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?)
Best Answer
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.