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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.) – Pierre-Yves Gaillard Feb 11 '12 at 13:43
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! – harlekin Feb 11 '12 at 14:39

The following are very similar:

  1. Integral of function $f$ with respect to measure $|dx|$.
  2. Integral of $n$-form $f(x)\,dx$.

But there are some differences:

  1. is independent of orientation of the region of integration.
  2. changes sign if the orientation is changed

Or, put another way:

  1. has the change of variables formula involving the absolute value of the Jacobian
  2. has the change of variables formula involving the Jacobian itself

One can infer from the above that the integral $\int_a^b f(x)\,dx$ encountered in calculus ($n=1$) is the integral of second kind, with $f(x)\,dx$ being a $1$-form.

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