# Why do differential forms and integrands have different transformation behaviours under diffeomorphisms?

Let $f$ be a diffeomorphism, say from $\mathbb R^n$ to $\mathbb R^n$ , such as the transition map between two coordinate charts on a differentiable manifold.

A differential $n$-form (or rather its coefficient function which is obtained by using the canonical one-chart atlas on $\mathbb R^n$) then transforms essentially by multiplication with $\mathrm{det}(Df)$, while integrals transform essentially by multiplication of the integrand with $\lvert\mathrm{det}(Df)\rvert$.

(This is the reason for the necessity to choose an orientation in order to define the integral of a top form on a differentiable manifold.)

Question: What is an intuitive or conceptional reason for these different transformation behaviours of forms and integrands?

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You say «A differential form then transforms essentially by multiplication with det(Df)», but that is quite not right: the differential form does not transform in anyway: what transforms is the coefficient in coordinate expressions. (This is not a minor nit-pick but the whole point of defining differential forms!) – Mariano Suárez-Alvarez Aug 21 '10 at 22:44

Transforming by a diffeomorphism $f$ shouldn't make an (unsigned) integral negative. For instance, say that the integral of $f$ over a subset $A$ of $\mathbb{R}^n$ is just the usual Lebesgue integral, no worries about differential forms of orientation. So shifting between $f$ and $f \circ \alpha$ for $\alpha$ some map cannot possibly change the sign. This is why the absolute value signs appear.

One can consider a "volume element:" this is something that transforms via $|Det f|$ for a diffeomorphism $f$. Any differential form leads to a volume element (take its absolute value). But the opposite is not always true. One can still define (unsigned) integration with respect to a volume element. However, note that the cancellations in signed integration are useful. These cancellations imply that the integral of an exact form on a compact oriented manifold is zero, for instance. So restricting to volume elements loses interesting and useful information.

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The absolute values are not introduced: they are a consequence of definitions (and in fact the statement that integrals transform as they do is a theorem). Likewise, the way differential forms transform follows from their definition, it is not that we make them transform in any way so as to keep them smooth. – Mariano Suárez-Alvarez Aug 21 '10 at 22:38
Yeah, that was a bad use of language: I was trying to explain why one would choose to define them that way. I've revised the post. – Akhil Mathew Aug 22 '10 at 15:56

A manifold is not necessary orientable. But the orientation bundle always exists. You can construe it as a real line bundle. Then the density bundle is obtained by tensoring the bundle of top forms with the orientation bundle. By definition, a nonvanishing section of the orientation bundle is an orientation, and a section of the density bundle is a density. Compactly supported densities can be integrated.

Locally, orientations "transform" by the sign of the determinant (of the jacobian matrix), top forms by the determinant, and densities by the absolute value of the determinant.

It suffices to understand the case of the real line.

EDIT. I've just realized that, in your question, you mention manifolds only in a parenthesis, and concentrate on $\mathbb R^n$. Open subsets of $\mathbb R^n$ are of course always orientable. But the key point, again, is to understand what happens on the real line.

For the real line, it's natural to write a 1-form as $f\ dx$, and a density as $f\ |dx|$ (say with $f$ continuous). If you have a compact interval $I$, then you can integrate $f\ |dx|$ over $I$, and $f\ dx$ from one bound to the other.

EDIT 2. You can write the standard orientation of $\mathbb R$ as $|dx|/dx$, and define the integral of the form $f\ dx$ over the interval $I$ equipped with the orientation $\pm|dx|/dx$ as the integral over the nonoriented interval $I$ of the density $\pm f\ |dx|$ equal to the orientation times the form.

So you only need to define the integral of a density over a nonoriented (compact) interval. [See for instance de Rham's book on differentiable manifolds.]

EDIT 3. In the same spirit, to define the $L^2$ space of a manifold, you consider half-densities. More precisely, you complete the pre-hilbert space of compactly supported half-densities. The half-densities "transform" by the square root of the determinant.

The general principle is this:

If $M$ is an $n$-manifold, then each action of $G:=GL_n(\mathbb R)$ on a manifold $F$ induces a bundle over $M$ with fiber $F$, called the associated bundle. If the action is an $r$-dimensional $\mathbb R$-linear representation, then the associated bundle is a rank $r$ vector bundle.

The most basic example is the tangent bundle, associated to the natural representation on $\mathbb R^n$.

Let $\chi$ be the characters of $G$ given by the inverse of the determinant. The bundles of $n$-forms, orientations, densities, and half-densities, are the line bundles respectively associated to the following characters of $G$: $\chi$, the sign of $\chi$, its absolute value, and the square root of its absolute value.

For the definition of the associated bundle, see

http://en.wikipedia.org/wiki/Associated_bundle#Fiber_bundle_associated_to_a_principal_bundle

In our case, the principal bundle is the frame bundle:

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