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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 $f: SO(3) \to R^n$?

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$SO(3) \cong \mathbb{R}P^3$, and I think the minimal $n$ for $\mathbb{R}P^k$ is $k+2$, but I can't remember. – Paul Siegel Mar 24 '14 at 21:45
But the question is for isometric embeddings for the bi-invariant metric. – ThiKu Mar 25 '14 at 7:06

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.

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12-dimensional $\leadsto$ 12-element, no? – Mariano Suárez-Alvarez Mar 25 '14 at 1:36
@Mariano: Yes, of course, you are right. I'll fix that. – Robert Bryant Mar 25 '14 at 6:37

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

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But of course there are lots of relations - elements above the diagonal determine the elements below the diagonal, determinant is $1$, etc. So $9$ is certainly not minimal. – Paul Siegel Mar 24 '14 at 21:42
This is the same as the Veronese embedding $\mathbb{R}\mathrm{P}^3\hookrightarrow \mathbb{R}^9$. – Anton Petrunin Mar 25 '14 at 0:11
It it easy to see that we can't do better if we want the action of $SO(3) \times SO(3)$ to extend linearly to $\mathbb{R}^n$: Irreducible representations of $SO(3)$ have dimension $2m+1$, so irreps of $SO(3) \times SO(3)$ have dimension $(2m+1) \times (2n+1)$, so the smallest representation of the product on which neither $SO(3)$ acts trivially is dimension $3 \times 3=9$. But it isn't clear to me that the symmetries have to extend to the ambient space. – David Speyer Mar 25 '14 at 0:31
@VítTuček: No. In their paper, they explicitly point out that the embedding into $\mathbb{R}^6$ that one gets for $r_3=0$ is not isometric for the bi-invariant metric. (Of course, it is isometric for a different left-invariant metric, just not the bi-invariant one.) – Robert Bryant Mar 25 '14 at 8:47
@VítTuček: However, they compute the induced metric embedding parameters explicitly and relate them to the $r_\mu$ in their equations $(55)$ and $(56)$, and it's obvious from these formulae that they don't get the bi-invariant metric when $r_3=0$. That's what I mean by 'explicitly point out'. – Robert Bryant Mar 25 '14 at 11:10

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.

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But the question is about isometric embedding for a specific metric. A flat torus embeds as a submanifold in 3-space but not isometrically, for example. – Mariano Suárez-Alvarez Mar 24 '14 at 23:50
@MarianoSuárez-Alvarez OP did not specify the class of embeddings, if it is $C^1$ then by Nash--Kuiper there is one, for $C^\infty$-embeddings 9 might be the best. – Anton Petrunin Mar 25 '14 at 0:04
@AntonPetrunin, yeah, sure. We all default to $C^\infty$, though :-) – Mariano Suárez-Alvarez Mar 25 '14 at 0:05
@DavidSpeyer: Actually, no. The generic orbit in this $\mathbb{R}^5$ is not $\mathrm{SO}(3)$; it is a quotient $\mathrm{SO}(3)/(\mathbb{Z}_2\times\mathbb{Z}_2)$. This is because the generic $3$-by-$3$ symmetric matrix with trace zero and distinct eigenvalues is stabilized by the subgroup of order $4$ that fixes the eigenspaces. Moreover, none of the metrics induced on these orbits is bi-invariant. – Robert Bryant Mar 25 '14 at 1:03
If you want to embed $\mathrm{SO}(3)$ into $\mathbb{R}^5$ (non-isometrically), the simplest way I know is to first embed it into $\mathbb{R}^6$ by projecting onto the first two columns of the matrix element in $\mathrm{SO}(3)$. Then, note that the image of this embedding lies in a hypersphere $S^5\subset \mathbb{R}^6$. Remove a point of this $S^5$ that does not lie in the image and then stereographically project this complement onto $\mathbb{R}^5$. This gives a non-isometric, non-equivariant embedding of $\mathrm{SO}(3)$ into $\mathbb{R}^5$. – Robert Bryant Mar 25 '14 at 6:30

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.

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yep, that's the one I cited (but only found the paywall link) – Carlo Beenakker Mar 25 '14 at 7:23
However, as I pointed out above, this paper does not prove that there is no isometric embedding of $\mathrm{SO}(3)$ with its bi-invariant metric into a lower dimensional space than $\mathbb{R}^9$. What the authors show is that their constructions don't yield such an embedding into any lower dimension. – Robert Bryant Mar 25 '14 at 8:29
@RobertBryant Yes, that bears out my impression, though since the authors are physicists, it is a little hard to figure out what they mean... – Igor Rivin Mar 25 '14 at 11:34

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?

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An element of $SO(3)$ is determined by an element of $S^2$ and a rotation around that pole, not an element of $RP^2$. As a bundle over $RP^2$, it has disconnected fibers. So your argument suggests an embedding in $\mathbb{R}^4$, which is unlikely. – Ben McKay Mar 25 '14 at 14:45
@ Ben: Not sure I understand. Why should it be $S^2$? For a unit vector $v\in \mathbf{R}^3$ and $\theta\in [0,2\pi)$ is not the 3d-rotation corresponding to the pair $(v,\theta)$ the same as that for $(-v,\theta)$? I guess it should be $\mathbf{RP}^2$. – P Vanchinathan Mar 25 '14 at 16:11
This description in terms of $(\nu,\theta)$ is very messy; you have to identify $(\nu,\theta)$ with $(-\nu,2\pi-\theta)$, but you also have to collapse every element $(\nu,0)$ into the identity matrix, so topologically you get quite a monster. So your description is not as natural as thinking of $SO(3)$ as the unit tangent bundle of the sphere. – Ben McKay Mar 25 '14 at 18:42
I see your objection. If we banish $\theta=0$ the remaining part should be topologically a non-monster, and can the whole thing be regarded as one-point compactification of that thing? Assuming there is a connection between minimal embedding dimensions of a manifold and its one-point compactification this might be useful. Is this approach worthwhile? – P Vanchinathan Mar 26 '14 at 0:20

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