# When is this differential form harmonic?

Let $$(M^3, g)$$ be a (closed) Riemannian manifold and let $$u: M \to S$$ be a harmonic function, where $$S$$ is a closed orientable surface. If $$\omega$$ is a $$2$$-form on $$S$$, what are sufficient conditions on $$\omega$$ in order $$u^* \omega$$ to be a harmonic $$2$$-form on $$M$$?

The concrete case I am analyzing is the following: we have $$u_1, u_2 : M \to \mathbb{S}^1$$ harmonic functions, $$u = (u_1, u_2):M \to \mathbb{S}^1 \times \mathbb{S}^1$$ and $$\omega = d\theta_1 \wedge d\theta_2$$, where $$d\theta_i$$ denotes the volume form on each $$\mathbb{S}^1$$.

I’d be glad to have a solution in either case.

• I think this is probably false most of the time in the concrete case that you're describing. Just like a product of two harmonic forms is not necessarily harmonic, there's no good a priori reason that a wedge product of harmonic 1-forms (which is $u^*\omega$ in your case) is harmonic. – Rohil Prasad Apr 2 at 0:20

In your particular case, it seems clear to sufficiency two conditions: the first is that $$\nabla u^*(\omega) = 0,$$ i.e, $$u^*(\omega)$$ is parallel.

The last is quite technical so let me introduce some concepts before:

Recall the Weitzenbock formula: $$\Delta = \nabla^*\nabla + Q^p.$$

Since $$\Sigma$$ is closed we can intrdocute the $$L^2$$ inner product of $$\Omega^p(\Sigma)$$ in the following way

$$(\omega,\beta) := \int_{\Sigma}\omega\wedge\star\beta = \int_{\Sigma} g(\omega,\beta)d\mu_g.$$

Therefore, $$\Delta$$ is self-adjoint on this inner product and $$\nabla^*$$ is the dual of $$\nabla$$ on this inner product.

In the particular case of $$n=3$$, $$Q^p$$ is nothing but the Ricci tensor on $$2$$-forms, i.e, if $$R$$ denotes the Riemann tensor, then $$R(X,Y)\omega := \nabla_X\nabla_Y\omega - \nabla_Y\nabla_X\omega -\nabla_{[X,Y]}\omega,$$ $$\langle R(X,Y)\omega,\omega\rangle$$ the Ricci tensor corresponds to $$(\mathrm{Ric}~\omega)(X_1,X_2) = \sum_{s=1}^3\left(R(X_s,X_1)\omega(X_s,X_1) + R(X_s,X_2)\omega(X_s,X_2\right)),$$ where $$\{X_1,X_2,X_3\}$$ is an orthonormal pair.

Hence the Weitzenbock formula reduces to (for $$p=2$$)

$$\Delta = \nabla^*\nabla + \mathrm{Ric}.$$

Therefore, for any parallel $$2$$-form $$\beta$$ one has

$$(\Delta\beta,\beta) =\int_{\Sigma} \langle \Delta\beta,\beta\rangle =\int_{\Sigma} \langle \nabla^*\nabla\beta,\beta\rangle + \int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle = \int_{\Sigma}|\nabla\beta|^2 +\int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle = \int_{\Sigma}\langle\mathrm{Ric}~\beta,\beta\rangle.$$

Since $$(\Delta\beta,\beta) = (d\delta\beta,\beta) + (\delta d\beta) = |\delta\beta|^2 + |d\beta|^2,$$ onde thus concludes that $$u^*(\omega)$$ is closed and co-closed, hence, harmonic, provided if $$\int_{\Sigma}\langle\mathrm{Ric}~u^*(\omega),u^*(\omega)\rangle\leq 0,$$ and this is our second condition.

Note that in general, parallel forms are not harmonic, but in this particular that if $$g$$ is Ricci flat, we are done, since $$\Delta u^*(d\theta_i) = 0,$$ and for $$p=1$$, $$Q^p$$ is also the Ricci tensor, it follows that $$0 = \int_{\Sigma}\langle \mathrm{Ric}(u^*(d\theta_i)),u^*(d\theta_i)\rangle +\int_{\Sigma} |\nabla u^*(\theta_i))|^2 = \int_{\Sigma} |\nabla u^*(\theta_i))|^2,$$ and hence, $$u^*(\theta_i)$$ is parallel, hence, all of our conditions would be satisfied.