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Let M be a d-dimension smooth manifold,we denote $\mathcal{X}(M)$ all the smooth vector field on $M$.

We can show $\mathcal{X}(M)$ is a $C^\infty(M)-$module using the coordinate representation.Correct?

As a module it needs not to have basis,as this excellent solution shows.If it has basis it must be at most d dimension,which also has been proved in this post.using the fact that :

if taking $X_1,X_2,...,X_d$ as the subset of elements in the basis set,such that at point $p$ these d elements span the $T_pM$.Then exist a neiborhood of $p$ denoted it as $U_p$,such that $X_1,X_2,...,X_d$ as basis in the neiborhood of $p$.

I prove this fact as follows:Since $X_1,...,X_d$ are smooth map $M\to TM$,they must be continuous also,since $\det(X_1(p),...,X_d(p)) \ne 0$ if we take some connected neiborhood $U_p$ then by precomposing $\det$ with some continuous map $(X_1,X_2,...,X_d)$ this set of vectors must be invertible at this neiborhood.So this set is in fact a basis in some neiborhood is my interpretation correct?

I guess my interpretation is correct.But there are bit details that confuse me is $\det$ is a map that from $V\times V...\times V \to \Bbb{R}$ but the vector field $X_i :M\to TM$ ,and $TM$ is not a vector space $V$,how to compse them?

Do we need to use the local trivialization map $\Phi :\pi^{-1}(U)\to U\times \Bbb{R}^d$ here?I mean first $(X_1,...,X_d):M\to TM\times TM...\times TM$ which is smooth by maps into product lemma,if we restrict it to some connected neiborboohd (we may shrink it if necessary) into the domain of local trivilization then $(X_1,...,X_d) : U \to \pi^{-1}(U)\times ...\times \pi^{-1}(U)$ (which is smooth by restriction of domain and codomain lemma) now for each $\pi^{-1}(U) \to U \times \Bbb{R}^d\to \Bbb{R}^d$ is smooth,so we map it into vector space $\Bbb{R}^d \times...\times \Bbb{R}^d$

(Further more we may restrict the Ring $C^\infty(M)$ into field of constant function $\Bbb{R}\subset C^\infty(M)$ so \mathcal{X}(M) becomes a vector space over $\Bbb{R}$,which is infinite dimension using the existence of bump function correct?)

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    $\begingroup$ I would just use local coordinates. If $M$ is a $d$-manifold then $M$ embeds in $\mathbb{R}^{2d}$. There is an orthogonal projection onto the tangent space at each point. From this you can conclude that the rank of the module of vector fields over smooth functions is at most $2d$. $\endgroup$ Apr 5, 2021 at 12:01
  • $\begingroup$ @Charlie Frohman thanks ,Can you elaborate more on "There is an orthogonal projection onto the tangent space at each point." ? $\endgroup$
    – yi li
    Apr 5, 2021 at 12:09

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Suppose that $M\leq \mathbb{R}^{2n}$ is a regular submanifold. This means for each $p\in M$ you can identify $T_pM$ with a subspace of $T_p\mathbb{R}^{2n}$. The tangent bundle of $\mathbb{R}^{2n}$ is parallelizable with global frame $\{\frac{\partial}{\partial x^i}\}$. Hence you can understand any tangent vector to any point $p\in \mathbb{R}^{2n}$ as a column vector. Hence the elementary definition of orthogonal projection gives rise to a family of maps, $$ \pi_p:\mathbb{R}^{2n}\rightarrow T_pM .$$ By looking at coefficients in a local parametrization you can see that the vector fields $\pi_p(\vec{e}_i)$ are smooth on $M$. Since at each $p\in M$ there is an open neighborhood $U$ and a choice of smooth functions $a_i$ so that any smooth vector field can be written $$ \sum_i a_i \pi_p(\vec{e}_i)$$ using a partition of unity subordinate to a cover of $M$ made up of these neighborhoods $U$, any smooth vector field is a smooth linear combination of the smooth vector fields $\pi_p(\vec{e}_i)$. Therefore the smooth vector fields on $M$ have rank less than or equal to $2n$ as a module over the smooth functions on $M$. Since by Whitney's hard embedding theorem every smooth $n$-manifold can be realized by a regular submanifold of $\mathbb{R}^{2n}$ we have a global bound on the rank of the module of vector fields on an $n$-manifold.

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