# Action of a finite group on a finite factor

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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• Minor correction: you should at least assume the action to be non-trivial. – Sebastien Palcoux 3 hours ago

## 1 Answer

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.