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}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$

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 $Tomega$ 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:

The form $omega$ is constant.

$f$ 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
$$
|f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
$$
thus
$$
left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
$$
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
$$
begin{split}
& |f^* omega-f_n^* omega|(p)= \
&left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
&left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
end{split}
$$

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:

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

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

edited Jan 15 at 16:29

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

add a comment |

$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$

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

The form $omega$ is constant.

$f$ is continuous.

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

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:

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

edited Jan 15 at 16:29

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

add a comment |

$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$

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

The form $omega$ is constant.

$f$ is continuous.

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

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:

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

edited Jan 15 at 16:29

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$

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

The form $omega$ is constant.

$f$ is continuous.

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

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:

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

dg.differential-geometry ap.analysis-of-pdes sobolev-spaces regularity exterior-algebra

edited Jan 15 at 16:29

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

edited Jan 15 at 16:29

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

edited Jan 15 at 16:29

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

asked Jan 13 at 13:33

Asaf Shachar

2,2691947

add a comment |

1 Answer
1

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This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. 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. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.

edited Jan 13 at 15:45

Martin Sleziak

2,97532028

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

$begingroup$
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 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 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 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 at 16:47

add a comment |

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1 Answer
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edited Jan 13 at 15:45

Martin Sleziak

2,97532028

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

$begingroup$
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 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 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 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 at 16:47

add a comment |

edited Jan 13 at 15:45

Martin Sleziak

2,97532028

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

$begingroup$
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 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 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 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 at 16:47

add a comment |

edited Jan 13 at 15:45

Martin Sleziak

2,97532028

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

edited Jan 13 at 15:45

Martin Sleziak

2,97532028

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

edited Jan 13 at 15:45

Martin Sleziak

2,97532028

edited Jan 13 at 15:45

Martin Sleziak

2,97532028

edited Jan 13 at 15:45

Martin Sleziak

2,97532028

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

answered Jan 13 at 15:40

Piotr Hajlasz

8,65142967

$begingroup$
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 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 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 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 at 16:47

add a comment |

$begingroup$
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 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 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 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 at 16:47

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 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 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 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 at 16:47

add a comment |

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