vector with trivial stabilizer in $ SO_3 $ representation

Question

Consider the standard representation of $ SU_2 $ on $ \mathbb{C}^2 $. For any nonzero vector $ v \in \mathbb{C}^2 $, $ v $ has trivial stabilizer.

Finding such a representation and such a vector seems natural and I would like to do it for $ SO_3(\mathbb{R}) $.

By contrast, in the standard representation of $ SO_3(\mathbb{R}) $ on $ \mathbb{R}^3 $ every nonzero vector has stabilizer $ SO_2(\mathbb{R}) $.

EDIT: Lspice points out that every element in the standard representation of $ SO_3(\mathbb{R}) $ has stabilizer a maximal torus ( just a circle $ SO_2(\mathbb{R} $) and intersecting two maximal tori here will give the center (which is trivial). So if we consider the 6d reducible rep of $ SO_3(\mathbb{R}) $ on $ \mathbb{R}^3 \oplus \mathbb{R}^3 $ (just two copies of the standard rep) then the stabilizer of a vector like $ (e_1,e_3) $ must be trivial. This is not totally surprising because $ \mathbb{R}^6 $ is known to contain a copy of $ \mathbb{R}P^3 $ realized as the unit tangent bundle of the sphere $ S^2 $.

I am looking for an irreducible representation $ \pi$ of $ SO_3(\mathbb{R}) $ on a finite dimensional vector space $ V $ and a vector $ v \in V $ such that the stabilizer of $ v $ is trivial.

The five dimensional irrep of $ SO_3(\mathbb{R}) $ will not help since we can identify it with the action by conjugation on the traceless symmetric $ 3 \times 3 $ real matrices, which are all orthogonally diagonalizable, and then by inspecting the diagonal form conclude that every element has stabilizer $ S(O_1(\mathbb{R}) \times O_1(\mathbb{R}) \times O_1(\mathbb{R})) \cong O_1(\mathbb{R}) \times O_1(\mathbb{R}) $ if all eigenvalues are distinct or stabilizer $ S(O_2(\mathbb{R}) \times O_1(\mathbb{R})) \cong O_2(\mathbb{R}) $ if there is a repeated eigenvalue.

To summarize, the stabilizers from the 3d, and 5d irreps are $ SO_3(\mathbb{R}), SO_2(\mathbb{R}),O_1(\mathbb{R}) \times O_1(\mathbb{R}), O_2(\mathbb{R}) $. None are trivial.

Maybe the 7d irrep has a vector with a trivial stabilizer? So far I know the 7d irrep has vectors with stabilizer: three element cyclic $ C_3 $, six element dihedral $ D_6 $, and the tetrahedral subgroup $ A_4 $.

I think it is very likely that the 9d irrep has a vector with a trivial stabilizer. For example it is a necessary condition that $ \mathbb{R}^9 $ contains an equivariantly isometrically embedded $ \mathbb{R}\mathbb{P}^3 $ and that is known to be the case for example by looking at $ SO_4(\mathbb{R}) $ acting by conjugation on the nine dimensional vector space of traceless symmetric $ 4 \times 4 $ real matrices and then considering the orbit of a matrix with a triple repeated eigenvalue.

I also know this must be possible for some finite dimensional rep of $ SO_3(\mathbb{R}) $ by Mostow-Palais theorem (since $ SO_3(\mathbb{R}) $ compact and trivial group is closed).

Answer 1

Let $\mathcal{P}_3$ denote the space of homogeneous cubics in three variables, and let $\mathcal{H}_3$ denote the (seven-dimensional) subspace of harmonic cubics, an irreducible $\mathrm{SO}(3)$-module. Let $H : \mathcal{P}_3 \to \mathcal{H}_3$ denote the orthogonal projection map. Then it turns out that any unit norm $p \in \mathcal{H}_3$ can be factorized as $p = H( \ell_1 \ell_2 \ell_3),$ where the $\ell_i$ are linear functions on $\mathbb{R}^3$ of unit norm. The linear factors $\ell_i$ are unique up to permutation and replacing two of them at a time by their negatives. Denote this equivalence relation by $\sim.$

The resulting map $(S^2 \times S^2 \times S^2) / {\sim} \to S^6$ is $\mathrm{SO}(3)$ equivariant, where $\mathrm{SO}(3)$ acts diagonally on $(S^2 \times S^2 \times S^2) / {\sim}.$ Therefore, to find an element $v \in \mathcal{H}_3$ with trivial $\mathrm{SO}(3)$-stabiliser, it suffices to find a collection of three points on $S^2$ for which there are no nontrivial $\mathrm{SO}(3)$ elements sending these elements to another member of their equivalence class under $\sim.$ The generic triple of points on $S^2$ should fit this requirement.