Let $\Pi$ be a nondegenerate tangent plane to $M$, a semi-Riemannian manifold, at $p$. If $P$ is a small enough neighborhood of 0 in $\Pi$. What is the Gaussian curvature at $p$ of $\exp_p(P)$?
Edit:
There was no full unambgous answer when this question was originally asked.
I am trying to solve the same problem:
I know that for a neighbourhood $U$ of $0$ in $\Pi\subset T_pM$, then $\exp_p$ is a diffeomorphism onto its image in $M$.
So the diffeomorphism would preserve the semi-Riemannian manifold structure so that $\exp_p(P)$ is a submanifold of M.
So $T_pP$ and $T(\exp(P))$ are ismorphic.
Consider the following vectors of $T_p(\exp(P))$ $$v=\alpha^{\prime}(0)$$
$$w=\beta^{\prime}(0)$$
$$x=\gamma^{\prime}(0)$$
$$y=\lambda^{\prime}(0)$$
The Gauss equation is given by:
$$\left \langle\bar{R}_{v,w} x,y \right\rangle= \left \langle{R}_{v,w} x,y \right \rangle - \left \langle\Pi(v,x)\Pi(w,y) \right \rangle- \left \langle \Pi(w,x),\Pi(v,y) \right \rangle$$
Replacing the derivatives of the geodesics. I suppose the $\Pi=0$ since we are dealing with geodesics but that is only true if $\exp(P)$ is totally geodesic. However, $P$ is not assumed to be total geodesic. I do not if the fact of P being a tangent plane, could imply that it is totally geodesic.
How would I proceed?
Question:
Is what I have done correct? How do I finish my proof?
Thanks in advance.
At least for Riemannian manifolds, the Gaussian curvature of $exp(P)$ is the sectional curvature of the plane $\Pi$ and it follows from the Gauss formula. A good exposition of this fact can be found in doCarmo's "Riemannian Geometry", especifically at the chapter on Isometric Immersions. There is actually a short commentary there claimming that this is possibily the best geometrical interpretation of the sectional curvature (as the gaussian curvature os a small embedded totally geodesic 2-manifold). I'm not quite sure if this is what you were looking for. Hope this can be usefull.