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I'm still trying to get some feeling about this question...
Given Jesse Peterson's answer to this question (he showed that $\phi\circ\phi\sim\phi$ is impossible), I suspect that the following is also impossible. But I'm unable to generalize his argument.

Let $M$ be a type III factor, and let $\phi:M\to M$ be an irreducible endomorphism (the relative commutant of $\phi(M)$ in $M$ is trivial). Let $v_1$, $v_2\in M$ be isometries with orthogonal ranges summing up to $1$
($v_1^*v_1=v_2^*v_2=v_1v_1^*+v_2v_2^*=1$). Define $$\phi\oplus\phi:m\mapsto v_1\phi(m)v_1^*+v_2\phi(m)v_2^*.$$

Question: Is it possible to have $\phi\circ\phi$ conjugate to $\phi\oplus\phi$?
$$ \phi ( \phi(m)) = u \Big( v_1 \phi (m) v_1^* + v_2 \phi (m) v_2^* \Big) u^* $$

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  • $\begingroup$ In the case $\phi$ has a conjugate endomorphism so that one can define it's dimension (see e.g. Longo & Roberts, "A theory of dimension", this implies the condition $d(\phi)^2 = 2 d(\phi)$, which at least imposes some restrictions on $\phi$. $\endgroup$
    – Pieter Naaijkens
    Commented Nov 23, 2010 at 9:10
  • $\begingroup$ @Peter: Ah! I now realize that the conditions I've imposed do not exclude the possibility of $\phi$ having a conjugate... (I was trying to cook up something that would make that impossible). Unrelated: I really like the paper "A theory of dimension". $\endgroup$
    – André Henriques
    Commented Nov 23, 2010 at 18:35
  • $\begingroup$ Replacing $v_i$ by $uv_i$, you may assume that $u=1$, if that makes computations any easier. $\endgroup$
    – Andreas Thom
    Commented Nov 29, 2010 at 14:41

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I think the situation you describe is impossible: Let $\bar{\phi}$ be the conjugate endomorphism to $\phi$. From the equation $d(\phi)^2 = 2d(\phi)$ we get $d(\phi) = 2$. Denote by $\langle \rho, \sigma \rangle$ the dimension of the intertwiner space between $\rho$ and $\sigma$. By Frobenius reciprocity and the irreducibility of $\phi$ we now have

$$ \langle \bar{\phi} \circ \phi, \phi \rangle = \langle \phi, \phi \circ \phi \rangle = \langle \phi, \phi \oplus \phi \rangle = 2. $$

Thus, $\bar{\phi} \circ \phi$ contains two copies of $\phi$ and a copy of the identity. Therefore

$$ 4 = d(\phi)^2 = d(\phi)\cdot d(\bar{\phi}) = d(\bar{\phi} \circ \phi) \geq d(id \oplus \phi \oplus \phi) = 1 + 2d(\phi) = 5 $$

which is a contradiction. Note that if you drop the assumption that $\phi$ is irreducible, there should be examples: Suppose $M$ carries an involution $\alpha \colon M \to M$, i.e. an action of $\mathbb{Z} / 2\mathbb{Z}$. Consider $\phi = id \oplus \alpha$ with the definition of the sum similar to the one in your question. Then

$$ [\phi \circ \phi] = [id \oplus \alpha] \circ [id \oplus \alpha] = [id \oplus \alpha \oplus \alpha \oplus \alpha^2] = [id \oplus \alpha \oplus id \oplus \alpha] = [\phi \oplus \phi] $$

where the brackets mean unitary equivalence classes of endomorphisms.

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  • $\begingroup$ Excellent! So you have excluded the possibility $d(\phi)<\infty$ (which I wanted to exclude anyways -- see my comment above). What I'm really interested in is techniques to show that $\phi^2=2\phi$ and $d(\phi)=\infty$ is impossible. $\endgroup$
    – André Henriques
    Commented Nov 5, 2011 at 7:52

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