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Is the group of translations on a spherical topology tied to the group of rotations of that sphere?

More concretely, suppose we define some set of orthonormal basis $e^{i}$ where i runs from 1 to n on a sphere $S^{n}$ (yes, I realize n=1, 3, and 7 are the only parellizable cases).

For example suppose we consider $S^{3}$, we can then translate some point (consider perhaps a vector) around on the three-sphere. However, we could have just rotated the sphere itself and achieved the same thing. Now I get that this all seems rather evident but I'm interested in something a bit more. We can say that the principle bundle of (oriented now) orthonormal frames is $P_{OF}(S^{3})=S^{3}\times SO(4)$ (which is a trivial bundle). Due to a special isomorphism we know $S^{3}=SU(2)$, so we have $P_{OF}(S^{3})=SU(2)\times SO(4)$.

Back to our orthonormal basis, each $e^{i}$ can be chosen to identify with a generator of the Lie algebra $su(2)$. Of course the lie algebra of $su(2)$ is the same as that of $so(3)$ (the former being the double cover of the latter).

Contrast this with the topology $\mathbb{R}^{3}$, whose principle oriented orthonormal frame bundle is $P_{OF}(\mathbb{R}^{3})=\mathbb{R}^{3}\times SO(3)$. In this latter case, the two (frame translations and rotations) are entirely divorced from one another, yet in the former a translation on the three sphere is representable by rotations of the three-sphere. I find myself wondering if the two are “soldered” in some fashion? It seems like a translation on the surface of the sphere is equivalent to a rotation in the fiber of the frame bundle of that sphere. For instance, if we consider the 1-forms dual to our basis, we call these the soldering forms. Are these then tied to our rotations in the fibers? Put another way, are the horizontal and vertical sub-bundles tied to one another? can someone please elaborate?

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  • $\begingroup$ I don’t quite understand what you mean by “translation” here but are you referring to parallel transport? en.m.wikipedia.org/wiki/Parallel_transport $\endgroup$ Commented Sep 16, 2020 at 16:24
  • $\begingroup$ @QiaochuYuan. Yes, by translation, that is what I mean. Apologies if the terminology is sloppy, I study Physics generally. $\endgroup$
    – R. Rankin
    Commented Sep 16, 2020 at 16:27
  • $\begingroup$ It seems that you are asking about the fact that the holonomy group $Hol_p$ of $S^n$ equals the stabilizer of $p$ in $SO(n)$ (which is isomorphic to $SO(n-1)$, of course). In contrast, the holonomy group of the Euclidean space $E^n$ is trivial. My best guess is that you are asking for what is responsible for this phenomenon (of equality of the holonomy and isotropy groups). See my answer here. Other than that, I do not understand your question at all. $\endgroup$ Commented Sep 16, 2020 at 19:43

1 Answer 1

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Let me first get the terminology straight. Suppose that $M$ is a connected oriented Riemannian manifold. Given a path $c$ in $M$ connecting a point $p$ to a point $q$, one defines the parallel transport $\Pi_c: T_pM\to T_qM$ along $c$, which is an isometry of tangent spaces. But this set of parallel transports does not form a group: You can hardly ever compose parallel transports: In order to compose $\Pi_{c_2}\circ \Pi_{c_1}$ it is necessary and sufficient to have that the terminal point of $c_1$ is the initial point of $c_2$. Instead of a group, parallel transports define a groupoid: All the group axioms hold but the binary operation is only partially defined. Therefore, it is meaningless to ask about "the group of translations." However, one can define a group based on parallel transports, it is called the holonomy group $Hol_p$: It consists of parallel transports along loops which start and end at $p$. Such parallel transports one can, of course, always compose, hence, $Hol_p$ is a group.

It makes sense to ask questions like:

  1. Suppose that $g$ is an (orientation-preserving) isometry of $M$, $g(p)=q$. Is there a path $c$ from $p$ to $q$ such that $\Pi_c$ equals the derivative $dg_p: T_pM\to T_qM$?

  2. Suppose that $c$ is a path from $p$ to $q$. Is there an isometry $g$ of $M$ which carries $p$ to $q$ such that $dg_p=\Pi_c$?

What you observed in the case of the Euclidean space is that 1 fails (rather badly) while 2 holds (for a rather trivial reason). In the case when $M$ is a unit sphere (regardless of its dimension), both 1 and 2 have positive answers. The same applies to hyperbolic spaces. The proof is a small variation on my answer given here. In terms of groups, the clean statement is that for each point $p\in S^{n-1}$, the group $G_p\cong SO(n-1)$ (the isotropy subgroup of $G=SO(n)$ fixing the point $p$) is naturally isomorphic to the holonomy group $Hol_p$ based at $p$. The isomorphism is given by $$ g\mapsto dg_p $$

The same holds (with a minor qualification) for a larger class of Riemannian manifolds, called compact Riemannian symmetric spaces as well as for Riemannian symmetric spaces of noncompact type. Round spheres are examples of the former while hyperbolic spaces are examples of the latter. (Euclidean spaces are symmetric spaces but they are neither of compact nor of noncompact type.) The precise statement is that for these classes of Riemannian manifolds, the isotropy groups $G_p$ and holonomy groups $Hol_p$ have naturally isomorphic Lie algebras. (The above map $g\mapsto dg_p$ defines an isomorphism of Lie algebras.)

A proof of this result, due to E.Cartan, (regarding symmetric spaces) can be found in

J.-H. Eschenburg, Lecture Notes on Symmetric Spaces, Theorem 7.2.

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  • $\begingroup$ This was extremely helpful! In truth I'm working with Lorentzian spacetimes ( so Pseudo/semi Riemannian spaces) Can I expect this to hold there as well? $\endgroup$
    – R. Rankin
    Commented Sep 17, 2020 at 4:29
  • $\begingroup$ @R.Rankin: I do not know, but it is likely true for semi-Riemannian symmetric spaces as well. $\endgroup$ Commented Sep 17, 2020 at 9:40

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