If we have an exact sequence of finite-dimensional vector spaces $$0\rightarrow E'\rightarrow E\rightarrow E''\rightarrow0$$ then an orientation of any two induces an orientation of the third. I have just read that apparently this follows as well in the case of vector bundles $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$ where A is an $n$-plane bundle, C is an $m$-plane bundle, and B is an $(n+m)$-plane bundle. I think I understand why this is true given the basic definition of an orientation (in terms of the orientation of each fiber and local compatibility), but from trying to read Milnor-Stasheff's book on characteristic classes, it seems that for the oriented n-plane bundle A, one obtains a unique generator for $n^{th}$ cohomology of the pair $(A,A-\{0\})$ that restricts to the chosen orientation of each fiber through the Thom isomorphism. Can anyone explain what is happening with the $n,m,$ and $n+m$-dimensional cohomology classes of $(A,A-\{0\})$, $(B,B-\{0\})$, and $(C,C-\{0\})$ respectively?


1 Answer 1


A cleaner way of thinking about an orientation of an $n$-dimensional vector bundle $V$ for the purposes of doing this is that it's a trivialization of the top exterior power $\Lambda^{\dim V} V$. This makes the argument for vector bundles almost identical to the one for vector spaces: show that if

$$0 \to U \to V \to W \to 0$$

is a short exact sequence of vector bundles then there is an isomorphism of line bundles

$$\Lambda^{\dim V} V \cong \Lambda^{\dim U} U \otimes \Lambda^{\dim W} W$$

which clearly shows that trivializations of any two of these line bundles induce trivializations of the third.


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