Today I will discuss the Riemann curvature tensor. This is the other main invariant of a connection, along with the torsion. It turns out that on Riemannian manifolds with their canonical connections, this has a nice geometric interpretation that shows that it generalizes the curvature of a surface in space, which was defined and studied by Gauss. When {R \equiv 0}, a Riemannian manifold is flat, i.e. locally isometric to Euclidean space.

Rather amusingly, the notion of a tensor hadn’t been formulated when Riemann discovered the curvature tensor. 

Given a connection {\nabla} on the manifold {M}, define the curvature tensor {R} by

\displaystyle R(X,Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z.

There is some checking to be done to show that {R(X,Y)Z} is linear over the ring of smooth functions on {M}, but this is a straightforward computation, and since it has already been done in detail here, I will omit the proof.

The main result I want to show today is the following:

Proposition 1

Let {M} be a manifold with a connection {\nabla} whose curvature tensor vanishes. Then if {s: U \rightarrow M} is a surface with {U \subset \mathbb{R}^2} open and {V} a vector field along {s}, then\displaystyle \frac{D}{\partial x} \frac{D}{\partial y} V = \frac{D}{\partial y} \frac{D}{\partial x} V.


In other words, there is a kind of symmetry that arises in this case. This too can be proved by computing in a coordinate system.

More conceptually, here is a different argument. Assume first that {s} is an immersion at some point {p \in U}, and extend {V} locally to the vector field {\bar{V}} in a neighborhood of {s(p)}. Now

\displaystyle \frac{D}{\partial x} \frac{D}{\partial y} V = (\nabla_{X} \nabla_Y \bar{V} ) \circ s

where {X,Y } are at least locally {s}-related to {\frac{\partial }{\partial x}, \frac{\partial }{\partial y}}. Similarly,

\displaystyle \frac{D}{\partial y} \frac{D}{\partial x} V = (\nabla_{Y} \nabla_X \bar{V} ) \circ s

Since {R \equiv 0}, their difference is

\displaystyle \nabla_{[X,Y]} V \circ s = ( \nabla_0 \bar{V} ) \circ s,

since {[X,Y]} is {s}-related (at least in a neighborhood of {p}) to {0}. 

This was a short post to tie some things up.  Tomorrow’s should be longer.