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

About embeddings, I don't know, but there is an isometric immersion of $\mathrm{SO}(3)$ with its biinvariant 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 12element 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 leftinvariant 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 leftinvariant metric is fully right invariant and hence is a multiple of the biinvariant 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 biinvariant 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. 


the nine matrix elements of $SO(3)$ represent a vector in $R^9$, see Isometric Embedding for Homogeneous Compact 3Manifolds (1996). 


Apparently, Hopf proved in [H] that the projective space $\mathbb{R}P^3$ embedds into $\mathbb{R}^5$ and Hantzsche [Ha] proved that it cannot be embedded into $\mathbb{R}^4$. Since this projective space is isomorphic to $\mathrm{SO}(3)$ as was noted by Paul Siegel, this at least gives a lower bound. Let me just add that $\mathrm{Spin}(3)$ is (isomorphic to) the unit sphere in quaternions and hence this double cover of $\mathrm{SO}(3)$ is embeddable into $\mathbb{R}^4$. [H] H.Hopf, "Systeme symmetrischer Bilinearformen und euklidische Modelle der projektiven Raume", Vierteljschr Naturforsch Gesellschaft Zurich 85 (1940) 165177. [Ha] W.Hantzsche, "Einlagerung von Mannigfaltigkeiten in euklidische Raume", Math Zeit 43 (1938) 3858. 


This paper (I. Oszvath and B. Schuking, 1996) seems to construct the embedding into $\mathbb{R}^9$ and seem to be claiming that there is not one int $\mathbb{R}^6,$ or any lowerthan9 dimensional space. 


SO(3), a 3dimensional rotation is determined by a choice of axis, an element of $\mathbf{RP}^2$, and the amount of rotation, a number in $[0,2\pi)$. So should it not be possible with a number 1 more than that of projective plane? 

