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Is the wedge product of two harmonic forms on a compact Riemannian manifold harmonic? I'm looking for a counter-example that the textbooks say exists.
I would like to see a counter example that is on a complex manifold, Ricci-flat (or Einstein) manifold or both, if it is at all possible.
In general, I'm trying to understand the interaction between the wedge product, Hodge star and the Laplacian on forms and it's eigen-vectors, references will be much appreciated.

flag	asked Nov 6 at 4:08 Boris Ettinger 101●4

There is a nice article today on arxiv, called REMARKS ON THE PRODUCT OF HARMONIC FORMS by LIVIU ORNEA AND MIHAELA PILCA. You may find usefull references there. arxiv.org/abs/1001.2129 – Dmitri Jan 14 at 9:56

Dmitri · Accepted Answer · 2009-11-06 16:24:05Z UTC

It is easy to construct examples on Riemann surfaces of genus >1. Take any surface like this. Let A and B be two harmonic 1-forms, that are not proportional. Then A \wegde B is non-zero, but it vanishes at some point, since both A and B have zeros. At the same time a harmonic 2 form on a Rieamann surface is constant. Explicite examples of 1-forms on Rieamann surfaces can be obtained as real parts of holomorphic 1-forms.

Note of course that the above example is complex, and Einstein just take the standard metric of curvature -1. If you want an example on a Ricci flat manifold you should take a K3 surface. It is complex and admits a Ricci flat metric. Now, its second cohomology has dimesnion 22. Now it should be possible to find two anti-self-dual two-forms whose wedge product vanishes at one point on K3 but is not identically zero. This is because the dimesnion of the space of self dual forms is 19 which is big enought to get vanishing at one point

José Figueroa-O'Farrill · Answer 2 · 2009-11-06 07:28:46Z UTC

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Generically, the wedge product of two harmonic forms will not be harmonic. It is harder to find examples than counter-examples. For example, on compact Lie groups with a bi-invariant metric or, more generally, on riemannian symmetric spaces, harmonic forms are invariant and invariance is preserved by the wedge product. In general, though, this is not the case.

According to Kotschick (see, e.g., this paper) manifolds admitting a metric with this property are called geometrically formal and their topology is strongly constrained. He has examples, already in dimension 4, of manifolds which are not geometrically formal.

answered Nov 6 at 7:28

José Figueroa-O'Farrill
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Nitpick: this is not true on all symmetric spaces (witness the example of hyperbolic surfaces above), only those whose isometry group admits a bi-invariant metric. – Tom Church Nov 6 at 18:54

Thank you for the link, I'll study it carefully, although my interest is more about the spectral aspect and the question was about the zero eigen-modes of the Laplacian. – Boris Ettinger Nov 6 at 22:02

@Tom: I thought that compact riemannian symmetric spaces are such that the transvection group always admits a bi-invariant metric. I am confused by your comment about the hyperbolic surfaces. Which surfaces are you talking about? No compact manifold with constant negative curvature can be symmetric, since by Bochner's Lemma, it cannot have any Killing vector fields. Probably I am misunderstanding your comment. – José Figueroa-O'Farrill Nov 7 at 15:58

I imagine that the difference is terminological. I would call a factor space of a general group a homogeneous space rather then symmetric space. – Boris Ettinger Nov 7 at 20:31

So would I. There is no reason to believe that in a general homogeneous space (one admitting a transitive action by isometries) the wedge product of harmonic forms should again be harmonic. – José Figueroa-O'Farrill Nov 7 at 21:28

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Tamas Hausel · Answer 3 · 2009-11-07 15:26:55Z UTC

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Interestingly in (24) of hep-th/9603176 it is mistakenly claimed that the wedge product of harmonic forms is automatically harmonic. Because it is false we still do not know the predicted existence of those middle dimensional $L^2$ harmonic forms on these non-compact complete hyperkahler manifolds.

answered Nov 7 at 15:26

Tamas Hausel
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	Interesting. Is this also the case for Taub-NUT? In 0902.0948, Witten attributes the construction of the middle-dimensional $L^2$ harmonic form on Taub-NUT to three papers, of which the one you mentioned is the latest. The other two are older papers by Eguchi+Hanson and Pope from the late 1970s. – José Figueroa-O'Farrill Nov 7 at 17:54
	No, the individual 2-forms are harmonic, and on Taub-Nut is is in $L^2$ as well. One can show that there are no other ones as in arxiv.org/abs/math/9909002 and arxiv.org/abs/math/0207169 – Tamas Hausel Nov 7 at 18:39

ian cy shipman · Answer 4 · 2009-11-07 18:25:59Z UTC

Using homological perturbation theory, one can repair this defect. More precisely, on the space of harmonic forms, there is an $A_\infty$ structure with no differential whose 2-ary operation (multiplication) is constructed by wedging two harmonic forms then projecting the result back to the space of harmonic forms. See "Strong homotopy algebras of Kahler manifolds" by S.A. Merkulov (Int. Math. Res. Lett. no. 3 153--164) for details of the construction.

EDIT: Also, if the manifold is compact then the natural inclusion of harmonic forms into arbitrary forms becomes an equivalence of $A_\infty$ algebras, where the space of all forms has its usual dg-algebra structure.

Is the wedge product of two harmonic forms harmonic?

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