Plane of symmetry and geodesics in surfaces Let $(M,n)$ be an oriented regular surface of $\mathbb{R}^{3}$ such that the intersection of $M$ with the plane
$\Pi:=\{(x,y,z) \in \mathbb{R}^{3}:z=0\}$ is the image of a curve parametrized by arc length $\alpha$.
I am trying to prove that if $\varphi:M \rightarrow M$ defined by $\varphi (x,y,z)=(x,y,−z)$ is an isometry of M then the curve $\alpha$ must be a geodesic of M.
I would like to prove that the normal vector $n \circ \alpha(t)$ is orthogonal to the vector $e_3 = (0,0,1)$ (for any $t \in \operatorname{dom}(\alpha)$).
I know that the vector $n \circ \alpha(t)$ is an eigenvector of the linear transformation $R: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ given by  $R(x,y,z)=(x,y,−z)$, but I don't know how to prove that the tangent space $T_{\alpha(t)}M$ can't be the vector space $\operatorname{span}\{e_1,e_2\}$.
I also know that the hypothesis $M \cap \Pi$ equal to the image of a curve is important and necessary.
I have tried many things but I'm not making any progress. So I would really appreciate any help.
 A: The reason that the tangent plane $T_{\alpha(t)}$ can not be $span\{e_1, e_2\}$ is that in that case the projection to the $x-y$ plane would be a chart near $\alpha(t)$, i.e. $M$ would be locally a graph of $z=f(x,y)$; but the only graph invariant w.r. to $R$ is the graph of the zero functions $z=0$. However, in such a case the intersection of $M$ with $\Pi$ would not be a curve (a smooth curve can not have an image that fills in any open set of a plane). So this is excluded.
Now, (as you have perhaps already figured out) the derivative of any map $\varphi$ mapping $M_1$ to $M_2$ sends the tangent plane $T_p M_1$ to the tangent plane $T_{\varphi(p)} M_2$. In our case, when $p$ is on the curve $\alpha$ we have $\varphi(p)=p$ and $D\varphi$, being the linearization of $R$, which is already linear and fixes $p$, is $R$. Thus the tangent plane is invariant, and so the normal vector is invariant (up to sign), i.e. an eigenvector with eigenvalue $\pm 1$, and the case of it being $\pm e_3$ is excluded, so it lies in the $x-y$ plane. Now the acceleration vector of $\alpha$ is also in the $x-y$ plane and, since $\alpha$ is unit speed it is orthogonal to the velocity vector of $\alpha$, hence parallel to the normal vector to $M$, making $\alpha$ a geodesic.
