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

$$\newcommand{\R}{\mathbb R}$$ $$\newcommand{\N}{\mathbb N}$$ $$\newcommand{\de}{\delta}$$ $$\newcommand{\sig}{\sigma}$$ $$\newcommand{\Average}{\left\langle#1\right\rangle}$$ $$\newcommand{\IP}{\Average{#1,#2}}$$

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: Jan 15, 2019 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,... Jan 15, 2019 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). Jan 15, 2019 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. Jan 15, 2019 at 16:47