Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear Experts,

On a compact, boundless, Riemannian manifold, the dimension of the space of harmonic k-forms is equal to the k-th Betti number. Is this correct (by Hodge theory)? For example, on the surface of a unit sphere S^2 and any equivalent topology, the 0-th Betti number is 1, which is consistent with the fact that any harmonic function on a compact, boundless, Riemannian manifold is a constant. More importantly to me, what about the 1st Betti number? I think the only harmonic 1-form on the topological class of S^2 is zero --- you can't comb every hair on a sphere towards the same direction --- right?

Basically, it is easy to prove both results (1st betti number = 0 and the only harmonic 1-form is zero) on the ideal case of S^2. But if I want to go further, some topological tool has to be involved?

I am a PDE guy, so if there is any PDE-based proof to the above question, it'd be highly appreciated!

Thank you for your valuable time!

Bin Cheng

share|improve this question
    
Some geometric and topological ideas are needed to set things up (e.g de Rham's theorem), but the bulk of the argument is PDE as has been pointed out already. Perhaps the proof in Wells' Differential Analysis on Complex Manifolds, which uses pseudodifferential operators and applies to elliptic complexes, might be to your taste. –  Donu Arapura Nov 5 '10 at 12:25
add comment

3 Answers

up vote 4 down vote accepted

Just to add some details to Paul's answer:

It is indeed correct that the $k$-th Betti number of a compact Riemannian manifold $M$ without boundary is equal to the dimension of the space of harmonic $k$-forms. This is the Hodge isomorphism theorem. A good proof of this may be found in Demailly's book (available for free here: http://tinyurl.com/2dcnycm) at the beginning of chapter 6.

Basically what happens is this: we use a Riemannian metric to define an inner product on the space $E^k := C^\infty(M, \bigwedge^k T_M^*)$ of smooth $k$-forms on $M$. The Riemannian metric also gives us the Laplacian $\Delta$, which turns out to be an elliptic differential operator on $E^k$. Thus we get an orthogonal direct sum decomposition

$$C^\infty(M, \bigwedge^k T_M^*) = \mathcal H^k(M) \oplus Im \Delta = \mathcal H^k(M) \oplus Im d \oplus Im d^{\*}$$

where $\mathcal H^k(M) := Ker \Delta$ is the space of harmonic $k$-forms on $M$, and $d^\*$ is the formal adjoint of the exterior derivative $d$. The subspace of $d$-closed forms of $E^k$ is exactly $\mathcal H^k(M) \oplus Im d$, and thus we obtain an isomorphism between $\mathcal H^k(M)$ and the $k$-th De Rham cohomology group $H^k(M,\mathbb{R})$, whose dimension is equal to the $k$-th Betti number of $M$.

Morally speaking this isomorphism is interesting because it gives a link between the topological and geometric structures of the manifold. The Betti numbers only depend on the topology of $M$, while the space of harmonic forms is defined by a Riemannian metric. With this theorem we can interpret the $k$-th Betti number as counting the number of linearly independent harmonic $k$-form on the original manifold. In particular, as you say, if a $k$-th Betti number is zero, then the only harmonic $k$-form on $M$ is the zero form.

share|improve this answer
add comment

The Hodge theorem (The space of harmonic $p$ forms on a closed Riemannian manifold $M$ is isomorphic to $H^p(M;R)$) is a PDE theorem! See e.g. Warner's book. Your "comb the hair" comment is a bit off, though. You can't comb the hair on a genus g surface when g>1 either, but the space of harmonic 1-forms is 2g dimensional. Harmonic p-forms can have zeros when p>0.

share|improve this answer
add comment

If I understood your question correctly, it is: how can I compute the dimension of the space of harmonic forms? There is one class of Riemann manifolds where it is possible to write down the harmonic forms explcitly, without topology. These are the symmetric spaces (for example spheres, projective spaces, compact Lie groups, tori, Grassmann manifolds, flag manifolds). The space of harmonic forms on the symmetric space $M=G/H$ is the same as the invariant forms, which is $\Lambda^* (\mathfrak{g}/\mathfrak{h})^G$. And you can compute this algebraically. I guess that each explicit computation of harmonic forms, including your example on $S^2$, is based on that principle (maybe for Riemann surfaces and nonsymmetric homogeneous spaces there are exceptions to this rule). As pointed out by Gunnar, the dimension of the space of harmonic forms on $M$ does not depend on the metric. So you can conclude that there is no harmonic 1-form on $S^2$, in any wild metric. Of course, most manifolds do not carry a symmetric metric. Here, if you want to compute the dimension of the space of harmonic forms, it might be easiest to compute first the cohomology and then appeal to the Hodge theorem.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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