## The Riemann curvature tensor November 9, 2009

Posted by Akhil Mathew in differential geometry, MaBloWriMo.
Tags: ,

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.
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