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15

2

$begingroup$

Let $Msubset mathbbR^n$ be a compact smooth manifold embedded in $mathbbR^n$, we define $$mathfrakX(M) := X: M to mathbbR^n; Xmbox is smooth and X(p) in T_p M subset mathbbR^n, forall p in M .$$

Choosing an atlas $(varphi_i,U_i)_i=1^n$, and compacts $K_i subset U_i$, such that $$bigcup_i=1^n K_i = M,$$ we define the $|cdot |_r$ norm as

beginalign*|cdot|_r : mathfrakX(M)&to mathbbR\
X &to max_substackiin1,...,n \ jin 0,...,rleftsup_x in varphi^-1_i(K_i)left,
endalign*

then we named $mathfrakX^r(M)$ as the complete Banach space $(mathfrakX(M),|cdot|_r)$ (it is possible to prove that the topology of $mathfrakX^r(M)$ does not depend on the selected atlas).

My Question: Let $X in mathfrakX(M)$ and $Y$ be a smooth vector field on $M$ defined just in a compact $K subset M$ such that
$$max_substackiin1,...,n \ jin 0,...,rleft textd^jleft( Xcircvarphi_i right) - textd^jleft( Ycircvarphi_i right) right<varepsilon,$$
is it possible extend $Y$ to a vector field $tildeY$ such that

1) $left.tildeYright|_K = Y$,

2) $|X-tilde Y|_r < Acdotvarepsilon$ , where $A $ is a constant that depends only on the manifold $K$ ?

The compact $K$ is a connected submanifold with boundary of $M$, such that $dim K = dim M$.

Edit: I changed $|X-tilde Y|_r < varepsilon$ to $|X-tilde Y|_r < Acdotvarepsilon$ after Moishe Kohan's comment.

---

My ideas

First, I extend $Y$ by a smooth vector field $Z$ $in mathfrakX(M)$, by the continuity of $Z$, so there exists a neighborhood $U$ of $K$, such that
$$max_substackiin1,...,n \ jin 0,...,rleft textd^jleft( Xcircvarphi_i right) - textd^jleft( Zcircvarphi_i right) right<varepsilon,$$

and then choosing a partition of unity $ phi_1, phi_2$ subordinate to the cover $U,Msetminus K$ we can define
$$tildeY = phi_1 Z + phi_2 X, $$
however I could not guarantee that $|X - tildeY|_r < varepsilon$, because I can not control de derivatives of $phi_1$ and $phi_2$. Does anyone know how should I proceed?

real-analysis analysis differential-geometry manifolds differential-topology

share|cite|improve this question

edited Apr 1 at 15:24

Matheus Manzatto

asked Mar 23 at 23:45

Matheus ManzattoMatheus Manzatto

1,2991626

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a form of Whitney extension theorem/problem. Take a look here: annals.math.princeton.edu/wp-content/uploads/… I suspect you cannot keep the same $epsilon$: Fefferman's result (and some follow up papers) show that you can find an extension with an error $Aepsilon$ where $A$ is some constant depending only on dimension.
  $endgroup$
  – Moishe Kohan
  Mar 28 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thx for the reference. Once the same $A$ holds for every function. It would be enough to solve my problem. The only complication I am seeing right know it is the fact that this result is for $mathbbR^n $ and not manifolds. Do you know how to generalize to manifolds this result?
  $endgroup$
  – Matheus Manzatto
  Mar 28 at 9:13
  

  
  
  
  

add a comment | 

15

2

$begingroup$

Let $Msubset mathbbR^n$ be a compact smooth manifold embedded in $mathbbR^n$, we define $$mathfrakX(M) := X: M to mathbbR^n; Xmbox is smooth and X(p) in T_p M subset mathbbR^n, forall p in M .$$

Choosing an atlas $(varphi_i,U_i)_i=1^n$, and compacts $K_i subset U_i$, such that $$bigcup_i=1^n K_i = M,$$ we define the $|cdot |_r$ norm as

beginalign*|cdot|_r : mathfrakX(M)&to mathbbR\
X &to max_substackiin1,...,n \ jin 0,...,rleftsup_x in varphi^-1_i(K_i)left,
endalign*

then we named $mathfrakX^r(M)$ as the complete Banach space $(mathfrakX(M),|cdot|_r)$ (it is possible to prove that the topology of $mathfrakX^r(M)$ does not depend on the selected atlas).

My Question: Let $X in mathfrakX(M)$ and $Y$ be a smooth vector field on $M$ defined just in a compact $K subset M$ such that
$$max_substackiin1,...,n \ jin 0,...,rleft textd^jleft( Xcircvarphi_i right) - textd^jleft( Ycircvarphi_i right) right<varepsilon,$$
is it possible extend $Y$ to a vector field $tildeY$ such that

1) $left.tildeYright|_K = Y$,

2) $|X-tilde Y|_r < Acdotvarepsilon$ , where $A $ is a constant that depends only on the manifold $K$ ?

The compact $K$ is a connected submanifold with boundary of $M$, such that $dim K = dim M$.

Edit: I changed $|X-tilde Y|_r < varepsilon$ to $|X-tilde Y|_r < Acdotvarepsilon$ after Moishe Kohan's comment.

---

My ideas

First, I extend $Y$ by a smooth vector field $Z$ $in mathfrakX(M)$, by the continuity of $Z$, so there exists a neighborhood $U$ of $K$, such that
$$max_substackiin1,...,n \ jin 0,...,rleft textd^jleft( Xcircvarphi_i right) - textd^jleft( Zcircvarphi_i right) right<varepsilon,$$

and then choosing a partition of unity $ phi_1, phi_2$ subordinate to the cover $U,Msetminus K$ we can define
$$tildeY = phi_1 Z + phi_2 X, $$
however I could not guarantee that $|X - tildeY|_r < varepsilon$, because I can not control de derivatives of $phi_1$ and $phi_2$. Does anyone know how should I proceed?

real-analysis analysis differential-geometry manifolds differential-topology

share|cite|improve this question

edited Apr 1 at 15:24

Matheus Manzatto

asked Mar 23 at 23:45

Matheus ManzattoMatheus Manzatto

1,2991626

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a form of Whitney extension theorem/problem. Take a look here: annals.math.princeton.edu/wp-content/uploads/… I suspect you cannot keep the same $epsilon$: Fefferman's result (and some follow up papers) show that you can find an extension with an error $Aepsilon$ where $A$ is some constant depending only on dimension.
  $endgroup$
  – Moishe Kohan
  Mar 28 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thx for the reference. Once the same $A$ holds for every function. It would be enough to solve my problem. The only complication I am seeing right know it is the fact that this result is for $mathbbR^n $ and not manifolds. Do you know how to generalize to manifolds this result?
  $endgroup$
  – Matheus Manzatto
  Mar 28 at 9:13
  

  
  
  
  

add a comment | 

$begingroup$

Let $Msubset mathbbR^n$ be a compact smooth manifold embedded in $mathbbR^n$, we define $$mathfrakX(M) := X: M to mathbbR^n; Xmbox is smooth and X(p) in T_p M subset mathbbR^n, forall p in M .$$

Choosing an atlas $(varphi_i,U_i)_i=1^n$, and compacts $K_i subset U_i$, such that $$bigcup_i=1^n K_i = M,$$ we define the $|cdot |_r$ norm as

beginalign*|cdot|_r : mathfrakX(M)&to mathbbR\
X &to max_substackiin1,...,n \ jin 0,...,rleftsup_x in varphi^-1_i(K_i)left,
endalign*

then we named $mathfrakX^r(M)$ as the complete Banach space $(mathfrakX(M),|cdot|_r)$ (it is possible to prove that the topology of $mathfrakX^r(M)$ does not depend on the selected atlas).

My Question: Let $X in mathfrakX(M)$ and $Y$ be a smooth vector field on $M$ defined just in a compact $K subset M$ such that
$$max_substackiin1,...,n \ jin 0,...,rleft textd^jleft( Xcircvarphi_i right) - textd^jleft( Ycircvarphi_i right) right<varepsilon,$$
is it possible extend $Y$ to a vector field $tildeY$ such that

1) $left.tildeYright|_K = Y$,

2) $|X-tilde Y|_r < Acdotvarepsilon$ , where $A $ is a constant that depends only on the manifold $K$ ?

The compact $K$ is a connected submanifold with boundary of $M$, such that $dim K = dim M$.

Edit: I changed $|X-tilde Y|_r < varepsilon$ to $|X-tilde Y|_r < Acdotvarepsilon$ after Moishe Kohan's comment.

---

My ideas

First, I extend $Y$ by a smooth vector field $Z$ $in mathfrakX(M)$, by the continuity of $Z$, so there exists a neighborhood $U$ of $K$, such that
$$max_substackiin1,...,n \ jin 0,...,rleft textd^jleft( Xcircvarphi_i right) - textd^jleft( Zcircvarphi_i right) right<varepsilon,$$

and then choosing a partition of unity $ phi_1, phi_2$ subordinate to the cover $U,Msetminus K$ we can define
$$tildeY = phi_1 Z + phi_2 X, $$
however I could not guarantee that $|X - tildeY|_r < varepsilon$, because I can not control de derivatives of $phi_1$ and $phi_2$. Does anyone know how should I proceed?

real-analysis analysis differential-geometry manifolds differential-topology

share|cite|improve this question

edited Apr 1 at 15:24

Matheus Manzatto

asked Mar 23 at 23:45

Matheus ManzattoMatheus Manzatto

1,2991626

$endgroup$

share|cite|improve this question

edited Apr 1 at 15:24

Matheus Manzatto

asked Mar 23 at 23:45

Matheus ManzattoMatheus Manzatto

1,2991626

share|cite|improve this question

edited Apr 1 at 15:24

Matheus Manzatto

asked Mar 23 at 23:45

Matheus ManzattoMatheus Manzatto

1,2991626

asked Mar 23 at 23:45

Matheus ManzattoMatheus Manzatto

1,2991626

asked Mar 23 at 23:45

Matheus ManzattoMatheus Manzatto

1,2991626