[Math] Isometric embedding of SO(3) into an euclidean space

Question

Consider $SO(3)$ with its bi-invariant metric and $R^n$ the euclidean space of dimension $n$. What is the minimal value of $n$ such that there exists an isometric embedding $f: SO(3) \to R^n$?

Answer 1

About embeddings, I don't know, but there is an isometric immersion of $\mathrm{SO}(3)$ with its bi-invariant metric into $\mathbb{R}^7$.

To see this, consider the natural representation $\rho_3:\mathrm{SO}(3)\to\mathrm{SO}\big({\mathcal{H}}_3\bigr)$, where $\mathcal{H}_3$ is the $7$-dimensional space consisting of the harmonic cubic polynomials on $\mathrm{R}^3$. This is an irreducible representation, so up to multiples there is a unique inner product on $\mathcal{H}_3$ that is invariant under this $\mathrm{SO}(3)$ action. Endow $\mathcal{H}_3$ with this inner product.

The stabilizer of the element $h = x_1x_2x_3\in\mathcal{H}_3$ is a 12-element discrete subgroup $A$ (isomorphic to $A_4$). The metric induced on $\mathrm{SO}(3)$ by the immersion $\iota:\mathrm{SO}(3)\to \mathcal{H}_3$ defined by $\iota(a) = \rho_3(a)h$ is clearly left-invariant and it is also invariant under right multiplication by elements of $A$. Since conjugation by elements of $A$ acts irreducibly on the Lie algebra of $\mathrm{SO}(3)$, it follows that this induced left-invariant metric is fully right invariant and hence is a multiple of the bi-invariant metric. Replacing $h$ by any nonzero multiple of $h$, we can scale the induced metric arbitrarily, so we can get any (positive) multiple of the bi-invariant metric that we want.

Note, however, that $\iota$ is an isometric embedding of $\mathrm{SO}(3)/A$, not $\mathrm{SO}(3)$ itself. It seems very likely to me that this isometric immersion can be isometrically perturbed to an isometric embedding, but I haven't tried to check that yet.

Actually, I suspect that there is an isometric embedding into $\mathbb{R}^6$, but there is certainly not an equivariant one, and, if it does exist, it might be hard to find.