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).

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