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Let $(M,g)$ be a riemannian manifold, $X$ a vector field of $M$ and $\gamma \colon M \to \mathbb{R}$ an integral curve of $X$ that is $\gamma'(t) = X(\gamma(t))$. Assume also that the Levi-Civita connection $\nabla_X X = 0$ where

Are the integral curves of a $X$ geodesics?

I believe is true. A curve $\gamma$ of $M$ is a geodesic if and only if the covariant derivative $\nabla_{\gamma'(t)}\gamma'(t) = 0$. But since $\gamma$ is an integral curve of $X$, just plugging $X(\gamma(t))$ in the connection and using the hypothesis that $\nabla_X X = 0$ gives the result.

Is this correct?

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$X(X,X)=0$ so that $|X|$ is constant on an integral curve $c(t)$. Since $c$ is a constant speed, so by an assumption, $c$ is a geodesic.

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  • $\begingroup$ I am having a hard time seeing why $\vert X \vert$ constant implies the integral curves of $X$ are a geodesics. For example, consider the vector field X = $(X_x, X_y) = (-y/\sqrt{x^2 + y^2}, x/\sqrt{x^2 + y^2})$ on $\Bbb R^2 \setminus \{(0, 0)\}$; we have $\vert X \vert = 1$ everywhere but the integral curves of $X$ are circles surrounding the origin. Mos' def' not geodesic in the usual, Euclidean metric on the plane. $\endgroup$ – Robert Lewis May 29 '18 at 2:21
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    $\begingroup$ Frankly, I do not understand your point. Would you show me your solution ? $\endgroup$ – HK Lee May 29 '18 at 2:42
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    $\begingroup$ Your example does not satisfy the assumption $\nabla_XX=0$. $\endgroup$ – HK Lee May 29 '18 at 3:13
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    $\begingroup$ No, nor do I claim such. I merely claim that constancy of $\vert X \vert$ is not sufficient to force $\nabla_X X = 0$, that is, to allow the conclusion that the integral curves of $X$ are geodesic. $\endgroup$ – Robert Lewis May 29 '18 at 3:17
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    $\begingroup$ I agree. I add more explantion. $\endgroup$ – HK Lee May 29 '18 at 3:23
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The answer is yes. I believe we can prove it with the pullback connection. On any curve $\alpha$ we have the induced pullback connection $\alpha^\ast \nabla$. The geodesic equation becomes with it $$\alpha^\ast \nabla_{d/ds} \alpha' = 0$$

We show this is satisfied. Let $\gamma$ be integral curve. Thus we have that $X \circ \gamma = \gamma'$. It turns out that this means that $\gamma' = \gamma^\ast X$, so that it is a pullback section. Thus $$\gamma^\ast \nabla_{d/ds} \gamma' =\gamma^\ast \nabla_{d/ds} \gamma^\ast X, $$

and by definition of the pullback connection $$\gamma^\ast \nabla_{d/ds} \gamma' =\gamma^\ast \nabla_{\gamma_\ast d/ds} X, $$

but $\gamma_\ast d/ds = X \circ \gamma$ and since the connection is $C^\infty$ linear on the below slot we have $$\gamma^\ast \nabla_{d/ds} \gamma' =\gamma^\ast \nabla_{X} X, $$

so when your condition is satisfied the curve is a geodesic.

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