# Why does homogeneity imply $d P^{\mu} / dX^{i} = 0$ along geodesic?

I'm reading the Cosmology lecture notes Daniel Baumann and there they describe the path of a freely-falling particle along a geodesic, which is denoted by the curve $$X^{\mu}(\tau)$$, $$\tau$$ being proper time and $$\mu$$ denoting a spacetime coordinate. The geodesic equation is given by $$\frac{d^2 X^{\mu}}{d \tau^2} = - \Gamma^{\mu}_{\alpha \beta} \frac{dX^{\alpha}}{d\tau} \frac{dX^{\beta}}{d\tau},$$ with $$\Gamma^{\mu}_{\alpha \beta}$$ the Cristoffel symbols. Defining $$U^{\mu} := dX^{\mu}/d\tau$$ and $$P^{\mu} := m U^{\mu}$$, they say that due to the homogeneity of the universe, one has $$d P^{\mu}/dX^{i} = 0$$ (the $$i$$ representing a spatial coordinate). This seems to be equivalent to saying that $$\frac{d^2 X^{\mu}}{dX^{i} d \tau} = 0$$, however I don't see how this follows from homogeneity. I would say $$X^{\mu}(\tau)$$ is only a function of $$\tau$$, hence the derivative would vanish. However, it seems that this reasoning is too simple and there is more at play here. What am I missing in my understanding?

First, $$\frac{d}{dX^{i}}= 0$$ can be considered as "generator" of displacement in $$X^i$$ direction so the statement $$d P^{\mu}/dX^{i} = 0$$ implies the velocity/momentum is independent of displacement along the $$X^i$$ axis, which is just the definition of homogeneity. I used quote when saying generator since Hamiltonian formulation of GR is bit involved. But the essence is what I just said.
Second, the fact that $$P^{i}$$ is only a function of $$t$$ or $$x^0$$ comes not just from homogeneity but also from isotropy of space. It's another way to say Hubble's law while excluding some parts.