# Must the electromagnetic 2-form be harmonic in vacuum?

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?

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

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.
• 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? – Sanjay Prabhakar Nov 15 '18 at 16:22