0
$\begingroup$

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?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ How do you integrate a 0-form over anything other than a 0-dimensional manifold? $\endgroup$
    – Michael Seifert
    Commented 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? $\endgroup$
    – Dr. user44690
    Commented 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. $\endgroup$
    – Richard Myers
    Commented Jun 10, 2021 at 17:37
  • 1
    $\begingroup$ @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
    Commented Jun 10, 2021 at 23:34

2 Answers 2

Reset to default
4
$\begingroup$

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

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

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ I believe that this is equivalent to the OP's version of the Lagrangian up to a boundary term. $\endgroup$
    – Michael Seifert
    Commented Jun 9, 2021 at 20:15
  • 3
    $\begingroup$ @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
    Commented 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? $\endgroup$
    – Dr. user44690
    Commented Jun 10, 2021 at 6:47
  • 1
    $\begingroup$ @user44690 done $\endgroup$
    – user21299
    Commented Jun 10, 2021 at 16:37
1
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.