# Completing The Space Sections in a Vectorbundle

Hi there.

Assume $(M,g)$ is a Riemanian manifold and $E\to M$ is a vector bundle with a bundle metric $\langle\cdot,\cdot\rangle$. We then have the pre-Hilbert space $H_0:=\Gamma_c^\infty(E)$ of compactly supported smooth sections with $(s_1,s_2):=\int_M\langle s_1,s_2\rangle dV_g$ . In a paper i'm currently working on it says that the completion $H$ of $H_0$ is (in a rather concrete Situation for $E$) given by the space of square integrable sections $L^2(M,E):=\{s\in\Gamma(E)\mid \int_M|\langle s,s\rangle|^2dV_g<\infty\}$, identifying sections being equal almost everywhere. Is this true in general and if yes, do you know a book where this is worked out?

Greetings, Robert Rauch

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This seems more appropriate for math.stackexchange.com than for MathOverflow. I also suggest that you try to work out the details yourself. It's really the same as how you define $L^2(R)$ as the completion of compactly supported smooth functions on $R$. –  Deane Yang Nov 30 '11 at 4:59

This is true in general. I don't know a reference for the statement, but it is pretty simple just to work it out. The point is that $L^2(M,E)$ is a Hilbert space which contains $H_0$ as a dense linear subspace, so it must be the completion.
I guess I should add that $H_0$ sits isometrically inside $L^2(M,E)$. –  MTS Nov 29 '11 at 22:01
What you are essentially saying is, that i should check the definition of the completition. The point is, its not that obvious to me that $L^2(M,E)$ is a Hilbertspace. I mean, is there a simpler argument than "go along the proof for the $L^p(X,\mu)$ spaces and change everything such that it fits our needs"? –  Robert Rauch Nov 29 '11 at 22:20
I also thought about this in the beginning, but we are not going to define an integral for sections in $E$ (in which space should $\int_M s dV$ lie for $s\in\Gamma(E)$?!), but a bilinear form $(\cdot,\cdot)$ on a certain subspace of $\Gamma(E)$, which involves only integration of, say $\mathbb{C}$-valued functions on the measure space $(M,dV)$. –  Robert Rauch Nov 29 '11 at 23:34