4
$\begingroup$

Let $\mathcal{M}$ be a diffuse semi-finite von Neumann algebra acting on a Hilbert space $\mathcal{H}$, equipped with a faithful normal semi-finite trace $\varphi$. Let $\Phi : \mathcal{M} \to \mathcal{M}$ be a $*$-automorphism of $\mathcal{M}$ satisfying the following conditions:

  • The set of non-zero projections $p \in \mathcal{M}$ for which $\Phi(p) = p$ is $\{1\}$, i.e., the only projection fixed by $\Phi$ is the identity projection.
  • For every non-zero projection $p \in \mathcal{M}$, there exists some $k \geq 1$ such that $$ \Phi^k(p) \mathcal{H} \cap p \mathcal{H} \neq \{0\}. $$
  • The automorphism $\Phi$ preserves the trace, i.e., $$ \varphi \circ \Phi(x) = \varphi(x) \quad \text{for all} \quad x \in \mathcal{M}. $$

Question: Is there a non-zero projection $q \in \mathcal{M}$ such that $$ \Phi(q)\mathcal{H} \cap q^\perp \mathcal{H} \neq \{0\}? $$

Let $\mathcal{K}_q = \Phi(q)\mathcal{H} \cap q^\perp \mathcal{H}$ for a non-zero projection $q \in \mathcal{M}$, and let $r_q$ denote the projection from $\mathcal{H}$ onto $\mathcal{K}_q$.

We aim to show that $r_q = 0$ for all non-zero projections $q$ in $\mathcal{M}$.

However, the following question arises: Is it possible to construct a non-zero projection $q \in \mathcal{M}$ such that $r_q \neq 0$?
Could you please help me to solve this problem? I am unable to proceed from here. Your assistance would be greatly appreciated. Thank you.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ If $\Phi$ is unital and $\Phi(q)\le q$ for every projection $q$, then $\Phi(1-q) \le 1-q$ as well, which implies $\Phi(q)=q$ for every $q$. $\endgroup$
    – Narutaka OZAWA
    Commented Nov 28 at 0:11
  • 1
    $\begingroup$ Pretty sure $\mathcal{M} = L^\infty(0, 1)^{\otimes \mathbb{Z}}$ and $\Phi$ being the Bernoulli shift should be an example? (It might be far harder to get $r_q = 0$ for all $q$, imo.) $\endgroup$
    – David Gao
    Commented Nov 28 at 2:13
  • $\begingroup$ @NarutakaOZAWA Sir, if $\Phi(q) = q$ for every $q$, then by condition 1, $q = 1$, which implies $\mathcal{M} = \mathbb{C} \mathbb{1}$. This seems to contradict the fact that $\mathcal{M}$ is diffuse. I hope I am understanding the argument correctly, and I would appreciate any clarification on this point. $\endgroup$
    – DenOfZero
    Commented Nov 28 at 6:12
  • 2
    $\begingroup$ @DenOfZero: I misunderstood your question. Please forget what I said. $\endgroup$
    – Narutaka OZAWA
    Commented Nov 28 at 9:14
  • $\begingroup$ BTW, Connes has proved that for every ergodic (or properly outer) $\Phi$ there is a nonzero projection $q$ such that $\| q\Phi(q) \|<0.1$, but you are asking for a nonzero $q$ with $q\Phi(q)=0$. $\endgroup$
    – Narutaka OZAWA
    Commented Nov 29 at 0:59

0

Reset to default

You must log in to answer this question.