The theorem of Hodge decomposition is on the compact Kahler manifold, is it generally true for the non-compact kahler manifold or are there examples to show the failure?
Here is my Hodge decomposition theorem:
$\begingroup$
$\endgroup$
4
-
3$\begingroup$ Your question is a bit imprecise. I presume that you are asking for a noncompact Kahler manifold where the $(p,q)$ decomposition including symmetry fails. In that case, take $\mathbb{C}^*$. Or do you mean something else? $\endgroup$– Donu ArapuraCommented Sep 7, 2018 at 22:32
-
$\begingroup$ @Donu Arapura, i have modified. And can you explain why $\mathbb{C}^*$ fails? $\endgroup$– 6666Commented Sep 7, 2018 at 23:21
-
1$\begingroup$ It would help for you to specify exactly what you mean by the Hodge decomposition. $\endgroup$– Keerthi MadapusiCommented Sep 8, 2018 at 2:24
-
$\begingroup$ @KeerthiMadapusiPera added $\endgroup$– 6666Commented Sep 8, 2018 at 2:47
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
8
The kernel and cokernel of an operator on a non-compact manifold could have infinite dimension. So the classical Hodge decomposition no longer holds. For a simple example, consider the unit open interval, or in complex setting the Riemann sphere removing the origin and the point $\infty$. There are various ways you may "regularize" the continuous spectrum depending on the boundary condition given. For example you may "glue" two manifold with boundary to a manifold without boundary.
One way to view these regularization methods is that they essentially `cut-off' the divergent part of heat kernel to restore the trace class property, and you get exact sequence in cohomology with boundary terms involved. However I am not entirely sure if there is a consensus on how to best approach these problems (the motivation is often from gauge theory or number theory). I think Rafe Mazzeo's PhD thesis is on Hodge theory for non-compact manifolds, which you can probably look up.
Update: Apparently due to my ignorance, I fail to recognize that the subject is still an ongoing research topic by experts in geometric analysis. See https://arxiv.org/pdf/1812.11764.pdf, for example.
answered Sep 7, 2018 at 23:53
-
$\begingroup$ I think "kernel and cokernel of an operator on a non-compact manifold could have infinite dimension" only shows we can not use $\mathcal{H}^p(X)\cong H^p(X,C)$, but we don't have to use harmonic forms to prove Hodge decomposition, so I wonder if the classical Hodge decomposition fails $\endgroup$– 6666Commented Sep 8, 2018 at 2:10
-
$\begingroup$ in any case, the example of $\mathbb C ^*$ shows that the first Betti number is not even so Hodge decomposition fails $\endgroup$ Commented Sep 8, 2018 at 3:28
-
$\begingroup$ @Venkataramana what's the definition $C^*$? $\endgroup$– 6666Commented Sep 8, 2018 at 3:56
-
1$\begingroup$ @6666 $\Bbb C$ minus the origin. This is homotopy equivalent to the circle. $\endgroup$– mmeCommented Sep 8, 2018 at 4:05
-
1$\begingroup$ @6666: Thanks for your comment. If my memory serves, there does exist a Hodge theory for singular varieties. The base field also does not need to be $\mathbb{C}$. The starting point is the filtration associated with a Hodge structure. However I am not very familiar with the subject ("mixture type theory?"). Other experts on the site may help you more. $\endgroup$ Commented Sep 8, 2018 at 20:03
$\begingroup$
$\endgroup$
Now that it's clear what you want, let me turn my (and various other) comment(s) into an answer. Any manifold with odd first Betti number will give a counterexample; $\mathbb{C}^*$ is the simplest such. Let me add that the reason why there was some confusion initially, is that "Hodge decomposition" can mean different things to different people. It could mean the $(p,q)$ decomposition, or it could simply mean the fact that on a compact manifold every cohomology class has a unique harmonic representative. BM's answer seems to be using this last interpretation. For noncompact manifolds, things can go badly wrong with this as well.
I'm just leaving this for the record.