# Do non-continuous Sobolev maps pull back closed forms to weakly closed forms?


Let $$n,d \in \mathbb{N}$$, and let $$\Omega \subseteq \R^n$$ be open. Let $$f \in W^{1,k}(\Omega,\R^d)$$. Let $$\omega \in \Omega^k(\R^d)$$ be a smooth closed $$k$$-form, such that $$\omega$$ and its derivative $$T\omega$$ are both uniformly bounded globally; Is it true that $$f^*\omega$$ is weakly closed? i.e. does $$\int_{\Omega} \IP{f^* \omega}{\de \sig}=0$$ hold for every compactly-supported $$k+1$$-form $$\sigma \in \Omega^{k+1}(\Omega)$$?

I have read that this should be true, but I am only able to show this in two special cases:

1. The form $$\omega$$ is constant.
2. $$f$$ is continuous.

Does this hold for non-continuous Sobolev maps in general?

Here is the problem as I see it: We approximate $$f$$ via smooth functions; suppose that $$f_n \in C^{\infty}(\Omega,\R^d)$$ satisfy $$f_n \to f$$ in $$W^{1,k}(\Omega,\R^d)$$. Then $$\int_{\Omega} \IP{f^* \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{ f_n^* \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{d f_n^* \omega}{ \sig}=0.$$ Now, we need to justify the passage to the limit $$\int_{\Omega} \IP{f^* \omega}{\de \sig}=\lim_{n \to \infty} \int_{\Omega} \IP{ f_n^* \omega}{\de \sig}$$.

If $$\omega$$ is constant, that is $$\omega_q= \alpha$$ independently of $$q \in \R^d$$, where $$\alpha$$ is a fixed element in $$\bigwedge^k (\R^d)^*$$, then we have $$|f^* \omega-f_n^* \omega| \le |\alpha| \, \left|\bigwedge^{k} df-\bigwedge^{k} df_n\right|_{op},$$ thus $$\left|\int_{\Omega} \IP{f^* \omega}{\de \sig}- \IP{ f_n^* \omega}{\de \sig}\right| \le |\alpha| \|\de \sig\|_{\sup} \int_{\Omega} \left|\bigwedge^{k} df-\bigwedge^{k} df_n\right|.$$ and The RHS tends to zero since Sobolev approximation lifts to exterior powers.

When $$\omega$$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is $$\begin{split} & |f^* \omega-f_n^* \omega|(p)= \\ &\left|\omega_{f(p)} \circ \bigwedge^{k} df_p -\omega_{f_n(p)} \circ \bigwedge^{k} (df_n)_p \right| \le \\ &\left|\big(\omega_{f(p)}-\omega_{f_n(p)}) \circ \bigwedge^{k} df_p +\omega_{f_n(p)} \circ \big( \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p) \right| \le \\ &|\omega_{f(p)}-\omega_{f_n(p)}| \, \, \cdot \, \, \left| \bigwedge^{k} df_p\right| +|\omega_{f_n(p)} | \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \end{split}$$

So, we we need to estimate $$\omega_{f(p)}-\omega_{f_n(p)}$$. When $$f$$ is continuous, we can take $$f_n$$ which converges uniformly to $$f$$, and thus we can continue the estimate:

$$\begin{split} &|\omega_{f(p)}-\omega_{f_n(p)}| \, \, \cdot \, \, \left| \bigwedge^{k} df_p\right| +|\omega_{f_n(p)} | \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \le \\ &|T\omega|_{sup} \, \cdot \, |f(p)-f_n(p)| \, \cdot \, \left| \bigwedge^{k} df_p\right| +|\omega |_{sup} \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \le \\ &|T\omega|_{sup} \, \cdot \, |f-f_n|_{sup} \, \cdot \, \left| \bigwedge^{k} df_p\right| +|\omega |_{sup} \, \, \cdot \, \, \left| \bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p \right| \stackrel{L^1}{\to} 0, %&|\alpha \circ \brk{\bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p} | \le |\alpha| \cdot |\bigwedge^{k} df_p-\bigwedge^{k} (df_n)_p|_{op}. \end{split}$$

So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?

This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf which I state below. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal.

Theorem. Let $$\mathcal{M}$$ be a smooth, $$k$$-dimensional oriented manifold with or without boundary.

• If $$f\in W^{1,1}_{\rm loc}(\mathcal{M},\mathbb{R}^m)$$, then $$f^\ast (\omega \wedge \eta) = f^\ast\omega \wedge f^\ast\eta$$ holds pointwise a.e.
• If $$f\in W^{1,p}_{\rm loc}(\mathcal{M},\mathbb{R}^m)$$, $$p\geq\ell+1$$, $$0\leq\ell\leq k-1$$, and $$\eta\in C^\infty(\bigwedge\nolimits^\ell\mathbb{R}^m)\cap W^{1,\infty}$$ (i.e. $$\eta$$ and $$|\nabla \eta|$$ are bounded), then $$d (f^\ast\eta) = f^\ast (d\eta)$$ holds in the weak sense, i.e. $$\int_{\mathcal{M}} f^*\eta\wedge d\varphi = (-1)^{\ell+1}\int_{\mathcal M} f^*(d\eta)\wedge \varphi$$ for all $$\varphi\in C_0^\infty(\bigwedge\nolimits^{k-\ell-1}\mathcal M)$$.
• If $$\eta\in W^{1,p}_{\rm loc}(\bigwedge\nolimits^{\ell_1}\mathcal M)$$, $$\omega\in W^{1,p}_{\rm loc}(\bigwedge\nolimits^{\ell_2}\mathcal M)$$, $$\ell_1+\ell_2\leq k-2$$, $$p\geq 2$$, then $$d(\eta\wedge d\omega)=d\eta\wedge d\omega$$ weakly in the sense that $$\int_{\mathcal{M}} \eta\wedge d\omega\wedge d\varphi = (-1)^{\ell_1+\ell_2}\int_{\mathcal{M}} d\eta\wedge d\omega\wedge\varphi$$ for all $$\varphi\in C_0^\infty(\bigwedge\nolimits^{k-\ell_1-\ell_2-2}\mathcal M)$$.

The proof is very easy. We simply approximate $$f$$ by smooth mappings $$f_\epsilon$$ (convolution type approximation). Then the corresponding formulas are clearly true with $$f$$ replaced by $$f_\epsilon$$ and we pass to the limit as $$\epsilon\to 0$$.

• Thanks. This looks similar to my question; Our week formulations of $df^*\omega$ are a bit different though-I used duality with the adjoint of the exterior derivative (which depends on the Riemannian structure, in general) while you used a more metric-free approach (which indeed seems more natural). I need to convince myself that one can move between these two pictures in such a weak setting. However, my main issue of concern is different: – Asaf Shachar Jan 15 '19 at 16:19
• As I described in the question, the problem lies within the "approximation step" (in the question it is phrased as $n \to \infty$, in your formulation you take $\epsilon \to 0$). As I see it, since the "point of evaluation" of the pullback form "moves with the function used to pull back" , we need uniform convergence of the mollifiers to the limiting map. (or so it seems to me). You can see how I elaborated on this point in my question. I don't see how your argument (which is essentially identical to mine) extends above the continuous setting,... – Asaf Shachar Jan 15 '19 at 16:21
• if the form that being pulled back is not constant. To conclude, I did not understand if you think that continuity is truly needed here or not. Thanks again for all your help. (By the way, I also assume that $\omega$ and its derivative are uniformly bounded globally. When the Sobolev map is assumed to be continuous, this is not need, since the statement can be reduced to the local case). – Asaf Shachar Jan 15 '19 at 16:22
• Finally, note that it suffices to require here $f \in W^{1,k}$ (instead of $f \in W^{1,k+1}$), since we are only discussing pull-backs of closed forms, and not arbitrary commutation of the exterior derivative and pull-backs. This is note crucial here, but turns out to be important in some applications. – Asaf Shachar Jan 15 '19 at 16:47