@ACuriousMind Is the Hodge theorem here the same one as Hodge decomposition theorem $\Omega^p=d\Omega^{p-1}\oplus d^\dagger\Omega^{p+1}\oplus \text{Harm}^p$?
@0celo7 Yes, I think "Hodge's theorem for functions" from Mike there is a special case of the general Hodge decomposition. The vanishing of the integral there shows there's no component of $\rho$ in the $\mathrm{d}^\dagger$-part, I think
@Danu Heh, yeah. I'm not advocating to take that stance in general, but there has to be a certain effort/usefulness ratio for it to be worth understanding such a proof.
@StanShunpike Since that'd be equivalent to the map being constant on a larger interval, yes, unless you demand the map to be injective (which isn't done in general).
My dream is to become a theoretical cosmologist. I love math and physics and I am incredibly curious about the mysteries of our universe. What is it like to be one (specifically research)?
@0celo7 Ah, c'mon keep answering stuff. Perhaps look at a few questions which don't require much formulae at all - those usually get many votes, and need not be boring.
@Danu Well, you can also do symmetric. But no, I don't know of any interesting consequences of it, it's just something you need to be aware of when quantizing a theory.
@ACuriousMind I know, I just didn't want to type it out and thought it was obvious from my previous comment :P
@ACuriousMind I do think there are very interesting things to be said about this. I think the entire program of "geometric quantization" is based on the idea that it provides a good motivation for one choice. Or at least that's what one of my professors told me.
@ACuriousMind I was still asking about the analogy with the derivatives btw. But you don't think it's more than just superficial, then?
My dream is to become a theoretical cosmologist. I love math and physics and I am incredibly curious about the mysteries of our universe. What is it like to be one (specifically research)?
