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I am interested in Lovelock actions in the metric-affine (or Palatini) formalism. It is well-known that the metric version (starting from the Levi-Civita curvature) of the Lovelock lagrangian of order $k$ is a topological term in $2k$ dimensions. For example, the case $k=2$ is the Gauss-Bonnet term.

I have also read in texts on Poincaré gauge gravity (now the connection contains torsion but not non-metricity) that this is also true for torsionful-Lovelock actions.

But some people extrapolate this to the general metric-affine-Lovelock action (with torsion and non-metricity, i.e. a completely general affine connection), but I cannot find the proof anywhere.

Is this true in general? Is it easy to prove?


If it were true:

What does it mean to be "topological" in this particular context? [See Two definitions of topological terms in field theory ] The metric theory can be written as a total derivative; but I think there is no way to do so in the metric-affine case (expanding the connection 2-form using the Cartan's structure equation and trying to extract one of the exterior derivatives), so it must be topological in the other sense. I am confused.

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This result was not known, and for that reason we investigated about it and we got a quite surprising answer: they are NOT boundary terms. And the responsible for this is the non-metricity tensor or, in particular, its traceless part (since the trace is undetermined by these theories due to the projective symmetry).

The idea behind the proof is that for boundary terms Lagrangians, the equations of motion are identically satisfied for any configuration of the fields (they are indeed trivial identities, $0=0$). And, for metric-affine (or Palatini) Lovelock theories in their critical dimension (when $D=2k$), it is always possible to find configurations of the fields that violate at least one equation of motion. Here we provide the counterexamples: https://arxiv.org/abs/1907.12100

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