2

I'm currently reading about symmetric spaces, and I've come across the following fact (exercise A1 chpt 2 in Helgason's book): if $(G,H)$ is a symmetric pair with respect to the involution $\sigma$, and we write $o$ for the point $eH\in G/H$, and $s_0$ denotes the involution of $G/H$ given by $gH\mapsto \sigma(g)H$, then there exists a unique affine connection $\nabla$ invariant under $s_0$ and the action of $G$.

Now this affine connection has a very nice form for its curvature tensor at the point $o$, namely $R_0(X,Y)(Z)=-[[X,Y],Z]$. Further, we have the fact that for every vector field $V$, $\nabla_V(R)=0$.

My question is, what does it 'mean' that $\nabla_V(R)=0$ for all vector fields? This seems like a really strong condition, but I don't know how to interpret it geometrically, or how it might be useful.

I realize the question is vague, but any answer would be appreciated.

Note: I don't know that much differential/Riemannian geometry. :)

freeRmodule
  • 1,852

1 Answers1

3

An affine connection $\nabla$ on a manifold $M$ defines a notion of "parallelism" on $M$.

Thinking $\nabla$ as a "covariant derivative" operator: For a tensor field $T$ on $M$, the tensor field $\nabla_VT$ is a sort of directional derivative of $T$ in the direction $V$. That is, $\nabla_V T = 0$ means that $T$ "does not change" in the direction $V$, meaning that $T$ is parallel in the direction $V$.

Thus: Saying that "$\nabla_V R = 0$ for all $V$" means that the curvature tensor is parallel in all directions $V$. Thus, the "covariant derivative of $R$ is zero," and we sometimes describe this condition by saying that "$R$ is covariantly constant."

So: "$\nabla_VR = 0$ for all $V$" is a type of "constant curvature" condition, in a sense. It is indeed quite strong.

Jesse Madnick
  • 31,524