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A homogeneous metric space $(X,d)$ is one where the isometry group $Iso(X)$ acts transitively on $X$. (See this answer to a previous question of mine; note that this is a special case of the notion of homogeneous space.) One consequence of being a homogeneous metric space is that the "local isometry groups", denoted $Iso(X,x)$ (the subgroup of $Iso(X)$ which fixes $x \in X$) for each point are isomorphic, that is, for any distinct $x_1, x_2 \in X$, one has that $Iso(X,x_1) \cong Iso(X,x_2)$. (See the same answer I mentioned previously for a proof of this fact.)

Question: Given a homogeneous metric space, does there always exist a subgroup $T(X) \subseteq Iso(X)$ such that one has, for an arbitrary $x \in X$, $$Iso(X)/T(X) \cong Iso(X,x)\,? $$ $$\text{or }\quad Iso(X)/Iso(X,x) \cong T(X)? $$

This is motivated by analogy to the case of Euclidean space, where $Iso(X)$ is the Euclidean group, $T(X)$ is the translation group, $Iso(X,x)$ is the orthogonal group, and the Euclidean group is actually the semi-direct product of the orthogonal group and the translation group.

Note: I am sure that this question can be asked and answered in the full generality of an arbitrary homogeneous space, not just a homogeneous metric space. However, I am most familiar with the metric space case, and that is the category in which I am interested in applying the result, if true, hence why I am asking at this level of specificity. If you can answer in greater generality, and if you want to, then please do not hesitate to do so.

Attempt: Is the sphere $\mathbb{S}^2$ a counterexample?

Its isometry group is isomorphic to $O(3)$, so for any point $p \in \mathbb{S}^2$, the only isometries fixing $p$ correspond to the elements of $O(3)$ which are rotations through the line connecting the origin and $p$ (using the standard embedding of $\mathbb{S}^2$ into $\mathbb{R}^3$) (I think, I am not really sure). So each $Iso(\mathbb{S}^2, p)$ is isomorphic to $O(2)$ or $SO(2)$. I am not really certain what $O(3)/O(2)$ or $O(3)/SO(2)$ is isomorphic to, and whether it corresponds to a translation group on the sphere or not. (Apparently, since $SO(3)/SO(2) \cong \mathbb{S}^2$, I would expect that $O(3)/O(2) \cong \mathbb{S}^2$ as well. But I don't know.)

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    $\begingroup$ Why not look instead at $Iso(X)/Iso(X,o)$ for some basepoint $o \in X$ ? $\endgroup$ – reuns Jul 17 '17 at 8:23
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    $\begingroup$ And what happens for the sphere ? $\endgroup$ – reuns Jul 17 '17 at 8:43
  • $\begingroup$ @reuns I don't know. I really only know the case of Euclidean space at all. $\endgroup$ – Chill2Macht Jul 17 '17 at 8:50
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    $\begingroup$ @Chill2Macht Well, try to work out the case of the 2-d sphere and see what you get. $\endgroup$ – Moishe Kohan Jul 17 '17 at 8:59
  • $\begingroup$ @MoisheCohen I assume you are implying that this is a counterexample, but I am not really sure how, since I am not sure exactly what $Iso(\mathbb{S}^2, p)$ would be, or what a "translation" is on the sphere. See my edit to the post if you're interested. $\endgroup$ – Chill2Macht Jul 17 '17 at 10:47
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The possibility of $\operatorname{Iso}(X)/\operatorname{Iso}(X,x) \cong T(X)$ is ruled out by the fact that $\operatorname{Iso}(X,x) $ is generally not a normal subgroup of $\operatorname{Iso}(X)$. For example, consider $X=S^2$ with $x$ being the North Pole. Rotations around the North-South axis fix $x$. But if such a rotation is conjugated by an isometry that moves $x$ to the equator, the result is a map that does not fix $x$.

Neither do we have a subgroup $T(X)$ such that $\operatorname{Iso}(X) / T(X) \cong \operatorname{Iso}(X,x)$ in general. Let $X$ be the vertices of an octahedron. Then $\operatorname{Iso}(X)$ is the binary octahedral group of order 48. The subgroup $T(X)$ would have to be of order $6$, considering that $X$ has $6$ elements. But the normal subgroups of $\operatorname{Iso}(X)$ are of orders $1, 2, 4, 8, 24, 48$.

Aside: Lie groups and homogeneous spaces

Focusing on $\mathbb{R}^n$ as an example of a homogeneous space is misleading because it's not just a homogeneous space but a Lie group, the group structure being the translation. So the underlying question could be "is every homogeneous space a Lie group?" to which the sphere $S^2$ is probably the simplest counterexample.

More specifically: fix a point $x\in S^2$ and a tangent vector $v\in T_x(S^2)$. If a "translation subgroup" $T(S^2)$ exists, then its elements push $v$ around, creating a nonvanishing vector field $F$ on $S^2$. If $T(S^2)$ is a closed subgroup of $\operatorname{Iso}(S^2)$, then it's also a Lie group, hence $F$ is a continuous nowhere vanishing vector field on $S^2$, contradicting the Hairy ball theorem.

The above does not preclude the possibility of constructing some non-closed subgroup $T(S^2)$ (probably using the Axiom of Choice) that could pass for "translations", but if this is what you want to do, you're on your own.

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  • $\begingroup$ I don't understand the premise of the answer -- why does the homogeneous metric space have to be a Lie group? What do you mean by a translation subgroup pushing $v$ around? Is there a definition of translation you have in mind of which I'm not aware? I literally don't know how to define what a translation is except in Euclidean $n$-space. You seem to have an intuition for "translation" which extends beyond that, however, which seems to be informing the content of your answer. Since I am not on the same page about that, I do not understand the answer as written unfortunately. $\endgroup$ – Chill2Macht Jul 17 '17 at 16:36
  • $\begingroup$ Also, why do we need to talk about Lie groups to show that $T(S^2)$ would create a non-vanishing vector field? Didn't you already argue/show that in the previous sentence, allowing one to apply the Hairy Ball theorem and get a contradiction? ("then its elements push $v$ around, creating a nonvanishing vector field $F$ on $S^2$" ) Finally, Lie groups seem kind of less general than arbitrary homogeneous metric spaces -- is there a reason why we have to restrict the generality to differentiable manifolds? $\endgroup$ – Chill2Macht Jul 17 '17 at 16:40
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    $\begingroup$ Alright, I moved the Lie stuff to "aside". The two paragraphs before the Aside should serve as an answer to your question. $\endgroup$ – user357151 Jul 17 '17 at 16:59
  • $\begingroup$ Thank you for taking the time to answer this question -- I apologize for being so grossly ignorant about the geometric significance of Lie groups, but even that part has also helped me understand this issue better. I appreciate it. $\endgroup$ – Chill2Macht Jul 17 '17 at 17:26

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