# Does the hyperfinite II_1 factor admit two irreducible representations that are not unitarily equivalent?

Regarding the hyperfinite $II_{1}$ factor $R$ as $C^{*}$-algebra, is it known whether any two irreducible representations of $R$ are unitarily equivalent? If it is known that there exists a pair of irreducible representations that are not unitarily equivalent, please provide a reference.

(Note: I am not requiring that the representations be normal!!)

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Don't want to see this qn ignored so here's an idea. $R$ is prime and antiliminal so the pure states of $R$ are weak*-dense in the state space (Glimm/Tomiyama/Takesaki). If any two pure states can be intertwined by a unitary in $R$ then any two states can be 'almost intertwined'. Let $R$ act as a vN-alg on a Hilbert space $H$. Let $P$ be a projection in the commutant of $R$. Take vectors $\xi$ and $\eta$ in $H$ with $P\xi=\xi$ and $P\eta=0$. Take the vector states on $R$ generated by $\xi$ and $eta$. Then surely any unitary in $R$ will have a job intertwining these...can anyone finish this? –  Douglas Somerset Mar 31 '12 at 19:04
@douglas: Normal representations, and hence normal states (in particular vector states for normal representations) will never give you different irreducible representations. Every normal representation of a type II$_1$ factor $M$ is given by left multiplication $\pi_p$ on the Hilbert space $H=(\ell^2(I)\otimes L^2(M,\tau))p$. Here $p$ is a projection in $N=B(\ell^2(\IN))\otimes M$ (acting by right multiplication). These representations are never irreducible (the relative commutant is a type II factor), but more importantly, $\pi_p$ is isomorphic to a subrepresentation of $\pi_q^{\oplus I}$. –  Steven Deprez Apr 4 '12 at 11:16
@steven: Does the fact that $\pi_p$ is isomorphic to a subrepresentation of $\pi_q^{\oplus I}$ mean that $\pi_p$ and $\pi_q$ can be approximately intertwined, i.e. approximated by unitarily equivalent irreducible representations? –  Douglas Somerset Apr 5 '12 at 21:01

There are $2^c$ mutually non-equivalent irreducible representations. Since $\ell_\infty(N)$ has $2^c$ many pure states (there are $2^c$ many ultrafilters on $N$), any $C^*$-algebra containing $\ell_\infty(N)$ has at least as much pure states. Since $R$ has $c$ many unitary elements, there are $2^c$ many mutually non-equivalent pure states. This is a very old observation, but I don't know a specific reference. (It was noted in our paper too. http://xxx.lanl.gov/abs/math.OA/0110152)