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This is undoubtedly a very easy question, but perhaps there are some subtleties. Under what circumstances can we integrate by parts over a non-compact Riemannian manifold? I am aware that having bounded curvature is sufficient (is there a reference for this?), but can this be weakened?

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Integrate "what" by parts? Stokes theorem will hold provided whatever you are integrating converges "at infinity" in a suitable sense (see Deane's answer below), and does not require a Riemannian structure. So you can have a suitable equality of the form $\int \alpha \wedge \mathrm{d}\beta = \epsilon \int \mathrm{d}\alpha\wedge\beta$. For functions on a manifold you don't necessarily have natural partial derivatives. If you are thinking about directional derivatives relative to a vector field then you need the vector field to be parallel, other wise you get an error term. –  Willie Wong Feb 29 '12 at 13:49
    
The question could be rephrased "under what curcumstances do we have the formula [ \int_{\mathcal{M}} f \Delta g \, dV = \int_{\mathcal{M}} g \Delta f \, dV ]" –  T-' Feb 29 '12 at 19:31
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1 Answer

up vote 11 down vote accepted

I know of two ways to approach this in general. One common way is to exhaust the manifold with a sequence of compact domains with smooth boundary and show that when you integrate by parts on the compact domain, the two integrals converge and the boundary term converges to zero.

I, however, usually prefer the second approach, which is to use a sequence of cut-off functions, i.e. compactly supported smooth functions that converge to 1 on any compact domain. There is no boundary term, but, when you integrate by parts, you get an extra term involving the gradient of the cut-off function. You then show that this term converges to zero, given a suitable sequence of cut-off functions and suitable assumptions about the growth or decay rate of the volume form of the manifold and the integrand.

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Willie, many thanks for fixing the typos! –  Deane Yang Feb 29 '12 at 15:21
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