Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be the hyperfinite type $III_1$ factor, and let $Aut(R)$ be its group of automorphisms, equipped with the $u$-topology (topology of pointwise convergence on the predual). An automorphism $\alpha\in Aut(R)$ is called inner if it is of the form $\alpha(x)=uxu^\*$ for some unitary $u\in R$.

I've heard that inner automorhpisms are dense in $Aut(R)$.

Can someone give me an explicit example of a sequence of inner automorphisms that converges to an automorphism that is not inner?


PS: An answer to the same question for the hyperfinite $II_1$ factor would also be interesting.

share|improve this question
    
Is the $u$-topology given by point-norm or point-weak convergence on the predual? –  Aaron Tikuisis Jul 10 '12 at 10:19
    
Yes: point-norm. –  André Henriques Jul 11 '12 at 13:50

1 Answer 1

up vote 8 down vote accepted

It is relatively easy to give an explicit sequence of inner automorphisms that converges to the flip automorphism $\sigma$. First of all, realize $R$ as an infinite tensor product of matrix algebras $(R,\varphi)=\bigotimes_n (M_{k_n}(\mathbb{C}),\varphi_n)$. For every $n\in \mathbb{N}$, we find a unitary $u_n\in M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$ that implements the flip automorphism on $M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$, i.e. $u_n(x\otimes y)u_n^\ast=y\otimes x$. (This follows from the general observation that every automorphism of a type I factor is inner, but it is a good exercise to find the $u_n$ explicitly).

Now it follows that the sequence of inner automorphisms $\sigma_n=Ad_{u_1\otimes\ldots\otimes u_n\otimes 1\ldots}$ converges to the flip automorphism $\sigma$: Let $\psi$ be any ultraweakly continuous functional on $R\otimes R$. Consider $R\otimes R$ to be represented on the infinite tensor product space $H=\bigotimes_n L^2(M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C}), \varphi_n\otimes\varphi_n)$. We know that $\psi$ is of the form $\psi(x)=\sum_k\langle\xi_k,x\eta_k\rangle$ for some $\ell^2$-summable sequences $\xi_k,\eta_k$ in $H$. Since the finite sequences are dense in the $ell^2$-summable ones, we can assume that $\psi(x)=\langle\xi,x\eta\rangle$. Because the finite tensor products are dense in the infinite ones, we can assume that $\xi,\eta\in H_N=\bigotimes_{n=1}^N L^2(M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C}), \varphi_n\otimes\varphi_n)$ for some $N\in\mathbb{N}$. Now it follows that $\psi(\sigma_n(x))=\langle \xi,\sigma_n(x)\eta\rangle=\langle\sigma_n^{-1}(\xi),x\sigma_n^{-1}(\eta)\rangle=\langle\sigma^{-1}(\xi),x\sigma^{-1}(\eta)\rangle=\psi(\sigma(x))$.

share|improve this answer
2  
I was thinking of the same example, but couldn't see how to prove the convergence. Can you explain this part? –  Aaron Tikuisis Jul 10 '12 at 10:46
    
I added a proof of convergence –  Steven Deprez Jul 11 '12 at 8:50
    
Thanks! I take it from your proof that $u$-convergence does mean point-norm convergence on the predual? –  Aaron Tikuisis Jul 11 '12 at 11:04
    
One more question: how can we prove that the flip isn't inner? –  Aaron Tikuisis Jul 11 '12 at 11:12
1  
Consider an endomorphism $\rho$ of $R$ (I'm thinking $R=III_1$; for the analogous argument with $II_1$ factors, just translate into $R$-$R$-bimodules), and its dual endomorphism $\bar \rho$. Then $(\rho\otimes 1)\circ(1\otimes \bar\rho)$ is irreducible, but $(\rho\otimes 1)\circ(\bar\rho\otimes 1)$ is not irreducible, so they cannot be conjugate to each other. But $(\rho\otimes 1)\circ(1\otimes \bar\rho)=(\rho\otimes 1)\circ\tau\circ(\bar\rho\otimes 1)\circ\tau$. So it $\tau$ was inner, then those two endomorphisms of $R\otimes R$ would be conjugate; contradiction. –  André Henriques Jul 11 '12 at 13:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.