# Linearly independence and Smooth Local Section

Let $\pi:E\rightarrow M$ a smooth vector bundle, $U$ a neighborhood of $p$ in $M$ and $\sigma_i:U\rightarrow E, i=1,\dots n$ smooth local sections of $E$. If $\{ \sigma_1(p), \dots, \sigma_n(p)\}$ are linearly independent (as elements of the vector space $E_p$), then by continuity they remain linearly independent in some neighborhood of $p$. Can someone please explain to me why this is true?

I assume that the context (vector bundle, sections etc) doesn't play any role, but I am not sure, so I put it in the context I read it.

My attempt was the following: Define $f_{(a_1,\dots, a_n)}(q)=a_1\sigma_1(q)+\dots+a_n\sigma_n(q)$ on $U$. For a fixed $(a_1,\dots, a_n)\neq 0$, since $f_{(a_1,\dots, a_n)}(p)\neq 0$, we have that there is a neighborhood of $p$ such that $f_{(a_1,\dots, a_n)}(q)\neq 0$ in this neighborhood. But there are infinetely many n-tuples $(a_1,\dots, a_n)\neq 0$ and if I take then intersection of all such neighborhoods there is no guarantee that I will end up with an open neighborhood.

Any ideas?

Think about $\sigma = \sigma_1 \wedge \dots \wedge \sigma_n$ as a smooth section of $\wedge^n E$. Saying that $\sigma_1(p), \dots, \sigma_n(p)$ are linearly independent is the same as saying that $\sigma(p)$ is non-vanishing. By continuity arguments, the set on which $\sigma$ is non-vanishing is open.
Equivalently, write the trivialised forms of $\sigma_1 \dots \sigma_n$ as a big $r \times n$ matrix of smooth functions, where $r$ is the rank of the vector bundle. Saying that $\sigma_1(p) \dots \sigma_n(p)$ are linearly independent is the same as saying that the determinant of at least one $n \times n$ submatrix within your big matrix is non-vanishing at $p$. Then use continuity of these determinants.
• Ignore the first paragraph if you are unfamiliar with exterior products because the second paragraph is totally equivalent. If $E$ is a rank $r$ bundle and if $U$ is a trivialising neighbourhood, then over $U$, $E$ looks like $U \times \mathbb R^r$. So over $U$, you can think of a section as a smooth map from $U$ to $\mathbb R^r$, i.e. as a column vector of $r$ smooth functions. This column vector is what I mean by the trivialised form of the section. – Kenny Wong Feb 26 '17 at 2:27