% LaTeX Article Template - customizing header and footer \documentstyle[12pt, oldlfont]{article} \input{amssymb.def} \input{amssymb.tex} %\def\Ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\sim$}}}{}} %\def\ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\scriptscriptstyle\sim$}}}{}} \setlength{\oddsidemargin}{0.in} \setlength{\textwidth}{7.in} \pagestyle{myheadings} \markright{G. Hjorth} % Set the beginning of a LaTeX document \begin{document} \def\Ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\sim$}}}{}} \def\ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\scriptscriptstyle\sim$}}}{}} \def\R{{\Bbb R}} \def\V{{\Bbb V}} \def\N{{\Bbb N}} \def\Q{{\Bbb Q}} \title{Non-smooth infinite dimensional group representations} % Enter your title between curly braces \author{Greg Hjorth} % Enter your name between curly braces \date{\today} % Enter your date or \today between curly braces \maketitle {\bf $\S$0 When it's bad it's worse} It is known from \cite{thoma} and \cite{glimm} that the infinite dimensional irreducible representations of a countable group do not posses a {\it nice} Borel structure unless $G$ is abelian-by-finite. Roughly speaking, the equivalence relation of conjugacy, or isomorphism, interprets the equivalence relation of eventual agreement on infinite sequences of integers, and thus is {\it non-smooth}: the irreducibles considered up to conjugacy do not form a standard Borel space, there is no Borel selector, there is no reasonable way to assign reals as complete invariants. This note asks when the classification problem for the irreducibles is no {\it worse than} that of infinite sequences considered up to eventual agreement. Surprsingly, the equivalence relation is {\it always} more complicated when non-smooth. Thus there is no reasonable way to assign complete invariants in sequences mod finite when $G$ is not abelian-by-finite. Equivalently, the cosets of ${\Bbb Q}$ in ${\Bbb R}$ can not provide complete invariants -- that is to say, there is no Borel function $\theta$ from the space of irreducible representations to ${\Bbb R}$ such that representations $\tau_1$ and $\tau_2$ are conjugate if and only if $\theta(\tau_1)-\theta(\tau_2)\in {\Bbb Q}$. Worse: there is no {\it countable Borel lacunary section} (a Borel subset of the space of irreducibles meeting each equivalence class in exactly $\aleph_0$ many points). We cannot assign countable sets of reals as complete invariants. And strongest of all -- and this is the real theorem -- for $G$ not abelian-by-finite, we cannot complete a classification by countable discrete structures considered up to isomorphism. \newpage \noindent{\bf $\S$1 Some facts} \medskip From here on let $G$ be a countable discrete group, $H$ a separable infinite dimensional Hilbert space, $U_{\infty}=U(H)$ the unitary group of $H$ -- the group consisting of all bounded linear operators $T:H\rightarrow H$ with $T^*=T^{-1}$. \medskip \noindent{\bf 1.1 Definition} Define Irr$(G,H)$ to be the space of {\it irreducible representations} -- that is the set of all homorphisms $\tau:G\rightarrow U_{\infty}$ such that there are no non-trivial topologically closed, linearly closed, $\tau[G]$- invariant subspaces of $H$. For $\tau\in$ Irr$(G,H)$ and $T\in U_{\infty}$, define $T\cdot \tau\in$ Irr$(G,H)$ by \[(T\cdot \tau)(g)=T\circ \tau(g)\circ T^{-1}.\] Let $\tau_1\approx\tau_2$ denote that there is some $T\in U_{\infty}$ with $T\cdot \tau_1=\tau_2$. \medskip Note that Irr$(G,H)$ is a $G_{\delta}$ subset of $H^G$, and hence Polish (in the product topology). $U_{\infty}$ is a Polish group and the action of this group on Irr$(G,H)$ is continuous. \medskip The following is certainly implicit in \cite{effros}. \medskip \noindent{\bf 1.2 Lemma} (Effros, Mackey) Let $\tau_1, \tau_2\in$ Irr$(G,H)$, $T_i\in U_{\infty}$, $e,\hat{e}\in H$, with \[T_i\cdot \tau_1 \rightarrow \tau_2,\] \[{\rm {liminf}}_{j\rightarrow \infty}(\langle T_j(e), \hat{e})\rangle\neq 0.\] Then $\tau_1\approx\tau_2$. \noindent Proof. Without loss of generality there is a bounded linear operator $T\in {\cal B}(H)$ such that \[T_j(e')\rightarrow T(e')\] weakly (in the sense that $\langle T_j(e'), e''\rangle \rightarrow \langle T(e'), e''\rangle$ all $e''\in H$) for all $e'\in H$. Note then by replacing $\tau_2$ by suitable $T'\cdot \tau_2$ and $T_i$ by $T'T_i$ we may assume $T_j(e)\rightarrow \delta e$ weakly for some $\delta>0$. \noindent Claim(1): For all $e'\in H$, \[T_j^*(e')\rightarrow T^*(e').\] \noindent Proof of claim. Since for all $e' , e'' \in H$ we have that \[\langle e', T^*(e'')\rangle=\langle T(e'), e''\rangle\] \[={\rm {lim}}_{i\rightarrow\infty}\langle T_i(e'), e''\rangle\] \[={\rm {lim}}_{i\rightarrow\infty}\langle e', T_i^*(e'')\rangle.\] \hfill ($\Box$Claim(1)) \noindent Claim(2): For all $g\in G$ \[T\circ \tau_1(g)=\tau_2(g)\circ T.\] \noindent Proof of claim: Since for all $e' , e''\in H$ \[\langle T\circ \tau_1(g)(e'), e''\rangle={\rm {lim}}\langle T_i\circ \tau_1(g)(e'), e''\rangle\] \[={\rm {lim}}_{i\rightarrow\infty}\langle \tau_2(g)\circ T_i(e'), e''\rangle ={\rm {lim}}_{i\rightarrow\infty}\langle T_i(e'), \tau_2(g)^*(e'')\rangle\] \[\langle T(e'), \tau_2(g)^*(e'')\rangle\] \[=\langle \tau_2(g)\circ T(e'), e''\rangle.\] \hfill ($\Box$Claim(2)) \noindent Claim(3): For all $g\in G$ \[T^*\circ \tau_2(g)=\tau_1(g)\circ T^*.\] \noindent Proof of claim: Since for all $e'\in H$ and $j\in {\Bbb N}$ we have \[T_j^*\circ \tau_2(g)(T_j(e'))=T_j^*\circ T_j\circ \tau_1(e')\] \[=\tau_1(e')=\tau_1\circ T_j^*(T_j(e')),\] we can use claim(1) to just repeat the proof of claim(2). \hfill ($\Box$Claim(3)) \noindent Claim(4): For all $g\in G$ \[T^*\circ T\circ \tau_1(g)=\tau_1(g)\circ T^*\circ T.\] \noindent Proof of claim: Follows by claim(2) and claim(3). \hfill ($\Box$Claim(4)) So now let $B\subset {\cal B}(H)$ be the $C^*$-algebra generated by the operator $T^*\circ T$. Since $T^*\circ T$ is self adjoint we have that the polynomials in $T^*\circ T$ are dense in $B$, and hence that every $\tau_1(g)$ commutes with every element of $B$. Following 3.5, 3.6, 3.11VIII of \cite{conway}, if we let $M_0$ be the orthogonal complement of the kernel of $T$ and let $M_1$ be the range of $T$, we may find an isometry \[U: M_0\rightarrow M_1\] and positive $S\in B$ with \[U\circ S=T\] \[S^2=T*T.\] Since {\it every} element of $B$ commutes with $\tau_1(g)$ we have that \[U\circ \tau_1(g)\circ S=U\circ S\circ\tau_1(g)\] \[=T\circ \tau_1(g)=\tau_2(g)\circ T=\tau_2(g)\circ U\circ S\] and thus $U\circ \tau_1(g)$ equals $\tau_2(g)\circ U$ on range of $S$. Since Ran$S$ is $\tau_1[G]$-invariant we have Ran$S=\emptyset, H$ by assumption of irreducibility. So we will be done once we see that $S$ is non-degenerate. However, $T*T\neq 0$ by assumption \[\langle T*\circ T(e), {e}\rangle=\langle T(e), T{e}\rangle\] \[=\langle \delta e, \delta e\rangle=\neq 0.\] \hfill $\Box$ \medskip {\bf 1.3 Corollary} Every $\approx$-equivalence calss is $F_{\sigma}$. \newpage \noindent {\bf $\S$2 A lemma} \medskip From \cite{hjorth}: \medskip \noindent {\bf 2.1 Theorem} Let $G_0$ be a Polish group. Let $S_{\infty}$ be the infinite symmetric group, let $G_1$ be a closed subgroup of $S_{\infty}$. Suppose $X, Y$ are Polish spaces on which $G_0$ and $G_1$ respectively act continuously. Suppose that \leftskip 0.5in \noindent (i) every $G_0$-orbit in $X$ is dense; \noindent (ii) every $G_0$-orbit in $X$ is meager; \noindent (iii) for all $x\in X$, {\it the local orbits of $x$} are somewhere dense; that is to say, if $V$ is an open neighbourhood of $1_{G_0}$, $U$ is an open set containing $x$, and if $O(x,U,V)$ is the set of all $\hat{x}\in [x]_{G_0}$ such that there is a finite sequence $(x_i)_{i\leq k}\subset U$ such that $x_0=x$, $x_k=\hat{x}$, and each $x_{i+1}\in V\cdot x_i$, then the closure of $O(x,U,V)$ contains an open set. \leftskip 0in {\it Then} there is no Borel function \[\theta: X\rightarrow Y\] such that for all $x_1, x_2\in X$ \[\exists g\in G_0(g\cdot x_1=x_2)\Leftrightarrow \exists g'\in G_1 (g'\cdot \theta(x_1)=\theta(x_2)).\] \medskip \noindent {\bf 2.2 Lemma} Let $G$ be a countable group that is not abelian-by-finite -- that is to say, there is no abelian subgroup $G_00$, $A\subset G$ of (finite) size $N$, and $n\in{\Bbb N}$ be such that \[\{\tau\in C: \langle \tau(g)(e_i), e_j\rangle - \langle \tau_1(g)(e_i), e_j\rangle|<\delta {\rm {\: for\: all\: }} i, j\leq n, g\in A\}\subset U\] and let \[U_0=\{\tau\in C: \langle \tau(g)(e_i), e_j\rangle - \langle \tau_1(g)(e_i), e_j\rangle|<\delta/2 {\rm {\: for\: all\: }} i, j\leq n, g\in A\}\subset U\] and \[\{T\in U_{\infty}: \langle T(e_i) , e_i\rangle >1-\delta {\rm {\: all \:}}i\leq n\}\subset V.\] Choose $U_1\subset U_0$ non-empty. We seek $O(\tau_1, U, V)\cap U_1\neq \emptyset$. Choose $\tau_2\in U_1$ with $\tau_1\not\approx \tau_2$. Choose $T_i\in U_{\infty}$ with \[T_i\cdot \tau_1\rightarrow \tau_2.\] Let $\{f_1,...,f_k\}$ be an orthonormal basis for the subbasis generated by $\{(\tau_1^*(g))(e_i), (\tau_1(g))(e_i), e_i: g\in A, i\leq n\}$. By 1.2, for all $j_1, j_2\in{\Bbb N}$ there is some $\epsilon>0$ such that there are only finitely many $i\in{\Bbb N}$ with \[|\langle e_{j_1}, T_i(f_{j_2})\rangle>|\epsilon.\] Thus for $\delta_0>0$ arbitrarily small we may find $i$ such that for all $j_1\leq n, j_2\leq k$ and $g\in A$ \[|\langle e_{j_1}, T_i(f_{j_2})\rangle| <\delta_0/k,\] \[T_i\cdot \tau_1\in U_1.\] Claim: Let $f'_1,..., f'_{l}$ be a set of unit vectors in $H$, and let $H_0$ be a finite dimensional subspace of $H$ with orthonormal basis $\hat{f}_1,...,\hat{f}_{k'}$. If for each $i\leq k'$ \[\sum_{j\leq l}|\langle f'_i, \hat{f}_j\rangle| <\epsilon,\] then there is $T\in U_{\infty}$ such that for each $\langle f'_i, T(f'_i)\rangle$ is bigger than $1-\epsilon$ and each $|\langle \hat{f}_i, T(f'_i)\rangle|=0$. Proof of claim. Let $\{\hat{f}_i:i\in{\Bbb N}\}$ be some orthonormal basis of $H$ extending the orthonormal basis of $H_0$ and such that $\{\hat{f}_i:i\leq k'+l\}$ provides a basis for the subspace generated by $H_0\cup\{f'_1,..., f'_{k'}\}$. Then choose $T$ so that it interchanges $\{\hat{f}_{1},...,\hat{f}_{k'}\}$ with $\{\hat{f}_{1+l+k'}, ...,\hat{f}_{k'+l+k'}\}$ and use that \[\sum_{j\in{\Bbb N}}|\langle f'_i, \hat{f}_j\rangle|^2=1.\] \hfill ($\Box$Claim) Applying the claim to the orthonormal set $\{T_i(f_j):j\leq k\}$ we can find $T'\in U_{\infty}$ such that \leftskip 0.5in \noindent (i) for all $j_1, j_2\leq n$ and $g\in A$ \[\langle T'(e_{j_1}), T_i((e_{j_2}))\rangle =0,\] \[\therefore \langle T_i^{-1}( T'(e_{j_1})), e_{j_2}\rangle =0,\] \[\therefore \langle \tau_1(g)(T_i^{-1}( T'(e_{j_1}))), \tau_1(g)(e_{j_2})\rangle =0,\] \[\langle T'(e_{j_1}), T_i(\tau_1(g)(e_{j_2}))\rangle =0,\] \[\therefore \langle T_i^{-1}( T'(e_{j_1})), \tau_1(g)(e_{j_2})\rangle =0,\] \[\langle T'(e_{j_1}), T_i(\tau_1^*(g)(e_{j_2}))\rangle =0\] \[\therefore\langle \tau_1(g)\circ T_i^{-1}\circ T'(e_{j_1}), e_{j_2}\rangle =0;\] \noindent (ii) for all $j\leq n$ \[|\langle T'(e_j), e_j\rangle|>1-\delta_0.\] \leftskip 0in \noindent From (ii) we obtain that \leftskip 0.5in \noindent (iii) for all $j\leq n$ \[|\langle (T')^{-1}(e_j), e_j\rangle|>1-\delta_0.\] \leftskip 0in \noindent and so $ (T')^{-1}\in V$. %Then \[\langle ((T')^{-1}\circ T_i\cdot(\tau_1))(g)(e_{j_1})), e_{j_2}\rangle - %\langle T'\cdot\tau_1(g))((T')^{-1}(e_{j_1})), e_{j_2}\rangle|\] %\[=T_i(\tau_1(g)((T')^{-1}(e_{j_1})), T'(e_{j_2})\rangle - %\langle \tau_1(e_{j_1}), e_{j_2}\rangle|< %(1+(n+1))\delta_0) |T_i(\tau_1(g)(e_{j_1}), T'(e_{j_2})\rangle - %\langle \tau_1(e_{j_1}), e_{j_2}\rangle|\] %since $\tau_1(g)$ is an isometry and (iii), %\[< (1+(n+1))\delta_0)^2 |T_i(\tau_1(g)(e_{j_1}), e_{j_2}\rangle - %\langle \tau_1(e_{j_1}), e_{j_2}\rangle|\] %by (ii). Thus $(T')^{-1}\circ T_i\cdot(\tau_1)\in U_0$. For future notation, let $\tau_3= (T')^{-1}\circ T_i\cdot(\tau_1)$. Now note that for all $j_1, j_2\leq n$ \[\langle \tau_3(g)(e_{j_1}), e_{j_2}\rangle\] \[=\langle ((T')^{-1}\cdot( T_i\cdot(\tau_1)))(g)(e_{j_1})), e_{j_2}\rangle\] \[=\langle ((T')^{-1}(T_i\cdot(\tau_1)))(g))(T'(e_{j_1}))), e_{j_2}\rangle\] \[=\langle (T_i\cdot \tau_1(g))(T'(e_{j_1})), T'(e_{j_1})\rangle =\] \[\langle (T_i\cdot \tau_1(g))(\delta_1 e_{j_1}), \delta_2 e_{j_2}\rangle +\langle (T_i\cdot \tau_1(g))(\delta_3 f), \delta_2 e_{j_2}\rangle \] \[+\langle (T_i\cdot \tau_1(g))(\delta_1 e_{j_1}), \delta_4 f'\rangle +\langle (T_i\cdot \tau_1(g))(\delta_3 f), \delta_4 f'\rangle \] where \[1>\delta_1=|\langle T'(e_{j_1}), e_{j_1})\rangle|>1-\delta_0,\] \[1>\delta_2=|\langle T'(e_{j_2}), e_{j_2})\rangle|\surd(1-\delta_0),\] and \[(\delta_1)^2+(\delta_3)^2=1,\] \[(\delta_2)^2+(\delta_4)^2=1\] \[f\bot e_{j_1},\] \[f'\bot e_{j_2}.\] Thus, since each $T_i\cdot \tau_1(g)$ is an isometry, for suitably small $\delta_0$ we can obtain \[\langle ((T')^{-1}\cdot( T_i\cdot(\tau_1)))(g)(e_{j_1})), e_{j_2}\rangle - \langle T_i\cdot\tau_1(g))((e_{j_1})), e_{j_2}\rangle|<\delta/2,\] and thus $\tau_3=(T')^{-1}\circ T_i\cdot(\tau_1)\in U$. Since $T_i\cdot(\tau_1)\in O((T')^{-1}\circ T_i\cdot(\tau_1), U, V)$, it suffices to show that $(T')^{-1}\circ T_i\cdot(\tau_1)\in O(\tau_1, U, V)$. Let $\{f_j:j\in {\Bbb N}\}$ be an orthonormal basis extending $\{e_j, (T_j)^{-1}\circ T'(e_j): j\leq n\}$. For $\varphi\in [0,2\pi]$ define $S_{\varphi}\in U_{\infty}$ by the requirement that for all $j\leq n$ \[S_{\varphi}(e_j)={\rm {cos}}(\varphi)(e_j)- {\rm {sin}}(\varphi)((T_i)^{-1}\circ T'(e_j)),\] \[S_{\varphi}((T_i)^{-1}\circ T'(e_j))={\rm {sin}}(\varphi)(e_j)+ {\rm {cos}}(\varphi)((T_i)^{-1}\circ T'(e_j)),\] while for $f_j$ outside $\{e_j, (T_i)^{-1}\circ T'(e_j): j\leq n\}$ we just have \[S_{\varphi}(f_j)=f_j.\] Since $(S_{\pi/2})^{-1}(e_j)=(T_i)^{-1}\circ T'(e_j)$ for each $j\leq n$ it follows that $(T')^{-1}\circ T_i\circ (S_{\pi/2})^{-1}\in V$. %$\tau_3(=_{df} (T')^{-1}\circ T_i\cdot(\tau_1))$ is in %$O(S_{\pi/2}\cdot \tau_1, U, V)$. Thus if we can only show $S_{\pi/2}\cdot \tau_1\in O(\tau_1, U, V)$ then we obtain $\tau_3(=_{df} (T')^{-1}\circ T_i\cdot(\tau_1))$ is in $O(S_{\pi/2}\cdot \tau_1, U, V)$, and hence $T_i\cdot\tau_1\in O(\tau_3, U, V)$ and the transitivity of this relation $O(\cdot, U, V)$ yields $T_i\cdot\tau_1\in O(\tau_1, U, V)$ as required. Then for all $j_1 ,j_2\leq n, g\in A$ we have %\[|(\langle S_{\varphi}\cdot (\tau_1))(g) (e_j), e_i\rangle - %\langle (\tau_1(g)) (e_{j_1}), e_{j_2}\rangle|= %|\langle\tau_1(g) (S_{\varphi}^{-1}(e_{j_1})), %S_{\varphi}^{-1}(e_{j_2})\rangle - \langle (\tau_1)(g) (e_{j_2}), e_{j_1}\rangle|\] \[|\langle\tau_1(g) (S_{\varphi}^{-1}(e_{j_1})), S_{\varphi}^{-1}(e_{j_2})\rangle |\] \[=|\langle \tau_1(g)({\rm {cos}}(-\varphi)(e_{j_1})- {\rm {sin}}(-\varphi)((T_i)^{-1}\circ T'(e_{j_1})),{\rm {cos}}(-\varphi)(e_{j_2})- {\rm {sin}}(-\varphi)((T_i)^{-1}\circ T'(e_{j_2})\rangle|\] %- \langle (\tau_1)(g) (e_{j_1}), e_{j_2}\rangle| \] \[=|\langle{\rm {cos}}^2(-\varphi)\langle \tau_1(g)(e_{j_1}), e_{j_2}\rangle +{\rm {sin}}^2(-\varphi)\langle \tau_1(g)((T_i)^{-1}\circ T'(e_{j_1})), (T')^{-1}\circ T_i(e_{j_2})\rangle|\] by orthogonality of the sets $\{e_j, (\tau_1(g))(e_j):j\leq n, g\in A\}$ and $\{(T_i)^{-1}\circ T'(e_j), (\tau_1(g))(T_i)^{-1}\circ T'(e_j)):j\leq n, g\in A\}$ as expressed at (i) above. Thus \[|\langle (S_{\varphi}\cdot (\tau_1))(g) (e_j), e_i\rangle - \langle (\tau_1(g)) (e_{j_1}), e_{j_2}\rangle|=\] \[=|\langle{\rm {cos}}^2(-\varphi)\langle \tau_1(g)(e_{j_1}), e_{j_2}\rangle +{\rm {sin}}^2(-\varphi)\langle \tau_1(g)((T_i)^{-1}\circ T'(e_{j_1})), (T_i)^{-1}\circ T'(e_{j_2})\rangle\] \[- \langle (\tau_1)(g) (e_{j_1}), e_{j_2}\rangle|\] But this final inequality is bounded by an expression of the form \[\lambda|\langle \tau_1(g)((T_i)^{-1}\circ T'(e_{j_1})), (T_i)^{-1}\circ T'(e_{j_2})\rangle - \langle (\tau_1)(g) (e_{j_1}), e_{j_2}\rangle|\] \[+(1-\lambda) |\langle \tau_1(g)(e_{j_1}), e_{j_2}\rangle - \langle (\tau_1)(g) (e_{j_1}), e_{j_2}\rangle|,\] where $\lambda\in[0,1]$, and hence must be less than or equal to the bound $\delta$ obtained at the extreme points. Thus each $S_{\varphi}\cdot (\tau_1)$ is in $U$, and so we have traced out a continuous path of group elements connecting $\tau_1$ to $S_{\pi/2}\cdot\tau_1$ inside $U$, and are therefore done. \hfill $\Box$ \medskip The further non-classifiability results alluded to in the introduction are now a consequence of 2.1 and standard facts from the theory of equivalence relations -- as may be found in \cite{beckerkechris}. \newpage \begin{thebibliography}{99} %\bibitem{becker}H. Becker, {\it Vaught's conjecture for complete left invariant %groups}, handwritten notes, University of North Carolina at Columbia, 1996. \bibitem{beckerkechris} H. Becker and A.S. Kechris, {\bf The descriptive set theory of Polish group actions}, London Mathematical Society Lecture Notes Series, Cambridge, 1996. \bibitem{conway} J.B. Conway, {\bf A course in functional analysis,} Graduate Texts in Mathematics 96, Springer-Verlag, New-York, 1990. \bibitem{effros} E. 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