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?


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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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