@timmdeeg @Dale, Let me put it more carefully by superposition. To linear order the local inertial frame gets two contributions: the bucket's own stress-energy, plus the integrated effect of all distant matter (Lense–Thirring summed over the cosmos):
##\Omega_{\text{local inertial}} = \Omega_{\text{bucket}} + \int (\text{distant matter contribution})##
The bucket term is there regardless. Distant matter only changes the conclusion if that integral is nonzero. So the whole Machian question reduces to a single well-defined quantity: does ##\int \frac{G}{c^2}\frac{\rho}{r} \,dV## vanish or not?
I shouldn't have claimed "removing the stars changes nothing", that presupposes the integral is negligible, which I can't show. In fact Sciama's estimate suggests it's of order unity for the observed density, which would support Mach. So I'll retract that.
But there's a point that holds regardless of that integral, and it's the core one: the spinning bucket has one Riemann tensor, the static bucket has another, and changing the reference frame does not turn one into the other. The Riemann tensor is a tensor: its components transform covariantly, but the curvature invariants (e.g. the Kretschmann scalar ##R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}##) are the same numbers in every coordinate system. Hence "stars rotating around a static bucket" and "bucket at rest" are the same physical system in two coordinate charts, same Riemann, same invariants. "Bucket physically spinning" is a different stress-energy tensor, different Riemann, different invariants, real gyroscope precession, curved surface. No change of frame can map one into the other, because no change of frame can alter an invariant scalar.
That distinction stands whether or not the Machian integral vanishes; whether the inertial frame itself is fixed locally or by the cosmic integral is the separate open question, and Timmdeeg is right that GP-B / Lense–Thirring is exactly the term that could make that integral matter.
If rotation could be removed by a change of frame, the Kerr metric would be diffeomorphic to Schwarzschild, but it isn't. They have different curvature invariants (the Kretschmann scalar depends on ##a##), Kerr has an ergosphere and frame-dragging that Schwarzschild lacks, and angular momentum ##J## is an invariant charge, not a coordinate artifact. A rotating source and a static one are genuinely different spacetimes.
).