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A vector $\boldsymbol{r}$ in $\mathbb{R}^3$ transforms under rotation $\boldsymbol{A}$ to $\boldsymbol{r}'=\boldsymbol{Ar}$. It is equivalent to an SU(2) "rotation" as $$\left( \boldsymbol{r}'\cdot\boldsymbol{\sigma} \right) = \boldsymbol{h} \left( \boldsymbol{r}\cdot\boldsymbol{\sigma} \right) \boldsymbol{h}^{-1},$$ where $\boldsymbol{h}$ is the counterpart of $\boldsymbol{A}$ in SU(2) given by the homomorphism between these two groups.

Now the question is, what would be the equivalent transformation in SU(2) of the rotation of a matrix in $\mathbb{R}^3$? In other words, what is the equivalent in SU(2) of $\boldsymbol{M}'=\boldsymbol{A}\boldsymbol{M}\boldsymbol{A}^{-1}$.

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    I am a little confused about what you're asking. You presumably know that there is a double cover $\phi : \text{SU}(2) \to \text{SO}(3)$. Given _any_ action of $\text{SO}(3)$ on any kind of thing, pulling back along $\phi$ gives you an action of $\text{SU}(2)$ on that thing (an element $g \in \text{SU}(2)$ acts by however $\phi(g)$ acts).2012-06-20
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    @QiaochuYuan: I didn't learn the concept of "pullback". Can you elaborate?2012-06-20
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    All I mean is that if $S$ is a set (e.g. a vector space) and $\rho : \text{SO}(3) \to \text{Aut}(S)$ is an action of $\text{SO}(3)$ on that set (e.g. a linear representation on a vector space) then $\rho \circ \phi : \text{SU}(2) \to \text{Aut}(S)$ is the corresponding action of $\text{SU}(2)$.2012-06-20
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    @Qiaochu: I think the OP is asking for an explicit formula for $\rho \circ \phi$ where $S = \mathcal{M}_{3\times 3}$ and $\rho(A)M = AMA^{-1}$.2012-06-20

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