Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$?

2010-11-23T08:22:09Z

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 $\phi\oplus\phi$?
$$ \phi ( \phi(m)) = u \Big( v_1 \phi (m) v_1^* + v_2 \phi (m) v_2^* \Big) u^* $$

Answer by Ulrich Pennig for Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$?

2011-11-04T09:59:27Z

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.

Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$? - MathOverflow

Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$?

Answer by Ulrich Pennig for Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$?