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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.

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  • $\begingroup$ 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. $\endgroup$ – Rohil Prasad Apr 2 at 0:20
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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.

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