# Quotient of $\mathbb{R}^n$ by a subgroup of $\mathrm{SO}(n)$

$$\DeclareMathOperator\SO{SO}$$ 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.

• 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 Apr 11, 2021 at 12:52
• 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. Apr 11, 2021 at 12:53
• 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 $n\ge 3$.
– YCor
Apr 11, 2021 at 13:10
• 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 $\mathbb{R}^2/pm$? Apr 11, 2021 at 13:20
• 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 $2\pi$.
– YCor
Apr 11, 2021 at 13:27

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.
• 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. Apr 14, 2021 at 13:18