Let $N\subset M$ be a be factors acting on a Hilbert space $H$.
Denote by $\mathrm{Cyc}(A)\subset H$ the set of cyclic vectors for $A = M,N$.


I am interested in the equality case of the inclusion $\mathrm{Cyc}(N)\subseteq \mathrm{Cyc}(M)$?


If, e.g., $M= M\_2(\mathbb C) \otimes N$, then it is easy to find vectors vectors which are cyclic for $M$ but not for $N$ (any product vector with respect to the induced splitting $H = \mathbb C^2\otimes H\_0$ works).
I was not able to find counterexamples for cases where the relative commutant is trivial, so I want to ask:


Does $\mathrm{Cyc}(M)=\mathrm{Cyc}(N)$ hold if the relative commutant is trivial (i.e., $N'\cap M=\mathbb C$)?


Any help is much appreciated



Cyclic vectors and subfactor inlcusion
No. Let $N$ be any $\rm II\_1$ factor, and let $\alpha: G\to \operatorname{Aut}(N)$ be an outer action. Then $M:= N\rtimes\_\alpha G$ is again a $\rm II\_1$ factor, and $N\subset M$ is irreducible, i.e., $N'\cap M = \mathbb{C}$. Now let $\Omega$ be the image of $1$ in $L^2M$, the GNS space of the trace. Then $\Omega$ is cyclic for $M$, but not for $N$.


This is true for any irreducible $\rm II\_1$ subfactor; $\Omega$ will always be cyclic for $M$ but not for $N$ unless $N=M$. The closure of $N\Omega$ is unitarily isomorphic to $L^2N$, and the projection $e\_N$ onto this subspace is $1$ if and only if $M=N$.


