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Let $M$ be a manifold with a circle action, i.e. with a 1-parameter group of diffeomorphisms $\phi_t:M\to M$ of period 1.

I think of the average of a differential form $\omega \in \Omega^n(M)$ with respect to the $S^1$ action as the $n$-form $A(\omega)$ defined pointwise by $A(\omega)_p(X_1,...,X_n)=\int_0^1 \phi_t^*(w)_p(X_1,...,X_n) dt$ for any tangent vectors $X_1,...,X_n \in T_pM$. Note that $\phi_t^*(w)_p(X_1,...,X_n)$ is a real valued smooth function of $t$ so the integral makes sense.

However, you can think of $A(\omega)=\int_0^1 \phi_t^*(w) dt$ as an integral in the Hilbert space of differential forms. Can someone help me in making this more precise? What is the norm that makes $\Omega^*(M)$ into a Hilbert space? Can we do this without choosing a metric for $M$?

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$\Omega^{\ast}(M)$ probably cannot be made into a complete inner product space in a natural way. If you have a Riemannian metric on $M$ you may at least be able to get an inner product, though. Anyway, I don't understand why you need an inner product in the first place. –  Qiaochu Yuan Aug 4 '12 at 4:08
    
Well, how would you make sense of the integral $\int_0^1 \phi_t^*(\omega)dt$ in the space of forms? –  Manuel Aug 4 '12 at 4:13
    
Pointwise. ${}{}$ –  Qiaochu Yuan Aug 4 '12 at 4:16

1 Answer 1

up vote 3 down vote accepted

On an oriented Riemannian manifold $M$ one has the natural inner product on each $\Omega^n(M)$, defined by $\langle \alpha, \beta\rangle = \int_M \alpha\wedge *\beta $ where $*$ is the Hodge star. It is also natural to let $\Omega^*(M)=\bigoplus_{n=0}^{\dim M}\Omega^n(M)$; the direct sum of inner product space also becomes an inner product space, simply by adding the products. Nothing can stop you now from completing $\Omega^*(M)$ with respect to the inner product to get a Hilbert space; the elements of the resulting space are the forms with $L^2$ coefficients. Which, of course, raises the issue of the definition of operator $d$. One considers the smaller Hilbert space of forms with $H^1$ coefficients (Sobolev space) on which $d$ is defined, but takes values in the larger space of forms with $L^2$ coefficients. Now we can dualize and say that $d$ also maps $L^2$ forms into $H^{-1}$ forms, where $H^{-1}$ is the Sobolev space of negative order, defined as the dual of $H^1$. Note that $H^{1}\subset L^2\subset H^{-1}$, so we have a rigged Hilbert space.

A classical reference for the above is Chapter 7 of Multiple integrals in the calculus of variations by Morrey. If you don't have access to it, take a look, e.g., at this paper. These two references lean into analysis. A geometric point of view is presented in Chapter 6 of Warner's Foundations of differentiable manifolds and Lie groups. Warner works with the inner product on $\Omega^*(M)$ defined above, but avoids taking the completion as long as he can. By the time the Sobolev spaces appear, the computation is moved to a Euclidean space.

This said, I agree with @Qiaochu that you don't need an inner product. Completeness is essential, of course, but there are other options:

But if you don't want to choose a metric on $M$, then you'll have to carefully define an LCS (locally convex topological vector space) of smooth forms using the family of seminorms given by the integrals against all smooth vector fields. And then deal with completeness and integration issues. Looks like an unnecessary effort.

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