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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 $\exp_p(P)$?

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It should be $K(\Pi)$ where $K$ is the sectional curvature of $M$??. –  Ivan Lurie Dec 12 '11 at 13:08

1 Answer 1

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.

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