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How can I show that following: If $F\subseteq TM$ is a smooth distribution then $F$ is vector bundle and the inclusion $F\hookrightarrow TM$ is a morphism of vector bundles?

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Using @Steve 's idea I did a sketch but I don't know if that's all right: Suppose $F$ is rank $k$ distribution. Consider the projection $\pi:F\rightarrow M$, $F_p\mapsto p$, where $F_p\subseteq T_pM$. We will show $(\pi, F, M)$ is a rank $k$ vector bundle:

(i) The fiber $\pi^{-1}(p)=F_p$ is a $k$-vector space for $F$ is a rank $k$ distribution.

(ii) Since $F$ is smooth given $p\in M$ there is an open neighbourhood $U\subseteq M$ containing $p$ such that $$F_q=\sum_{i=1}^k\alpha_iX_i(q),$$ for smooth vector fields (which are smooth sections of $TM$) $$X_1, \ldots, X_k\in \mathfrak{X}(U),$$ and scalars $\alpha_1, \ldots, \alpha_k$. Define, $$\phi:\pi^{-1}(U)\rightarrow U\times \mathbb R^k,$$ setting $$E_q=\sum_{i=1}^k\alpha_iX_i(q)\mapsto (q, x),$$ where $x=(\alpha_1, \ldots, \alpha_k)\in\mathbb R^k$. I guess $\phi$ defined this way will furnish me a diffeomorphism, but I didn't prove that yet, what you think @Steve and @Federica?

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Modulo some notational issues (e.g., the equality involving $F_q$ should have a set on the RHS) this looks ok to me. You will also have to check that the transition maps are linear. Then the fact that the inclusion of $F$ into $TM$ is a vector bundle map is immediate (it is linear on each fiber by definition). – Stephen Apr 22 '13 at 20:39
@Steve you're right, I guess $$\displaystyle F_q=\sum_{i=1}^k\alpha_iX_i(q)$$ is not good because I meant $$F_q=\textrm{span}\{X_1(q), \ldots, X_k(q)\}$$ so I can't fix the scalars $\alpha_i$ and write that equality.. – PtF Apr 22 '13 at 20:50

$F$ being a smooth distribution means by definition (according to most standard references):

first, the fiber $F(m)$ is a $k$-dimensional subspace of the tangent space $TM(m)=T_mM$, for some global integer $k$,

and second, that given a point $m \in M$ there is a neighborhood $U$ of $M$ and smooth sections $f_1,\dots,f_p$ of $TM$ such that the fiber of $F(u)$ over any $u \in U$ is spanned by $f_1(u),\dots,f_p(u)$.

Fix $m \in M$. Choose $f_1,\dots,f_k$ such that $f_1(m),\dots,f_k(m)$ are a basis for the fiber $F(m)$. In coordinates, the condition that $f_1(u),\dots,f_k(u)$ are linearly independent is expressed in terms of a determinant which is continuous in $u$. This means that they are linearly independent on a neighborhood of $m$ (since they are so at $m$) and therefore give a basis of $F(u)$ for each point $u$ in some neighborhood of $m$. Cover $M$ by neighborhoods of this type, and use the corresponding $f_1, \dots, f_k$ to construct a chart on each such neighborhood.

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I'll think about it, thanks.. – PtF Apr 22 '13 at 20:10
Can I write the distribution $F$ as $\displaystyle F=\bigcup_{p\in M}F_p$? If so I can use a procedure very similar for showing the tangent bundle is a vector bundle.. – PtF Apr 22 '13 at 22:37
@PtF, yes, since $F \subseteq TM$ by definition. – Stephen Apr 22 '13 at 23:34

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