# Extension of Vector Field in the $\mathcal{C}^r$ topology.

Let $$M\subset \mathbb{R}^n$$ be a compact smooth manifold embedded in $$\mathbb{R}^n$$, we define $$\mathfrak{X}(M) := \{X: M \to \mathbb{R}^n;\ X\mbox{ is smooth and }\ X(p) \in T_p M \subset \mathbb{R}^n,\ \forall\ p \in M \}.$$

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

\begin{align*}\|\cdot\|_r : \mathfrak{X}(M)&\to \mathbb{R}\\ X &\to \max_{\substack{i\in\{1,...,n\} \\ j\in \{0,...,r\}}}\left\{\sup_{x \in \varphi^{-1}_i(K_i)}\left\| \text{d}^{j}\left( X\circ\varphi_i \right) \right\|\right\}, \end{align*}

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

My Question: Let $$X \in \mathfrak{X}(M)$$ and $$Y$$ be a smooth vector field on $$M$$ defined just in a compact $$K \subset M$$ such that $$\max_{\substack{i\in\{1,...,n\} \\ j\in \{0,...,r\}}}\left\{\sup_{x \in \varphi^{-1}_i(K_i\cap K)}\left\| \text{d}^{j}\left( X\circ\varphi_i \right) - \text{d}^{j}\left( Y\circ\varphi_i \right) \right\|\right\}<\varepsilon,$$ is it possible extend $$Y$$ to a vector field $$\tilde{Y}$$ such that

1) $$\left.\tilde{Y}\right|_{K} = Y$$,

2) $$\|X-\tilde Y\|_r < A\cdot\varepsilon$$ , 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 $$\|X-\tilde Y\|_r < A\cdot\varepsilon$$ after Moishe Kohan's comment.

## My ideas

First, I extend $$Y$$ by a smooth vector field $$Z$$ $$\in \mathfrak{X}(M)$$, by the continuity of $$Z$$, so there exists a neighborhood $$U$$ of $$K$$, such that $$\max_{\substack{i\in\{1,...,n\} \\ j\in \{0,...,r\}}}\left\{\sup_{x \in \varphi^{-1}_i(K_i\cap U)}\left\| \text{d}^{j}\left( X\circ\varphi_i \right) - \text{d}^{j}\left( Z\circ\varphi_i \right) \right\|\right\}<\varepsilon,$$

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

• 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 $A\epsilon$ where $A$ is some constant depending only on dimension. – Moishe Kohan Mar 28 at 3:38
• 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 $\mathbb{R}^n$ and not manifolds. Do you know how to generalize to manifolds this result? – Matheus Manzatto Mar 28 at 9:13