Galois theory for subfactors I’m aware that there is some Galois theory established for subfactors but could not find anything related to the following:
Suppose that $N$ is a subfactor of the II$_1$ factor $M$. Let $G$ denote the group of automorphisms of $M$ fixing $N$ point wise and let $Q$ be the elements in $M$ fixed by all elements of $G$. Clearly $Q$ contains $N$. Are there sufficient conditions which ensure that $Q$ is a factor?
If it helps I’m willing to assume that $N$ has spectral gap in $M$ or infinite index.
 A: (This "answer" is actually not an answer at all, since I am looking here at the Galois property, i.e. the property that the fixed point algebra under $G$ is again $N$. See the comments below.)
When $N \subset M$ is a "generic" subfactor (and I am actually not specifying what this is supposed to mean), the group $G$ of automorphisms of $M$ fixing $N$ pointwise would generically be as small as $\{\operatorname{Ad} u \mid u \in \mathcal{U}(N' \cap M)\}$. In particular, for irreducible subfactors, the Galois property should be "rare" and only hold in the presence of special structure.
The following illustrates this. Assume that $\Gamma$ is a countable group and $\Gamma \curvearrowright^\alpha N$ is an action of $\Gamma$ by outer automorphisms of $N$. Then put $M = N \rtimes \Gamma$ so that $N \subset M$ is an irreducible subfactor. The group $G$ of automorphisms of $M$ that fix $N$ pointwise coincides with the group of characters
$$\hat{\Gamma} = \{\omega : \Gamma \rightarrow \mathbb{T} \mid \omega \;\text{is a group homomorphism}\;\}$$
by associating to every $\omega \in \hat{\Gamma}$ the automorphism $\theta_\omega$ of $M$ given by $\theta_\omega(a) = a$ if $a \in N$ and $\theta_\omega(u_g) = \omega(g) u_g$ if $g \in \Gamma$.
It then follows that the Galois property holds if and only if the group $\Gamma$ is abelian.
Another family of irreducible subfactors with the Galois property arises by taking the free product $M = N * P$ of two II$_1$ factors $N$ and $P$. For every unitary $u \in \mathcal{U}(P)$, one can define the automorphism $\theta_u$ of $M$ by $\theta_u(a) = a$ for all $a \in N$ and $\theta_u(b) = ubu^*$ for all $b \in P$. The only elements of $M$ that are fixed by all these automorphisms are the elements of $N$.
A: One sufficient condition, as I mentioned in the comment is that $N' \cap M= \mathbb C$. I guess you are looking for something less trivial than that. One other sufficient condition that I could think of is the following situation:
Suppose $M=N \bar{\otimes} P$, where $P$ is a $\rm II_1$  factor. Note that $G$ contains $\{Ad(u): u \in \mathcal U(P)\}$. Moreover, using Ge's result, $Q=N \bar{\otimes} P_0$, where $P_0 \subseteq P$ is a subalgebra. Since $uxu^*=x$ for all $x \in P_0$, and for all $u \in \mathcal U(P)$, we conclude that $P_0=\mathbb C$ ( as $\mathcal U(P) \curvearrowright P$ is ergodic).
More generally, if $N=(N'\cap M)' \cap M$, then $N=Q$. (Because of your condition, $Q \subseteq (N'\cap M)' \cap M$, and hence the more general condition yields that $Q=N$.)
On the other hand, suppose that $Out(M)$ is trivial. Then $G$ may be identified with unitaries in $N' \cap M$. If $N' \cap M$ is nontrivial, abelian, then $Q$ won't be a factor.
