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Let $M_2(\mathbb{C})$ be the algebra of $2\times 2$ complex matrices and $\mathbb{S}^1$ the unit circle.

How many actions of $\mathbb{S}^1$ on $M_2(\mathbb{C})$ exist (up to isomorphism)? And on $M_n(\mathbb{C})?$

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    $\begingroup$ What kind of actions are you considering? Homeomorphism? Diffeomorphism? $\endgroup$ Jul 7, 2015 at 17:33
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    $\begingroup$ It's known that every automorphism of $M_n(\mathbb{C})$ is inner; it follows that the automorphism group is $\text{PGL}_n(\mathbb{C})$, so equivalently the question is to classify the projective representations of $S^1$. $\endgroup$ Jul 7, 2015 at 18:04
  • $\begingroup$ @MarcoGolla, it is by automorphism [algebraically]. $\endgroup$
    – Paulo Boff
    Jul 7, 2015 at 22:53
  • $\begingroup$ @QiaochuYuan, it's by automorph.. For $n=2$, I claim there exist just an only action $\alpha$ of $\mathbb{S}^1$ on $M_2(\mathbb{C})$. It would be given by $\alpha_z(a)=\sum_{n\in \mathbb{Z}}a_n z^n $, where $z \in \mathbb{S}^1$ and $a \in M_2(\mathbb{C})$ is such that $a= a_{-1} + a_ {0} + a_{1}$, $a_{-1}= \begin{pmatrix} 0 & 0 \\ \ast & 0 \end{pmatrix}$, $a_{0}= \begin{pmatrix} \star & 0 \\ \ 0 & \star \end{pmatrix}$, $a_{1}= \begin{pmatrix} 0 & \star \\ \ 0 & 0 \end{pmatrix}$ and $a_{n}= \begin{pmatrix} 0 & 0 \\ \ 0 & 0 \end{pmatrix}$ $\forall n \in \mathbb{Z} \backslash \{ -1, 0, 1 \} $. $\endgroup$
    – Paulo Boff
    Jul 7, 2015 at 22:54

1 Answer 1

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As Qiaochu mentioned in a comment, every automorphism of $M_n(\mathbb{C})$ is inner (this is easy to show by considering what happens to matrices with only one nonzero entry), so you are asking to classify continuous homomorphisms $S^1\to PGL_n(\mathbb{C})$. These are easier to think about by passing to the universal covers, so let's think about homomorphisms $f:\mathbb{R}\to SL_n(\mathbb{C})$. Every such homomorphism is of the form $f(t)=\exp(2\pi i tA/n)$ for a unique $A\in \mathfrak{sl}_n(\mathbb{C})$ (the factor of $2\pi i/n$ is included because it will make the final answer nicer). When does such a homomorphism descend to a map $S^1\to PGL_n(\mathbb{C})$ (thinking of $S^1$ as $\mathbb{R}/\mathbb{Z})$)? Well, this happens iff $f(1)=\exp(2\pi i A/n)$ is a scalar matrix that is an $n$th root of unity. Explicitly, this happens iff $A$ is a diagonalizable matrix whose eigenvalues are integers that are all equal mod $n$.

So to sum up, actions of $S^1$ on $M_n(\mathbb{C})$ are naturally in bijection with traceless diagonalizable matrices $A$ whose eigenvalues are integers that are all equal mod $n$. Given such a matrix, the action of $e^{2\pi i t}\in S^1$ on $M_n(\mathbb{C})$ is given by conjugation by $\exp(2\pi i tA/n)$. For $n=2$, such matrices $A$ have a particularly simple description: they are matrices with eigenvalues $k$ and $-k$ for some integer $k$.

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  • $\begingroup$ In other words, they are given by and almost classified by Laurent polynomials (in $x$) $f$ such that all the coefficients are nonnegative integers, and $f(1) =n$ (that is, a character of the circle of degree $n$). Two such yield equivalent actions iff one is a power of $x$ times the other. $\endgroup$ Jul 8, 2015 at 1:25

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