Must the action be a coordinate scalar?

Question

I know that an action must be locally-Lorentz invariant based on physical reasons, but is there any requirement for it to be a coordinate pseudo-scalar (up to surface terms)? In particular, would an action of the following form be permissible, $$ S=\int \sqrt{-g} L \, d^4x \ , $$ where $L$ is not a coordinate scalar (i.e. has no free indices but explicitly depends on coordinates) but is still a local Lorentz scalar. An example of such a Lagrangian that isn't a coordinate scalar would be using a contraction of the Christoffel symbols $$ L= g^{\mu \nu}\Gamma_{\mu \lambda}^{\rho} \Gamma^{\lambda}_{\nu \rho} \ . $$

Will this still give a well-defined variation principle $\delta S = 0$ despite $S$ being non-covariant and depending on the choice of coordinates?

A similar question but regarding Lorentz invariance is asked here Must the action be a Lorentz scalar? but I'm unsure if the argument also applies here too (with diffeomorphism invariance instead of Lorentz invariance).

Also note I do not care if the equations of motion resulting from $\delta S = 0$ are not covariant.

-- On further thought, rather than Lagrangian wrote above one could just consider something simple like $$ L = \partial_{\mu}\partial_{\nu}g^{\mu \nu} $$ which is probably easier to work with. One potential problem I see is that the action could vanish or diverge to plus or minus infinity in some coordinate systems - is this a problem for the variation principle $\delta S= 0$?

Answer 1

When we write an integral like $$ \int f\sqrt{g}\mathrm{d}^4x$$ then $f : M \to \mathbb{R}$ is a scalar function on our spacetime $M$. Coordinate-dependent expressions do not define such functions, when you write something like $$ L = g^{\mu \nu}\Gamma_{\mu \lambda}^{\rho} \Gamma^{\lambda}_{\nu \rho}, $$ then that actually doesn't define anything - the expression on the r.h.s. does not have a single value, but infinitely many, depending on what coordinate system you choose.

Of course you can define a function if you declare that $L$ always takes the value that the r.h.s. has in one fixed coordinate system, but since we usually do not want to fix a coordinate system this is rather pointless - there is no naturally "preferred" coordinate system on manifold so it is hard to see how any quantity of interest could ever be of this form.