1
$\begingroup$

Given an single vector field $A_\mu(x)$ is it possible to make a diffeomorphism invariant action in 4 dimensions? In the same way that General Relativity is diffeomorphism invariant?

My first guess would be:

$$ S = \int \left( \det_{ij}(\partial_i A_j(x))\right)^{1/2} dx^4 $$

or the same question with a single scalar field if we set $A_\mu(x)=\partial_\mu \phi(x)$ ? (But then I think we'd end up with zero).

I think this might be wrong because it doesn't use covariant derivatives so won't be diffeomorphism invariant. Do we always require a metric tensor in 4 dimensions?

(I know its possible to do it with 4 vector fields since we can define a metric tensor as $g_{\mu\nu}(x)=\eta_{ij}A^i_\mu(x) A^j_\nu(x)$. This is just the vierbien. But can it be done with just one?)

$\endgroup$
  • $\begingroup$ Probably not, since everyone knows about Chern-Simmons theory on 3D but I've never heard of something in 4D. But of course that's not a proof. $\endgroup$ – Javier Sep 15 '17 at 2:40
  • $\begingroup$ In 3D for a U(1) vector field. Would that simply be $S=\int\varepsilon^{ijk}A_i(x)\partial_j A_k(x) dx^3$ in 3D? $\endgroup$ – zooby Sep 15 '17 at 21:33
  • $\begingroup$ Yes, that's the Chern-Simmons action. You can write it more compactly as $\int A \wedge dA$ if you want. $\endgroup$ – Javier Sep 15 '17 at 21:45

Your Answer

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

Browse other questions tagged or ask your own question.