1
$\begingroup$

I am trying to prove the following statement:

Let $(M, g)$ an Riemannian manifold an $X \in \Gamma(TM)$ a parallel vector field, i. e. $\nabla^g_Y X = 0$ for all $Y \in \Gamma(TM)$. Then all integral curves are geodesics.

My "proof" looks like that:

Let $I \subseteq \mathbb R$ and a $\gamma: I \to M$ an arbitrary integral curve, i. e. $\gamma'(t) = X(\gamma(t))$ for all $t \in I$. Since $X$ is parallel we have that

$$\nabla^g_{\partial/\partial t} \gamma ' = \nabla^g_{\partial/\partial t} X \circ \gamma = \nabla^g_{\gamma_*(\partial/\partial t)} X = \nabla^g_{\gamma'} X = 0,$$ where $\nabla^g$ denotes the Levi-Civita connection of $g$ on $TM$. Hence $\gamma$ ia a geodesic. $\square$

I am wondering if this solution is correct because I feel it is too short. I appreciate your help :)

$\endgroup$
4
$\begingroup$

It looks good, but I would write it as

$$ \frac{D\gamma'}{dt}(t) = \frac{D(X \circ \gamma)}{dt}(t) = (\nabla_{\gamma'(t)} X)(\gamma(t)) = 0. $$

The point is that $\gamma \colon I \rightarrow M$ is a curve and $\gamma' \colon I \rightarrow TM$ is a vector field along $\gamma$, not a vector field on $M$. To check whether $\gamma$ is a geodesic, you need to use the induced covariant derivative along $\gamma$ which I denote by $\frac{D}{dt}$. Since $\gamma$ is an integral curve of $X$, it turns out that $\gamma'$ is the "restriction" of a globally defined vector field $X$ on $M$ (that is, $\gamma' = X \circ \gamma$) and for such fields, we know that the covariant derivative along $\gamma$ is related to the regular covariant derivative of vector fields by the relation

$$ \frac{D(X \circ \gamma)}{dt}(t) = (\nabla_{\gamma'(t)} X)(\gamma(t)) $$

where the right hand side should actually be interpreted as $(\nabla_{Y_t} X)(\gamma(t))$ where $Y_t \in \mathcal{X}(M)$ is (any) vector field on $M$ for which $Y_t(\gamma(t)) = \gamma'(t)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.