endomorphism of factor: can it be idempotent up to congugacy? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:50:23Z http://mathoverflow.net/feeds/question/45102 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45102/endomorphism-of-factor-can-it-be-idempotent-up-to-congugacy endomorphism of factor: can it be idempotent up to congugacy? André Henriques 2010-11-06T21:41:45Z 2010-11-07T02:41:05Z <p>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.</p> <p>Is it possible to have $\phi\circ \phi$ conjugate to $\phi$?<br> In other words, is it possible to have an endomorphism $\phi$, and a unitary $u\in M$, such that $$\phi(\phi(x))=u\phi(x)u^*,\quad\forall x\in M.$$</p> <p>If this is possible, I would like to see an example. <hr> <em>Note:</em> an answer to the above question would also settle <a href="http://mathoverflow.net/questions/41327/subfactor-of-finite-rank-but-infinite-index-is-this-possible" rel="nofollow">this question</a>.</p> http://mathoverflow.net/questions/45102/endomorphism-of-factor-can-it-be-idempotent-up-to-congugacy/45131#45131 Answer by Makoto Yamashita for endomorphism of factor: can it be idempotent up to congugacy? Makoto Yamashita 2010-11-07T02:36:04Z 2010-11-07T02:36:04Z <p>There is Thompson's group $$F=\langle x_0, x_1, \ldots \mid x_i^{-1} x_n x_i = x_{n+1}, 0 \le i &lt; n \rangle.$$ 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.</p> http://mathoverflow.net/questions/45102/endomorphism-of-factor-can-it-be-idempotent-up-to-congugacy/45132#45132 Answer by Jesse Peterson for endomorphism of factor: can it be idempotent up to congugacy? Jesse Peterson 2010-11-07T02:41:05Z 2010-11-07T02:41:05Z <p>This is not possible. If it were, then using the notation above, given any $x \in \phi(M)$, we would have <code>$x u^* = u^* \phi(x)$</code>, and $\phi(x) u = u x$. Hence, for any $x \in \phi(M)$ we have <code>$$x u^* \phi(u^*) u^2 = u^* \phi(x u^*) u^2$$</code> <code>$$= u^* \phi(u^*) \phi \circ \phi (x) u^2 = u^* \phi(u^*) u^2 x.$$</code></p> <p>Hence <code>$u^* \phi(u^*)u^2 \in \phi(M)' \cap M = \mathbb C$</code> and so <code>$\phi(u^*) \in \mathbb C \cdot u^*$</code>. Then, for any $y \in M$ we would have that <code>$$\phi \circ \phi (y) = u \phi(y) u^*$$</code> <code>$$= \phi(u y u^*).$$</code> Since $\phi$ is injective we then have <code>$\phi(y) = u y u^*$</code>, and hence $\phi$ is invertible.</p> <p>If you don't require that $\phi(M)$ be irreducible then this is possible.</p>