Diffeomorphism invariance of metric determinant

[Pacopag]

Hi;
I am pretty sure that
sqrt(-g) is diffeomorphism-invariant.

I am wondering if all powers of this are diffeo-invariant too. For example, are
-g, g^2, etc. all invariants too?

 

[bcrowell]

If you want to use less cumbersome terminology, you can just say "is a scalar" rather than "is diffeomorphism-invariant." That's what "scalar" means in GR.

The answer to your question is yes, because any function of a scalar is also a scalar.

[--D'oh -- that was incorrect -- sorry! --]

 

[Pacopag]

Thank you for your reply. This is good news for me. To be sure, a valid action may take the form
$$S = \int d^4x \sqrt{-g}\left(R+\sqrt{-g}\phi^2\right)$$
where
$$\phi^2$$
is a scalar
Is this correct?

 

[Stingray]

The metric determinant is not a scalar. Think of the volume elements associated with (say) spherical and Cartesian coordinates in flat space.

 

[Pacopag]

Yes. You are right. The second term in my action does not work.

 

[bcrowell]

D'oh, thanks for the correctoin, Stingray. I was obviously not awake this morning when I posted #2 :-)