The Maxwell equations in vacuum are $dF=0$ and $d*F=0$. Is this not the same as saying $F$ is both closed and co-closed, and hence harmonic? But Hodge's theorem says the space of harmonic $p$-forms on a manifold is isomorphic to its $p$th cohomology, which seems to imply that the only solution to Maxwell's equations in vacuum on $\mathbb{R}^4$ is $F=0$? Presumably I have missed something here?

  • $\begingroup$ Maybe it would be good to add what $F$, coclosure and Hodge's theorem are. $\endgroup$ – ahemmetter Nov 15 '18 at 15:18

You've just forgotten the space has to be compact.

Think about the simple case $p = 0$. This is the set of harmonic functions $\nabla^2 f = 0$. Indeed, on a compact connected space the zeroth cohomology group vanishes, and there are no nontrivial solutions for $f$. But on $\mathbb{R}^n$ there are plenty of solutions, one example being $e^z$ on the plane.

If you require $F$ to vanish at infinity, we may compactify $\mathbb{R}^n$ to $S^n$, and the theorem does apply.

  • $\begingroup$ With this caveat this seems like a really nice result - is there anything I can do with the knowledge that there are exactly $b^2$ independent solutions on a compact manifold/that vanish at infinity? $\endgroup$ – Sanjay Prabhakar Nov 15 '18 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.