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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$\begingroup$ $SO(3) \cong \mathbb{R}P^3$, and I think the minimal $n$ for $\mathbb{R}P^k$ is $k+2$, but I can't remember. $\endgroup$– Paul SiegelCommented Mar 24, 2014 at 21:45
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$\begingroup$ But the question is for isometric embeddings for the bi-invariant metric. $\endgroup$– ThiKuCommented Mar 25, 2014 at 7:06
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$\begingroup$ $SO(3)$ is the unit tangent bundle to $S^2$. Since $S^2$ has an isometric embedding into Euclidean 3-space $\mathbb{R}^3$, and the tangent space to $\mathbb{R}^3$ is isometric to $\mathbb{R}^6$, we get an isometric embedding of $SO(3)$ into $\mathbb{R}^6$. However, I'm not sure if the induced metric is the bi-invariant metric (I suspect not, given Robert's comments). $\endgroup$– Ian AgolCommented Jul 2, 2016 at 16:15
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6 Answers
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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.
answered Mar 25, 2014 at 1:24
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$\begingroup$ How about taking two copies of the standard representation of $ SO_3(\mathbb{R}) $? The six dimensional (reducible) $ SO_3(\mathbb{R}) $ representation $ \mathbb{R}^3 \oplus \mathbb{R}^3 $ has the property that for any pair of linearly independent vectors $ v,w \in \mathbb{R}^3 $ the vector $ (v,w) \in \mathbb{R}^3 \oplus \mathbb{R}^3 $ has trivial stabilizer and so orbit $ \mathbb{R}P^3 $. (if $ v,w $ are linearly dependent but not both 0 the orbit is a sphere $ S^2 $ and the orbit of $ (0,0) $ is trivial of course). Is this embedding isometric? It at least has nice equivariance properties. $\endgroup$ Commented Jan 27, 2022 at 23:25
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1$\begingroup$ @IanGershonTeixeira: That always gives you a left-invariant metric on $\mathrm{SO}(3)$, but it is never a biïnvariant metric on $\mathrm{SO}(3)$. $\endgroup$ Commented Jan 28, 2022 at 11:11
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$\begingroup$ Ok that's kind of what I expected. Is it possible for the metric on this orbit $ \mathbb{R}P^3 \hookrightarrow \mathbb{R}^6 $ to be round? Or is it always squashed in one direction like the Berger sphere $ S^3 $? I guess a biinvariant $ \mathbb{R}P^3\cong S^3/-I $ would have to be round since the (orientation preserving) isometry group is $ SO_4/-I\cong SO_3 \times SO_3 $. So for my embedding of $ \mathbb{R}P^3 $ in $ \mathbb{R}^6 $ would it have to be isometric to the unit tangent bundle of round $ S^2 $ and have isometry group $ O_3 $? $\endgroup$ Commented Jan 28, 2022 at 18:32
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1$\begingroup$ @IanGershonTeixeira: The actual metric depends on the choice of $v$ and $w$ in $\mathbb{R}^3$. If $v$ and $w$ are an orthonormal pair, then, yes, the induced metric on the orbit will be isometric to the unit tangent bundle of round (unit) S^2. The isometry group of that guy is actually 4-dimensional: The obvious $\mathrm{SO}(3)$ plus the $S^1$ action on the circle fibers. $\endgroup$ Commented Jan 28, 2022 at 19:55
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$\begingroup$ I am interesting in perturbing this to be an actual embedding of $ \mathbb{R}P^3 $ into $ \mathbb{R}^7 $ as you suggested above. In other words, finding a harmonic cubic polynomial with trivial stabilizer (rather than $ A_4 $ stabilizer for $ xyz$ ). Maybe $ xyz+x^3+3x^2y-3xy^2-y^3 $? I asked a question about this on MSE and Gavin Ball said you personally communicated a factorization result for these polynomials math.stackexchange.com/questions/4362071/…. I was wondering if you could expand on that or if you had any references. $\endgroup$ Commented Jun 1, 2022 at 17:48
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the nine matrix elements of $SO(3)$ represent a vector in $R^9$, see Isometric Embedding for Homogeneous Compact 3-Manifolds (1996).
answered Mar 24, 2014 at 21:26
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$\begingroup$ 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. $\endgroup$ Commented Mar 24, 2014 at 21:42
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4$\begingroup$ This is the same as the Veronese embedding $\mathbb{R}\mathrm{P}^3\hookrightarrow \mathbb{R}^9$. $\endgroup$ Commented Mar 25, 2014 at 0:11
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2$\begingroup$ 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. $\endgroup$ Commented Mar 25, 2014 at 0:31
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2$\begingroup$ @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.) $\endgroup$ Commented Mar 25, 2014 at 8:47
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2$\begingroup$ @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'. $\endgroup$ Commented Mar 25, 2014 at 11:10
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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.
answered Mar 24, 2014 at 23:44
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5$\begingroup$ 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. $\endgroup$ Commented Mar 24, 2014 at 23:50
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3$\begingroup$ @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. $\endgroup$ Commented Mar 25, 2014 at 0:04
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4$\begingroup$ @AntonPetrunin, yeah, sure. We all default to $C^\infty$, though :-) $\endgroup$ Commented Mar 25, 2014 at 0:05
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5$\begingroup$ @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. $\endgroup$ Commented Mar 25, 2014 at 1:03
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6$\begingroup$ 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 $\mathbb{R}^5$. $\endgroup$ Commented Mar 25, 2014 at 6:30
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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.
answered Mar 25, 2014 at 1:47
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$\begingroup$ yep, that's the one I cited (but only found the paywall link) $\endgroup$ Commented Mar 25, 2014 at 7:23
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$\begingroup$ 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. $\endgroup$ Commented Mar 25, 2014 at 8:29
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$\begingroup$ @RobertBryant Yes, that bears out my impression, though since the authors are physicists, it is a little hard to figure out what they mean... $\endgroup$ Commented Mar 25, 2014 at 11:34
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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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$\begingroup$ 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. $\endgroup$ Commented Mar 25, 2014 at 14:45
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$\begingroup$ @ 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 $\mathbf{RP}^2$. $\endgroup$ Commented Mar 25, 2014 at 16:11
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$\begingroup$ 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. $\endgroup$ Commented Mar 25, 2014 at 18:42
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$\begingroup$ 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? $\endgroup$ Commented Mar 26, 2014 at 0:20
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(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.
answered Jul 1, 2016 at 22:53
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$\begingroup$ I don't want to edit your answer, but receipt -> recipe . To make your recipe work for $SO(n)$ (or even $SO(3)$ you need to have an explicit polynomial in the matrix entries which gives the (squared) distance between two matrices. It is plausible that such a polynomial exists for $S^3,$ but less clear that it does for $SO(3).$ $\endgroup$ Commented Jul 1, 2016 at 23:05
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$\begingroup$ @IgorRivin Thank you, I fixed the spelling. Now I see what is the problem with my recipe: there are infinitely_many functions depending only on the rotation angle of $AB^{-1}$, which define a metric. $\endgroup$ Commented Jul 2, 2016 at 5:51
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1$\begingroup$ @IgorRivin: The formula for the bi-invariant Riemannian distance $d(A,B)$ between two matrices $A,B\in\mathrm{SO}(3)$ is $$1+2\cos\bigl(d(A,B)\bigr) = \mathrm{tr}(AB^T).$$ There is no local coordinate system on $\mathrm{SO}(3)$ in which the function $d(A,B)$ itself is algebraic, let alone polynomial. (This latter statement is also true for the standard Riemannian metric on $S^3$.) $\endgroup$ Commented Jul 2, 2016 at 14:51
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$\begingroup$ @RobertBryant Thank you. But in fact, the Cayley-Menger thing is only relevant for the chordal distance between $A, B$ - I think part of the confusion in this answer is exactly the confusion between the two notions of isometric embedding (one, isometric as metric space, the other, isometric as inducing the same metric on the submanifold. For $S^3$ the chordal distance is $2\sin \theta/2,$ where $\theta$ is the actual distance, and the square of that is a polynomial function of the cosine locally. $\endgroup$ Commented Jul 2, 2016 at 15:03
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$\begingroup$ I am bit confused. Do we consider isometric embedding of metric spaces or embedding as a manifold which defines a bi-invariant Riemannian structure? $\endgroup$ Commented Jul 2, 2016 at 15:41