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

Let $M$ be a factor, and let $\phi:M\to M$ be an irreducible endomorphism ("irreducible" means that the relative commutant of $\phi(M)$ in $M$ is trivial). Let's also assume that $\phi$ is not invertible.

Is it possible to have $\phi\circ \phi$ conjugate to $\phi$?
In other words, is it possible to have an endomorphism $\phi$, and a unitary $u\in M$, such that $$\phi(\phi(x))=u\phi(x)u^*,\quad\forall x\in M.$$

If this is possible, I would like to see an example.


Note: an answer to the above question would also settle this question.

share|improve this question
add comment

2 Answers

up vote 7 down vote accepted

This is not possible. If it were, then using the notation above, given any $x \in \phi(M)$, we would have $x u^* = u^* \phi(x)$, and $\phi(x) u = u x$. Hence, for any $x \in \phi(M)$ we have $$ x u^* \phi(u^*) u^2 = u^* \phi(x u^*) u^2 $$ $$ = u^* \phi(u^*) \phi \circ \phi (x) u^2 = u^* \phi(u^*) u^2 x. $$ `

Hence $u^* \phi(u^*)u^2 \in \phi(M)' \cap M = \mathbb C$ and so $\phi(u^*) \in \mathbb C \cdot u^*$. Then, for any $y \in M$ we would have that $$ \phi \circ \phi (y) = u \phi(y) u^* $$ $$ = \phi(u y u^*). $$ Since $\phi$ is injective we then have $\phi(y) = u y u^*$, and hence $\phi$ is invertible.

If you don't require that $\phi(M)$ be irreducible then this is possible.

share|improve this answer
    
Very nice. That's a proof. But it's not the kind of proof that you just stumble upon by randomly playing with symbols... You must have known from before that this was not possible. Could you maybe share some of your intuition? –  André Henriques Nov 7 '10 at 11:46
    
I think it's easier to work with the map Ad`$(u^*) \circ \phi = \theta$` instead of $\phi$. Conjugating by $u$ then moves the image of $\theta$ into the fixed points of $\theta$, both of these subfactors being irreducible. From here the above proof is fairly natural. I didn't want to introduce extra notation above though. Also, by working with the original map $\phi$ you get to see explicitly what role $u^*\phi(u^*)u^2$ plays in the proof. –  Jesse Peterson Nov 7 '10 at 14:00
add comment

There is Thompson's group $$ F=\langle x_0, x_1, \ldots \mid x_i^{-1} x_n x_i = x_{n+1}, 0 \le i < n \rangle. $$ If you let $M = LF$ and $\phi$ $M \rightarrow M$ be the extension of $\phi(x_i) = x_{i+1}$, then $u = x_0^{-1}$ will satisfy your condition.

share|improve this answer
1  
Jesse seems to be right. I messed up with irreducibility. so this is not really an example. –  Makoto Yamashita Nov 7 '10 at 3:02
add comment

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.