Prove that background independence and diffeomorphism invariance of a spacetime theory are equivalent

Question

A theory's equations can generally be derived from an action along with a principle of least action ($\delta S=0$). The action is given by:

$$ S[f_1, f_2, ...]=\int_M \mathcal{L}(f_1, f_2, ..., g_1, g_2, ...) $$

where $S$ is a functional of $f_1, f_2, ...$, and is not a functional of $g_1, g_2, ...$. I call the $g_1, g_2, ...$ background objects. I have defined a theory to be background independent if the action $S$ used to derive its field equations cannot be defined in such as way as as to depend on no objects for which is it not a functional (i.e. it has no background objects $g_1, g_2, ...$).

A model of a theory is an ordered set $<$$M, A_1, A_2, ...$$>$ which represents a possible solution to the equations of the theory. A theory is diffeomorphism invariant if, for any solution to the equations of the theory $<$$M, A_1, A_2, ..., A_n, \rho$$>$, $<$$M, f^{\ast}A_1, f^{\ast}A_2, ..., f^{\ast}A_n, f^{\ast}\rho$$>$ is also a solution for any diffeomorphism $f$ ($f^{\ast}A_i$ is the drag-along under $f$ of $A_i$). The $A_1, A_2, ...$ are particular values of the $f_1, f_2, ... g_1, g_2, ...$ which solve the equations of the theory (i.e. for which $\delta S=0$ for infinitesimal variations around the $A_1, A_2, ...$ values).

I want to show that the background independence and diffeomorphism invariance of a theory are equivalent (i.e. that a theory is background independent if and only if it is diffeomorphism invariant). Here is my current attempt, but I am not sure if it is a rigorous proof (or if this statement is even definitely true!):

It is clear that, as I have defined the terms, a theory which is not background independent will generally not be diffeomorphism invariant, because a general diffeomorphism $f$ will not leave the absolute background objects invariant. This will only be the case for the subclass of diffeomorphisms for which $f^{\ast}A_i=A_i$ for all background objects $A_i$. Perhaps less obvious is the fact that a background independent theory will be diffeomorphism invariant. To see this, take some model of a background independent theory $<$$M, A_1, A_2, ..., A_n, \rho$$>$ (where all of the $A_i$ must not be background objects) and some diffeomorphism $h$. The action $S$ used to derive the field equations of the theory is given by:

$$ S[f_1, f_2, ...]=\int_M \mathcal{L}(f_1, f_2, ...) $$

where $A_1, A_2, ..., A_n, \rho$ are particular values of $f_1, f_2, ...$ that satisfy $\delta S=0$. This means that the value of $S$ doesn't change for infinitesimal variations around the $A_1, A_2, ..., A_n, \rho$ values. Since $f^{\ast}A_i(f(p)) \equiv A_i(p)$, and since $S$ is given by an integral over the entire manifold $M$, $S[A_1, A_2, ...]=S[f^{\ast}A_1, f^{\ast}A_2, ...]$. Moreover, since a diffeomorphism is smooth, an infinitesimal change to $f^{\ast}A_i$ corresponds to an infinitesimal change to $A_i$, and therefore, the value of $S$ will not change for infinitesimal variations around $f^{\ast}A_1, f^{\ast}A_2, ..., f^{\ast}A_n, f^{\ast}\rho$ either, and so these values for $f_1, f_2, ...$ also satisfy $\delta S=0$.

Any help would be much appreciated here!

Answer 1

I would have considered showing that diffeomorphism invariance implies background independence to be the harder direction to prove. The reason is you could be given an action that is written in terms of a lot of background structures, but is secretly diffeomorphism-invariant. The question becomes whether the action then be written in a manifestly diffeomorphism-invariant way. One example of such an action would be the $3+1$ split of the GR action: everything gets written in terms of the spatial geometry of the foliation (intrinsic and extrinsic curvatures), and hence would seem to depend on the foliation as a background structure, although it of course does not because it came from a manifestly background-independent action.

There is a paper by Iyer and Wald that addressed this question directly: https://arxiv.org/abs/gr-qc/9403028, see Lemma 2.1. They show that it is indeed true that a diffeomorphism-invariant action can always be written in a manifestly diffeomorphism-invariant (i.e. background-independent) way.

The implication that background-independence implies diffeomorphism-invariance I would have regarded as the more obvious one. It seems from your definition of background objects, you would say that there absence means that the Lagrangian functional satisfies \begin{equation}L[f^* A_i] = f^* L[A_i] \qquad (*)\end{equation} which gives $S[A_i] = S[f^* A_i]$ upon integration (up to boundary terms which I'll just ignore for this question. The boundary terms have important consequences, but it seems like this question is more about the local equivalence of diff-invariance and background-independence). Varying the Lagrangian gives $$\delta L[A_j] = E^i[A_j] \delta A_i$$ again dropping boundary terms of the form $d\theta[A_j, \delta A_j]$. But then varying ($*$), we derive $$E^i[f^* A_j] \delta(f^* A_i) = f^*(E^i[A_j]\delta A_i) = f^*E^i[A_j] f^*\delta A_i,$$ which uses that $\delta$ commutes with $f^*$. There certainly exist diffeos $f$ that satisfy $[\delta, f^*]=0$, and it is enough to restrict attention to them. The above equation is true for all variations, allowing us to conclude that $E^i[f^* A_j] = f^*E^i[A_j]$ and hence $A_j$ define a solution (E[A_j]=0) iff $f^* A_j$ do.