📂Geometry

Sectional Curvature of Differential Manifolds

Theorem¹

Let $\sigma \subset T_{p}M$ be a 2-dimensional subspace of the tangent space $T_{p}M$. Suppose that $x, y \in \sigma$ are linearly independent. Then, the following $K$ does not depend on the choice of $x, y$.

$$K(x, y) = \dfrac{R(x,y,x,y)}{\left\| x \times y \right\|^{2}}$$

Here, $R$ is the Riemann curvature tensor.

Explanation

According to the above theorem, if $\sigma$ is given, the value of $K$ is the same for any basis of $\sigma$. Therefore, $K$ is defined as follows.

Definition

For a point $p \in M$ on the differentiable manifold $M$ and a 2-dimensional subspace of the tangent space $\sigma \subset T_{p}M$,

$$K(\sigma) = K(x, y)$$

is called the sectional curvature of $\sigma$ at $p$, where $\left\{ x, y \right\}$ is any basis of $\sigma$.

Proof

Consider the following transformations that change the basis $\left\{ x, y \right\}$ of $\sigma$ to another basis $\left\{ x^{\prime}, y^{\prime} \right\}$.

$$\begin{align*} \left\{ x, y \right\} &\to \left\{ y, x \right\} \\ \left\{ x, y \right\} &\to \left\{ \lambda x, y \right\} \\ \left\{ x, y \right\} &\to \left\{ x + \lambda y, y \right\} \end{align*}$$

Then, $K$ remains invariant under these transformations. Due to the linearity and symmetry of $R$,

$$K(y, x) = \dfrac{R(y,x,y,x)}{\left\| y \times x \right\|^{2}} = \dfrac{R(x,y,x,y)}{\left\| x \times y \right\|^{2}} = K(x, y)$$

$$K(\lambda x, y) = \dfrac{R(\lambda x,y,\lambda x,y)}{\left\|\lambda x \times y \right\|^{2}} = \dfrac{\lambda^{2} R(x,y,x,y)}{\lambda^{2}\left\| x \times y \right\|^{2}} = \dfrac{R(x,y,x,y)}{\left\| x \times y \right\|^{2}} = K(x, y)$$

Since $R$ is symmetric, $R(y,y,x,y) = R(x,y,y,y) = 0$ and because of $y \times y = 0$,

$$\begin{align*} K(x + \lambda y, y) &= \dfrac{R(x + \lambda y,y,x + \lambda y,y)}{\left\| (x + \lambda y) \times y \right\|^{2}} \\[1em] &= \dfrac{R(x,y,x,y)}{\left\| x \times y \right\|^{2}} \\[1em] &= K(x, y) \end{align*}$$

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