# Is a geodesic in the 4d spacetime still a geodesic after projection onto the 3d space?

But what we can observe is the projected path of the object onto the 3d space, which is itself a Riemannian manifold, with the inherited Riemannian metric from the spacetime.

Not true. To get a 3d space, you have to pick a surface of simultaneity. GR doesn't in general have a preferred surface of simultaneity.

Is the projected geodesic onto the 3d space still a geodesic?

No. For example, the earth orbits the sun, and the sun's field is approximately static, so that there is a preferred time-slicing. With this time-slicing, space is nearly flat, and the earth's orbit is an ellipse, which is not a geodesic.

The spatial metric associated to some familly of local observers is the following (in some coordinates system. I'm using signature $\eta = (1, -1, -1, -1)$): $$\tag{1} h_{\mu \nu} = u_{\mu} \, u_{\nu} - g_{\mu \nu}, $$ where $u^{\mu}$ are the components of the 4-velocity of the local observer. Components (1) define a projector: $$\tag{2} h_{\mu \lambda} \, h^{\lambda}_{\; \nu} = h_{\mu \nu}. $$ Also: $h_{\mu \nu} \, u^{\nu} \equiv 0$ and obviously $h_{\mu}^{\; \lambda} \, g_{\lambda \kappa} \, h^{\kappa}_{\; \nu} \equiv h_{\mu \nu}$. The 3D spatial section defined with this lower metric depends on the familly of observers you select.

You could then compute the 3D Riemann tensor on that lower space. The *spacelike* geodesics of that lower space have nothing to do with the *timelike* geodesics of the full 4D metric, despite that (1) is a *projector*.

Notice that the lower metric (1) isn't compatible with the full connection: $$\tag{3} \nabla_{\mu} \, h_{\lambda \kappa} = \nabla_{\mu} (u_{\lambda} \, u_{\kappa}) \ne 0. $$ To define your lower Riemann tensor and also the 3D geodesics, you'll need a new connection $\tilde{\Gamma}_{\mu \nu}^{\lambda}$ from $h_{\mu \nu}$ such that $$\tag{4} \tilde{\nabla}_{\mu} \, h_{\lambda \kappa} = 0. $$