In mathematics, a uniformly bounded representation T {\displaystyle T} of a locally compact group G {\displaystyle G} on a Hilbert space H {\displaystyle H} is a homomorphism into the bounded invertible operators which is continuous for the strong operator topology, and such that sup g ∈ G ‖ T g ‖ B ( H ) {\displaystyle \sup _{g\in G}\|T_{g}\|_{B(H)}} is finite. In 1947 Béla Szőkefalvi-Nagy established that any uniformly bounded representation of the integers or the real numbers is unitarizable, i.e. conjugate by an invertible operator to a unitary representation. For the integers this gives a criterion for an invertible operator to be similar to a unitary operator: the operator norms of all the positive and negative powers must be uniformly bounded. The result on unitarizability of uniformly bounded representations was extended in 1950 by Dixmier, Day and Nakamura-Takeda to all locally compact amenable groups, following essentially the method of proof of Sz-Nagy. The result is known to fail for non-amenable groups such as SL(2,R) and the free group on two generators. Dixmier (1950) conjectured that a locally compact group is amenable if and only if every uniformly bounded representation is unitarizable.
Statement Let G be a locally compact amenable group and let Tg be a homomorphism of G into GL(H), the group of an invertible operators on a Hilbert space such that
for every x in H the vector-valued gx on G is continuous; the operator norms of the operators Tg are uniformly bounded. Then there is a positive invertible operator S on H such that S Tg S−1 is unitary for every g in G. As a consequence, if T is an invertible operator with all its positive and negative powers uniformly bounded in operator norm, then T is conjugate by a positive invertible operator to a unitary.
Proof By assumption the continuous functions
f x , y ( g ) = ( T g − 1 x , T g − 1 y ) , {\displaystyle \displaystyle {f_{x,y}(g)=(T_{g}^{-1}x,T_{g}^{-1}y),}}
generate a separable unital C* subalgebra A of the uniformly bounded continuous functions on G. By construction the algebra is invariant under left translation. By amenability there is an invariant state φ on A. It follows that
( x , y ) 0 = φ ( f x , y ) {\displaystyle \displaystyle {(x,y)_{0}=\varphi (f_{x,y})}}
is a new inner product on H satisfying
M − 1 ‖ x ‖ ≤ ‖ x ‖ 0 ≤ M ‖ x ‖ {\displaystyle \displaystyle {M^{-1}\|x\|\leq \|x\|_{0}\leq M\|x\|}}
where
M = sup g ‖ T g ‖ < ∞ . {\displaystyle \displaystyle {M=\sup _{g}\|T_{g}\|<\infty .}}
So there is a positive invertible operator P such that
( x , y ) 0 = ( P x , y ) . {\displaystyle \displaystyle {(x,y)_{0}=(Px,y).}}
By construction
( T g x , T g y ) 0 = ( x , y ) 0 . {\displaystyle \displaystyle {(T_{g}x,T_{g}y)_{0}=(x,y)_{0}.}}
Let S be the unique positive square root of P. Then
( S T g x , S T g y ) = ( P T g x , T g y ) = ( P x , y ) = ( S x , S y ) . {\displaystyle \displaystyle {(ST_{g}x,ST_{g}y)=(PT_{g}x,T_{g}y)=(Px,y)=(Sx,Sy).}}
Applying S−1 to x and y, it follows that
( S T g S − 1 x , S T g S − 1 y ) = ( x , y ) . {\displaystyle \displaystyle {(ST_{g}S^{-1}x,ST_{g}S^{-1}y)=(x,y).}}
Since the operators
U g = S T g S − 1 {\displaystyle \displaystyle {U_{g}=ST_{g}S^{-1}}}
are invertible, it follows that they are unitary.
Examples of non-unitarizable representations
SL(2,R) The complementary series of irreducible unitary representations of SL(2,R) was introduced by Bargmann (1947). These representations can be realized on functions on the circle or on the real line: the Cayley transform provides the unitary equivalence between the two realizations. In fact for 0 < σ < 1/2 and f, g continuous functions on the circle define
( f , g ) σ = 1 4 π 2 ∫ − π π ∫ − π π f ( s ) g ( t ) ¯ k σ ( s − t ) d s d t , {\displaystyle \displaystyle {(f,g)_{\sigma }={1 \over 4\pi ^{2}}\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }f(s){\overline {g(t)}}k_{\sigma }(s-t)\,ds\,dt,}}
where
k σ ( s ) = ( 1 − cos s ) σ − 1 / 2 . {\displaystyle \displaystyle {k_{\sigma }(s)=(1-\cos s)^{\sigma -1/2}.}}
Since the function kσ is integrable, this integral converges. In fact
( f , g ) σ ≤ ‖ f ‖ ⋅ ‖ g ‖ , {\displaystyle \displaystyle {(f,g)_{\sigma }\leq \|f\|\cdot \|g\|,}}
where the norms are the usual L2 norms. The functions
f m ( t ) = e i m t {\displaystyle \displaystyle {f_{m}(t)=e^{imt}}}
are orthogonal with
( f m , f m ) σ = ∏ i = 1 | m | i − 1 / 2 − σ i − 1 / 2 + σ = Γ ( 1 / 2 + σ ) Γ ( | m | + 1 / 2 − σ ) Γ ( 1 / 2 − σ ) Γ ( m + 1 / 2 + σ ) . {\displaystyle \displaystyle {(f_{m},f_{m})_{\sigma }=\prod _{i=1}^{|m|}{i-1/2-\sigma \over i-1/2+\sigma }={\Gamma (1/2+\sigma )\Gamma (|m|+1/2-\sigma ) \over \Gamma (1/2-\sigma )\Gamma (m+1/2+\sigma )}.}}
Since these quantities are positive, (f,g)σ defines an inner product. The Hilbert space completion is denoted by Hσ. For F, G continuous functions of compact support on R, define
( F , G ) σ ′ = ∫ − ∞ ∞ ∫ − ∞ ∞ F ( x ) G ( y ) ¯ | x − y | 2 σ − 1 d x d y . {\displaystyle \displaystyle {(F,G)_{\sigma }^{\prime }=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }F(x){\overline {G(y)}}|x-y|^{2\sigma -1}\,dx\,dy.}}
Since, regarded as distributions, the Fourier transform of |x|2σ – 1 is Cσ|t|−2σ for some positive constant Cσ, the above expression can be rewritten:
( F , G ) σ ′ = C σ ∫ − ∞ ∞ F ^ ( t ) G ^ ( t ) ¯ | t | − 2 σ d t . {\displaystyle \displaystyle {(F,G)_{\sigma }^{\prime }=C_{\sigma }\int _{-\infty }^{\infty }{\widehat {F}}(t){\overline {{\widehat {G}}(t)}}|t|^{-2\sigma }\,dt.}}
Hence it is an inner product. Let H'σ denote its Hilbert space completion. The Cayley transform gives rise to an operator U:
U f ( x ) = 2 σ / 2 − 3 / 4 π − 1 | x + i | 1 − 2 σ f ( x − i x + i ) . {\displaystyle \displaystyle {Uf(x)=2^{\sigma /2-3/4}\pi ^{-1}|x+i|^{1-2\sigma }f\left({x-i \over x+i}\right).}}
U extends to an isometry of Hσ onto H 'σ. Its adjoint is given by
U ∗ F ( e i t ) = 2 3 / 4 − σ / 2 π | 1 − e i t | 1 − 2 σ F ( 1 + e i t 1 − e i t ) . {\displaystyle \displaystyle {U^{*}F(e^{it})=2^{3/4-\sigma /2}\pi |1-e^{it}|^{1-2\sigma }F\left({1+e^{it} \over 1-e^{it}}\right).}}
The Cayley transform exchanges the actions by Möbius transformations of SU(1,1) on S1 and of SL(2, R) on R. The operator U intertwines corresponding actions of SU(1,1) on Hσ and SL(2,R) on H 'σ. For g in SU(1,1) given by
g = ( α β β ¯ α ¯ ) , {\displaystyle \displaystyle {g={\begin{pmatrix}\alpha &\beta \\{\overline {\beta }}&{\overline {\alpha }}\end{pmatrix}},}}
with
| α | 2 − | β | 2 = 1 , {\displaystyle \displaystyle {|\alpha |^{2}-|\beta |^{2}=1,}}
and f continuous, set
π σ ( g − 1 ) f ( z ) = | β ¯ z + α ¯ | 1 − 2 σ f ( α z + β β ¯ z + α ¯ ) . {\displaystyle \displaystyle {\pi _{\sigma }(g^{-1})f(z)=|{\overline {\beta }}z+{\overline {\alpha }}|^{1-2\sigma }f\left({\alpha z+\beta \over {\overline {\beta }}z+{\overline {\alpha }}}\right).}}
For g' in SL(2,R) given by
g ′ = ( a b c d ) , {\displaystyle \displaystyle {g^{\prime }={\begin{pmatrix}a&b\\c&d\end{pmatrix}},}}
with ad – bc = 1, set
π σ ′ ( ( g ′ ) − 1 ) F ( x ) = | c x + d | 1 − 2 σ F ( a x + b c x + d ) . {\displaystyle \displaystyle {\pi _{\sigma }^{\prime }((g^{\prime })^{-1})F(x)=|cx+d|^{1-2\sigma }F\left({ax+b \over cx+d}\right).}}
If g ' corresponds to g under the Cayley transform then
U π σ ( g ) U ∗ = π σ ′ ( g ′ ) . {\displaystyle \displaystyle {U\pi _{\sigma }(g)U^{*}=\pi _{\sigma }^{\prime }(g^{\prime }).}}
Polar decomposition shows that SL(2,R) = KAK with K = SO(2) and A the subgroup of positive diagonal matrices. K corresponds to the diagonal matrices in SU(1,1). Since evidently K acts unitarily on Hσ and A acts unitarily on H 'σ, both representations are unitary. The representations are irreducible because the action of the Lie algebra on the basis vectors fm is irreducible. This family of irreducible unitary representations is called the complementary series. Ehrenpreis & Mautner (1955) constructed an analytic continuation of this family of representations as follows. If s = σ + iτ, g lies in SU(1,1) and f in Hσ, define
π s ( g − 1 ) f ( z ) = | β ¯ z + α ¯ | 1 − 2 s f ( α z + β β ¯ z + α ¯ ) . {\displaystyle \displaystyle {\pi _{s}(g^{-1})f(z)=|{\overline {\beta }}z+{\overline {\alpha }}|^{1-2s}f\left({\alpha z+\beta \over {\overline {\beta }}z+{\overline {\alpha }}}\right).}}
Similarly if g ' lies in SL(2,R) and F in H 'σ, define
π s ′ ( ( g ′ ) − 1 ) F ( x ) = | c x + d | 1 − 2 s F ( a x + b c x + d ) . {\displaystyle \displaystyle {\pi _{s}^{\prime }((g^{\prime })^{-1})F(x)=|cx+d|^{1-2s}F\left({ax+b \over cx+d}\right).}}
As before the unitary U intertwines these two actions. K acts unitarily on Hσ and A by a uniformly bounded representation on H 'σ. The action of the standard basis of the complexification Lie algebra on this basis can be computed:
π s ( L 0 ) f m = m f m , π s ( L − 1 ) f m = − ( m + 1 / 2 + s ) f m + 1 , π s ( L 1 ) f m = − ( m − 1 / 2 − s ) f m − 1 . {\displaystyle \displaystyle {\pi _{s}(L_{0})f_{m}=mf_{m},\,\,\pi _{s}(L_{-1})f_{m}=-(m+1/2+s)f_{m+1},\,\,\pi _{s}(L_{1})f_{m}=-(m-1/2-s)f_{m-1}.}}
If the representation were unitarizable for τ ≠ 0, then the similarity operator T on Hσ would have to commute with K, since K preserves the original inner product. The vectors Tfm would therefore still be orthogonal for the new inner product and the operators
L i ′ = T L i T − 1 {\displaystyle \displaystyle {L_{i}^{\prime }=TL_{i}T^{-1}}}
would satisfy the same relations for
f m ′ = T f m = λ m f m . {\displaystyle \displaystyle {f_{m}^{\prime }=Tf_{m}=\lambda _{m}f_{m}.}}
In this case
[ L m ′ , L n ′ ] = ( m − n ) L m + n ′ , ( L i ′ ) ∗ = L − i ′ . {\displaystyle \displaystyle {[L_{m}^{\prime },L_{n}^{\prime }]=(m-n)L_{m+n}^{\prime },\,\,(L_{i}^{\prime })^{*}=L_{-i}^{\prime }.}}
It is elementary to verify that infinitesimally such a representation cannot exist if τ ≠ 0. Indeed, let v0 = f '0 and set
v 1 = L − 1 ′ v 0 . {\displaystyle \displaystyle {v_{1}=L_{-1}^{\prime }v_{0}.}}
Then
L 1 ′ v 1 = c v 0 {\displaystyle \displaystyle {L_{1}^{\prime }v_{1}=cv_{0}}}
for some constant c. On the other hand,
‖ v 1 ‖ 2 = ( L − 1 ′ v 0 , v 1 ) = ( v 0 , L 1 ′ v 1 ) = c ¯ ‖ v 0 ‖ 2 . {\displaystyle \displaystyle {\|v_{1}\|^{2}=(L_{-1}^{\prime }v_{0},v_{1})=(v_{0},L_{1}^{\prime }v_{1})={\overline {c}}\|v_{0}\|^{2}.}}
Thus c must be real and positive. The formulas above show that
c = 1 4 − s 2 = 1 4 − σ 2 + τ 2 − 2 i σ τ , {\displaystyle \displaystyle {c={1 \over 4}-s^{2}={1 \over 4}-\sigma ^{2}+\tau ^{2}-2i\sigma \tau ,}}
so the representation πs is unitarizable only if τ = 0.
Free group on two generators The group G = SL(2,R) contains the discrete group Γ = SL(2,Z) as a closed subgroup of finite covolume, since this subgroup acts on the upper half plane with a fundamental domain of finite hyperbolic area. The group SL(2,Z) contains a subgroup of index 12 isomorphic to F2 the free group on two generators. Hence G has a subgroup Γ1 of finite covolume, isomorphic to F2. If L is a closed subgroup of finite covolume in a locally compact group G, and π is non-unitarizable uniformly bounded representation of G on a Hilbert space L, then its restriction to L is uniformly bounded and non-unitarizable. For if not, applying a bounded invertible operator, the inner product can be made invariant under L; and then in turn invariant under G by redefining
( x , y ) 1 = ∫ H ∖ G ( g x , g y ) d g . {\displaystyle \displaystyle {(x,y