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> **Question**: Is the center of the automorphism group of a von Neumann algebra $\mathscr{M}$ trivial (=$\{\mathrm{id}\}$) whenever $\mathscr{M}$ is a factor (=$\mathscr{M}$ has center $\{\lambda I; \lambda \in \mathbb{C}\}$)?
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It is true in the *finite dimensional* case. A finite dimensional factor is $M\_n(\mathbb{C})$ for some $n \in \mathbb{N}$.


We proceed in two steps. First, every automorphism of $M\_n$ is of the form $a \mapsto u^\*a u$ for some unitary $u$. We will show that if an automorphism is in the center of $\mathop{\mathrm{Aut}}$, then $u$ is in the center of its unitary group. Finally, we show that the center of the unitary group is "trivial".


1. There is a surjective group homomorpism $\varphi\colon U(n) \to \mathop{\mathrm{Aut}}(M\_n)$ with $\varphi(u)(a)=u^\*au$.
The kernel of $\varphi$ are the unitaries in the center of $M\_n$. Thus $\ker \varphi = \{\lambda I;\ |\lambda|=1\}$. Now, given any $u$ such that $\varphi(u)$ is in the center of $\mathop{\mathrm{Aut}}(M\_n)$. Then for any other $v$ we have $vuv^{-1}u^{-1} \in \ker\varphi$. Thus $vuv^{-1}u^{-1}=\lambda I$ for some $\lambda \in \mathbb{C}$ with $|\lambda|=1$. In particular, when $u=v$ we find $I=\lambda I$. Apparently $vuv^{-1}u^{-1}=I$. Thus $u$ is in the center of $U(n)$.
2. The embedding $e\colon U(n) \to GL(n)$ is an irreducible group representation. A unitary $u$ in the center of $U(n)$, is an intertwining map from this representation to itself. Thus, by Schur's lemma, it is $\lambda I$ for some $\lambda \in \mathbb{C}$. Clearly $|\lambda| = 1$. Thus the center of $U(n)$ is the kernel of $\varphi$.



Is the center of the automorphism group of a von Neumann algebra M trivial whenever M is a factor?
Suppose that $\phi $ is in the centre of $\text {Aut}(M) $. Fix a unitary $u\in\mathcal M $. Then $$\tag {1}\phi (uxu^\*)=u\phi(x)u^\*$$ for all $x $. In particular, when $x=u $, we have $\phi ( u)=u\phi (u)u^\*,$ or $\phi (u)u=u\phi (u)$. As $u $ is unitary, this also implies that $u^\*\phi (u) =\phi (u)u^\*$.


Replacing $x $ with $xu $ in $(1) $, we get $\phi (ux)=u\phi (xu)u^\*$, so 
$$u^\*\phi (u)\phi (x)=\phi (x)u^\*\phi (u). $$ As $\phi $ is onto, we deduce that $u^\*\phi (u) $ is in the centre of $\mathcal M $.


Since $\mathcal M $ is a factor, $\phi (u)=\lambda\_u u $ for some $\lambda\_u \in\mathbb C $. It is easy to see that the map $u\longmapsto\lambda\_u $ is a group homomorphism.


*(Thanks to Jesse Peterson for the following argument)* From $\phi(u)=\lambda\_u u$, we get that $\sigma(u)=\lambda\_u\,\sigma(u)$, since $\phi$ preserves the spectrum. So, if the spectrum of $u$ has no rotational symmetry, then $\lambda\_u=1$. The set of unitaries with no rotational symmetry is dense in the set of unitaries of $\mathcal M$ (via the Spectral Theorem, since we can use unitaries with finite spectrum and tweak the eigenvalues slightly so that there is no rotational symmetry). By continuity, it turns out that $\phi$ is the identity on all the unitaries, and so $\phi$ is the identity. 


