(V(g)^{-1}=V(g)^*\), and therefore its orthogonal complement is invariant under every \(V(g)\). Thus \(N=S^{-1}((SM)^\perp)\) is a closed subspace, invariant under every \(\pi(g)\), with \(M\cap N=0\) and \(M+N=H\oplus H\), because \(SM\) and \((SM)^\perp\) have these properties and \(S\) is a linear homeomorphism.

Let \(P_1,P_2:H\oplus H\to H\) be the coordinate projections. The restriction \(P_2|_N:N\to H\) is injective, since its kernel is \(M\cap N\), and surjective, since \((0,\eta)=m+n\) with \(m\in M\), \(n\in N\) gives \(P_2n=\eta\). By the inverse mapping theorem it has a bounded inverse, and \(B=P_1(P_2|_N)^{-1}\in B(H)\) satisfies \(N=\{(B\eta,\eta):\eta\in H\}\). Invariance of \(N\) under \(\pi(g)\) says that the vector \(\pi(g)(B\eta,\eta)\), which is \((U(g)B\eta+D(g)U(g)\eta,\ U(g)\eta)\), lies in \(N\), that is, \[ \begin{gathered} U(g)B\eta+D(g)U(g)\eta=BU(g)\eta\\ (\eta\in H). \end{gathered} \] Substituting \(\eta=U(g)^{-1}\zeta\) gives \(D(g)=B-U(g)BU(g)^{-1}\). \(\square\)

The converse holds as well (Exercise 5.2).

3. Subgroups

Lemma 3.1 (induction). Let \(K\) be a subgroup of \(G\) and \(\sigma:K\to GL(E)\) a uniformly bounded representation. There is a uniformly bounded representation \(\Pi\) of \(G\) on \(\ell^2(G/K;E)\) with \(|\Pi|=|\sigma|\). If \(\Pi\) is unitarizable, so is \(\sigma\). If \(G\) is countable and \(E\) separable, then \(\ell^2(G/K;E)\) is separable.

Proof. Let \(X=G/K\) be the set of left cosets and choose representatives \(s:X\to G\), \(s(x)\in x\), with \(s(K)=e\). For \(g\in G\) and \(x\in X\) put \(c(g,x)=s(gx)^{-1}g\,s(x)\). Since \(g\,s(x)\in gx=s(gx)K\), we have \(c(g,x)\in K\). Directly, \[ \begin{aligned} &c(gh,x)\\ &\quad =s(ghx)^{-1}g\,s(hx)\cdot s(hx)^{-1}h\,s(x)\\ &\quad =c(g,hx)\,c(h,x),\\ &c(e,x)=e. \end{aligned} \] For \(\xi=(\xi_x)_{x\in X}\) in the Hilbert direct sum \(\ell^2(X;E)\) define \(\Pi(g)\xi\) by \((\Pi(g)\xi)_{gx}=\sigma(c(g,x))\xi_x\). As \(x\mapsto gx\) is a bijection of \(X\), \(\|\Pi(g)\xi\|^2=\sum_x\|\sigma(c(g,x))\xi_x\|^2\), which is at most \(|\sigma|^2\|\xi\|^2\). The cocycle identity gives \[ \begin{aligned} &(\Pi(g)\Pi(h)\xi)_{ghx}\\ &\quad =\sigma(c(g,hx))\sigma(c(h,x))\xi_x\\ &\quad =\sigma(c(gh,x))\xi_x=(\Pi(gh)\xi)_{ghx}, \end{aligned} \] and \(\Pi(e)=1\), so \(\Pi\) is a representation with \(|\Pi|\le|\sigma|\).

Let \(j:E\to\ell^2(X;E)\) be the isometric inclusion at the coset \(K\). For \(k\in K\) we have \(kK=K\) and \(c(k,K)=s(K)^{-1}k\,s(K)=k\), so \(\Pi(k)j=j\sigma(k)\). Hence \(\|\sigma(k)\|\le\|\Pi(k)\|\), and \(|\Pi|=|\sigma|\).

If \(\Pi\) is unitarizable, Proposition 1.1 gives \(Q\) with \(a\le Q\le b\) and \(\Pi(g)^*Q\Pi(g)=Q\). The compression \(Q_K=j^*Qj\) satisfies \(a\le Q_K\le b\), and for \(k\in K\), using \(\Pi(k)j=j\sigma(k)\), \[ \begin{aligned} \sigma(k)^*Q_K\sigma(k)&=j^*\Pi(k)^*Q\Pi(k)j\\ &=j^*Qj=Q_K. \end{aligned} \] By Proposition 1.1, \(\sigma\) is unitarizable. If \(G\) is countable, \(X\) is countable, and \(\ell^2(X;E)\) is separable when \(E\) is. \(\square\)

Corollary 3.2. Every subgroup of a unitarizable group is unitarizable. \(\square\)

Lemma 3.3. If every finitely generated subgroup of \(G\) is amenable, then \(G\) is amenable. Consequently every nonamenable group has a finitely generated, hence countable, nonamenable subgroup.

Proof. Let \(\mathcal M\) be the set of means on \(\ell^\infty(G)\). A mean has norm \(m(1)=1\): for \(|f|\le1\) and a scalar \(\theta\) with \(|\theta|=1\) and \(\theta m(f)=|m(f)|\), we have \(|m(f)|=m(\operatorname{Re}\theta f)\le m(1)\). The set \(\mathcal M\) is weak* closed in the unit ball of \(\ell^\infty(G)^*\), which is weak* compact by the Banach–Alaoglu theorem; so \(\mathcal M\) is weak* compact.

For a finitely generated subgroup \(K\), choose a left invariant mean \(m_K\) on \(\ell^\infty(K)\) and put \(\mu_K(f)=m_K(f|_K)\). For \(k\in K\) and \(x\in K\), \((L_kf)(x)=f(k^{-1}x)\) with \(k^{-1}x\in K\), so \((L_kf)|_K=L_k(f|_K)\) and \(\mu_K(L_kf)=\mu_K(f)\). Let \(C_K\subseteq\mathcal M\) be the weak* closure of \(\{\mu_{K'}:K'\supseteq K\text{ finitely generated}\}\). Each \(C_K\) is nonempty, and \(C_{K_1}\cap\dots\cap C_{K_n}\supseteq C_{\langle K_1,\dots,K_n\rangle}\neq\varnothing\). By compactness some \(\mu\) lies in every \(C_K\). For \(g\in G\) and \(f\in\ell^\infty(G)\), the set \(\{\nu\in\mathcal M:\nu(L_gf)=\nu(f)\}\) is weak* closed and contains every \(\mu_{K'}\) with \(K'\supseteq\langle g\rangle\); hence it contains \(C_{\langle g\rangle}\ni\mu\). So \(\mu\) is a left invariant mean on \(\ell^\infty(G)\). \(\square\)

4. The plan of the proof

By Lemma 3.3 and Lemma 3.1, it suffices to produce, for every countable nonamenable group \(G\) and every \(\varepsilon>0\), a representation \(\pi\) with \(|\pi|\le1+\varepsilon\) that is not unitarizable. By Proposition 2.2 it suffices to find a unitary representation \(U\) and a cocycle \(D\) with \(\sup_g\|D(g)\|\le\varepsilon\) that is not inner. The representation \(U\) will be a direct sum of left regular representations on \(\ell^2(G;\mathbb C^{k_j})\), and \(D\) a direct sum of cocycles \(D_{T_j}\) as in Example 2.1. The operators \(T_j\) have uniformly bounded conjugation differences \(T_j-U(g)T_jU(g)^{-1}\), but every operator implementing their direct sum would have to be far from the commutant of \(U\) in every summand by an amount that grows with \(j\).

5. Exercises

Exercise 5.1 (easy). Let \(\pi:G\to GL(H)\) be a representation, not assumed uniformly bounded, that preserves an inner product \([\cdot,\cdot]\) with \(a\|\xi\|^2\le[\xi,\xi]\le b\|\xi\|^2\). Show that \(|\pi|\le(b/a)^{1/2}\).

Exercise 5.2 (easy). In Proposition 2.2, suppose \(D(g)=B-U(g)BU(g)^{-1}\) with \(B\in B(H)\). Show that \(R=\begin{pmatrix}1&-B\\0&1\end{pmatrix}\) satisfies \(R\pi(g)R^{-1}=U(g)\oplus U(g)\), so \(\pi\) is unitarizable.

Exercise 5.3 (medium). Show that a finite group is unitarizable directly, by averaging \(\langle\pi(g)\xi,\pi(g)\eta\rangle\) over the group, and compute the constants \(a,b\) of the resulting equivalent inner product in terms of \(|\pi|\).

Exercise 5.4 (medium). For \(j\in\mathbb N\) let \(U_j\) be a unitary representation on \(H_j\) and \(D_j\) a bounded operator cocycle for \(U_j\), with \(\sup_{j,g}\|D_j(g)\|<\infty\). Show that \(D=\bigoplus_jD_j\) is a bounded operator cocycle for \(U=\bigoplus_jU_j\), and that if \(D\) is inner with implementer \(B\), then every \(D_j\) has an implementer \(B_j\) with \(\|B_j\|\le\|B\|\).

6. Solutions

5.1. If \(a\|\xi\|^2\le[\xi,\xi]\le b\|\xi\|^2\) and \([\pi(g)\xi,\pi(g)\xi]=[\xi,\xi]\), then \(a\|\pi(g)\xi\|^2\le[\pi(g)\xi,\pi(g)\xi]\), which is \([\xi,\xi]\le b\|\xi\|^2\), so \(\|\pi(g)\|\le(b/a)^{1/2}\) for every \(g\).

5.2. Here \(R^{-1}=\begin{pmatrix}1&B\\0&1\end{pmatrix}\). The matrix \(R\pi(g)\) has diagonal entries \(U(g)\), lower left entry \(0\) and upper right entry \(D(g)U(g)-BU(g)\); multiplying on the right by \(R^{-1}\) keeps the diagonal and the lower left entry and gives the upper right entry \(U(g)B+D(g)U(g)-BU(g)\). Since \(D(g)U(g)=BU(g)-U(g)B\), the upper right entry vanishes.

5.3. Put \([\xi,\eta]=\frac1{|G|}\sum_{g\in G}\langle\pi(g)\xi,\pi(g)\eta\rangle\). Replacing \(g\) by \(gh\) permutes the terms, so the form is invariant, and as in the proof of Theorem 1.2, \(|\pi|^{-2}\|\xi\|^2\le[\xi,\xi]\le|\pi|^2\|\xi\|^2\); so \(a=|\pi|^{-2}\), \(b=|\pi|^2\). This is the proof of Theorem 1.2 with the uniform average in place of the mean.

5.4. The direct sum \(D(g)=\bigoplus_jD_j(g)\) is a bounded operator of norm \(\sup_j\|D_j(g)\|\), and (2.1) holds summand by summand. Let \(I_j:H_j\to\bigoplus_iH_i\) be the inclusion and \(B_j=I_j^*BI_j\), so \(\|B_j\|\le\|B\|\). Each summand reduces \(U\): \(U(g)I_j=I_jU_j(g)\) and \(I_j^*U(g)=U_j(g)I_j^*\). Hence \(I_j^*U(g)BU(g)^{-1}I_j=U_j(g)B_jU_j(g)^{-1}\), and compressing \(D(g)=B-U(g)BU(g)^{-1}\) by \(I_j\) gives \(D_j(g)=B_j-U_j(g)B_jU_j(g)^{-1}\). The compression does not require \(B\) to preserve the summands. This step is used in the lesson Unitarizability implies amenability.

References