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$\DeclareMathOperator\SO{SO}$ Let $\mathcal{M}$ be an open subset of $\mathbb{R}^n$ endowed with the Euclidean metric and $\mathcal{N}$ be a Riemannian manifold. Assume that $G$ is a Lie subgroup of $\SO(n)$ such that $\mathcal{N} = \mathcal{M}/G$ and that the cannonical projection is a Riemannian submersion.

My question is: can $\mathcal{N}$ have a non null curvature? (Why?)

Thank you for your help.

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  • $\begingroup$ Not many of those will be manifolds, because the origin is fixed. In general there is o neills formula that relates the curvature of a manifold to the quotient $\endgroup$
    – Thomas Rot
    Apr 11, 2021 at 12:52
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    $\begingroup$ Quotients $\mathbb{R}^n/G$ will rarely be manifolds. One example is $\mathbb{R}^2/C_n=\mathbb{C}/C_n\simeq\mathbb{C}$ by $z\mapsto z^n$, and one can get other examples by taking products of examples of that type. Of course all the resulting quotients are just $\mathbb{R}^n$. I don't know any other examples. $\endgroup$ Apr 11, 2021 at 12:53
  • $\begingroup$ Even when the quotient is a topological manifold (e.g. $\mathbf{R}^2/\pm$) it's not Riemannian. $\mathbf{R}^n/\pm$ is not even a topological manifold for $n\ge 3$. $\endgroup$
    – YCor
    Apr 11, 2021 at 13:10
  • $\begingroup$ I didn't mean that it is always a manifold, I am interested in manifolds that arise in this way, I will edit the question. @YCor, what do you mean $\mathbb{R}^2/pm$ is not Riemannian? The $\mathbb{R}^2$ metric does not descends to $\mathbb{R}^2/pm$? $\endgroup$
    – Chevallier
    Apr 11, 2021 at 13:20
  • $\begingroup$ On $R^2/\pm$, there's no quotient metric at the singular point. Still there's a quotient distance. The total angle around the singular point is $\pi$ rather than $2\pi$. $\endgroup$
    – YCor
    Apr 11, 2021 at 13:27

1 Answer 1

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Yes, it can have nonzero curvature, moreover, it will typically have nonzero curvature. Consider, for example, the quotient of $\mathbb{R}^4$ by the standard Hopf $S^1$ action. The quotient will be a cone over a round sphere $S^2$ (it is easy to see from the fact that the quotient of $S^3$ by Hopf action is $S^2$, and standard dilation $x \rightarrow \lambda x$ acts conformally on the metric and commutes with action, you can also calculate it by hands).

Now, such cone only has $0$ curvature if it is actually a Euclidean space. Lets check its parameters. The standard radius $1$ sphere $S^3$ has riemannian $3$-volume $2\pi^2$, the length of every orbit is $2\pi$, which means that the quotient $2$-sphere has the volume $\pi$, which gives the radius $1/2$.

That shows that this cone is not flat - because it has the sphere $S^2$ of radius $1/2$ on the distance $1$ from the origin.

edit: I happened to play a lot with such quotients (also of $S^n$ and some other homogeneous compact spaces), so if you have some additional questions maybe I can answer.

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  • $\begingroup$ Very nice example! $\endgroup$
    – Chevallier
    Apr 13, 2021 at 8:38
  • $\begingroup$ I would like to see a proof or disproof of the fact that it should have non-negative sectional curvature, I don't know any examples when this fails. Maybe it is even more general, can riemannian quotient of non-negatively curved space have negative sectional curvature anywhere? (: $\endgroup$ Apr 13, 2021 at 16:28
  • $\begingroup$ I would be interested in the answer. I was at first motivated by the Riemannian Wasserstein geometry of Gaussian measures, which is given by submersion of the Euclidean geometry of $GL(n)$ by the map $A^TA$. The sectional curvatures are all positive or null, you can check the papers of Asuka Takatsu if you are interested. I feel like taking this submersion is equivalent to take the quotient by the action of rotation by left multiplication. I was a bit bothered by the fact that this introduces curvature, I thought I didn't get the construction right, but your example is convincing. $\endgroup$
    – Chevallier
    Apr 14, 2021 at 13:18

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