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.^[1]

In fact for 0 < σ < 1/2 and f, g continuous functions on the circle define

${\displaystyle \displaystyle {(f,g)_{\sigma }={1 \over 4\pi ^{2}}\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }f(s){\overline {g(t)}}k_{\sigma }(s-t)\,ds\,dt,}}$

where

${\displaystyle \displaystyle {k_{\sigma }(s)=(1-\cos s)^{\sigma -1/2}.}}$

Since the function k_σ is integrable, this integral converges. In fact

${\displaystyle \displaystyle {(f,g)_{\sigma }\leq \|f\|\cdot \|g\|,}}$

where the norms are the usual L² norms.

The functions

${\displaystyle \displaystyle {f_{m}(t)=e^{imt}}}$

are orthogonal with

${\displaystyle \displaystyle {(f_{m},f_{m})_{\sigma }=\prod _{i=1}^{|m|}{i-1/2-\sigma \over i-1/2+\sigma }={\Gamma (1/2+\sigma )\Gamma (|m|+1/2-\sigma ) \over \Gamma (1/2-\sigma )\Gamma (m+1/2+\sigma )}.}}$

Since these quantities are positive, (f,g)_σ defines an inner product. The Hilbert space completion is denoted by H_σ.

For F, G continuous functions of compact support on R, define

${\displaystyle \displaystyle {(F,G)_{\sigma }^{\prime }=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }F(x){\overline {G(y)}}|x-y|^{2\sigma -1}\,dx\,dy.}}$

Since, regarded as distributions, the Fourier transform of |x|^{2σ – 1} is C_σ|t|^−2σ for some positive constant C_σ, the above expression can be rewritten:

${\displaystyle \displaystyle {(F,G)_{\sigma }^{\prime }=C_{\sigma }\int _{-\infty }^{\infty }{\widehat {F}}(t){\overline {{\widehat {G}}(t)}}|t|^{-2\sigma }\,dt.}}$

Hence it is an inner product. Let H'_σ denote its Hilbert space completion.

The Cayley transform gives rise to an operator U:

${\displaystyle \displaystyle {Uf(x)=2^{\sigma /2-3/4}\pi ^{-1}|x+i|^{1-2\sigma }f\left({x-i \over x+i}\right).}}$

U extends to an isometry of H_σ onto H '_σ. Its adjoint is given by

${\displaystyle \displaystyle {U^{*}F(e^{it})=2^{3/4-\sigma /2}\pi |1-e^{it}|^{1-2\sigma }F\left({1+e^{it} \over 1-e^{it}}\right).}}$

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 H_σ and SL(2,R) on H '_σ.

For g in SU(1,1) given by

${\displaystyle \displaystyle {g={\begin{pmatrix}\alpha &\beta \\{\overline {\beta }}&{\overline {\alpha }}\end{pmatrix}},}}$

with

${\displaystyle \displaystyle {|\alpha |^{2}-|\beta |^{2}=1,}}$

and f continuous, set

${\displaystyle \displaystyle {\pi _{\sigma }(g^{-1})f(z)=|{\overline {\beta }}z+{\overline {\alpha }}|^{1-2\sigma }f\left({\alpha z+\beta \over {\overline {\beta }}z+{\overline {\alpha }}}\right).}}$

For g' in SL(2,R) given by

${\displaystyle \displaystyle {g^{\prime }={\begin{pmatrix}a&b\\c&d\end{pmatrix}},}}$

with ad – bc = 1, set

${\displaystyle \displaystyle {\pi _{\sigma }^{\prime }((g^{\prime })^{-1})F(x)=|cx+d|^{1-2\sigma }F\left({ax+b \over cx+d}\right).}}$

If g ' corresponds to g under the Cayley transform then

${\displaystyle \displaystyle {U\pi _{\sigma }(g)U^{*}=\pi _{\sigma }^{\prime }(g^{\prime }).}}$

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 H_σ 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 f_m 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.^[2] If s = σ + iτ, g lies in SU(1,1) and f in H_σ, define

${\displaystyle \displaystyle {\pi _{s}(g^{-1})f(z)=|{\overline {\beta }}z+{\overline {\alpha }}|^{1-2s}f\left({\alpha z+\beta \over {\overline {\beta }}z+{\overline {\alpha }}}\right).}}$

Similarly if g ' lies in SL(2,R) and F in H '_σ, define

${\displaystyle \displaystyle {\pi _{s}^{\prime }((g^{\prime })^{-1})F(x)=|cx+d|^{1-2s}F\left({ax+b \over cx+d}\right).}}$

As before the unitary U intertwines these two actions. K acts unitarily on H_σ 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:^[3]

${\displaystyle \displaystyle {\pi _{s}(L_{0})f_{m}=mf_{m},\,\,\pi _{s}(L_{-1})f_{m}=-(m+1/2+s)f_{m+1},\,\,\pi _{s}(L_{1})f_{m}=-(m-1/2-s)f_{m-1}.}}$

If the representation were unitarizable for τ ≠ 0, then the similarity operator T on H_σ would have to commute with K, since K preserves the original inner product. The vectors Tf_m would therefore still be orthogonal for the new inner product and the operators

${\displaystyle \displaystyle {L_{i}^{\prime }=TL_{i}T^{-1}}}$

would satisfy the same relations for

${\displaystyle \displaystyle {f_{m}^{\prime }=Tf_{m}=\lambda _{m}f_{m}.}}$

In this case

${\displaystyle \displaystyle {[L_{m}^{\prime },L_{n}^{\prime }]=(m-n)L_{m+n}^{\prime },\,\,(L_{i}^{\prime })^{*}=L_{-i}^{\prime }.}}$

It is elementary to verify that infinitesimally such a representation cannot exist if τ ≠ 0.^[4]

Indeed, let v₀ = f '₀ and set

${\displaystyle \displaystyle {v_{1}=L_{-1}^{\prime }v_{0}.}}$

Then

${\displaystyle \displaystyle {L_{1}^{\prime }v_{1}=cv_{0}}}$

for some constant c. On the other hand,

${\displaystyle \displaystyle {\|v_{1}\|^{2}=(L_{-1}^{\prime }v_{0},v_{1})=(v_{0},L_{1}^{\prime }v_{1})={\overline {c}}\|v_{0}\|^{2}.}}$

Thus c must be real and positive. The formulas above show that

${\displaystyle \displaystyle {c={1 \over 4}-s^{2}={1 \over 4}-\sigma ^{2}+\tau ^{2}-2i\sigma \tau ,}}$

so the representation π_s 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.^[5] The group SL(2,Z) contains a subgroup of index 12 isomorphic to F₂ the free group on two generators.^[6] 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

${\displaystyle \displaystyle {(x,y)_{1}=\int _{H\backslash G}(gx,gy)\,dg.}}$

As in the previous proof, uniform boundedess guarantees that the norm defined by this inner product is equivalent to the original inner product. But then the original representation would be unitarizable on G, a contradiction. The same argument works for any discrete subgroup of G of finite covolume. In particular the surface groups, which are cocompact subgroups, have uniformly bounded representations that are not unitarizable.

There are more direct constructions of uniformly bounded representations of free groups that are non-unitarizable: these are surveyed in Pisier (2001). The first such examples are described in Figà-Talamanca & Picardello (1983), where an analogue of the complementary series is constructed.

Later Szwarc (1988) gave a related but simpler construction, on the Hilbert space H = ${\displaystyle \ell }$²(F₂), of a holomorphic family of uniformly bounded representations π_z of F₂ for |z| < 1; these are non-unitarizable when 1/√3 < |z| < 1 and z is not real. Let L(g) denote the reduced word length on F₂ for a given set of generators a, b. Let T be the bounded operator defined on basis elements by

${\displaystyle \displaystyle {Te_{1}=0,\,\,Te_{g}=e_{g^{\prime }},}}$

where g ' is obtained by erasing the last letter in the expression of g as a reduced word; identifying F₂ with the vertices of its Cayley graph, a rooted tree,^[7] this corresponds to passing from a vertex to the next closest vertex to the origin or root. For |z| < 1

${\displaystyle \displaystyle {\pi _{z}(g)=(I-zT)^{-1}\lambda (g)(I-zT)}}$

is well-defined on finitely supported functions. Pytlik & Szwarc (1986) had earlier proved that it extends to a uniformly bounded representation on H satisfying

${\displaystyle \displaystyle {\|\pi _{z}(g)\|\leq {1+|z| \over 1-|z|}.}}$

In fact it is easy to check that the operator λ(g)Tλ(g)⁻¹ – T has finite rank, with rangeV_g, the finite-dimensional space of functions supported on the set of vertices joining g to the origin. For on any function vanishing on this finite set, T and λ(g)Tλ(g)⁻¹ are equal; and they both leave invariant V_g, on which they acts as contractions and adjoints of each other. Hence if f has finite support and norm 1,

${\displaystyle \displaystyle {\|\pi _{z}(g)f\|=\|\lambda (g)f+\sum _{n=0}^{\infty }z^{n+1}T^{n}[T,\lambda (g)]f\|\leq 1+2\sum _{n=0}^{n}|z|^{n+1}={1+|z| \over 1-|z|}.}}$

For |z| < 1/√3, these representations are all similar to the regular representation λ. If on the other hand 1/√3 < |z| <1, then the operator

${\displaystyle \displaystyle {D=\pi _{z}(a)+\pi _{z}(a^{-1})+\pi _{z}(b)+\pi _{z}(b^{-1})}}$

satisfies

${\displaystyle \displaystyle {Df=(3z+z^{-1})f}}$

where f in H is defined by

${\displaystyle \displaystyle {f(1)=1,\,\,f(g)={3 \over 4}(3z)^{-L(g)}\,\,(g\neq 1).}}$

Thus, if z is not real, D has an eigenvalue which is not real. But then π_z cannot be unitarizable, since otherwise D would be similar to a self-adjoint operator.