44
Noncommutative
boundaries
of q-deformations
Sergey Neshveyev
Introduction
The study of random walks on duals ofcompact quantum groups was initiated by Masaki Izumi
in [II]. The motivation was to compute the relative commutant of the fixed point algebra for
product-type actions of compact quantum groups. In the classical case such actions are always
minimal, that is, the commutant is trivial. For quantum groups this is not so. It turns$\mathrm{n}\mathrm{s}$ out that
the relative
commutant
can be interpretedas
the algebra of bounded measurable functions onthePoissonboundary of the dual discrete quantum group. The general theory ofnoncommutative
Poisson boundariesdevelopedin [I1]wasillustratedbythecomputation of theboundaryof$\overline{SU_{q}(2}$),
whichwas shown to be isomorphic to the quantum sphere $s_{q}^{2}$
.
In thepresent notewe
discuss thework ofIzumi, Tuset andthe author [INT] , where wecomputed the Poissonboundary ofthe dual
of$SU_{q}(n)$ for arbitrary$n\geq 2$.
This note is basedonthe talk given by the author at the RIMS Symposium“Recent
develop-ments in classification problems in Operator Algebras”, January 24-26, 2005, Kyoto.
1
Main
result
Let $G$ be a compact quantumgroup [W]. Inother words, we are given a unital C’-algebra $C(G)$
and aunital homomorphism $\Delta:C(G)arrow C(G)\otimes$$C(G)$ such that (A$($& $\iota)\Delta=(\iota\otimes\Delta)\Delta$ and that
both$\Delta(C(G))(C(G)\otimes 1)$ and$\Delta(C(G))(1\otimes C(G))$
are
dense in$C(G)\otimes C(G)$. Thenwealso haveadual discrete quantumgroup $G^{\Lambda}$
suchthat thealgebra$c_{0}(\hat{G})$ of functionson$\hat{G}$vanishingat infinity
is the group $\mathrm{C}^{*}$ algebra C’(G) of$G$, so
$c_{0}( \hat{G})=\bigoplus_{s\in \mathrm{I}\mathrm{r}\mathrm{r}(G)}B(H_{s})$,
where the sumis over theset $\mathrm{I}\mathrm{r}\mathrm{r}(G)$ of equivalence classes ofirreducible representations of
$G$.
Givenanormalstate $\phi$ on$\ell^{\infty}(G)=W^{*}(G)$, consider the convolutionoperator $P\psi$ on $\ell^{\infty}(\hat{G})$,
$P_{\phi}(x)=(\phi\otimes\iota)\hat{\Delta}(x)$
.
Then consider the space
$H^{\infty}(\hat{G}, \phi)=\{x\in\ell^{\infty}(\hat{G})|P\phi(x)=x\}$.
of $P\phi$-harmonic elements. A priori this is just a weakly operator closed operator system. But it
has aunique vonNeumann algebrastructure, whichis explicitly given by
The algebra $H^{\infty}(\hat{G}, \phi)$ should be thought of as the algebra ofbounded measurable functions on
the Poisson boundaryof$\hat{G}$ definedby
$\phi$ $[\mathrm{I}\mathrm{I}]$
.
This algebra has a right action of the quantum group $\hat{G}$
coming from the right action by
translations of$\hat{G}$ on itself.
There exists a left adjoint action of the quantum group $G$ on $W^{*}(G)$
.
We shall only considerstates $\phi$whichareinvariant underthis action. This givesus anadditional symmetry of the Poisson
boundary, so$H^{\infty}(\hat{G}, \phi)$ becomes a vonNeumann algebrawitha left action of$G$ andaright action
of$\hat{G}$. Notice that the action of$\hat{G}$ is always ergodic. It turns out that the right action of$G$ is also
ergodic,if the fusionalgebra, or the representation ring, of$G$is commutative,
Consider now the group $G=SU_{q}(n)$, $q\in[-1,1]$. By definition the algebra $C(SU_{q}(n))$ is
generated by$n^{2}$ elements
$uij$, $1\leq \mathrm{i}$,$j\leq n$, satisfyingthe relations
ikUjk $=qujkuik$, ukiukj $=qukjUki$ for $\mathrm{i}<j$,
uuujk $=ujku_{il}$ for $\mathrm{i}<j$, $k<l$,
$u_{ik}u_{il}-u_{gl}u_{ik}=(q-q^{-1})u_{jk}u_{il}$ for$\mathrm{i}<j$, $k<l$, $\det_{q}(U)=1$,
where $U=(u_{ij})_{i,j}$ and $\det_{q}(U)=\sum_{w\in S_{n}}(-q)^{\ell(w)}u_{w(1)1}\ldots$$u_{w(n)n}$, with $\ell(w)$ being the number of
inversions in$w\in S_{n}$. Theinvolution on $C(SU_{q}(n))$ is given by
$u_{ij}^{*}=(-q)^{i-i}\det_{q}(U_{\hat{j}}^{\hat{i}})$,
where $U_{\hat{j}}^{\hat{i}}$ is the matrix obtained from $U$ by removing the
$\mathrm{i}\mathrm{t}\mathrm{h}$ row
and the $j\mathrm{t}\mathrm{h}$ column. The
comultiplication is given by
$\Delta(u_{ij})=\sum_{k}u_{ik}\otimes u_{kj}$.
Denote by $T$ the maximal torus in $SU_{q}(n)$ and consider the homogeneous space $SU_{q}(n)/T$.
More explicitly, for $(t1, \ldots, t_{n})\in \mathrm{T}^{\tau l}$ such that $t_{1}\ldots$$t_{n}=1$ define an automorphism of$C(SU_{q}(n))$
by
$u_{ij}\mapsto t_{j}u_{ij}$
.
Thisway weget an actionof$T\cong \mathbb{T}^{n-1}$on$C(SU_{q}(n))$, and $C(SU_{q}(n)/T)$ isthe fixedpoint algebra
for thisaction.
Wecan now formulate ourmainresult,
Theorem
If
$0<q<1$, thenfor
anyleft
$SU_{q}(n)$-invariant normal generating state on$\ell^{\infty}(S\overline{U_{q}(n}))$,the Poisson boundary
of
$S\overline{U_{q}(n}$) is$SU_{q}(n)$- and$S\overline{U_{q}(n}$)-equivariantly isomorphic to the quantum
flag
manifold
$SU_{q}(n)/T$.
Here the left action of$SU_{q}(n)$ on $SU_{q}(n)/T$ is the action by translations. The rightaction of
$S\overline{U_{q}(n})$ comes from the right adjoint action of $S\overline{U_{q}(n}$) on $C(SU_{q}(n))=C^{*}(S\overline{U_{q}(n}))$. However, a
more correct way ofthinking ofthis action is to consider it
as
a quantum analogue of dressingtransformations, see e.g. $[\mathrm{K}\mathrm{o}\mathrm{S}]$
.
Inthe classical case the orbits ofthe action by dressingtransfor-mationsareleaves of the canonical Poissonstructure. The flagmanifold has one open denseleave,
the Schubert cell, so that the action is ergodic. This makes it easier to believe that the action of
$S\overline{U_{q}(n})$ onthe quantumflag manifold is ergodic, which should be the caseifourresult is true.
The above theorem says in particular that the Poisson boundary of $S\overline{U_{q}(n}$) does not depend
onthe generating state. This in fact can be shown without actually computing the boundary: if
the fusion algebra of$G$ is commutative, then the space $H^{\infty}(\hat{G}, \phi)$ is the same for any G-invariant
2
Poisson integral
and
Berezin transform
For anycompactquantum group$G$thereexists auniquenormalunital G- and$\hat{G}$-equivariant map$\mathrm{O}-$
from $L^{\infty}(G)$ into $\ell^{\infty}(\hat{G})$. Explicitly,
$\Theta$$=(\varphi\otimes\iota)\hat{\Phi}$,
where$\varphi$ istheHaarstateon$L^{\infty}(G)$ and
$\hat{\Phi}:L^{\infty}(G)arrow L^{\infty}(G)\otimes\ell^{\infty}(\hat{G})$ is the rightadjointaction of
$\hat{G}$
on$G$, which we discussed above. This mapwas introduced in [I1] to showthat anisomorphism
between $s_{q}^{2}$ and the Poisson boundary of
$\overline{SU_{q}(2}$) is $\overline{SU_{q}(2}$)-equivariant. It was also shown in [I1]
that forany $G$ and any normal left $G$-invariant stateon $\ell^{\infty}(\hat{G})$ theimage of this map is contained
in$H^{\infty}(\hat{G}, \phi)$
.
Thus ifwe
expectthe Poissonboundary of$\hat{G}$to beahomogeneous space $G/H$, thenthis map shouldbe anisomorphism of$L^{\infty}(G/H)$ onto $H^{\infty}(\hat{G}, \phi)$. But the only thing we can say
in general is that this map is completely positive. We call this mapthe Poissonintegral.
Recall that given von Neumann algebras $N_{1}$ and $N_{2}$, normal faithful states $\nu_{1}$ and $\nu_{2}$ on $N_{1}$
and $N_{2}$, respectively, and a normal unital completely positive map$\theta:N_{1}arrow N_{2}$ suchthat $\nu_{2}\theta=\nu 1$
and $\sigma_{t}^{12}\theta J=\theta\sigma_{t}^{\nu_{1}}$, there exists a normal unital completelypositive map
$\theta^{*}:$$N_{2}arrow N_{1}$ such that
$\nu_{2}(\theta(x_{1})x_{2})=\nu_{1}(x_{1}\theta^{*}(x_{2}))$ for$x_{1}\in N_{1}$, x2 $\in N_{2}$
.
Then by [P] an element $x\in N_{1}$ isin the multiplicative domain of$\theta$ ifandonly if$(\theta^{*}\theta)(x)=x$.
We want to apply this criterion to our map $\Theta$
.
For thiswe
need to compute $\Theta^{*}$.
Let$\Theta_{s}$:$L^{\infty}(G)arrow B(H_{s})$, $s\in$Irr(G),be the components of
0.
Explicitly,$\Theta_{s}(a)=(\varphi\otimes\iota)(U^{s}(a\otimes 1)U^{s*})$,
where$U^{s}\in C(G)\otimes B(H_{s})$ is
a
fixed representative of theequivalenceclass$s\in$Irr(G) of irreduciblerepresentations of$G$. The representation$U^{s}$defines twoadjointactions of$G$
on
$B(H_{s})$. There existauniqueleft-invariant state$\phi_{s}$ and auniqueright-invariant state$\omega_{s}$on$B(H_{s})$ (in the classical case
they bothcoincide with thenormalized$\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}$). Then itisnot difficult to checkthatthe adjoint $\Theta_{s}^{*}$
for $\Theta_{s}$:$(L^{\infty}(G), \varphi)\mapsto(B(H_{s}), \phi_{s})$ is givenby
$\Theta_{s}^{*}(x)=(\iota$$($&$\omega_{s})(U^{s*}(1\otimes x)U^{s})$.
On $H^{\infty}(\hat{G}, \phi)$ there is a canonical state$\nu_{0}$ given by the restriction ofthe counit
$\hat{\epsilon}$, that is, by
“the evaluation at the unit of$\hat{G}$
”. Thenit can be shownthat the adjoint $\Theta^{*}$ to $\Theta$:$(L^{\infty}(G), \varphi)arrow$
$(H^{\varpi}(\hat{G}, \phi)$,$\iota/_{0})$ is
$\Theta^{*}(x)=s^{*}-\lim_{narrow\infty}$ $\sum$ $\phi^{n}(I_{s})\Theta_{s}^{*}(x)$,
$s\in \mathrm{I}\mathrm{r}\mathrm{r}(\mathrm{G})$
where$I_{s}$ is the unit in $B(H_{s})$ and $\phi^{n}$is the nth convolution power of$\phi$
.
Thus, ifwedenote $\Theta_{s}^{*}\Theta_{S}$by$B_{s}$,
an
element $a\in L^{\infty}(G)$liesin themultiplicativedomain ofthePoisson integral if andonlyif $\sum$ $\phi^{n}(I_{s})B_{s}(a)arrow a$.
$s\in \mathrm{I}\mathrm{r}\mathrm{r}(\mathrm{G})$
It turns out that the operators $\Theta_{s}$ and $\Theta_{s}^{*}$
are
analogues of well-known classicalconstruc-tions [Be, Per]. Let forthe moment $G$be a compact Lie group, $U:Garrow B(H)$ a finite dimensional
unitary representation. Fix a vector $\xi\in H$, $||\xi||=1$
.
Let $T\subset G$ be the stabilizer of the line $\mathbb{C}\xi$.For an operator $S\in B(H)$, its covariant Berezinsymbol$\sigma(S)$ is defined by$\sigma(S)(g)$ $=(SU_{g}\xi, U_{g}\xi)$.
The covariant symbol aisa$G$-equivariantmap from$B(H)$ into $C(G/T)$
.
Consider the innerprod-uctson $C(G/T)$ and$B(H)$ givenbythe$G$-invariant probability
measure
andthenormalized$\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{e}$,respectively. Then there exists an adjoint $\dot{\sigma}$:$C(G/T)$ $arrow B(H)$ of
$\sigma$. Explicitly,
where$d=\dim H$ and $P_{\xi}$ is the projection onto
C4.
A function $f$ is called a contravariant Berezinsymbol of$\dot{\sigma}(f)$
.
The map $B=\sigma\dot{\sigma}$ is called theBerezin transform.Ifwe consider $U$ as acorepresentationof$C(G)$, then the definition of$\sigma$can be writtenas
$\sigma(S)=(\iota \mathrm{C}\otimes\omega_{\xi})(U^{*}(1\otimes S)U)$,
where$\omega\xi=(\cdot\xi, \xi)$is thevector-statedefinedby
4.
Thuswe see that ouroperator$\Theta_{s}^{*}$ isjust awith$\omega\xi$ replaced by $\mathrm{c}\mathrm{u}_{\mathrm{s}}$. Then$\Theta_{s}=(\Theta_{s}^{*})^{*}$ is an analogueof $\dot{\sigma}$.
Suppose now that $G$ is a semisimple Lie group, $U=U^{\lambda}:Garrow B(H\lambda)$ an irreducible
represen-tation with highest weight $\lambda$, $\langle$ $=\xi\lambda$ a highest weightvector. Let $B_{\lambda}$be thecorrespondingBerezin
transform. It is proved in [D] that the sequence $\{B_{n\lambda}\}_{n=1}^{\infty}$ converges to the identity on $C(G/T)$
as $narrow\infty$. This is a key step to show that the full matrix algebras $B(H_{n\lambda})$, $n\in \mathrm{N}$, provide a
quantization of$C(G/T)$, see $[\mathrm{L}, \mathrm{R}]$. In other words, in the classical case the Berezin transforms
converge to theidentity operator on the flagmanifold along any ray in theWeyl chamber. While
to prove multiplicativityofthe Poissonintegralwe have to show that certainconvexcombinations
of quantum Berezin transforms defined by a random walk on the Weyl chamber converge to the
identity operatoron thequantum flag manifold.
Theproofin the classical caseisbased ontheobservation that the operators$B_{n\lambda}$ are given by
convolution with measures whichare absolutelycontinuous with respect to the Haarmeasure and
such that the Radon-Nikodym derivatives are, up to normalization, powers ofasingle function$h$
such that $h(g)=1$ for$g\in T$ and $h(g)<1$ for $g\not\in T$
.
The proof of our $q$-analoguewillbe based onthe study of ergodicpropertiesof
an
auxiliary operator.Since the maps $B_{s}$ are $G$-equivariant, it suffices to check the convergence “at the unit element
of$G”$
.
More precisely, assuming that the counit $\in$iswell-defined on $C(G)$, that is, thegroup $G$ iscoamenable, a subalgebra$C(G/H)$ is in the multiplicativedomain of $\Theta$if and only if
$\sum_{s}\phi^{n}(I_{s})(\epsilon B_{s})(a)arrow\epsilon(a)$
forany $a\in C(G/H)$. We have
$(\epsilon B_{s})(a)=(\omega_{s}\Theta_{s})(a)=(\varphi\otimes \omega_{s})(U^{s}(a\otimes 1)U^{s*})$
.
Consider theoperator $A_{\omega_{s}}$:$C(G)arrow C(G)$ defined by
$A_{\omega_{\theta}}(a)=(\iota\otimes\omega_{s})(U^{s}(a\otimes 1)U^{s*})$.
Then$\epsilon B_{s}=\varphi A_{\omega_{S}}$, and wearriveat the following criterion,
Proposition Assume $G$ is coamenable. For
a
normalleft
$G$-invariant state $\phi$ considerthecorre-spondingnght$G$-invariant state$\omega$$= \sum_{s}\phi(I_{s})\omega_{s}$ and the operator$A_{\omega}$:$C(G)arrow C(G)$,
$A_{\omega}= \sum_{s}\phi(I_{s})A_{(v_{\mathit{8}}}$
.
Then the states $\varphi A_{d}^{n}$
‘ converge to a state on $C(G)$ as $narrow\infty$. For a subgroup $H\subset G$ the algebra
$C(G/H)$ is in the multiplicative domain
of
$\Theta:L^{\infty}(G)arrow H^{\infty}(\hat{G}, \phi)$if
and onlyif
the limit statecoincides with the counit$\epsilon$
on
$C(G/H)$.Consider
now
$G=SU_{q}(2)$.
Thenand the relations
can
bewritten as$\alpha^{*}\alpha+\gamma^{*}\gamma=1$, $\alpha\alpha^{*}+q^{2}\gamma^{*}\gamma=1$, $\gamma^{*}\gamma=\gamma\gamma^{*}$, $\alpha\gamma=q\gamma\alpha$, $\alpha\gamma^{*}=q’\gamma^{*}\alpha$.
The comultiplicationA is determined by theformulas
$\Delta(\alpha)=\alpha\otimes\alpha-q\gamma^{*}\otimes\gamma$, $\Delta(\gamma)=\gamma\otimes\alpha+\alpha^{*}\otimes\gamma$
.
Since the Poisson boundary does not depend on the generating state, it suffices to consider the
state $\phi$ correspondingto thefundamental representation $U$
.
So$\phi=\frac{1}{q+q^{-1}}\mathrm{R}$
(.
$(\begin{array}{ll}q 00 q^{-1}\end{array})$),
$\omega=\frac{1}{q+q^{-1}}\mathrm{B}($.
$(\begin{array}{ll}q^{-1} \mathrm{o}0 q\end{array})$$)3$and thus
$A_{\omega}(a)$ $=$ $\frac{1}{q+q^{-\mathrm{l}}}\mathrm{n}($$(\begin{array}{ll}\alpha -q\gamma^{*}\gamma \alpha^{*}\end{array})(\begin{array}{ll}a 00 a\end{array})(\begin{array}{ll}\alpha^{*} \gamma^{*}-q\gamma \alpha\end{array})(\begin{array}{ll}q^{-1} 00 q\end{array})$$)$
$=$ $\frac{1}{q+q^{-1}}(q^{-1}(\alpha a\alpha^{*}+q^{2}\gamma^{*}a\gamma)+q(\gamma a\gamma^{*}+\alpha^{*}a\alpha))$.
We see in particular that $A_{\omega}$ commutes with the left and the right actions of
$T\cong \mathrm{T}$ given by $\alpha\mapsto z\alpha$, $\gamma$ $\mapsto\overline{z}\gamma$ and $\alpha\mapsto z\alpha$, $\gamma\mapsto 27$, respectively. It folows that the limit ofthe states
$\varphi A_{\omega}^{n}$
is also invariant with respect to these actions. Though the counit is not invariant on the whole
group, notice that in general $\epsilon$ is invariant with respect to the left action of$H$
on
$G/H$, as wellas with respect to the right action of $H$ on $H\backslash G$
.
It follows that to prove multiplicativity ofthe Poisson integral on $C(SU_{q}(2)/T)$ it suffices to provethat thelimit state coincides with $\epsilon$ on
$C(T\backslash SU_{q}(2)/T)$
.
The latter algebra is generated by 77. It is known that the spectrum of $\gamma^{*}\gamma$is the set $I_{q^{2}}=\{0\}\mathrm{U}\{q^{2n}\}_{n=0}^{\infty}$
.
Thuswe
can identify $C(T\backslash SU_{q}(2)/T)$ with the algebra $C(I_{q^{2}})$ ofcontinuous functions
on
$I_{q^{2}}$.
Since thecounit isacharacter such that$\alpha\mapsto 1$ and$\gamma\mapsto 0$, under this
identification it isgiven by the evaluation at$0\in I_{q^{2}}$
.
Furthermore,using the identities$\alpha^{*}(\gamma^{*}\gamma)^{k}\alpha=q^{-2k}(\gamma^{*}\gamma)^{k}(1-\gamma^{*}\gamma)$, $\alpha(\gamma^{*}\gamma)^{k}\alpha^{*}=q^{2k}(\gamma^{*}\gamma)^{k}(1-q^{2}\gamma^{*}\gamma)$,
we
see
that theaction of$A_{\omega}$ on thefunctions on $I_{q^{2}}$ is given by$(A_{\omega}f)(t)= \frac{1}{q+q^{-1}}(q^{-1}((1-q^{2}t)f(q^{2}t)+q^{2}tf(t))+q(tf(t)+(1-t)f(q^{-2}t)))$.
In otherwords, $A_{\omega}$ is the Markov operator with transition probabilities$p(\mathrm{O}, 0)=1$,
$p(q^{2n}, q^{2(n-1)})$ $=$ $\frac{q-q^{2n+1}}{q+q^{-1}}$, $p(q^{2n}, q^{2n})$ $=$
$\underline{2q^{2n+1}}$
$q+q^{-1\}}$
$p(q^{2n}, q^{2(n+1)})$ $=$ $\frac{q^{-1}-q^{2n+1}}{q+q^{-1}}$
.
It is not difficult to show that the restriction of this random walk to $I_{q^{2}}\backslash \{0\}$ is transient. In
particular, $(A_{\omega}^{n}\delta_{x})(y)arrow 0$ for any $x,y\in I_{q^{2}}\backslash \{0\}$
.
Hence for any $f\in C(I_{q^{2}})$ the functions $A_{\omega}^{n}f$convergepointwiseto the constant$f(0)$
.
Thiscompletesthe proof of multiplicativityofthePoissonFor $n>2$ we prove that $\in$ is the only $A_{(p}$-invariant state on $C(SU_{q}(n)/T)$. The proofis by
induction on$n$, and it turns out that for the induction stepit suffices to check that $\epsilon$ is the only
$A_{\omega}$-invariant state on
$C(S(U_{q}(n-1)\mathrm{x} \mathbb{T})\backslash SU_{q}(n)/S(U_{q}(n-1)\mathrm{x}\mathrm{F}’))\cong C(T\backslash SU_{q}(2)/T)$,
whichis already established.
3
Random
walk
on
the
center
In the previous sectionwe sketchedanargument for multiplicativityof thePoisson integral. Since
the Haar state is faithful, we also automatically have injectivity of the Poisson integral on its
multiplicative domain. We shall next discuss surjectivity.
We need an estimate on the dimensionsof the spectral subspaces of$H^{\infty}(\hat{G},\phi)$. By a result of
Hayashi [H], whichwe have alreadymentioned,if the fusion algebra ofagroup $G$ is commutative,
theaction of$G$onthePoisson boundary is ergodic. This alreadyimpliesthat thespectral subspaces
of$H^{\infty}(\hat{G}, \phi)$ are finite dimensional.
To obtainabound ontheirmultiplicities, note that ergodicity of the actionof$G$ on$H^{\infty}(\hat{G}, \phi)$
is equivalent to triviality of the Poisson boundary of the restriction of$P\psi$ to the center $\ell^{\infty}(\hat{G})$,
since the center is exactly the fixed point algebra$\ell^{\infty}(\hat{G})^{G}$. Identify the center with$\ell^{\infty}(I)$, where
$I=$ Irr(G). For
a
fixed generatingstate $\phi$, let $\{p(s, t)\}_{s,t\in I}$ be the transition probabilities definedby the restriction of $P\phi$ to $\ell^{\infty}(I)$, so $P\emptyset(I_{t})I_{s}=p(s, t)I_{s}$
.
Let $(\Omega,\mathrm{P}_{0})$ be the path space of thecorrespondingrandom walk,
$\Omega=\prod_{n=1}^{\infty}I$, $\mathrm{P}_{0}(\{\underline{s}|s_{1}=t_{1}, \ldots, s_{n}=t_{n}\})=p(0, t_{1})p(t_{1}, t_{2})$. .
.
$p(t_{n-1}, t_{n})$.Denote by $\pi_{n}$ the projection $\Omegaarrow I$ onto the
$n\mathrm{t}\mathrm{h}$ factor. Let E.$\ell^{\infty}(\hat{G})arrow\ell^{\infty}(I)$ be the unique
$G$-equivariant conditional expectation, $E(x)(s)=\phi_{s}(x)$. If$x\in l^{\infty}(\hat{G})$ is $P\psi$-harmonic, then
$x^{*}x=P\phi(x)^{*}P\phi(x)\leq P\psi(x^{*}x)$,
whence $E(x^{*}x)\leq P\emptyset(E(x^{*}x))$
.
It follows that the sequence $\{\pi_{n}(E(x^{*}x))\}_{n}$ is a submartingale.Hence it converges$\mathrm{a}.\mathrm{e}$. But since the Poisson boundary of the center is trivial, the limit must be
aconstant. This constant is $\nu_{0}($$’ .$x)$. Thus we get the following result.
Proposition Let $\phi$ be a normal
left
$G$-invariant generating state on $\ell^{\infty}(\hat{G})$.
Assume that thePoisson boundary
of
the center is trivial, Thenfor
any $x$,$y\in H^{\infty}(\hat{G}, \phi)$ and almost every path$\underline{s}\in\Omega$, ate have $\phi_{s_{n}}(xy)arrow\nu_{0}(x\cdot y)$ as $narrow\infty$
.
In other words, the completely positive maps$(H^{\infty}(\hat{G}, \phi)$,$\nu_{0})arrow(B(H_{s_{n}})_{:}\phi_{s_{n}})$ are asymptotically isometric in $L^{2}$-norm.
In particular,
for
any $\mathrm{i}$ reducible representation $V$of
$G$ the multiplicityof
$V$ in $H^{\infty}(\hat{G}, \phi)$ isnotlargerthanthe supremum
of
the multiplicitiesof
$V$ in $\overline{U}\mathrm{x}$$U$for
allirreducible representations$U$
of
$G$.For the $q$-deformation $G$ of a compact connected semisimple Lie group the last estimate is
optimal, see e.g. $[\check{\mathrm{Z}}, \S 131]$
.
For example, for $SU_{q}(2)$ ifwe take thespin $s \in\frac{1}{2}\mathbb{Z}_{+}$representation$U^{s}$then $\overline{U}^{s}\cong U^{s}$, and $U^{s}\mathrm{x}$ $U^{s}$ is isomorphic to the direct sum of $U^{t}$, $t=0,1$,
.
.
.
,$2\mathrm{s}$, Onthe other
handweknow that only integer spin representationsappearin$C(SU_{q}(2)/T)$,eachwithmultiplicity
one.
Thus if $G$ is the $q$-deformation of a compact connected semisimple Lie group, $T\subset G$ the
maximal torus, $\phi$ a generating state, then the Poisson integral $\Theta:L^{\infty}(G/T)arrow H^{\infty}(\hat{G}, \phi)$ is an
4
Conclusion
Asignificant part of our considerations is validfor $q$-deformations ofarbitrary compact connected
semisimple Lie groups. The only pointwhere weused that thegroup was $SU_{q}(n)$ istheinduction
step in the proofofmultiplicativity of thePoisson integral. It isbased on a quantum analogue of
the fact that $S(U(n-1)\mathrm{x} \mathbb{T})\subseteq SU(n)$is aRiemarmian symmetric pairof rank
one.
Hence thereis hope that similar considerations could workfor SO(n), $Sp(n)$ and F4, see [He, Ch. $\mathrm{X}$, Table $\mathrm{V}$].
Nevertheless for the exceptional groups $E_{6}$,$E_{7}$,$E_{8}$ and $G_{2}$ this approach definitely requires
more
computations than was the case for An. So it would be desirable to find a universal method. A
more difficultandinteresting problemis to computethe Martinboundary.
The Poisson boundary of a noncompact semisimple Lie group $G$ was computed by
Fursten-berg [Pu], who showed that it is isomorphic to the flag manifold $B=G/P$, where $P$ is the
min-imal parabolic subgroup of $G$. The Martin boundary of $G$, or of the corresponding symmetric
space $S=G/K$, was computedby Olshanetsky, whoannounced the result more than thirtyyears
ago [Oil], but a detailedproofappeared only recently [012, GJT]. The result says that the Martin
compactification is the minimal compactiflcation which dominates both the visibility
compactifi-cation and theFurstenberg compactiflcation. The latter is defined as the closure of$S$ in $M_{1}(B)$,
the space ofprobability
measures
on$B=G/P$, under the map $\mathrm{g}\mathrm{x}0\mapsto gm$, where $m$ is the unique$K$-invariant probability
measure
on $B$ (notethat quiteconfusingly theboundaryof$S$in thiscom-pactification doesnotalways coincide withthe Furstenberg boundary $B$, but inthe rankone
case
they both do coincide with the sphere at infinity, which is the boundary in the visibility
com-pactification). Moreconcretelythe Furstenbergcompactiflcation
can
be obtainedbyembedding $S$in the projective space of matrices using certain irreducible representations of$G$
.
The results ofFurstenbergwere extended to arbitrary (connected) Lie groups by Raugi [R]. On the other hand,
the problemofcomputingMartinboundaries evenforsolvable Lie groupsremains unsettled.
The rank
one
caseconsidered in [NT1]isnot sufficient toformulatea
plausibleconjectureaboutthe Martin boundary of the dual of the $q$-deformation of a compact semisimple Lie group. The
works of Biane [B] andCollins [C]suggest that theminimalboundary isaquantization of the sphere
inthe dual of the Liealgebra. Sincewe nowknow that thePoissonboundary is thequantizationof
the Furstenberg boundary, it is natural to conjecture that tocompute thewholeMartinboundary
one should look for quantizations ofclassical Martin boundaries.
The basis for the above conjecture is that the Poisson boundary of$SU_{q}(n)$ is the quantization
ofthe Poisson boundaryof $5\mathrm{L}\mathrm{n}(\mathrm{C})$. For the moment
we
do not have a good explanation of thisphenomenon. To really understand it we need a better understanding of what happens when
$q=1$
.
The Poisson boundary of $\overline{SU(n}$), and in fact of thedual of any compact group, is trivial.However, the limit $qarrow 1$ should rather be considered in the category of$\underline{\mathrm{H}\mathrm{o}\mathrm{p}\mathrm{f}}$-Poisson algebras.
In other words, the classical limit of $S\overline{U_{q}(n}$) is not the Pontryagin dual $SU(n)$, but the Poisson
dual of$SU(n)$
.
Thus the question is whether Furstenberg’s results have dual Poisson analogues.This could also clarify the appearance of the Berezin transform in our work [INT], since to
our
knowledgeit has not played any role in theclassical picture.
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Mathematics Institute, UniversityofOslo, PB 1053Blindern, Oslo0316, Norway
