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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?

Thank you in advance

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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.

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  • $\begingroup$ Thank you very much for your response. It seems that the book I read hasn't developed this machinery until the point of my question. Could you please tell me what do you mean by the wedge of smooth sections? Also, what do you mean by the trivialized form of a section? Just give me a brief answer in order to see if indeed I haven't explored these notions yet or I have to insist more and understand your comment. Thaks again $\endgroup$ – perlman Feb 26 '17 at 2:18
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    $\begingroup$ 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. $\endgroup$ – Kenny Wong Feb 26 '17 at 2:27
  • $\begingroup$ Now everything is clear. Thank you very much for your help! $\endgroup$ – perlman Feb 26 '17 at 2:51

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