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This is a follow-up to a previous question, where it was concluded that at any point of Euclidean space, the "infinitesimal isometry group" is $O(n) \times \mathbb{R}^n$ and the "pointed infinitesimal isometry group" is $O(n)$ (precise definitions of each are to follow).

Questions (working from most specific to most general examples):

  1. Are the "infinitesimal (pointed) isometry groups" at each point of the 2-sphere $\mathbb{S}^2$ the same as those for the Euclidean plane? (They should be equal at each point since $\mathbb{S}^2$ is homogeneous.) Or are they different because of the sphere's positive curvature?

  2. For an arbitrary Riemannian $2$-dimensional manifold, are the "infinitesimal (pointed) isometry groups" at each point functions of the Gaussian curvature? Is the Gaussian curvature insufficient information to predict them? Or are they totally independent of it?

  3. For an arbitrary $n$-dimensional Riemannian manifold $M$, is it true that, at any point $p \in M$, one has that the "infinitesimal isometry group" is $O(n) \times \mathbb{R}^n$ and that the "pointed infinitesimal isometry group" is $O(n)$? I.e. does the curvature of $M$ at $p$ affect these groups?

Note: An answer to any of the above three questions will suffice for an answer. The first question seems like the easiest and the one I am trying to solve right now, but ultimately the final level of generality I am interested in is the third question. Again, please answer whichever you want.

(If the answer to any of the above questions turns out to be false, then these groups should not be called "infinitesimal isometry groups", since Riemannian manifolds are supposed to be "infinitesimally Euclidean". Hence the quotation marks around the terms.)

Definitions: Given an arbitrary metric space $(X,d)$ (not necessarily a Riemannian manifold),

  • A map $f: X \to X$ is called a local isometry at $x \in X$ if $x$ has two (possibly equal) neighborhoods $U_x$ and $V_x$, such that (the restriction of) $f$ maps $U_x$ isometrically onto $V_x$.

  • A map $f: X \to X$ is called a pointed local isometry at $x \in X$ if it is a local isometry at $x \in X$ and if $f(x)=x$, i.e. if $f$ fixes $x$.

  • The infinitesimal isometry group at $x \in X$ is the set of all germs of local isometries at $x$.

  • The pointed infinitesimal isometry group at $x \in X$ is the same as the above, except replacing "local isometries at $x$" with "pointed local isometries at $x$".

It was shown in this previous question that both of these sets are actually groups, and that for Euclidean space the former is the Euclidean group and the latter is the orthogonal group, i.e. the same as the "global (pointed) isometry groups".

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Chill2Macht
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    Not completely sure I understand your definitions, but it looks to me that the "infinitesimal isometry group" of a round $n$-sphere is $O(n + 1)$ and the "pointed infinitesimal isometry group" of a round $n$-sphere is $O(n)$. There are, respectively, the global isometry group and the isotropy group of a point. – Andrew D. Hwang Jul 22 '17 at 21:18
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    most surfaces with Riemannian metrics have no isometries at all other than the identity map. – Will Jagy Jul 22 '17 at 21:26
  • @AndrewD.Hwang How does one get the result that the "infinitesimal isometry group" of the sphere is the same as its global isometry group? I'm kind of confused since, while the sphere is compact, most of the neighborhoods of any given point are not -- so it is unclear to me why every local isometry would be the restriction of a global isometry and why every global isometry is the extension of a local isometry. https://math.stackexchange.com/a/2364201/327486 – Chill2Macht Jul 23 '17 at 06:43
  • @WillJagy Can you please explain how this is relevant to the question? I don't understand. I am talking about metric space isometries, not Riemannian metric isometries, although I thought that there wasn't a distinction between them for Riemannian manifolds. – Chill2Macht Jul 23 '17 at 06:44
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    A Riemannian isometry of a connected manifold is completely determined by its action on some orthonormal frame. (Intuitively, the action on a frame at $p$ determines the action in a neighborhood of $p$ via composition with the exponential map, and this determines the isometry throughout the entire path component.) If memory serves, Kobayashi's Transformation Groups in Differential Geometry contains details. – Andrew D. Hwang Jul 23 '17 at 13:14
  • @AndrewD.Hwang The Riemannian isometries coincide with the metric space isometries, correct? Also, I don't understand how global isometries are relevant here -- the question is about germs of local isometries. A local isometry, for example, can be arbitrarily badly behaved outside of a neighborhood of the point in question. – Chill2Macht Jul 23 '17 at 13:49
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    Yes, Riemannian isometries coincide with metric space isometries. The relevance to global isometries is that there's a natural bijection from local isometries of a round sphere (or Euclidean space) and global isometries: A local isometry (even a germ of a local isometry, i.e., an isometry of an orthonormal frame) extends uniquely to a global isometry, a global isometry restricts to a local isometry near an arbitrary point, and these associations are "morally inverse" to each other. – Andrew D. Hwang Jul 23 '17 at 16:20
  • @AndrewD.Hwang It took me a while, but I do get it now. I really appreciate you taking so much time to explain it to me. – Chill2Macht Jul 23 '17 at 16:43
  • You're very welcome. :) – Andrew D. Hwang Jul 23 '17 at 18:48
  • @AndrewD.Hwang Sorry, 2 follow-up questions I just thought of (1) to confirm, your argument shows that the group of germs of local isometries at $x$ is isomorphic to the global isometry group of $M$ for any point of any Riemannian manifold $M$, not just a point of the homogeneous $S^2$, correct? (2) It also shows that the group of germs of pointed local isometries at $x$ is isomorphic to the global pointed isometry group of $M$ for any point of of any Riemannian manifold $M$, correct? I.e., both parts of 3. have been shown false, correct?Do you not want to post that as an answer? – Chill2Macht Jul 23 '17 at 20:23
  • @AndrewD.Hwang Provided that the Riemannian manifold is connected, since "a Riemannian isometry of a connected manifold is completely determined by its action on some orthonormal frame"? Would the reference for that in Kobayashi's Transformation Groups in Differential Geometry be in the section about Killing Vector Fields? (Which he also calls infinitesimal isometries?) My point being that this seems to completely answer my question in full generality, so I am confused why you seemed to not want to post an answer to be upvoted and accepted. – Chill2Macht Jul 23 '17 at 20:25
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  • On a connected manifold, two global isometries having the same action on some orthonormal frame agree, but not every action on an orthonormal frame comes from an isometry. (As Will notes, a generic surface has no non-trivial isometries.) 2. The fact that "local isometries correspond to global isometries" is a rare property, held by complete manifolds of constant sectional curvature: Euclidean space, round spheres, and hyperbolic spaces. 3. I've been commenting instead of answering because I'm not sure I've fully understood all your definitions; when time permits I'll have another look.
    • – Andrew D. Hwang Jul 23 '17 at 22:03
  • I need to look into this further, but as a note to myself or other future readers, I think the "pointed infinitesimal isometry group" might be related to the notion of "holonomy group" (or at least the path component of that group containing the identity), cf. remark 3.33 and proof of theorem 3.32 (pp. 45-46) in Chapter I.3 by Haefliger in the book Metric Spaces of Non-Positive Curvature edited by Bridson. https://mathworld.wolfram.com/HolonomyGroup.html https://ncatlab.org/nlab/show/holonomy+group https://link.springer.com/chapter/10.1007/978-3-662-12494-9_3 – Chill2Macht Mar 06 '22 at 03:25