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We can write the Einstein-Hilbert action in first order formalism as follows:

$$S(g, \Gamma)=\int_M \sqrt{-g}g^{\mu \nu}(\Gamma^{\alpha}_{\mu \nu}\Gamma^{\lambda}_{\lambda \alpha}-\Gamma^{\alpha}_{\lambda \nu}\Gamma^{\lambda}_{\alpha \nu})+\int_M \partial_{\lambda}(w^{\lambda}\sqrt{-g})d^nx+$$ $$+\int_{M}\partial_{\nu}(\sqrt {-g}g^{\mu \nu})\Gamma^{\lambda}_{\lambda \mu}-\int_{M}\partial_{\lambda}(\sqrt {-g}g^{\mu \nu})\Gamma^{\lambda}_{\mu \nu}$$ where $$w_{\lambda}\equiv g^{\mu \nu}\Gamma^{\lambda}_{\mu \nu}-g^{\mu \lambda}\Gamma^{\nu}_{\nu \mu}$$ I kow that if $\Gamma$ is the the Levi civita connection, than the integral $$\int_M \partial_{\lambda}(w^{\lambda}\sqrt{-g})d^nx$$

can be reduced to an integral over the boundary $\partial M$ of the manifold M.

In fact, in this case we can use $$\partial_{\lambda}(\sqrt{-g})=\sqrt{-g}\, \Gamma^{\mu}_{\mu \lambda} \qquad (1)$$ so that $$\int_M \partial_{\lambda}(w^{\lambda}\sqrt{-g})d^nx=\int_M \nabla_{\lambda}w^{\lambda}\sqrt{-g}d^nx$$ Now we can use the Stokes theorem: $$\int_M \nabla_{\mu}V^{\mu}\sqrt{-g}d^nx=\int_{\partial_M}n_{\mu}V^{\mu}\sqrt{-\gamma}d^{n-1}x$$ where n is the 4-vector normal to $\partial M$ and $\gamma$ is the induced metric on $\partial M$

If $\Gamma$ is not the Levi-Civita connection, this proof breaks down because we cannot use (1) anymore.

In first order formalism, we treat $g$ and $\Gamma$ as two independent variables, so $\Gamma$ isn't the Levi-Civita connection until we solve the EOMs. Despite of this, we still consider the integral $\int_M \partial_{\lambda}(w^{\lambda}\sqrt{-g})d^nx$ as an integral on $\partial M$ in first order formalism, as far as I know. Why is it so?

  • Is $\int_M \partial_{\lambda}(w^{\lambda}\sqrt{-g})d^nx$ a boundary integral if $\Gamma$ is a general connection?
  • In particular, is this the case if $\Gamma$ is torsionless, but no metric compatibility is assumed?
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This post has a really bad case of physics math. It's not clear what $$\tag{$*$}\int_M \partial_\lambda(w^\lambda \sqrt{-g})\,d^nx$$ is supposed to be. You're doing a Lebesgue integral on $\Bbb R^n$, but the domain is $M$? So $M$ has to be an open set in $\Bbb R^n$. With appropriate conditions on the boundary, the Gauss-Green Theorem gives you what you want. Or, you could interpret that as $$\int_M \frac{1}{\sqrt{-g}}\partial_\lambda(w^\lambda \sqrt{-g})\,\sqrt{-g}d^nx.$$ This can be interpreted as $$\int_M\mathrm{div}(w)\,dv,$$ where $w$ is the vector field with components $w^\lambda$ in whatever coordinate systems were being used implicitly. This is then defined for any manifold, and is a boundary integral as explained in this PSE post. This does not require any kind of connection.

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