In the book vector bundle and K-theory of Allan Hatcher, proposition $1.3$ state that

If $p : E\to B$ is a vector bundle over a paracompact base $B$ and $E_{0} \subset E$ is a vector subbundle, then there is a vector subbundle $E_{0}^{\perp} \subset E$ such that $E_{0} \oplus E_{0}^{\perp} \cong E$

The basic idea is whenever $B$ is paracompact we can give it an inner product so it's natural to let $E_{0}^{\perp}$ be the subspace of $E$ which in each fiber consists of all vector orthogonal to vectors in $E_{0}$. If one can proves $E_{0}^{\perp}$ satisfies the local triviality condition then $E \cong E_{0} \oplus E_{0}^{\perp}$ via canonical map $(v,w) \to v + w$. Firstly, he assumed $E$ to be trivial and near each point $b_0 \in B$ ( a neighborhood ), the subbundle has $m$ independent local sections $s_1,s_2,.., s_m$ ( $m$ is dimension of bundle $E_{0} \to B$ ). Then he enlarged this $m$ sections to $n$ local sections of bundle $E \to B$ by choosing $s_{m+1},...s_n$ in each fiber $p^{-1}(b_0)$, then taking the same vectors for all nearby fibers and hence if $s_1,..s_n$ are independent at $b_0$, it will remain to be independent for all $b$ near $b_0$

I want to ask how to can taking the same vectors... and how it will be independent for nearby $b$. Maybe he explained so vaguely and brief?


1 Answer 1


Since $E$ is locally trivial, $E|_U \cong V \times U$ for some neigborhood $U$ of $b_0$ and some vector space $V$. This same congruence maps $E_0$ to subspaces $V_b$ of $V$ for each $b \in U$, which vary continuously with $b$. He picks sections that form a basis for each $V_b$ (potentially shrinking $U$ to a smaller neighborhood of $b_0$).

At $b_0$, he completes the basis for $V_{b_0}$ to full basis of $V$. The additional vectors can be considered constant sections of $V \times U$. Since the full set of sections give linearly independent vectors at $b_0$ and vary continuously, there must be some neighborhood where they remain linearly independent as you move away from $b_0$.

So when he is talking about keeping the same vectors, he means in the trivialization $V \times U$, not the original vector bundle $E$.


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