# Is a rotationally invariant surface necessarily invariant w.r.t. reflection?

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.
• 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 $\mathbb{R}^3$? – Sammyy Delbrin Apr 29 '17 at 18:18