Let be a Riemannian manifold with metric , Levi-Civita connection , and curvature tensor . Define Apparently people sometimes refer to this as the curvature tensor, though it is probably not too confusing.
Recall also the following three identities, proved here:
- (Bianchi identity)
I claim now that there is a type of symmetry:
This is in fact a general algebraic lemma.
Lemma 1 Let be a real vector space and let 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 . If we add and use skew-symmetry several times, we obtain
Now using gives
and suitably interchanging all the variables gives the result.
Notation as above, for a 2-dimensional subspace , define the sectional curvature as
if form an orthonormal basis for .I now claim this is well-defined.
Proposition 2 If span , then In particular, depends only on and is well-defined.
Indeed, write . Then this is a computation depending on the previous identities already proved. (Messy algebra and blogging do not mix well.)
Sectional curvature determines
The sectional curvature actually encodes all the information contained in . Indeed, if we had two Riemannian metrics on the same manifold with curvature tensors with the same sectional curvature in all two-dimensional planes, then
for all .
Consider the difference . Then it satisfies all the four identities in the first section of this post, along with (by skew-symmetry again). Also by the fourth identity, is symmetric in , so the skew-symmetry just proved implies
In particular, we get another skew-symmetric identity:
Applying this again gives
and the Bianchi identity clearly gives that .
I should say something about the geometric interpretation about all this. If , then for a small neighborhood of is a surface in . Then the Gauss curvature of that surface is .