Hilbert space on line bundle

Question

Suppose that $L$ is a complex line bundle on a manifold $M$ with measure $\mu$, How can we prove, $L^2(M,L,\mu)$ is Hilbert space?

Answer 1

You must be leaving something out of the question here - you need an inner product in order to say that something is a Hilbert space.

Probably the situation is something like this: you have a Hermitian metric on $L$, i.e. an inner product $\langle \cdot , \cdot \rangle_x$ on $L_x$ for each $x \in M$, such that the inner products vary smoothly with $x$.

Then you can define an inner product on smooth (or even continuous) sections of $L$ by $$ \langle \phi, \psi \rangle = \int_M \langle \phi(x), \psi(x) \rangle_x d \mu(x). $$ If $M$ is not compact then you will need to restrict to compactly supported sections in order for this integral to make sense.

Now the question is, how do you define $L^2(M,L,\mu)$? There are essentially two options: the first is that you define it to be the completion of the space of sections with respect to the inner product just defined, in which case it is automatically a Hilbert space; the other option is to define an equivalence relation on sections saying that two sections are equivalent if they are equal almost everywhere, then define $L^2(M,L,\mu)$ to be the set of equivalence classes of square-integrable sections, in which case the proof of completeness is essentially identical to that for $L^2(M,\mu)$.