Isometric embedding of SO(3) into an euclidean space 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 $f: SO(3) \to R^n$? 
 A: the nine matrix elements of $SO(3)$ represent a vector in $R^9$, see Isometric Embedding for Homogeneous Compact 3-Manifolds (1996).
A: 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) 165-177.
[Ha] W.Hantzsche, "Einlagerung von Mannigfaltigkeiten in euklidische
     Raume", Math Zeit 43 (1938) 38-58.
A: 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 lower-than-9 dimensional space.
A: 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.
A: SO(3), a 3-dimensional 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?
A: (Not an answer at all.)
If a metric space $X$ admits an isometric embedding into Euclidean $\mathbb{R}^n$, then all Caley--Menger determinants defined by $n+2$ points from $X$ are equal to 0. Moreover, for a metric space embeddable into some $\mathbb{R}^N$ with large $N$ (this is our case) the above property guarantees that the image of such embedding lies in an $n$-dimensional subspace. It allows to get the minimal dimension for any specific metric, but there are many bi-invariant metrics, this is the problem.
