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I want to get some deep understanding on closed orientable Riemannian manifolds admitting $k$-forms ($k\geq 2$) $\omega$ that satisfices the following conditions: $$\nabla \omega\neq 0,\quad \Delta\omega=0.$$ where $\Delta=(\delta +d)^2$ is the Laplace-Beltrami operator (Hodge Laplacian). What other useful properties can these forms have? Any reference would be greatly appreciated.

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There are too many of these for them to have any particularly interesting structure. For example, consider any metric $g$ on the $3$-torus $\mathbb{T}^3$. By the Hodge theorem, the space of $g$-harmonic $2$-forms has dimension $3$. However, if, say, all of them were $g$-parallel, then their duals would be a basis of $g$-parallel $1$-forms, which would imply that the metric $g$ is flat. Thus, any non-flat metric $g$ on the $3$-torus will have a $g$-harmonic $2$-form that is not $g$-parallel.

In fact, as long as $H^2_{\mathrm{deRham}}(M^3,\mathbb{R})\not=0$, where $M$ is an orientable compact $3$-manifold, the generic metric $g$ on $M$ will have a $g$-harmonic $2$-form that is not $g$-parallel, since having a nonzero $g$-parallel $2$-form would force $g$ to be locally a product metric, and the generic metric is not locally a product.

Unfortunately, I think that there not much 'deep' to say about such $2$-forms in general.

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  • $\begingroup$ Thanks professor. Do you know that these non-parallel forms are invariant under metric deformation? i.e. for a deformation of $g$ like $h$, $\nabla_h \omega\neq 0$ and $\Delta_h\omega=0$? (something like this post) $\endgroup$
    – C.F.G
    Commented Feb 23, 2021 at 6:25
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    $\begingroup$ @C.F.G: There's no reason to believe that a form of positive degree that is harmonic for one metric would be harmonic for a different metric. If $\omega$ is closed (which doesn't depend on the metric), you are asking whether $\mathrm{d}(\ast_h\omega) = 0$ for a metric $h$, which is a non-trivial first-order system of pde for $h$ as long as $\omega$ is nonzero, and it is generally weaker than the equation $\nabla_h\omega = 0$ (which is also a first order equation for $h$). $\endgroup$
    – Robert Bryant
    Commented Feb 23, 2021 at 10:34
  • $\begingroup$ That's seems reasonable but I compare it with homotopy(homology) groups like loops homotopy class that are invariant under homeo. but why these are not invariant under homeo intuitively? $\endgroup$
    – C.F.G
    Commented Feb 23, 2021 at 11:10
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    $\begingroup$ @C.F.G: I'm not sure what you mean. The deRham cohomology group is a topological invariant, but the choice of a form representing a given cohomology class is not a topological invariant, there is not even a choice of a form that is a diffeomorphism invariant, since the diffeomorphisms homotopic to the identity don't fix any form of positive degree other than the form that vanishes identically. $\endgroup$
    – Robert Bryant
    Commented Feb 23, 2021 at 12:21
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    $\begingroup$ @C.F.G: Well, I chose 3 dimensions just because it was easier, but there are certainly examples in dimension 4 as well. Let $M^4$ be a compact orientable 4-manifold that is simply-connected, not a product, does not support a Kähler metric, and yet does have non-trivial Betti number $b_2$. (It's easy to construct such examples, for example, the oriented connected sum of $\mathbb{CP}^2$ with itself.) Then every metric $g$ on it will have non-zero harmonic $2$-forms (by deRham's Theorem), but none of them can be $g$-paralell. $\endgroup$
    – Robert Bryant
    Commented Jul 7, 2022 at 0:32

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