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

Question

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.

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

Answer 1

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.

Answer 2

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.