Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T16:51:07Z http://mathoverflow.net/feeds/question/47071 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/47071/endomorphism-of-type-iii-factor-can-it-satisfy-phi-circ-phi-phi-oplus-phi Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$? André Henriques 2010-11-23T08:22:09Z 2011-11-05T07:54:44Z <p>I'm still trying to get some feeling about <a href="http://mathoverflow.net/questions/41327/subfactor-of-finite-rank-but-infinite-index-is-this-possible" rel="nofollow">this question</a>...<br> Given Jesse Peterson's answer to <a href="http://mathoverflow.net/questions/45102/endomorphism-of-factor-can-it-be-idempotent-up-to-congugacy" rel="nofollow">this question</a> (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.</p> <hr> <p>Let $M$ be a type <i>III</i> 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 $1$<br> ($v_1^*v_1=v_2^*v_2=v_1v_1^*+v_2v_2^*=1$). Define $$\phi\oplus\phi:m\mapsto v_1\phi(m)v_1^*+v_2\phi(m)v_2^*.$$</p> <blockquote> <p><b>Question:</b> Is it possible to have $\phi\circ\phi$ conjugate to $\phi\oplus\phi$?<br> $$\phi ( \phi(m)) = u \Big( v_1 \phi (m) v_1^* + v_2 \phi (m) v_2^* \Big) u^*$$</p> </blockquote> http://mathoverflow.net/questions/47071/endomorphism-of-type-iii-factor-can-it-satisfy-phi-circ-phi-phi-oplus-phi/80030#80030 Answer by Ulrich Pennig for Endomorphism of type III factor: can it satisfy $\phi\circ\phi = \phi\oplus\phi$? Ulrich Pennig 2011-11-04T09:59:27Z 2011-11-05T07:54:44Z <p>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</p> <p>$$\langle \bar{\phi} \circ \phi, \phi \rangle = \langle \phi, \phi \circ \phi \rangle = \langle \phi, \phi \oplus \phi \rangle = 2.$$</p> <p>Thus, $\bar{\phi} \circ \phi$ contains two copies of $\phi$ and a copy of the identity. Therefore </p> <p>$$4 = d(\phi)^2 = d(\phi)\cdot d(\bar{\phi}) = d(\bar{\phi} \circ \phi) \geq d(id \oplus \phi \oplus \phi) = 1 + 2d(\phi) = 5$$</p> <p>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</p> <p>$$[\phi \circ \phi] = [id \oplus \alpha] \circ [id \oplus \alpha] = [id \oplus \alpha \oplus \alpha \oplus \alpha^2] = [id \oplus \alpha \oplus id \oplus \alpha] = [\phi \oplus \phi]$$ </p> <p>where the brackets mean unitary equivalence classes of endomorphisms. </p>