I am dealing with Riemannian immersions and I am stuck on the following: Given a totally geodesic immerserd surface $S$ on a compact riemannian manifold $M$ with metric $g$, is there another metric $\tilde g \neq g$ on $M$ such that $S$ is isometric immersed (not necessarely totally geodesically) on $(M,\tilde g)$? In this case, what can I say about the second fundamental form of this second immersion? Can I conclude with assumptions that the induced connection on $S$ is the same on both immersions?
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3$\begingroup$ If the connection on $M$ is the Levi-Civita connection, then the induced connection of an immersion is the Levi-Civita connection for the induced metric. Thus in both cases, since the immersion is isometric, the induced connection is the same. $\endgroup$– Paul BryanOct 21, 2017 at 1:44
1 Answer
Suppose that $S$ is a pair of disjoint Euclidean 2-spheres, say of radius 1 and radius 2. Isometrically immerse to the 3-sphere $M$ as totally geodesic spheres of radius 1, intersecting along a totally geodesic curve $C$ lying on each of the 2-spheres. The preimage of $C$ in $S$ is a pair of circles of different lengths. No perturbation of the metric on $M$ can give that curve $C$ those two different lengths. So $S$ is not isometrically immersed for any metric on $M$.
Suppose that you do have two immersions, one totally geodesic, the other isometric, $S \to M$. If they agree at one point, and have the same Levi-Civita connection induced, they are both isometric immersions. So we can easily invent a counterexample. Take $S$ the plane with standard metric, immersed into 3-dimensional Euclidean space by an immersion which is isometric only at the origin, and which puts $S$ into the horizontal 2-plane, so totally geodesic. Then the Levi--Civita connection of the pullback metric on $S$ cannot be the Levi--Civita connection of the original metric on $S$.