Does the geometric version of Nakayama's lemma hold for smooth manifolds?

Consider the following geometric formulation of Nakayama's lemma.

Proposition. Let $$F$$ be a quasi-coherent sheaf locally of finite type on a scheme $$X$$. Consider the quotient map $$\pi:F_x\to F_x\otimes\Bbbk (x)$$. Given $$s_1,\dots ,s_n\in F_x$$, suppose their image generates $$F_x\otimes \Bbbk (x)$$. Then the $$s_i$$ extend to a neighborhood $$U\subset X$$ of $$x$$ on which they define a surjective arrow $$(\mathcal O _X|_U)^n\overset{(s_1,\dots ,s_n)}{\longrightarrow}F|_U\to \bf 0$$ on $$U$$. When this holds, we say $$s_1,\dots ,s_n$$ generate $$F$$ over $$U$$.

Let $$(M,\mathcal T_M)$$ be a manifold with the sheaf of sections of its tangent bundle. The $$x$$-fiber of $$\mathcal T_M$$ is the vector space of tangents at $$x$$. The $$x$$-stalk is the module of germs at $$x$$ of vector fields. Does "Nakayama" hold for $$(M,\mathcal T_M)$$?

Yes, this is true for any vector bundle $$E$$ on a smooth manifold $$M$$. Let $$s_1,\dots,s_n$$ be sections of $$E$$ in a neighborhood of $$x$$ whose values at $$x$$ span the fiber $$E_x$$. We may assume that the values $$s_1(x),\dots,s_n(x)$$ are also linearly independent (if not, take a maximal linearly independent subset), so $$n$$ is the rank of $$E$$. Now choosing a local trivialization of $$E$$, we can think of $$(s_1,\dots,s_n)$$ as a map $$A:U\to\mathbb{R}^{n\times n}$$ for some open neighborhood $$U$$ of $$x$$. Since $$s_1,\dots,s_n$$ are linearly independent at $$x$$, $$A(x)$$ is an invertible matrix. The set of invertible matrices is open in $$\mathbb{R}^{n\times n}$$, so $$A(y)$$ is invertible for all $$y$$ in some neighborhood $$V$$ of $$x$$.
Now let $$s$$ be any section of $$E$$ over $$V$$, which we think of as a map $$V\to\mathbb{R}^n$$ via our local trivialization. For all $$y\in V$$, $$s(y)=A(y)A(y)^{-1}s(y)$$. The columns of $$A(y)$$ are just $$s_1(y),\dots,s_n(y)$$, so this writes $$s(y)$$ as a linear combination $$\sum f_i(y)s_i(y)$$. Here $$f_i(y)$$ is the $$i$$th coordinate of the vector $$A(y)^{-1}s(y)$$, which is a smooth function of $$y$$ (here we use that matrix inversion is smooth). So, $$s=\sum f_i s_i$$ is a linear combination of the $$s_i$$ with coefficients in $$C^\infty(V)$$. That is, the $$s_i$$ generate the sections of $$E$$ over $$V$$ as a module over $$C^\infty(V)$$.
• Dear @Eric, I have a bit of a follow up question. Is there an example where $s_1(x),\dots ,s_n(x)$ are linearly independent yet $s_1,\dots ,s_n$ are not? – Arrow Dec 12 '18 at 23:58
• Sure. You could have sections that are linearly independent near $x$, but then on some open set away from $x$ they are identically $0$. The sections are then not linearly independent over $C^\infty(M)$ because they are annihilated by a bump function supported on the set where they're $0$. – Eric Wofsey Dec 13 '18 at 0:00
• What if we look at the $s_1,\dots ,s_n$ as germs at $x$ of vector fields? Wouldn't that prevent looking at opens "away" from $x$? – Arrow Dec 13 '18 at 0:06