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I am asked to compute the orbits and isotropy groups (a.k.a. stabilizers) of the Lie group action \begin{align*} \varphi\colon \operatorname{SO}(3)\times\operatorname{Sym}(3) & \rightarrow \operatorname{Sym}(3) \\ (R,A) &\mapsto R A R^t \end{align*} where $\operatorname{Sym}(3)$ is the space of symmetric $3\times 3$ matrices. If I did it well, for an element $A\in\operatorname{Sym}(3)$, its orbit $\mathcal{O}(A)$ and isotropy group $G_A$ are given by:

\begin{equation*} \mathcal{O}(A) = \{B\in\operatorname{Sym}(3) \ \big| \ \operatorname{Eigenvalues}(B)=\operatorname{Eigenvalues}(A) \} \end{equation*}

\begin{align*} G_A \cong \begin{cases} \operatorname{SO}(3) &\text{ if $A$ has 3 identical eigenvalues}\\ \operatorname{O}(2) &\text{ if $A$ has 2 identical eigenvalues}\\ F &\text{ if $A$ has no identical eigenvalue} \end{cases} \end{align*} where $F$ is the finite subgroup $\{I,\ \operatorname{diag}(1,-1,-1),\ \operatorname{diag}(-1,1,-1),\ \operatorname{diag}(-1,-1,1)\}$ of $\operatorname{SO}(3)$.

What I did is mostly to use standard linear algebra results for the case of $A$ diagonal and then extend to the non-diagonal case using that:

  • Any $A$ is always on the same orbit of a diagonal matrix with the same eigenvalues.
  • For two elements on the same orbit, their isotropy groups are conjugated and, therefore, isomorphic.

Now I am curious about if there is a more "sophisticated" way of computing these orbits and stabilizers using some results or techniques about representation theory. I don't know much about it, but I believe that the following map is a representation of $\operatorname{SO}(3)$: \begin{align*} \Pi \colon \operatorname{SO}(3) &\rightarrow \operatorname{GL}(\operatorname{Sym}(3)) \\ R &\mapsto \varphi_R \end{align*} where $\varphi_R(A) = \varphi(R,A) = R A R^t$.

Thanks a lot for your help!

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    $\begingroup$ Sometimes representation theory is just linear algebra. I think this is one of those cases. $\endgroup$
    – anon
    May 15, 2020 at 19:04
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    $\begingroup$ Sym(3) is isomorphic to the space of homogeneous polynomials in variable $ x,y,z $ by taking $ \mathbb{R}^3 \otimes \mathbb{R}^3 $ to be the space of $ 3 \times 3 $ real matrices. Then polynomials become symmetric matrices by the identification ($e_1,e_2,e_3$ standard basis) $ xy \to e_1 \otimes e_2 + e_2 \otimes e_1 $ and $ x^2 \to 2 e_1 \otimes e_1 $ etc... Here laplacian of polynomial=trace of matrix. So the standard 5d irrep of degree two homogeneous harmonic polynomials is identified with symmetric traceless matrices. But ultimately I agree with @runway44 your way is best to find orbits $\endgroup$ Jan 21, 2022 at 15:58

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