 Is the wedge product of two harmonic forms on a compact Riemannian manifold harmonic? I'm looking for a counterexample that the textbooks say exists.
 I would like to see a counter example that is on a complex manifold, Ricciflat (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 eigenvectors, references will be much appreciated.

$\begingroup$ 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 $\endgroup$ – Dmitri Panov Jan 14 '10 at 9:56
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 $A \wedge B$ is nonzero, but it vanishes at some point, since both $A$ and $B$ have zeros. At the same time a harmonic $2$form on a Riemann surface is constant. Explicit examples of $1$forms on Riemann 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 dimension $22$. Now it should be possible to find two antiselfdual twoforms whose wedge product vanishes at one point on $K3$ but is not identically zero. This is because the dimension of the space of self dual forms is 19 which is big enought to get vanishing at one point

$\begingroup$ Thank you Dmitri. The genus 2 case is obvious to me after your comment. I'll try to reinforce it with some computations. $\endgroup$ – Boris Ettinger Nov 6 '09 at 21:59
Generically, the wedge product of two harmonic forms will not be harmonic. It is harder to find examples than counterexamples. For example, on compact Lie groups with a biinvariant 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.

2$\begingroup$ Nitpick: this is not true on all symmetric spaces (witness the example of hyperbolic surfaces above), only those whose isometry group admits a biinvariant metric. $\endgroup$ – Tom Church Nov 6 '09 at 18:54

$\begingroup$ 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 eigenmodes of the Laplacian. $\endgroup$ – Boris Ettinger Nov 6 '09 at 22:02

$\begingroup$ @Tom: I thought that compact riemannian symmetric spaces are such that the transvection group always admits a biinvariant 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. $\endgroup$ – José FigueroaO'Farrill Nov 7 '09 at 15:58

$\begingroup$ I imagine that the difference is terminological. I would call a factor space of a general group a homogeneous space rather then symmetric space. $\endgroup$ – Boris Ettinger Nov 7 '09 at 20:31

1$\begingroup$ I think y
all are getting confused by
symmetric' vs. `locally symmetric'that is, having a universal cover which is symmetric. Const. negative curv. surfaces are locally symmetric. $\endgroup$ – Richard Montgomery Feb 2 '10 at 7:18
Interestingly in (24) of hepth/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 noncompact complete hyperkahler manifolds.

1$\begingroup$ Interesting. Is this also the case for TaubNUT? In 0902.0948, Witten attributes the construction of the middledimensional $L^2$ harmonic form on TaubNUT 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. $\endgroup$ – José FigueroaO'Farrill Nov 7 '09 at 17:54

$\begingroup$ No, the individual 2forms are harmonic, and on TaubNut 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 $\endgroup$ – Tamas Hausel Nov 7 '09 at 18:39
Here's a counterexample from the theory of nilmanifolds, which by their very nature are not formal. Take a compact quotient $H^3/\Gamma$ of the Heisenberg group. It admits invariant 1forms $e^1,e^2,e^3$ with $de^1=0=de^2$ and $de^3=e^1\wedge e^2$. Then $e^1,e^2$ are harmonic, but $e^1\wedge e^2$ is exact, so not harmonic. You can take a product with $S^1$ to get a complex (nonKähler) surface on which the same thing works, but not I am afraid Ricciflat or Einstein.

$\begingroup$ What do you mean by "by their very nature"? More interestngly: can you give an example of a space which is formal against its nature? :) $\endgroup$ – Mariano SuárezÁlvarez Jul 25 '10 at 22:59

1$\begingroup$ The differential graded algebra of invariant forms on a nilmanifold M is a minimal model for M in Sullivan's sense because the nilpotency makes d decomposable. Unless d kills everything invariant (like on a torus), M will not therefore be formal. $\endgroup$ – Simon Salamon Jul 25 '10 at 23:24


$\begingroup$ Dear Prof. Salamon, could you please give a bit more explanation on the phrase "exact, so not harmonic?" I looked at your arXiv pieces, there are a few items on nilmanifolds but nothing that quite explains this to me, or to the friend who asked me for clarification. Anyway, if this medium is not convenient, you can always email me, just search using my last name at ams.org/cml $$ $$ Granted, I think Dmitry (not the Dmitri above) was thinking about Riemann surfaces while I was talking about Riemannian manifolds. $\endgroup$ – Will Jagy Jul 26 '10 at 21:24
Here is a copout answer (sorry in advance). Consider the 0form $f(x)=x$ on $\mathbb{R}$. Clearly $f$ is harmonic, but $(f\wedge f)_{x}=x^2$ isn't. Since the wedge product is just the pointwise product on 0forms, this is a valid example!
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 2ary 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 153164) 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 dgalgebra structure.

$\begingroup$ Thanks Ian, but my interest is to analyze the product of two harmonic forms, rather than the general structures on harmonic forms. $\endgroup$ – Boris Ettinger Nov 7 '09 at 20:33