We know that $$\frac{dv}{d t}=\frac{d\left(v^{i} e_i\right)}{d t}=\partial_{j} v^{i} v^{j} e_{i}+v^{i} v^{j} \partial_{j} e_{i}$$ As $\partial_{j} e_{i}$ is another vector we can expand it in the same basis $$\partial_{j} e_{i}=C_{i j}^{k} e_{k}$$ then we obtain the condition for constant vector as $$\nabla_{j} v^{k}=0$$

By recognizing $C_{i j}^{k}=\Gamma_{i j}^{k}$ we can rewrite geodesic equation as $$U^{i}\nabla_{i} U^{j}=0$$ which implies $$\nabla_{i} U^{j}=0$$ Thus we can conclude that velocity of the particle in gravitational field is constant. But how can the velocity of particle in arbitrary gravitational field be constant?


Let us first consider the general set of equations $$\nabla_\mu V^\nu = 0$$ for $\nu,\mu=0,1,2,3$. This set of equations generally corresponds to 16 partial differential equations for the four components of $V^\nu$. In coordinate components and using Christoffel symbols $\Gamma^\nu_{\mu\kappa}$ they read $$\frac{\partial V^\nu}{\partial x^\mu} = - \Gamma^\nu_{\mu\kappa}V^\kappa$$ Now the question is: are these equations integrable? For that it must hold that partial derivatives commute $\partial^2 V^\nu/\partial x^\mu \partial x^\lambda=\partial^2 V^\nu/\partial x^\lambda \partial x^\mu$, which is, using the equation above $$\frac{\partial}{\partial x^\lambda}(- \Gamma^\nu_{\mu\kappa}V^\kappa) = \frac{\partial}{\partial x^\mu}(- \Gamma^\nu_{\lambda\kappa}V^\kappa)$$ which reduces to $$\left(\Gamma^\nu_{\mu\kappa,\lambda} - \Gamma^\nu_{\lambda\kappa,\mu} - \Gamma^\nu_{\mu\gamma}\Gamma^\gamma_{\lambda\kappa} + \Gamma^\nu_{\lambda\gamma}\Gamma^\gamma_{\mu\kappa}\right)V^\kappa = 0$$ One can now recognize the expression in the brackets as the Riemann curvature tensor and the integrability condition simply reads $$R^\nu_{\;\kappa \mu\lambda}V^\kappa =0$$ For a generic $R^\nu_{\;\kappa \mu\lambda}$ this equation has too many independent components for any $V^\nu$ to satisfy them. Indeed, I cannot recall any case of interest where a covariantly constant vector existed apart from flat space-time ($R^\nu_{\;\kappa \mu\lambda} = 0$ in any coordinate system).

Now for the geodesic equations, which read $$U^\mu \nabla_\mu U^\nu =0$$ I would like to stress that Einstein summation is being used here and there are only 4 equations (and only 3 of them non-trivial). Perhaps it would be helpful to write that by definition $$U^\mu \nabla_\mu U^\nu \equiv U^0 \nabla_0 U^\nu + U^1 \nabla_1 U^\nu + U^2 \nabla_2 U^\nu+ U^3 \nabla_3 U^\nu $$ This being set to zero of course does not imply that $\nabla_\mu U^\nu$ is zero for every choice of $\nu,\mu=0,1,2,3$ separately!

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