General Relativity: Definition of Locally Inertial Frame
seems to me that this should imply that a local observer standing on the earth (so not free falling at all) should be considered as an accelerating, non inertial frame.
Yes, an observer standing on the earth is not inertial in relativity. The definitive test is to have the observer carry a good accelerometer. In this case it will indicate an acceleration of 1 g upwards, conclusively showing that the observer is non-inertial.
Just a nitpick on language: an observer isn’t a reference frame, he or she has a reference frame, or even better there is a reference frame where he or she is at rest.
there is another, more geometrical, equivalent formulation of EEP: Locally spacetime looks like 4 This is not the precise formulation of the geometrical formulation, but it's good enough.
Agreed, it is good enough for present purposes.
This means that in every sufficiently small region of spacetime it's like being into a inertial special relativity frame, so no accelerating, no gravity, no shenanigans.
It does not mean that at all. You can certainly have accelerating reference frames with pseudo-gravitational forces in 4. All 4 means is that you cannot have any tidal effects.
4 is a flat spacetime manifold and can be equipped with an endless number of coordinate systems, including non-inertial ones. What “locally spacetime looks like 4 means is that all of the curvature tensors on spacetime are 0 over the local region, but it does not restrict you to a certain class of coordinate systems.
But the geometrical formulation states that every sufficiently small reference frame, myself included, should be like an inertial SR frame!
No, the observer is unambiguously non-inertial. The geometrical formulation does not contradict that at all. The geometrical formulation merely says that in a small region spacetime is flat, not that an observer is inertial. It is perfectly consistent to have non-inertial observers and reference frames in flat spacetime. Only tidal effects are forbidden.
Standing on the surface of the Earth, the reference frame at rest relative to yourself is certainly not a frame with Minkowski metric. Here is the proof: release an object, so that it is in free fall. There is relative acceleration between the object and the chosen frame. Hence the frame is not inertial and its metric is not Minkowskian.
To define a tangent space in general relativity it is not sufficient that the metric be Minkowskian just at one event. It must be Minkowskian AND have no first-order dependence on distance or time near that event. In other words the Christoffel symbols must all vanish. But since the released object is accelerating relative to the frame at rest on Earth, at least one of the Christoffel symbols is not zero.