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In his book Riemannian manifolds, John Lee states the following on pages 8-9:

The most fundamental fact about geodesics is that given any point $p\in M$ and any vector $V$ tangent to $M$ at $p$, there is a unique geodesic starting at $p$ with initial tangent vector $V$.

Here is a brief recipe for computing some curvatures at a point $p\in M$:

  1. Pick a 2-dimensional subspace $\Pi$ of the tangent space to $M$ at $p$.

  2. Look at all the geodesics through $p$ whose initial tangent vectors lie in the selected plane $\Pi$. It turns out that near $p$ these sweep out a certain 2-dimensional submanifold $S_{\Pi}$ of $M$, which inherits a Riemannian metric from $M$.

  3. Compute the Gaussian curvature $S_{\Pi}$ of $p$, which the Theorema Egregium tells us can be computed from its Riemannian metric. This gives a number, denoted $K(\Pi)$, called the sectional curvature of $M$ at $p$ associated with the plane $\Pi$.

Thus the curvature of $M$ at $p$ has to be interpreted as a map \begin{equation} K : \lbrace \text{2-planes in } T_pM\rbrace \rightarrow \Bbb{R} \end{equation}

My question:

Does Lee mean select a $2$-dimensional subspace of an $n$-dimensional tangent space $T_pM$ at $p$? In other words, what is the dimensionality of rhe manifold we are working with here? Is this method designed for curvature in higher dimensions?

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The tangent space $T_pM$ has some arbitrary dimensionality $n$, but the sectional curvature is always a function of some planar subspace of the tangent space.

Of course, in the case that the tangent space is itself two-dimensional, then the scalar curvature can be expressed as a single number.

Using exterior algebra, we can say $K: \bigwedge^2 T_pM \mapsto \mathbb R$. This is a natural consequence of how the Riemann tensor can be considered as a linear map $R: \bigwedge^2 T_pM \mapsto \bigwedge^2 T_pM$. Let $B \in \bigwedge^2 T_p M$. Then the sectional curvature is just $K(B) = \langle B, R(B)\rangle/ \langle B, B \rangle$, with the inner product extended to 2-vectors as usual.

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  • $\begingroup$ Does this "planar subspace" requirement apply to using the Ricci curvature tensor? $\endgroup$ – Stan Shunpike Jan 28 '15 at 4:20
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    $\begingroup$ No, the Ricci tensor would be viewed as acting on line-like subspaces--also known as common vectors. - The definition most like that of the Riemann I gave above would just be a linear operator, mapping vectors in $T_pM$ to vectors in $T_p M$. $\endgroup$ – Muphrid Jan 28 '15 at 4:21

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