Understanding twisted differential forms

Integral of a twisted differential $p$-form on a $p$-submanifold $\Sigma \subset M$ does not depend on the orientiation of $\Sigma$ (in fact $\Sigma$ can be non-orientable), but rather on external orientation (orientation of the normal bundle $\left. TM \right|_\Sigma / T \Sigma$). To give you a physical example of this, consider the current $j$ of certain substance (e.g. water or electric charge) in a container $M$ (manifold of dimension $D$). I claim that $j$ is geometrically a twisted $(D-1)$-form on $M$. Indeed, given any two-sided hypersurface $\Sigma \subset M$, the flux $\int_\Sigma j$ of the flowing substance through $\Sigma$ is well-defined. Clearly it should not depend on the orientation of $\Sigma$, but it should change sign if we switch which side of $\Sigma$ is regarded as "in" vs "out".

Of course twisted $D$-forms are even easier: they represent densities and can be integrated over volume, without having to choose orientation. One example is the volume form: certainly given a metric you can calculate the area of a Mobius strip! To extend my previous physical analogy, suppose that the substance under consideration has density profile $\rho$. Then $\rho$ is a twisted $D$-form. If the amount of substance is (pointwise) conserved, we will have a continuity equation $$ \frac{\partial}{\partial t} \rho + d j=0, $$ in which $d$ is the exterior derivative along the "spatial" manifold $M$ and $t$ is additional "time" parameter. This equation is the statement that the rate of change of amount of the substance in some volume $\Omega$ is given by the flux of the substance through boundary $\partial \Omega$.

This is not an answer to the questions at the end of the post, but rather an answer to the question marks in

Somehow this means that they can be used to integrate on non-orientable manifolds. (???)

Namely, I want to expand on the mathematical details of why exactly this property allows $n$-pseudoforms to be integrated on any smooth $n$-manifold without having to make additional choices. For this I need the machinery of associated bundles.

Associated bundles $\newcommand{\R}{\mathbb R}$ $\newcommand\id{\mathrm{id}}$ $\newcommand\sgn{\mathrm{sgn}}$

The frame bundle $LM$ on a manifold $M$ is a principal $GL(n)$-bundle: there is a fibrewise regular (i.e. free and transitive) right $GL(n)$-action on $LM$, a matrix $A\in GL(n)$ sends a frame $e=(e_i)$ to the new frame $e\triangleleft A=(\sum_j e_jA^j_i)$.

If $GL(n)$ acts on a vector space $V$ by a representation $\rho:GL(n)\to GL(V)$, then you can construct a new vector bundle $LM[V,\rho]\to M$ as $$LM[V,\rho] = (LM\times V)/\sim$$ where the relation $\sim$ is given by $(e\triangleleft A,v)\sim(e,\rho(A)v)$ or, equivalently, as $(e,v)\sim(e\triangleleft A,\rho(A^{-1})v)$. Write the equivalence class of $(e,v)$ as $[e,v]$.

This construction allows us to "tie a vector bundle to its components with respect to $LM$" as follows:

Since for any two $[e,v]$, $[e',v']$ in $LM[V,\rho]$ (over the same basepoint) there is a unique $A\in GL(n)$ such that $e\triangleleft A=e'$ (because the action is regular) it follows that, given a local section $X$ of $LM[V,\rho]$ and a local section $F$ of $LM$, for every point $p$ of (an open subset $U$ of) the manifold, you can find a unique vector $X_F(p)\in V$ such that $$X(p) = [F(p),X_F(p)].$$ Indeed, if $X(p)=[e,v]$, then let $X_F(p)=\rho(A^{-1})v$, where $A$ is the unique matrix in $GL(n)$ such that $e\triangleleft A=F(p)$. This gives us a function $X_F:U\to V$ whis is the "components of $X$ with respect to $F$". I'll write this as $$X = [F,X_F]$$ which holds at the domain of $F$.

If $F:U\to LM$ is the local frame associated to a coordinate frame $(U,x)$, then write $F=\partial/\partial x$.

Now you take $V=\R^n$ and the standard representation $\id:GL(n)\to GL(\R^n)$, then you can express a section $X$ of $LM[\R^n,\id]$ using a local coordinate frame $\partial/\partial x$ as $$\begin{align} X = \left[\frac{\partial}{\partial x},X_{(x)}\right] \end{align}$$ where $X_{(x)}:U\to\R^n$ are the "components of $X$ with respect to $\partial/\partial x$".

Notice the following: if you take a new coordinate patch $y$ on $U$, then you have $$ \frac{\partial}{\partial x} = \frac{\partial}{\partial y} \triangleleft \frac{\partial y}{\partial x} $$

$$\begin{align} X &= \left[\frac{\partial}{\partial x},X_{(x)}\right] \\ &= \left[\frac{\partial}{\partial y} \triangleleft \frac{\partial y}{\partial x},X_{(x)}\right] \\ &= \left[\frac{\partial}{\partial y}, \frac{\partial y}{\partial x}X_{(x)}\right] \\ \end{align}$$

but also $$ X = \left[\frac{\partial}{\partial y},X_{(y)}\right] $$

hence

$$ X_{(y)} = \frac{\partial y}{\partial x}X_{(x)} $$ or, in components, $$ X_{(y)}^i = \sum_j \frac{\partial y^i}{\partial x^j}X^j_{(x)} $$ which is precisely the "transformation behaviour" of a tangent vector field. Indeed, you can show that $LM[\R^n,\id]$ is isomorphic to the tangent bundle $TM$.

Case of interest

Now take an $n$-form $\omega$, i.e. a section of the bundle of $n$-forms $\bigwedge^n T^*M$.

If $(U,x)$ is a coordinate chart, you can express $\omega$ locally as $$\omega = \omega_{(x)} dx^1\wedge\dots\wedge dx^n.$$ where $\omega_{(x)}:U\to\R$ is a function of the manifold that depends on the chart map $x$. If you change coordinates to a new chart map $y$, then $\omega$ has the new expression $$ \omega = \omega_{(y)} dy^1\wedge\dots\wedge dy^n $$ which is related to the old one by $$\begin{align} \omega &= \omega_{(x)} dx^1\wedge\dots\wedge dx^n \\ &= \omega_{(x)} \frac{\partial x^1}{\partial y^{i_1}}dy^{i_1} \wedge\dots\wedge \frac{\partial x^n}{\partial y^{i_n}}dy^{i_n} \\ &=\omega_{(x)} \det\left(\frac{\partial x}{\partial y}\right)dy^1\wedge\dots\wedge dy^n. \end{align}$$ So $$\omega_{(y)} = \det\left(\frac{\partial x}{\partial y}\right) \omega_{(x)} = \det\left(\frac{\partial y}{\partial x}\right)^{-1} \omega_{(x)} .$$ What this is saying is that the bundle $\bigwedge^n T^*M$ can be obtained as an associated bundle to the principal bundle $LM$ via the representation $\rho:GL(n)\to GL(1)=\R^*$ given by $\rho(A)v = \det(A)^{-1}v$, because for this associated bundle $LM[\R,\rho]$ you have $$\begin{align} \left[\frac{\partial}{\partial x},\omega_{(x)}\right] &= \left[\frac{\partial}{\partial y}\triangleleft\frac{\partial y}{\partial x},\omega_{(x)}\right] \\ &= \left[\frac{\partial}{\partial y},\rho\left(\frac{\partial y}{\partial x}\right)\omega_{(x)}\right]. \end{align}$$ but also $$\begin{align} \left[\frac{\partial}{\partial x},\omega_{(x)}\right] &= \left[\frac{\partial}{\partial y},\omega_{(y)}\right] \\ &= \left[\frac{\partial}{\partial y},\det\left(\frac{\partial y}{\partial x}\right)^{-1}\omega_{(x)}\right]. \end{align}$$

Similarly, you can check that if you take the representation $\sigma:GL(n)\to GL(1)=\R^*$ given by $\sigma(A)=|\det A|^{-1}$, then the component functions $\mu_{(x)}$ and $\mu_{(y)}$ of a section $\mu$ of $LM[\R,\sigma]$ with respect to coordinate charts $x$ and $y$ obbey the transformation law $$\mu_{(y)} = \left|\det\frac{\partial x}{\partial y}\right| \mu_{(x)} .$$ Now you can define the integral of $\mu$ over $U$ as $$\int_U \mu = \int_{x(U)} \mu_{(x)}\circ x^{-1}$$ where the right hand side is the Lebesgue integral in $\R^n$. Now this is independent of the chart maps because if $y$ is any other chart map defined on $U$ whatsoever (in particular, it can be orientation reversing) then you have $y(U)=y\circ x^{-1}\circ x(U)$, so $$ \begin{align} \int_{y(U)} \mu_{(y)}\circ y^{-1} &= \int_{y\circ x^{-1}\circ x(U)} \mu_{(y)}\circ y^{-1} \\ &= \int_{x(U)} (\mu_{(y)}\circ y^{-1}\circ (y\circ x^{-1}))|J(y\circ x^{-1})| \\ &= \int_{x(U)} (\mu_{(y)}\circ x^{-1}) \left|\det\frac{\partial y}{\partial x}\circ x^{-1}\right| \\ &= \int_{x(U)} \left(\mu_{(y)}\left|\det\frac{\partial y}{\partial x}\right|\right)\circ x^{-1} \\ &= \int_{x(U)} \mu_{(x)}\circ x^{-1}. \end{align} $$ The bundle given by this representation $\sigma(A)=|\det A|^{-1}$ is called the $1$-density bundle or also the $n$-pseudoform bundle. Its sections are called $1$-densities or $n$-pseudoforms.

My comments about twisting the bundle of $n$-forms $\bigwedge^n T^*M$ with the bundle given by the representation $A\mapsto \sgn(\det A)$ is basically due to the fact that the representation $\sigma$ that gives rise to $n$-pseudoforms can be decomposed as $$ \sigma(A)=|\det A|^{-1} = \sgn(\det A)\det(A)^{-1} = \sgn(\det A)\rho(A) $$ where $\rho$ is the representation that gives rise to the $n$-form bundle $\bigwedge^n T^*M$.

Given a manifold $M$, differential forms over it are the sheaf of sections of the exterior cotangent bundle:

$\Omega(M) := \Gamma(\wedge T^*M)$

Given a vector bundle $E$ over the same manifold, we say the exterior tangent bundle is twisted by $E$ when we form:

$\wedge T^*M \otimes E$

Then again taking sections we define differential forms over $M$ twisted by $E$.

$\Omega(M,E) := \Gamma(\wedge T^*M \otimes E)$

These are also called differential forms valued in $E$. Now one important case is when $E$ is actually a line bundle. And an important such line bundle is:

$Or(M) := \wedge^m T^*M$

Where $m$ is the dimension of the manifold. Thos os the orientation bundle. Now differential forms twisted by $Or(M)$, that is $\Omega(M, Or( M))$, are called densitised forms. These are the forms you are interested in. Its this twisting by the orientation bundle that allows it to be integreted over non-orientable manifolds.