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Just like a gravitational action may be written as an integration over forms, namely $$S = \int_{\mathcal{M}} \star F_{ab}\wedge e^a \wedge e^b$$ (here $\star$ is not the hodge action but $\star F_{ab} = \epsilon_{abcd}F^{cd}$). I wonder if I can similarly write the curved space scalar action as an integration over a form. One immediate idea is $$S_1 = \int_{\mathcal{M}}\phi~ d^{\dagger}d~\phi$$ where $d^{\dagger}d$ is the Hodge Laplacian which just a fancy way to write the usual Laplacian in terms of exterior derivative operators. However, the above is not entirely correct as it doesn't give the correct $\sqrt{g}$ volume factor, hence, we are compelled propose the following $$S_2 = \int_{\mathcal{M}}*(\phi~ d^{\dagger}d~\phi)$$

However, mathematically both are valid as $S_1$ is an integration over a zero-form form while $S_2$ is an integration over a hodge-dual of a zero form. Both will give the same equation of motion. But physics-wise which is the correct curved-space action for the scalar field?

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    $\begingroup$ How do you integrate a 0-form over anything other than a 0-dimensional manifold? $\endgroup$ Jun 9, 2021 at 19:19
  • $\begingroup$ @MichaelSeifert Oh you can't ? I am sorry I wasn't aware of this. Can you explain why can we only integrate over an $m$-form in $m$-dimensions? $\endgroup$
    – user44690
    Jun 9, 2021 at 19:24
  • $\begingroup$ @user44690 this is definitional for forms. But even dimensionally, an $m$-form is like a product of $m$ differentials and hence would have "units" like an $m$-dimensional volume. $\endgroup$ Jun 10, 2021 at 17:37
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    $\begingroup$ @user44690 there's an actual reason for integrating top forms which is the Jacobian determinant in the formula for changes of variables in an integral (mathworld.wolfram.com/ChangeofVariablesTheorem.html). Briefly: top forms are, in particular, tensors, and the change of variables formula for top forms produces a determinant. $\endgroup$
    – user21299
    Jun 10, 2021 at 23:34

2 Answers 2

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You want $$ \int d\phi\wedge \star d\phi\,. $$

This appears in e.g. (2.2) of https://arxiv.org/pdf/0902.4032.pdf

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    $\begingroup$ I believe that this is equivalent to the OP's version of the Lagrangian up to a boundary term. $\endgroup$ Jun 9, 2021 at 20:15
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    $\begingroup$ @MichaelSeifert: as you pointed out, OP's S_1 is wrong. You are correct that S_2 is equivalent to what I wrote up to sign factors and boundary terms, but boundary terms matter. The form I wrote is also the one most commonly found in the literature and it manifestly reduces to the usual Klein Gordon action on flat spacetime. $\endgroup$
    – user21299
    Jun 9, 2021 at 20:34
  • $\begingroup$ @alexarvanitakis Can you please provide a reference to the literature where you found the scalar action written this way? $\endgroup$
    – user44690
    Jun 10, 2021 at 6:47
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    $\begingroup$ @user44690 done $\endgroup$
    – user21299
    Jun 10, 2021 at 16:37
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In general, you can only integrate an n-form over an n-dimensional manifold in a coordinate-independent way. Since the Hodge dual of a 0-form is an n-form, $S_2$ is a well-defined quantity (independent of coordinates), but $S_1$ is not.

As to why n-forms are the "natural" objects to integrate on n-dimensional manifold, I'm not an expert; hopefully someone will give a more thorough answer here. In addition, this answer and this answer over on Math.SE might help elucidate things.

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