$\DeclareMathOperator\SO{SO}$ 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.



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


