Is a circle action on M_n necessarily inner?

An action $\alpha$ of a locally compact topological group G on a unital $C^*$-algebra $A$ is called inner if there exists a continuous group homomorphism $u\colon G\to U(A)$ such that $\alpha_g(a)=u_gau_g^*$ for all $g\in G$ and all $a\in A$.

Question: let $\alpha\colon S^1\to Aut(M_n)$ be a continuous action. Does it follow that $\alpha$ is inner?

For every $\zeta\in S^1$, the automorphism $\alpha_\zeta$ is inner, say $\alpha_\zeta=Ad(u_\zeta)$ for some unitary $u_\zeta$ in $M_n$. This unitary is unique up to multiplication by elements of $S^1\cdot 1_n\subseteq M_n$ (the unitaries in the center of $M_n$). Is it known whether one can choose the map $\zeta\mapsto u_\zeta$ to be a continuous group homomorphism?

I suppose this may be known to some people but I could not find any references.

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If I'm not mistaken, the datum of your continuous action $\alpha : S^1 \to \operatorname{Aut}(M_n(\mathbb{C}))$ should be equivalent to the datum of a continuous projective unitary representation $\beta : S^1 \to PU(n)$, via $\alpha(\theta)(T) = \beta(\theta)T\beta(-\theta)$; your action $\alpha$ is inner if and only if $\beta$ lifts to a continuous unitary representation $\tilde{\beta} : S^1 \to U(n)$ of $S^1$ on $\mathbb{C}^n$. I don't know, though, off the top of my head, whether or not the obstruction to such a lift necessarily vanishes for $S^1$. –  Branimir Ćaćić Jul 15 '13 at 14:57
Thanks for your comment. What you say is definitely right. One can see, using algebraic topology, that there is a continuous lift, but it is not clear to me that one can choose this lift to be a group homomorphism. –  Eusebio Gardella Jul 15 '13 at 17:14
The obstruction to this lift being a group homomorphism, then, will be precisely a $U(1)$-valued $2$-cocycle on $S^1 (=U(1))$, so if you can figure out the group cohomology $H^2(S^1,U(1))$, you should have your answer one way or another. I'm afraid, though, that I know nothing about the computation of group cohomology. –  Branimir Ćaćić Jul 15 '13 at 17:22
If $G$ is a compact Lie group, the $U(1)$-valued 2-cocycles (for so-called Segal cohomolgoy) are the same as classes for $H^3(BG,\mathbb Z)$. For $G=U(1)$, that group vanishes. So yes: all actions of $U(1)$ are "inner". –  André Henriques Jul 15 '13 at 21:22
$H^3(BU(1),\mathbb Z)=0$ is easy: $CP^\infty$ has only even dimensional cells. For the first fact, the key point is that Segal cohomolgoy of a compact group G with coefficients in R vanishes (that's the middle term in the LES associated to the SES of coefficient groups $Z\to R\to S^1$). My recollection is that there's an explicit formula that you can write down, that uses the Haar measure on G: if you start with an R-valued cocycle $c$, then the explicit formula gives you a cochain whose differential is $c$. Finally, here's a reference: Corollary 97 of arxiv.org/pdf/0911.2483.pdf –  André Henriques Apr 8 '14 at 18:19