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?
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$\begingroup$ You should give your definition of $L^2(M,L,\mu)$. $\endgroup$– wildildildlifeJul 2, 2013 at 16:17
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$\begingroup$ The space of sections of $C^{\infty}(M,L)$, has a natural Hermitian pairing as I explained in one of my comments and so one can form $L^2(M,L,\mu)$ in natural way. $\endgroup$– MatiasJul 2, 2013 at 16:26
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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 $L^2(M,\mu)$.
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$\begingroup$ In your inner product , I think we should change, d$\mu$ to $\mu$ $\endgroup$– MatiasJul 2, 2013 at 15:59
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$\begingroup$ Also your integrand, $<\phi(x),\psi(x)>$ should be belong to trivial line bundle $M\times\mathbb{C}$ $\endgroup$– MatiasJul 2, 2013 at 16:04
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$\begingroup$ also, I guess, $<\phi(x),\psi(x)>=\eta(f,g)$ where $\eta:L\bigotimes_{\mathbb{C}} L\to M\times \mathbb{C}$ $\endgroup$– MatiasJul 2, 2013 at 16:09
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1$\begingroup$ $d \mu$ vs. $\mu$ is just a matter of taste. About your second and third comments - that is what I meant by a Hermitian metric on $L$. The smoothness condition ensures that the function $x \mapsto \langle \phi(x), \psi(x) \rangle_x$ is a smooth function on $M$. $\endgroup$– MTSJul 2, 2013 at 18:43
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1$\begingroup$ (Oops, posted too soon.) Hence this function can be integrated over $M$ with respect to $\mu$. To be precise, a Hermitian metric is conjugate-linear in the first variable, not linear, so if you want to write the metric as a bundle map, it should be something like $\eta : \overline{L} \otimes L \to M \times \mathbb{C}$, where $\overline{L}$ is the complex conjugate bundle to $L$. $\endgroup$– MTSJul 2, 2013 at 18:46