let $V$ be vector space, and let $\nabla$ be the Levi-Civita connection wrt. a constant metric on $V$ (in the sense that the metric is the same at every point of $V$).

Let $w$ be a vector in $V$, and consider it a constant vector field on V, along some curve $\gamma$ on V.

Does it hold that

$\nabla_{\dot{\gamma}_t}w = 0$


  • $\begingroup$ Yes, it does hold. $\endgroup$ – Berci Sep 27 at 10:57

Yes. This is the easiest possible case of a Riemannian manifold.

The Christoffel symbols $\Gamma^i_{jk}$ are 0 because all the derivatives of the metric are 0. So, in coordinates/components, $\nabla_j w^i = \dfrac{\partial w^i}{\partial x^j} + \Gamma^j_{ik}w^k = 0$ for every $i,j$ (and hence also for the vector $\dot\gamma$).


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