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

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?

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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 $\epsilon\to 0$.

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  • $\begingroup$ 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: $\endgroup$ – Asaf Shachar Jan 15 '19 at 16:19
  • $\begingroup$ 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,... $\endgroup$ – Asaf Shachar Jan 15 '19 at 16:21
  • $\begingroup$ 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). $\endgroup$ – Asaf Shachar Jan 15 '19 at 16:22
  • $\begingroup$ 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. $\endgroup$ – Asaf Shachar Jan 15 '19 at 16:47

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