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Question: Let $G$ be a finite group and let $P$ be a $\rm II_1$ factor. Assume that $G$ acts on $P$ in a trace-preserving manner, such that the crossed product algebra $P \rtimes G$ is a factor. Is $G \curvearrowright^{\sigma} P$ outer?

Motivation: It's mentioned in Example 2.3.3(b) in Jones-Sunder's book that the above result is true. However, the proof is not given. I tried to prove it myself, but got stuck. Below is my attempt.

Attempt: Suppose $G \curvearrowright^{\sigma} P$ is not outer. Then there exists $g \in G$, $g \neq e$ such that $\sigma_g$ is inner. Assume that $\sigma_g =id$. Let $v =\sum_{h \in G} u_h u_g u_{h^{-1}}$. Then, it's easy to verify that $v $ lies in the center of $P \rtimes G$, and hence is a scalar. Now, let $H=C_G(g)$. Then, $v= \sum_{h \in H} u_g + \sum_{h \in G \setminus H} u_{hgh^{-1}}=c \in \mathbb C$. Therefore, $|H| + \sum_{h \in G \setminus H} u_{hgh^{-1}g^{-1}}=cu_{g^{-1}} $. Taking traces on both sides, we get that $|H|=0$, which is a contradiction.

I tried a similar trick when $\sigma_g=ad(u)$ for some unitary $u \in P$. I took $v= \sum_{h \in G} u_h u^{\ast}u_{gh^{-1}}$. I can show that $v $ lies in the center of $P \rtimes G$, and hence is a scalar. However, I got stuck after that.

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  • $\begingroup$ Minor correction: you should at least assume the action to be non-trivial. $\endgroup$ – Sebastien Palcoux 3 hours ago
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I think that the claim in the question is false. One can construct a counterexample as follows. First assume in general that $G$ is a finite abelian group of order $n$ and that $\Omega : G \times G \to S^1$ is a bicharacter (i.e. a map that is multiplicative in both variables). Define the projective representation $U : G \to U(\ell^2(G))$ by $$U(g)e_k = \overline{\Omega(g,k)} e_{gk}$$ where $(e_k)_{k \in G}$ is the standard orthonormal basis of $G$. One checks that $$U(g) U(h) = \overline{\Omega(g,h)} U(gh)$$ for all $g,h \in G$. Identify $\ell^2(G) \cong \mathbb{C}^n$ and let $P_0$ be any $II_1$ factor. Define $P = M_n(\mathbb{C}) \otimes P_0$ and the action $G \curvearrowright^\sigma P$ by $\sigma_g = \operatorname{Ad}(U(g) \otimes 1)$. Then every automorphism $\sigma_g$ is inner by construction.

Consider the twisted group von Neumann algebra $L_\Omega(G)$ generated by canonical unitary operators $(v_g)_{g \in G}$ satisfying $$v_g v_h = \Omega(g,h) v_{gh} \; .$$ There then is a $*$-isomorphism $$\pi : P \rtimes_\sigma G \to P \otimes L_\Omega(G) : \pi(a u_g) = a(U(g) \otimes 1) \otimes v_g$$ for all $a \in P$ and $g \in G$.

To produce a counterexample, it now suffices to give an example where $L_\Omega(G) \cong M_n(\mathbb{C})$. This happens for instance when $\Gamma$ is a finite abelian group and $G = \Gamma \times \widehat{\Gamma}$ with $$\Omega : G \times G \to S^1 : \Omega((g,\omega),(h,\eta)) = \omega(h) \; .$$ Defining the unitary operators $W(g,\omega)$ on $\ell^2(\Gamma)$ by $$W(g,\omega)e_h = \omega(h) e_{gh} \; ,$$ it follows that $$\theta : L_\Omega(G) \to B(\ell^2(\Gamma)) \cong M_n(\mathbb{C}) : \theta(v_{(g,\omega)}) = W(g,\omega)$$ is a $*$-isomorphism.

So for instance the group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ admits an inner action on a $II_1$ factor such that the crossed product is a factor.

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