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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 $\{\lambda I; \lambda \in \mathbb{C}\}$)?

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 $\varphi$.

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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.

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  • $\begingroup$ Thanks. You have proven the fact for arbitrary unital C*-algebras with trivial center. $\endgroup$
    – westerbaan
    Feb 9, 2015 at 9:45
  • $\begingroup$ You do not need Fuglede-Putnam-Rosenblum: $au = ua \Rightarrow a = uau^* \Rightarrow u^* a = au^*$. $\endgroup$
    – westerbaan
    Feb 9, 2015 at 11:03
  • $\begingroup$ Here is an alternate way to finish the argument (for von Neumann algebras): Automorphisms preserve the spectrum, and so since $\phi(u) = \lambda_u u$ for any unitary it follows that $\phi$ must fix all unitaries whose spectrum has no rotational symmetry. By the spectral theorem such unitaries are dense in the set of all unitaries. $\endgroup$ Feb 9, 2015 at 21:55
  • $\begingroup$ By the way, $e^{ix} = e^{iy}$ doesn't imply $e^{itx} = e^{ity}$ for all $t$. $\endgroup$ Feb 9, 2015 at 21:57
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    $\begingroup$ Here is a way to find other examples. If $f \in L^\infty(X, \mu)$ is a real function then $e^{if} = 1$ if and only if $f(x)/2\pi \in \mathbb Z$, a.e. $x \in X$. Taking different masas in $\mathcal B(\mathcal H)$ you can then find $x, y \in \mathcal B(\mathcal H)$, self-adjoint, which don't commute, yet $e^{ix} = e^{iy} = 1$. $\endgroup$ Feb 10, 2015 at 0:35

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