📂Geometry

If the Sectional Curvature is the Same, the Riemannian Curvature is also the Same.

Theorem¹

Let $V$ be defined in a vector space of dimension $2$ or higher, and let $\left\langle \cdot, \cdot \right\rangle$ be the inner product defined on $V$. Let $R : V \times V \times V \times V \to V$ and $R^{\prime} : V \times V \times V \times V \to V$ be multilinear functions satisfying the conditions below.

$$R(x,y,z,w) = \left\langle R(x,y)z, w \right\rangle,\quad R^{\prime}(x,y,z,w) = \left\langle R^{\prime}(x,y)z, w \right\rangle$$

$$\begin{align*} R(x, y, z, w) + R(y, z, x, w) + R(z, x, y, w) &= 0 \tag{a}\\ R(x, y, z, w) &= - R(y, x, z, w) \tag{b}\\ R(x, y, z, w) &= - R(x, y, w, z) \tag{c}\\ R(x, y, z, w) &= R(z, w, x, y) \tag{d} \end{align*}$$

For two linearly independent vectors $x, y$, let $K, K^{\prime}$ be the following.

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

Here, $\sigma$ is a $2$-dimensional subspace with a basis of $\left\{ x, y \right\}$. If for all $\sigma \subset V$, $K(\sigma) = K^{\prime}(\sigma)$ holds, then $R = R^{\prime}$.

Explanation

The $R$ mentioned in the statement refers to the Riemann curvature tensor, and $K$ refers to the sectional curvature.

This theorem tells us that we can know the Riemann curvature tensor $R$ with just the information about the $2$-dimensional subspace.

Proof

The proof is not difficult; it just requires some computation effort.

Claim: $R(x,y,z,w) = R^{\prime}(x,y,z,w)\quad \forall x,y,z,w \in V$

First, it is certain that the following holds by the definition of $K, K^{\prime}$.

$$\begin{equation} K(\sigma) = K^{\prime}(\sigma) \implies R(x,y,x,y) = R^{\prime}(x,y,x,y) \end{equation}$$

Therefore, we obtain the following.

$$R(x+z, y, x+z, y) = R^{\prime}(x+z, y, x+z, y)$$

This is due to the linearity of $R, R^{\prime}$ as follows.

$$\begin{align*} R(x,y,x,y) + R(z,y,x,y)\quad &= R^{\prime}(x,y,x,y) + R^{\prime}(z,y,x,y) \\ \quad + R(x,y,z,y) + R(z,y,z,y) &\quad + R^{\prime}(x,y,z,y) + R^{\prime}(z,y,z,y) \end{align*}$$

The first and fourth terms of both sides can be eliminated by $(1)$.

$$R(z,y,x,y) + R(x,y,z,y) = R^{\prime}(z,y,x,y) + R^{\prime}(x,y,z,y)$$

Then, by changing the first term of each side according to $(d)$, it becomes the following.

$$\begin{align*} && R(x,y,z,y) + R(x,y,z,y) &= R^{\prime}(x,y,z,y) + R^{\prime}(x,y,z,y) \\ \implies && R(x,y,z,y) &= R^{\prime}(x,y,z,y) \tag{2} \end{align*}$$

From $(2)$, we obtain the following equation.

$$R(x, y+w, z, y+w) = R^{\prime}(x, y+w, z, y+w)$$

Similarly, due to linearity, it is broken down as follows.

$$\begin{align*} R(x,y,z,y) + R(x,w,z,y)\quad &= R^{\prime}(x,y,z,y) + R^{\prime}(x,w,z,y) \\ \quad + R(x,y,z,w) + R(x,w,z,w) &\quad + R^{\prime}(x,y,z,w) + R^{\prime}(x,w,z,w) \end{align*}$$

The first and fourth terms of both sides are mutually eliminated by $(2)$.

$$\begin{align*} && R(x,w,z,y) + R(x,y,z,w) &= R^{\prime}(x,w,z,y) + R^{\prime}(x,y,z,w) \\ \implies && R(x,y,z,w) -R^{\prime}(x,y,z,w) &= R^{\prime}(x,w,z,y) - R(x,w,z,y) \end{align*}$$

Using $(b), (d)$ on the two terms of the right side,

$$R(x,y,z,w) -R^{\prime}(x,y,z,w) = R(y,z,x,w) - R^{\prime}(y,z,x,w)$$

From this equation, we again obtain the following.

$$\begin{equation} \begin{aligned} R(x,y,z,w) -R^{\prime}(x,y,z,w) &= R(y,z,x,w) - R^{\prime}(y,z,x,w) \\ &= R(z,x,y,w) - R^{\prime}(z,x,y,w) \end{aligned} \end{equation}$$

Now, from $(a)$, we obtain the following.

$$R(x,y,z,w) + R(y,z,x,w) + R(z,x,y,w) = 0 \\ R^{\prime}(x,y,z,w) + R^{\prime}(y,z,x,w) + R^{\prime}(z,x,y,w) = 0$$

Subtracting the equation below from the above,

$$\left( R(x,y,z,w) - R^{\prime}(x,y,z,w) \right) + \left( R(y,z,x,w) - R^{\prime}(y,z,x,w) \right) \\ \ + \left( R(z,x,y,w) - R^{\prime}(z,x,y,w) \right) = 0$$

At this time, the three terms enclosed in parentheses are all the same by $(3)$. Therefore,

$$3\left( R(x,y,z,w) - R^{\prime}(x,y,z,w) \right) = 0 \\[1em] \implies R(x,y,z,w) = R^{\prime}(x,y,z,w)$$

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