Diffeormophism invariance of a non-local series possible? If I want to construct Lorentz invariant forms involving a scalar field $\phi(x)$ I could have non-local terms such as:
$$\int \phi(x)\frac{1}{|x-y|^2}\phi(y) dx^4 dy^4$$
or 'local' forms such as:
$$\int \partial_\mu \phi(x)\partial^\mu\phi(x) dx^4 $$
Now if I want to construct diffeomorphic invariant forms involving the metric $g^{\mu\nu}(x)$, I would like to know if I could construct them from a series of non-local terms. e.g.:
$$ a + \int g(x)b(x)dx^4 + \int \int g(x)g(y)c(x,y)dx^4dy^4 + \int \int \int g(x)g(y)g(z)d(x,y,z)dx^4dy^4dz^4 + ...$$
(hiding indices)
Can such a series be diffeomorphic invariant if the functions $b,c,d,..$ are non-local? (i.e. not derivatives of dirac delta functions, so that for instance $c(x,y)\neq 0$ when $x \neq y$).
Otherwise the only way I can think of to form diffeomorphic invariant form is in terms of local forms such as:
$$\int \sqrt{g}(a + b R + c R^2 + d R^3 + ....) dx^4$$
(hiding indices where $R^n$ are the various contractions of the curvature tensor). Is this true?
(The reason I am interested is thinking about solutions to Wheeler-de-Witt equation but that's not important for the question.)
 A: You have to be more precise about what you mean by "diffeomorphism invariance".
If it is coordinate invariance you seek, the answer is "yes", though you'll need an extra structure.
For example, as long as there's a function $f(x, y)$ that is scalar in both $x$ and $y$, you can have an integral
$$ \int d^4 x \sqrt{|g(x)|} \int d^4 y \sqrt{|g(y)|} \phi(x) \phi(y) f(x, y). $$
One such function is the heat kernel $K_t(x, y)$, that is, the solution of
$$ \frac{\partial}{\partial t} K_t(x, y) = g^{\mu \nu}(y) \nabla_{\mu} \nabla_{\nu} K_t(x, y) $$
where covariant derivatives are w.r.t. $y$, and the initial condition is
$$ K_0(x, y) = \frac{1}{\sqrt{|\det g(x)|}} \cdot \delta^{(4)}(x, y). $$
The heat kernel is completely determined by the choice of the metric, which is local.
If, on the other hand, by "diffeomorphism invariance" you meant what is really called "background independence", which is essentially the lack of background structures such as a metric tensor, a function $f(x, y)$, etc; then the answer is "no".
