# Neighbourhood of Linearly Independent Vector Fields $(X_1,\dots,X_n)$

Actually, this is a piece of some argument i need to prove some proposition. I've been thinking this all day so i hope someone can help me with this.

Suppose that we have smooth vector fields $(X_1,\dots,X_n)$ defined on some open subset $U\subset M$. At a point $p \in U$ we know that $(X_1|_p,\dots,X_n|_p)$ are linear independent, hence basis for $T_pM$. Can i find some neighbourhood $V$of $p$ in $U$ such that the $(X_1,\dots,X_n)$ are linear independent for all points in $V$ ?

My idea is to use the determinant argument. That at $p$ the determinant matrix with entries consist of components of $X_i$'s at $p$, that is $det (X_1|_p,\dots,X_n|_p) = det (X^i_j(p))$ is not zero, so we have an open neighbourhood of $p$ such that $X_1,\dots,X_n$ are l.i there.

Is this correct ? Any other idea to find such neighbourhood ? Thank you.

$\textbf{Update : }$

After a while, i think this is true. This also can be proved using contradiction too. Pretty much by the same route. Only this time we assume on the contrary that, there is no such neighbourhood for $p$ and then deriving a contradiction.

$\textbf{Proof }$ : Suppose $M$ is an $n$-dimensional smooth manifold and $(X_1,\dots,X_n)$ are smooth vector fields defined on some open subset $U\subseteq M$ such that for a point $p \in U$, $X_1(p),\dots,X_n(p)$ are linearly independent. We claim that there are no neighbourhood $V$ of $p$ such that for any point $x \in V$, then vectors $X_1(x),\dots,X_n(x)$ are linearly independent. Stated it differently, every neighbourhood of $p$ contain a point where the vectors $X_i$ are linearly dependent.

This will lead to contradiction as follows : Choose a smooth charts $(U',x^i)$ contain $p$. By shrinking $U'$, assume that $U'\subseteq U$. In this chart, the value of vector fields $X_i$ at any point $x \in U'$ is $$X_i(x) = X_i^j(x) \frac{\partial}{\partial x^j}\bigg|_x.$$

Define a map from $m : U' \to M(n\times n,\mathbb{R} )$ defined by $$m : x \mapsto \begin{pmatrix} X_1^1(x) & \cdots & X_n^1(x) \\ \vdots & \ddots & \vdots \\ X_1^n(x) & \cdots & X_n^n(x) \end{pmatrix} \in M(n\times n,\mathbb{R} ).$$ This is a smooth map since the entries are smooth functions on $U'$. By composing this with determinant map $\text{det} : M(n\times n, \mathbb{R}) \to \mathbb{R}$, we have a smooth function $f = \text{det} \circ m : U' \to \mathbb{R}$. This function is also smooth and at $p \in U'$, $f(p) \neq 0$ since $X_1(p),\dots,X_n(p)$ are linearly independent vectors.

Now, by assumption, any neighbourhood $V_1 \subseteq U'$ of $p$ contain a point $p_1$ such that $X_1(p_1),\cdots,X_n(p_1)$ are linearly dependent. Therefore $f(p_1) = 0$. By doing this repeatly, we have a sequence $(p_k)_{k=1}^{\infty}$ in $U'$ converging to $p$ such that $f(p_k) =0$ for all $k$. Since $f$ is smooth (hence continous), then as $p_k \to p$, then $f(p_k) \to f(p)$. But this is not happen since $f(p_k)= 0$ is a constant sequence and $f(p) \neq 0$. Therefore $f$ is not continous. Contradiction.

Your idea is correct. Just use local coordinates $x^1, \dots, x^n$ on $U$. If $\partial_1, \dots, \partial_n$ are the corresponding coordinate vector fields, then each $X_i$ can be written as a linear combination of the coordinate vector fields, $$X_i = \sum_{j=1}^n A_i^j\partial_j,$$ where each $A_i^j$ is a smooth function on $U$ and therefore the function $f = \det [A_i^j]$ is also a smooth function on $U$. Since $X_1(p), \cdots, X_n(p)$ are linearly independent, $\det [A_i^j(p)] \ne 0$. By continuity, $f^{-1}(\mathbb R\backslash\{0\})$ is open (and contains $p$).

This is too long for a comment, but disclaimer: this is not my area of expertise.

This seems mostly correct. And the idea is good. However, slight issue with the current argument as I see it, I'm not familiar with a definition of determinant for a collection of vectors of an abstract vector space. I'm familiar with the determinant of a linear transformation, and this definition can be used to give a definition of the determinant of a collection of vectors once a basis has been chosen. Namely if $e_i$ is the distinguished basis and $v_i$ is the collection of vectors, define $\det(v_i) = \det(T)$, where $T$ is the linear map $e_i\mapsto v_i$. Thus I think to adapt your argument, you need to reduce to a smaller neighborhood of $p$ diffeomorphic to an open subset of $\Bbb{R}^n$, and then take determinants with respect to the standard bases of the tangent spaces.

Anyway, please let me know if anyone spots anything wrong with this, but it seems like this fixes the problem I spotted anyway.

Actually, you don't really need to reduce to a smaller neighborhood, just cover $U$ with open sets diffeomorphic to an open subset of $\Bbb{R}^n$ and locally use the standard bases to compute the determinants, since the transition between different bases has nonzero determinants, whether or not $\det(X_i)$ is 0 will be the same regardless of the basis used, so everything will agree on the overlaps of the patches.

• Another adaptation, which is more coordinate-free: $v_1 \wedge \cdots \wedge v_n \in \bigwedge^n V$ is nonzero at $p$, so it's nonzero on some neighborhood of $p$. (And this one should work even if $\dim V > n$, I think.) Jan 6, 2018 at 0:26

The problem is to define $\det$ on a set of vectors, which is impossible unless you have a choice of standard basis like $\Bbb R^n$. But we could use a chart to transfer the problem to $\Bbb R^n$. Consider a chart $(U',\varphi)$ on $M$ with $p\in U'\subseteq U$. Note that $\varphi$ is a diffeomorphism from $U'$ to an open set $\varphi(U')$ of $\Bbb R^n$, so its differential at $p$, $d\varphi_p$, is a vector space isomorphism. Hence, the set of tangent vectors $\{d\varphi_p(X_1|_p),\dots,d\varphi_p(X_n|_p)\}$ is linearly independent.

In general, the vectors $d\varphi_q(X_i|_q)$ are simply elements in the tangent space $T_{\varphi(q)}\varphi(U')$, so we can regard them as column vectors in $\Bbb R^n$. Here, we used that every open set in $\Bbb R^n$ has trivial tangent bundle.

Now consider a matrix-valued function $f$ on $U'$ defined by that the $i$-th column of $f(q)$ is $d\varphi_q(X_i|_q)$. Of course $f$ is smooth and $f$ outputs square matrices. Now look at the composite function $\det\circ\ f$, which is continuous. By that $\{d\varphi_p(X_1|_p),\dots,d\varphi_p(X_n|_p)\}$ is linearly independent, $\det\circ\ f(p)$ is not zero. By continuity of $\det\circ\ f$, $\det\circ\ f$ is not zero on a neighbourhood $V$ of $p$, i.e. the set of vectors $\{d\varphi_q(X_1|_q),\dots,d\varphi_q(X_n|_q)\}$ are linearly independent for all $q\in V$, and so is $\{X_1|_q,\dots,X_n|_q\}$ linearly independent for all $q\in V$.

You may notice that we only need continuity of the vector fields, not smoothness.