Let $M$ be a smooth $n$-manifold with or without boundary. If $(X_1, \dots , X_n)$ is a linearly independent $n$-tuple of smooth vector fields along a closed subset $A \subset M$, then there exists a smooth local frame $(\tilde{X_1}, \dots, \tilde{X_n})$ on some neighborhood of $A$ such that $\tilde{X}_i |_A = X_i$ for $i = 1, \dots, n.$

I think I need to use the following theorem . Let $M$ be a smooth manifold with or without boundary, and let $A \subset M$ be a closed subset. Suppose $X$ is a smooth vector field along $A$. Given any open subset $U$ containing $A$, there exists a smooth global vector field $\tilde{X}$ on $M$ such that $\tilde{X}|A = X$ and supp$ \tilde{X} \subset U$.

Since there is no open set containing $A$ given, I need to construct such an open set somehow. Smooth vector field $X$ along a closed subset $A$ means for each $p \in A$, there is a neighborhood $W_p$ of $p$ in $M$ and a smooth vector field $\tilde{X}$ on $W_p$ that agrees with $X$ on $W_p \cap A$. So I thought about taking $U = \cup_{p \in A} W_p$ and extending each $X_i$ to $U$ using the theorem. But I cannot guarantee that the extended vector fields will still be linearly independent. I have been stuck on this problem for some time. I would greatly appreciate any help.


The key point is that linear independence is an open condition: if you have some collection of vector fields, the set of points at which they are linearly independent is open. Indeed, looking in local coordinates near any point, you can think of your vector fields as functions to $\mathbb{R}^n$, and they are linearly independent at a point iff the $n\times n$ matrix they form has nonzero determinant (here I assume you have $n$ vector fields as in your problem; if you had $k$ for some $k<n$ you would want some $k\times k$ minor of the $k\times n$ matrix to have nonzero determinant). Since the determinant of a matrix is a continuous function of its entries, the set of points where the determinant is nonzero is open.

So, you can just take any extensions of $X_1,\dots,X_n$ to $M$, and the set where they are linearly independent will be an open set containing $A$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.