Usually in differential geometry one proves the Stokes theorem and then obtains divergence theorem and Green's formulas as corollaries. However, divergence theorem is also valid for nonorientable riemannian manifolds when one replaces forms with densities. But then the Green's formulas should also be valid. Am I missing something? I haven't found a discussion about where precisely the orientability is needed.
Further supposing Green formulas in nonorientable manifolds one could define variational formulation for example for elliptic PDEs there. I wonder if the (non)orientability has some effect on the solutions of such PDEs?
Best Answer
Similar answer, a friend recently asked about twisted forms for non-orientable manifolds, what I found was pages 79-88 in Bott and Tu, "Differential forms in algebraic topology." I also think a twisted Stokes' Theorem is possible, as they present an entire twisted de Rham complex. Anyway, take a look:
http://books.google.com/books?id=S6Ve0KXyDj8C&pg=PA79&dq=bott+tu+twisted&cd=1#v=onepage&q&f=false
I googled "twisted Stokes theorem." My friend Dmitry asked originally based on some physics inquiries. It appears that these physics people give a pretty direct discussion, maybe it is enough. "Foundations of classical electrodynamics: charge, flux, and metric" By Friedrich W. Hehl, Yuri N. Obukhov, (2003) Birkhauser
http://books.google.com/books?id=48-hHXL-CYUC&pg=PA93&lpg=PA93&dq=twisted+Stokes+theorem&source=bl&ots=EQhcJBqPC2&sig=iK7FdNGL7xFeePoqjFhcnLbn3d4&hl=en&ei=3UcJTJOxFIiMNtDR2bUE&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBYQ6AEwAA#v=onepage&q=twisted%20Stokes%20theorem&f=false