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I'm working on a problem in do Carmo chapter 4. We have that a Riemannian manifold, M, is locally symmetric if $\nabla R = 0$, where R is the curvature tensor of M (i.e., $R(X,Y,Z,W) = \langle R(X,Y)Z, W \rangle$, and $R(X,Y)Z = {\nabla}_Y{\nabla}_XZ - {\nabla}_X{\nabla}_YZ - {\nabla}_{[X,Y]}Z$).

I'm trying to show that if M is locally symmetric, $\gamma: [0,l] \rightarrow M$ a geodesic, and $X,Y,Z$ are parallel vector fields along $\gamma$, then $R(X,Y)Z$ is a parallel field along $\gamma$.

So, Let $X,Y,Z,W$ be vector fields on M parallel along $\gamma$. Then since $\nabla R = 0$ we have in particular that $\nabla_\dot{\gamma}R = 0$. So then

$$0 = \dot{\gamma}(R(X,Y,Z,W)) - R(\nabla_\dot{\gamma}X, Y, Z, W) - R(X, \nabla_\dot{\gamma}Y, Z, W) - R(X, Y, \nabla_\dot{\gamma}Z, W) - R(X, Y, Z, \nabla_\dot{\gamma}W)$$

But since X,Y,Z,W are parallel along $\gamma$ this reduces to $$0 = \dot{\gamma}(R(X, Y, Z, W)) = \dot{\gamma}(\langle R(X,Y)Z, W \rangle) = \langle \nabla_{\dot{\gamma}}R(X,Y)Z, W\rangle + \langle R(X,Y)Z, \nabla_{\dot{\gamma}}W \rangle$$

And since W is parallel along $\gamma$, we have $$0 = \langle \nabla_{\dot{\gamma}}R(X,Y)Z, W\rangle$$

So I want to say this implies that $\nabla_{\dot{\gamma}}R(X,Y)Z = 0$ so that $R(X,Y)Z$ is parallel along $\gamma$, but I am thinking there is something else I need to show since there is the possibility, I think, that $\nabla_{\dot{\gamma}}R(X,Y)Z \neq 0$ but rather $\nabla_{\dot{\gamma}}R(X,Y)Z$ and $W$ are orthogonal. Is there any nice way to show this other than just by assumption?

Agathon
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    There is always a parallel orthonormal frame $(E_1,\dots,E_n)$ along $\gamma$. Your calculation shows that $\nabla_{\dot\gamma}R(X,Y)Z$ is orthogonal to $E_i$ for each $i$, and thus it's zero. – Jack Lee Nov 18 '15 at 20:01
  • Ah! That would do it, thanks! – Agathon Nov 18 '15 at 20:04

1 Answers1

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We cannot assume that $W$ is a parallel vector field, only that $X,Y,Z$ are. Instead, let $W$ be an arbitrary vector field along $\gamma$. Since $M$ is locally symmetric, $\nabla R=0$:

\begin{align} 0&=(\nabla R)(X,Y,Z,W,\gamma’)\\ &= (\gamma’)R(X,Y,Z,W) - R(\nabla_{\gamma’}X,Y,Z,W) - R(X, \nabla_{\gamma’} Y,Z,W) -R(X,Y, \nabla_{\gamma’} Z,W) - R(X,Y,Z, \nabla_{\gamma’} W)\\ &= (\gamma’)R(X,Y,Z,W) - R(X,Y,Z, \nabla_{\gamma’} W)\\ &= (\gamma’)\langle R(X,Y)Z,W\rangle - \langle R(X,Y)Z, \nabla_{\gamma’} W\rangle\\ &= \langle \nabla_{\gamma’}R(X,Y)Z,W\rangle + \langle R(X,Y)Z,\nabla_{\gamma’}W\rangle -\langle R(X,Y)Z,\nabla_{\gamma’}W\rangle \\ &= \langle \nabla_{\gamma’}R(X,Y)Z,W\rangle \end{align}

Since this holds for any choice of vector field $W$ along $\gamma$ we can conclude that $R(X,Y)Z$ is a parallel vector field.

mxnoqwerty
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