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Let $E \rightarrow X$ be a vector bundle with an inner product. If $F$ is a sub-bundle, we can define an orthogonal complement bundle $F^\perp$ (see for the construction, and the source of the problem). I am trying to show that $F^\perp$ is independent of the choice of inner product, up to isomorphism. I have tried to do a local construction, defining the isomorphism in each locally trivial neighborhood, but I do not see a way to patch these together into a global isomorphism. Any suggestions?

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Consider two inner products $\langle\cdot\rangle_1$, $\langle\cdot\rangle_2$, and the two orthogonal complement bundles $C_1,~C_2$ to $F$ they define. There are canonical isomorphisms as follows: consider the restriction to $C_2$ of the projection onto $C_1$ parallel to $F$. This gives you an isomorphism $C_2\rightarrow C_1$, and similarly in the other direction. These are actually inverse to one another. – Olivier Bégassat Jul 20 '12 at 4:06
up vote 4 down vote accepted

Hint: First show that for an inner product space $(V, \langle \cdot, \cdot \rangle)$ and a subspace $W \subseteq V$, $$W^\perp \cong V/W.$$ Then, using the quotient bundle construction of the previous exercise (exercise 2), show that for a Euclidean vector bundle $E \longrightarrow X$ with sub-bundle $F$, $$F^\perp \cong E/F.$$ Since $E/F$ is independent of the bundle metric, this will show that $F^\perp$ does not depend on the bundle metric (up to isomorphism).

Let me know if you want me to provide the details of the argument.

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Thanks. I think I need a little more detail; I tried this already, and was unable to get it to work. – user15464 Jul 20 '12 at 11:53

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