Suppose I have a riemannian manifold $(N,g^1)$ and a submanifold $i:M\hookrightarrow N$. Furthermore, assume that $M$ is totally geodesic with respect to $N$, therefore the second fundamental form vanishes. Now consider another riemannian metric $g^2$ on $N$ such that $g^2=e^{2f}g^1$ for a smooth function $f$. Now I was wondering if with respect to the metric $g^2$ will $M$ also be totally geodesic with respect to $N$. I belive the answer is yes but I wanted to make sure my proof was correct.
The goal is to check that the second fundamental form associated with $g^2$ vanishes. We will have $B^2(X,Y)=\tilde \nabla ^2_{\tilde X}\tilde Y- \nabla ^2_X Y$. But since we are in the same conformal class we know that , from this answer Levi-Civita connection between conformal metrics,
$$\tilde \nabla^2_{\tilde X}\tilde Y=\tilde \nabla^1_{\tilde X}\tilde Y+\tilde X(f)\tilde Y+\tilde Y(f)\tilde X- g^1(\tilde X,\tilde Y)grad(f)$$
and we get an analogous result for $\nabla^2_{X}Y$. Therefore using the fact that $\tilde \nabla^1_{\tilde X}\tilde Y- \nabla^1_{X}Y=0$ and that $\tilde X,\tilde Y$ are extensions of $X$ and $Y$ we get that $B^2$ will vanishe, and hence $(M,i^*(g^2))$ is totally geodesic with respect to $(N,g^2)$.
What do you think about this proof? Any insight is appreciated, thanks in advance.
Best Answer
The answer is no. Take $M$ to be the unit disk in $\Bbb R^2$ and consider the two metrics $g_1 = g_{\text{eucl}}$ the Euclidean metric and $g_2 = g_{\text{hyp}} = \frac{4}{(1-|x|^2)^2}g_1$ the hyperbolic metric.
Any straight line in $M$ is a geodesic in $(M,g_1)$, therefore a totally geodesic submanifold. But only those passing through the origin are geodesics in $(M,g_2)$.
You can adapt this example in any dimension: any affine hyperplane intersected with the unit ball is a totally geodesic euclidean submanifold while only those passing through the origin will be totally geodesic hyperbolic submanifolds.