Let ${M}$ be a Riemannian manifold with metric ${g}$, Levi-Civita connection ${\nabla}$, and curvature tensor ${X,Y,Z \rightarrow R(X,Y)Z}$. Define $\displaystyle R(X,Y,Z,W) := g\left( R(X,Y)Z, W\right).$ Apparently people sometimes refer to this as the curvature tensor, though it is probably not too confusing.

Some algebra

Recall also the following three identities, proved here:

• (Skew-symmetry) ${R(X,Y,Z,W) = -R(Y,X,Z,W)}$
• (Skew-symmetry) ${R(X,Y,Z,W) = -R(X,Y,W,Z)}$
• (Bianchi identity) ${R(X,Y,Z,W) + R(Z,X , Y, W) + R(Y, Z, X, W) = 0}$

I claim now that there is a type of symmetry: $\displaystyle \boxed{ R(X,Y,Z,W) = R(Z,W,X,Y).}$ This is in fact a general algebraic lemma.

Lemma 1 Let ${V}$ be a real vector space and let ${R:V \times V \times V \times V \rightarrow \mathbb{R}}$ be a quadrilinear map satisfying the three bulleted identities. Then it satisfies the boxed one.

The proof is some slightly messy algebra, which I’ll only sketch. There is a geometrical way of thinking about this that Milnor presents in Morse Theory.

Let the Bianchi identity as written above be denoted by ${Rel_{X,Y,Z,W}}$. If we add ${Rel_{X,Y,Z,W}, Rel_{W,Y,Z,X}, Rel_{X,W,Z,Y}, Rel_{X,Y,W,Z}}$ and use skew-symmetry several times, we obtain $\displaystyle R(W,Y,Z,X) + R(Z,W,Y,X) + R(X,W,Z,Y) = 0.$

Now using ${Rel_{W,Y,Z,X}}$ gives $\displaystyle R(X,W,Z,Y) = R(Y,Z,W,X)$

and suitably interchanging all the variables gives the result.

Sectional curvature

Notation as above, for a 2-dimensional subspace ${S \subset T_p(M)}$, define the sectional curvature ${K(S)}$ as $\displaystyle K(S) := R( E, F, F, E)$

if ${(E,F)}$ form an orthonormal basis for ${S}$.I now claim this is well-defined.

Proposition 2 If ${X,Y \in T_p(M)}$ span ${S}$, then $\displaystyle K(S) = \frac{ R(X,Y,Y,X) }{(X,X)(Y,Y) - (X,Y)^2}.$ In particular, ${K(S)}$ depends only on ${S}$ and is well-defined.

Indeed, write ${X = x_1 E + x_2 F, Y = y_1 E + y_2 F}$. Then this is a computation depending on the previous identities already proved. (Messy algebra and blogging do not mix well.)

Sectional curvature determines ${R}$

The sectional curvature actually encodes all the information contained in ${R}$. Indeed, if we had two Riemannian metrics on the same manifold with curvature tensors ${R,R'}$ with the same sectional curvature in all two-dimensional planes, then $\displaystyle R'(X,Y,Y,X) = R(X,Y,Y,X)$ for all ${X,Y \in T_p(M)}$.

Consider the difference ${R' - R}$. Then it satisfies all the four identities in the first section of this post, along with ${B(X,Y,X,Y) = 0}$ (by skew-symmetry again). Also by the fourth identity, ${B(X,Y,X,Y')}$ is symmetric in ${Y,Y'}$, so the skew-symmetry just proved implies $\displaystyle B(X,Y,X,Y') \equiv 0.$ In particular, we get another skew-symmetric identity: $\displaystyle R(X,Y,Z,W) = -R(Z,Y,X,W)=R(Y,Z,X,W) .$

Applying this again gives $\displaystyle R(X,Y,Z,W) = R(Z,X,Y,W) ,$ and the Bianchi identity clearly gives that ${B \equiv 0}$.

I should say something about the geometric interpretation about all this.  If $S \subset T_p(M)$, then $\exp_p(S \cap U)$ for $U$ a small neighborhood of $0 \in T_p(M)$ is a surface in $M$.  Then the Gauss curvature of that surface is $K(S)$.