## endomorphism of factor: can it be idempotent up to congugacy?

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.

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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.

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 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? – AndrĂ© Henriques Nov 7 2010 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 2010 at 14:00

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.

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Jesse seems to be right. I messed up with irreducibility. so this is not really an example. – Makoto Yamashita Nov 7 2010 at 3:02