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Suppose I have a Riemann Surface (i.e. an oriented manifold) and I have an integral that uses the volume density $|dx|$ instead of the volume form $dx = dx_1dx_2$ (here the $(x_1,x_2)$ are local real coordinates, i.e. we have the holomorphic coordinates $z = x_1 + ix_2$ and $\bar{z} = x_1 - ix_2$).

Here comes my question that I'd need help to proceed: Can I simply "ignore" the difference and write \begin{equation} |dx| = dx_1dx_2 \end{equation} in this case ? My guess is yes because I am working on an oriented manifold and on these one can identify 1-densities with n-forms. But I am not sure, my knowledge on differential forms is not solid enough to be confident. (The integral I am looking at is supposed to hold over general even-dimensional manifolds (not necessarily oriented), that's why the density is used. It's only a special case of mine that I have additional complex structure, forcing my manifold to be oriented.) Thanks for your help!

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    Dear harlekin: What you write looks correct to me. The choice of an orientation on an $n$-manifold enables you to identify densities to $n$-forms. (What's behind this is, I think, the notion of fiber bundle associated to a principal bundle, if you know these words.)2012-02-11
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    ok that is reassuring, I will have a look at the concept of a fibre bundle, I realize I have to learn a lot more to fully understand the theory. Thanks a lot for your help!2012-02-11

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