I have a question about using orientation in the integration of differential forms on a manifold $M$. I have seen two different "versions" of integration. Sorry if this is a little long.
One, as seen in Arnold's Mathematical Methods of Classical Mechanics, has to do with integration on chains. To keep things simple, we can look at integrating what he calls a "$k$-dimensional cell"; let's label it $\sigma$. It is given by a triple $\sigma=(D,f,\text{Or})$ consisting
- a convex polyhedron $D\subset\mathbb{R}^k$,
- a differentiable map $f\colon D\rightarrow M$, and
- an orientation on $\mathbb{R}^k$, denoted by $\text{Or}$.
The integral of a $k$-form $\omega$ on $\sigma$ is then defined by $$\int_\sigma\omega=\int_D f^*\omega.$$
There is no mention of orienting $M$ at all here; instead the orientation seems is given in $\mathbb{R}^k$. So, basically what we do (I hope I'm right) is pull back the form $\omega$ into something that is written in the form of $g(x)\,dx_1\wedge dx_2\wedge\cdots\wedge dx_k$. So if we give the standard basis on $\mathbb{R}^k$ positive orientation, then we simply take a normal Riemann integral of $g(x)\,dx_1dx_2\cdots dx_n$; if we gave the standard basis negative orientation then we would have the same integral but with a negative value.
What's interesting here to me is how the orientation is given in $\mathbb{R}^k$, not on the manifold itself. So, for example, if I integrate the $1$-form $dx\wedge dy$ on the upper hemisphere of the sphere $S^2$ (seen as embedded in $\mathbb{R}^3)$, one way to do this would be to paramaterize the region with a polyhedron in $\mathbb{R}^2$ (with the standard orientation) via $$f(u,v)=(\cos u\sin v,\sin u\sin v,\cos v),$$ where $u\in(0,2\pi)$ and $v\in(0,\pi/2)$. If you evaluate all of the pullbacks, etc., you get that the integral of this form is $-\pi$.
On the other hand, if you orient the upper hemisphere itself, say with an "upward" pointing unit normal, you would expect to get $\pi$ as the answer. So it seems like what I need to do get $\pi$ is to reverse the orientation on $\mathbb{R}^2$, if we're looking at Arnold's definition.
This whole operation then seems a little awkward. My question, then is, how does this relate to the usual definition of integration on an oriented manifold? To me, it seems like any form $\omega$ on $M$ is already been fixed, after all, it just eats in tangent vectors and spits out a number. If a positively oriented chart is used, then we would get $\pi$. Is this correct? I feel like I'm just confused in general about how Arnold doesn't even orient the manifold but instead orients $\mathbb{R}^n$, while in other places the orientation of the manifold (and I guess $\mathbb{R}^n$ as well) also matters.
You don't need to orient $M$ for this operation. Think about it as the specialization of the following scenario. $N$ an oriented differentiable manifold, $D \subset N$ an open subset and $M$ a differentiable manifold.Let $F:D \to M$ be a differential map and $\omega$ a differential form on $M$. Endow $D$ with the orientation given by $N$ then the following integral is well defined $$\int_D F^{*}\omega$$