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 $\exp_p(P)$?


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 $T_p(\exp(P))$ $$v=\alpha^{\prime}(0)$$




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?


Is what I have done correct? How do I finish my proof?

Thanks in advance.

  • $\begingroup$ It should be $K(\Pi)$ where $K$ is the sectional curvature of $M$??. $\endgroup$
    – Ivan Lurie
    Dec 12, 2011 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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