I have encountered an apparent inconsistency in my calculations when dealing with covariant derivatives of different spacetime points (we assume they are timelike separated, and there exists a geodesic connecting them) acting on the Dirac delta. We denote a primed index to represent $x'$, and unprimed for $x$, thus $\nabla_{\mu}$ is a covariant derivative acting at $x$, while $\nabla_{\rho'}$ acts at $x'$. Further we define the biscalar Dirac delta $\tilde{\delta}(x;x')=\frac{\delta^D(x-x')}{\sqrt{-g}}$, where we make use of metric compatibility to ensure that the $\sqrt{-g}$ doesn't interfere with our covariant derivatives.
We define a bitensor as an object that transforms as a tensor w.r.t. both position at two independant orders, e.g. if $f(x;x')$ is a biscalar then $\nabla_{\mu}f(x;x')$ transforms as a covector in $x$ and as a scalar in $x'$.
There are two properties I have for dealing with covariant derivatives of the Dirac delta, they are
- For $T(x;x')$ a bitensor, the order of the prime and unprimed covariant derivatives are independent of each other and only have to remember the order of prime, and order of unprimed (See Synge - Relativity: The General Theory Ch.2§2). That is to say they commute: $$[\nabla_{\mu},\nabla_{\rho'}]T(x;x')=0$$ The proof involves looking at Fermi coordinates, where the $\Gamma$'s along the geodesic connecting $x$ and $x'$ vanish, which reduces the above relation to partial derivates which then commute (in the case of a two (un)primed derivatives, then $\partial_.\Gamma^._{..}$ terms survive, and we don't have a vanishing commutator).
- We can change the position of a covariant derivative when it acts directly on the Dirac delta via: $$\nabla_{\mu}\delta^D(x-x') = -\nabla_{\mu'}\delta^D(x-x')$$ This can be shown by looking at evaluating $$\int d^Dx' \sqrt{-g'}\nabla_{\mu}(\delta^D(x-x'))f(x') =\nabla_{\mu}\bigg(\int d^Dx' \sqrt{-g'}\delta^D(x-x')f(x')\bigg) =\sqrt{-g}\nabla_{\mu}f(x)$$ and then in the primed case we discard the boundary term when integrating by parts: $$-\int d^Dx' \sqrt{-g'}\nabla_{\mu'}(\delta^D(x-x'))f(x')=\int d^Dx' \sqrt{-g'}(\delta^D(x-x'))\nabla_{\mu'}f(x')=\sqrt{-g}\nabla_{\mu}f(x)$$
Now for the inconsistency. We want to exchange the position of the outermost covariant derivative of $\nabla_{\rho}\nabla_{\mu}\nabla_{\nu}\tilde{\delta}(x;x')$. As we can freely commute the prime and unprimed derivatives, we can first change the inner most one, and then as the first two derivatives always commute on a (bi)scalar we don't pick up any additional terms:
$$\nabla_{\rho}\nabla_{\mu}\nabla_{\nu}\tilde{\delta}(x;x') = -\nabla_{\rho}\nabla_{\mu}\nabla_{\nu'}\tilde{\delta}(x;x')=\nabla_{\mu}\nabla_{\nu'}\nabla_{\rho'}\tilde{\delta}(x;x')=-\nabla_{\rho'}\nabla_{\mu}\nabla_{\nu}\tilde{\delta}(x;x')$$
Alternatively, we manually commute through the final derivative, picking up a Riemann curvature term:
$$\nabla_{\rho}\nabla_{\mu}\nabla_{\nu}\tilde{\delta}(x;x') = \nabla_{\mu}\nabla_{\nu}\nabla_{\rho}\tilde{\delta}(x;x') + [\nabla_{\rho},\nabla_{\mu}]\nabla_{\nu}\tilde{\delta}(x;x') = -\nabla_{\rho'}\nabla_{\mu}\nabla_{\nu}\tilde{\delta}(x;x') - R^\lambda{}_{\nu\rho\mu}\nabla_{\lambda}\tilde{\delta}(x;x')$$
I tried to move this to a Differential Geometry framework, but I don't have a lot of experience there, and wasn't able to get to any conclusions. To me it is the second property that is more strange, and the one I expect to be breaking something, but I don't see the flaw in the argument. However I need to be able to change the position like this in some capacity.
EDIT: In the case that anybody is reading this at a later stage, I should point out that property 2 is actually only valid in the regime where we have a scalar function multiplying our function. For a biscalar that proof breaks down. I have yet to come up with a new method for exchanging the position of my derivative when multiplying by a bitensor however.