0
$\begingroup$

Hodge's decomposition theorem saids that on a compact orientable Riemannian manifold any $k$-form can be decomposed uniquely as the sum of an exact, a co-exact and a harmonic form:

$\Omega^k(M)=d\Omega^{k-1}(M) \oplus \delta\Omega^{k+1}(M) \oplus \mathcal{H}^k(M)$

Is there a similar decomposition if the metric is indefinite? What is the closest such result we know? Are there any references on this?


Added:

I know that if the metric is indefinite Hodge's theorem doesn't hold. This is why my question is whether there is something similar (not identical) for the indefinite case.

$\endgroup$
  • $\begingroup$ Hodge decomposition uses orthogonality of the three listed subspaces under the $L^2$ inner product. The analogous scalar product on indefinite signature no longer gives you nice orthogonality conditions. $\endgroup$ – Willie Wong May 7 '11 at 16:39
  • 1
    $\begingroup$ Your question might be a little too general as is. One could plunge into the depths of Hodge theory and try and salvage as much as possible from an indefinite metric, but the ocean is deep and it would help enormously to have in idea of what one is looking for. So, what motivated your question? $\endgroup$ – Gunnar Þór Magnússon May 8 '11 at 18:56
11
$\begingroup$

It seems to be far from true. The first example I can think of is the torus $\mathbb R^2/\mathbb Z^2$ with indefinite metric $dxdy$. The Laplacian on functions is $\frac{\partial^2}{\partial x\partial y}$, so the space of harmonic functions is infinite dimensional.

I think the point is that in the definite case the Laplacian is an elliptic operator.

Edit: This example is atypical in one way. Substitute a generic lattice for $\mathbb Z^2$ and now the only harmonic functions are constant.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. Your example shows that Hodges decomposition does not apply to the indefinite case, but I am interested whether there is something instead. $\endgroup$ – Cristi Stoica May 8 '11 at 10:47
  • $\begingroup$ I don't know. But the fact that even on a compact manifold harmonic does not imply closed is not a good sign. $\endgroup$ – Tom Goodwillie May 8 '11 at 18:07

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.