Let $E \to M $ be a vector bundle of rank $k$. Suppose that for each $p \in M $ we are given a subspace $E'_p$ of $E_p$ and consider the set $\displaystyle E' = \bigcup_{p \in M} E'_p $.

Show that $E'$ is the total space of a rank $l$ subbundle if and only if for each $p\in M$ there is an open set $U$ of $p$ on which smooth sections $\sigma_1, \ldots, \sigma_l$ are defined such that for each $q \in M$ the set $\{\sigma_1(q), \ldots, \sigma_l(q)\}$ is a basis of the subspace $E'_q$.

Attempt of proof:

($\implies$) Suppose $E'$ is the total space of a rank $l$ subbundle then for every $p \in M$ there exists a VB-chart such that

$$\phi (\pi^{-1}(U) \,\,\cap\,\,E' ) = U \times V' \subseteq U \times V$$

This rises a chart of the triple $(E', \pi|_{E'}, M, V')$ if we define the $\phi'$ as $\phi|_{\pi^{-1}_{E'}(U)}$ (the restriction of $\phi$ to $\pi^{-1}_{E'}(U)$), where $\pi^{-1}(U) \cap E' = \pi^{-1}|_{E'}(U)$ and such that the following diagram commutes

$$\require {AMScd} \begin{CD}\pi^{-1}(U) \cap E' = \pi^{-1}|_{E'}(U) @>\phi'>> U \times V' \subseteq U \times V\\@V \pi|_{E'} VV @VVpr_1V\\U @>>id >U\end{CD}$$

Let $\{e_1, \ldots, e_l\}$ be a basis of $V'$. Since $\phi$ is a diffeomorphism we maydefine the sections of $E'$ as

$$\sigma_j (q) = \phi^{-1} (q, e_j) \, \, , j = 1, \ldots, l , \forall q \in U $$

Now the inclusion $\iota_j : U \to U \times V' $, $\iota_j (x) = (x, e_j)$ is a section of $U \times V'$ since $pr_1 \circ \iota_j = id_U$ and as $\phi' $ is a bundle isomorphism then $\sigma_j$ are sections of $E'$. Since the vectors $(q, e_j)$ form a basis of $\{q\}\times V'$ and $\phi'$ is an isomorphism then $\{\sigma_1(q), \ldots, \sigma_l(q)\}$ is a basis of $E'_q$, ( notice that $\sigma_j(q) \in E'_q, \forall j$) for each $q$ as wanted.

($\Longleftarrow$) Suppose there exists an open neighborhood $U$ of $p$ (taken arbitrarily) such that for each $q \in U$ the set $\{\sigma_1(q), \ldots, \sigma_l(q)\}$ is a basis of the subspace $E'_q$.

Define $\phi' : U \times V' \to E' \cap \pi^{-1}(U)$ by $\phi' (q, c_1, \ldots, c_l) = \displaystyle \sum_{j=1}^l c^j\sigma_j(q)$. As $\sigma_j$ are smooth then $\phi$ is smooth.

We have that $\pi(\phi'(q,c)) = q = pr_1 (q,c)$ and the restriction of $\phi'$ to $\{q\} \times V'$ is linear. Since $\sigma_1(q), \ldots, \sigma_l(q)$ are linearly independent the homomorphism $\phi'$ is injective and thus an isomorphism on each fiber.

We conclude that $\phi$ is a bundle isomorphism and $$\phi'^{-1}(E' \cap \pi^{-1}(U)) = U \times V'$$

and therefore $E'$ is a total space of a rank $l$ subbundle.

  • I would like to check if this is correct.

  • Do I have to show that $E'$ is a submanifold of $E$?


The proof basically looks fine to me. However, in the second part, you have not really shown that $E'$ is a smooth subbundle of $E$ but only that $E'$ is a vector bundle of rank $\ell$.

To prove that $E'$ is a subbundle, you should complete the sections $\sigma_1,\dots,\sigma_\ell$ to a local frame for $E$. To do this, just use a vector bundle chart $\psi$ for $E$ and for some point $x$, complete $\psi(\sigma_1(x)),\dots,\psi(\sigma_\ell(x))$ to a basis of $V$, say using vectors $v_{\ell+1},\dots, v_n$. Then for $i=\ell+1,\dots,n$ define $\sigma_i(y)$ as $\psi^{-1}(y,v_i)$ on the domain of $\iota$. Then you obtain local smooth sections $\sigma_1,\dots,\sigma_n$ whose values at $x$ form a basis for the fiber $E_x$. Hence there is an open neighbourhood $U$ of $x$ such that the values in each $y\in U$ form a basis for $E_y$. Then as in your proof you can construct a vector bundle chart for $E$ on $U$ in which $E'$ corresponds to a linear subspace.

  • $\begingroup$ I see, so I just need to complete the basis first and then carry on with my proof? $\endgroup$ – Aaron Maroja Jun 17 '16 at 13:17
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Andreas Cap Jun 17 '16 at 14:54
  • $\begingroup$ This $\psi$ composition seems to be incorrect. I guess it should be $(pr_2 \circ \psi) (\sigma_i (x))$. $\endgroup$ – Aaron Maroja Jun 17 '16 at 18:26
  • $\begingroup$ Yes you are right, it should be the second component of $\psi$. $\endgroup$ – Andreas Cap Jun 18 '16 at 10:43
  • $\begingroup$ I've completed the basis of $E'_q$ to $E_q$ (dimension $k$), already using smooth section (this can be done locally) then defined $\psi$ as $\phi^{-1}$ where $$\phi (q,c) = \sum_{j=1}^k c^j \sigma_j(q)$$ this seems to solve the question. $\endgroup$ – Aaron Maroja Jun 18 '16 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.