suppose we have a surface (2-dimensional submanifold of $\mathbb{R}^3$) equipped with a Riemannian metric which is invariant with respect to rotations about the $z$-axis (that is, all rotations about the z-axis are isometries of the surface).

I am wondering if the surface is necessarily invariant with respect to reflections at (Euclidean) planes containing the $z$-axis (e.g. the x-z-plane or y-z-plane).

Surfaces of revolution with the Riemannian metric induced by the space $\mathbb{R}^3$ have this property but I am not sure if it must hold in general.

Best wishes


Yes. For any point $p$ in your surface, its reflection $p'$ at a plane containing the $z$-axis can also be obtained by some rotation around the $z$-axis, hence by the assumption of rotational invariance (under arbitrary rotation angles) $p'$ is also in the surface.

  • $\begingroup$ Thank you for your help. Is it also possible to show that the reflection must be an isometry? Of course, if we equip the surface with the Riemannian metric coming from $\mathbb{R}^3$, then the reflection is an isometry. But what happens when we would equip the surface with a metric different than the one induced by the ambient $\mathbb{R}^3$? $\endgroup$ – Sammyy Delbrin Apr 29 '17 at 18:18
  • $\begingroup$ @SammyyDelbrin Just any Riemannian metric? Of course the reflection is not necessarily an isometry. It may become more interesting if you require your metric to be invariant under rotations. $\endgroup$ – Amitai Yuval Apr 29 '17 at 20:13
  • $\begingroup$ @AmitauYuval: Actually I require the metric to be rotationally invariant. Does it follow that the metric is also invariant to reflections as mentioned in my question above? $\endgroup$ – Sammyy Delbrin Apr 30 '17 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.