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I know that for a surface $S \subseteq \mathbb{R}^3$, the gauss curvature and the sectional curvature are equivalent.

I was wondering if there is any useful way we can understand sectional curvature in terms of gaussian curvature for higher dimensional Riemannian manifolds.

To be more concrete, let $M$ be a $m$ dimensional Riemannian manifold where $m>2$. For simplicity, assume that $M$ is geodesically completely and connected (any two points in the manifold are joined by a geodesic). Consider some point $x \in M$ and let $U$ be a small enough neighborhood around $x$. Let $T_{x}M$ be the tangent space of $M$ at $x$. Consider any two linearly independent elements of the tangent space, $u,v$. Let $k$ be the section curvature at $x$ of the 2-plane $\sigma$ spanned by $u,v$. I have the following question:

Consider the geodesics $\gamma_{u}$ and $\gamma_{v}$ emanating at $x$ in the directions of $u$ and $v$ such that $\gamma_{u}, \gamma_{v} \subseteq U$. Is there a way to use these geodesics to define some two dimensional surface whose gauss curvature equals $k$? Otherwise, what are some intuitive interpretations of gauss curvature relate to sectional curvature that you know?

Thanks in advance!

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  • $\begingroup$ For hypersurface in R^{n+1}, the tangent space is formed by an orthonormal basis $\{e_1,\cdots,e_n\}$, any two $e_i,e_j$ of them plus the normal vector forms a three dimensional space which has Guassian curvature (multiplication of two principal curvatures), i.e., sectional curvature $K(e_i,e_j)$. $\endgroup$ – H-H Jun 27 '18 at 17:17
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You can get the surface you desire using the exponential map $e : T_x M \to X$.

First let $P \subset T_x M$ be the 2-dimensional linear subspace spanned by the vectors $u,v$. Also choose $r>0$ so that $e$ is injective on the open ball $B_r \subset T_x M$ of radius $r$. The surface you want is simply $e(P \cap B_r)$.

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  • $\begingroup$ Thanks this is useful ! Could you give a hint for how to prove that this two dimensional surface has the desired curvature? $\endgroup$ – ace7047 Jun 28 '18 at 12:16
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    $\begingroup$ The sectional curvature of the plane $P$ in the tangent space $T_x M$ is equal to the sectional curvature of the surface $S = e(P \cap B_r)$ in the tangent space $T_x S$, and the latter equals the Gauss curvature of $S$ at $x$, by definition. $\endgroup$ – Lee Mosher Jun 28 '18 at 13:14

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