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When it comes to integration on manifolds, I speak two languages. The first is of course the language of differential forms, which is something I am relatively well acquinted with.

The second language is what is often used in general relativity books that follow a rather traditionalist viewpoint (eg. no explicit usage of modern differential geometry, just coordinate and index-based tensor-analysis with "funky coordinates"), which try to define "tensor densities" as indexed objects that pick up some power of the jacobian determinant during coordinate change, in addition to the usual jacobian matrices on the tensor indexes. It is then stated that these "densities" can be integrated invariantly.

These concepts generally do not come with any rigorous, or even semi-rigorous explanation, just stated and then used without much thought.

In time I came to understand "scalar densities" as the single independent component of a top-order differential form expressed in some coordinate basis, seeing as this component transforms by picking up the jacobian determinant, which comes directly from the skew-symmetric properties of the differential form, as well as the fact that a top-order form has one independent component only.

On the other hand, I have been told by a mathematician whose Lie-groups course I have been attending that densities are perfectly well defined mathematical objects that can be used to integrate on manifolds that aren't orientable.

I understand the general idea behind densities, but I have no idea how to rigorously define objects like these, neither do I know how any concrete density even "looks like".

I mean, if someone asked me the same thing I am asking here, but with differential forms, I would

1) Explain to them algebraic exterior forms on finite dimensional vector spaces, exterior products, etc.

2) Show them that on a tangent space of a differential manifold, any exterior form can be written as a linear combination of wedge products of coordinate differentials, eg. $\mathrm{d}x^\mu$.

3) Define fields of these objects by either showing what it means for a $p\mapsto\omega_p$ assignment ($p\in M$, $\omega_p\in\Lambda^kT^*_pM$) to vary smoothly or by "bundle-izing" exterior products of contangent spaces over the manifold.

4) Define the exterior derivative and show how it works.

5) Show that an integral of $k$-differential form over a $k$-dimensional submanifold of $M$ is independent of coordinates, basically by defining ($k=n$ for simplicity) $$ \int_\mathcal{D}\omega=\int_\mathcal{D}F\mathrm{d}x^1\wedge...\wedge\mathrm{d}x^n=\int_{\Psi(\mathcal{D})}F(x^1...x^n)dx^1...dx^n, $$ where $\Psi$ is the coordinate map, and the right side is a standard Riemann-integral or an integral against the standard Lebesgue-measure of $\mathbb{R}^n$ and $\omega=F\mathrm{d}x^1\wedge...\wedge\mathrm{d}x^n$, and then showing that if I change the coordinates, everything in the integral transforms in such way that the value of the integral stays the same.

Questions:

1) How do I (mostly) rigorously define and construct a density on a manifold and how does it look like?

(Eg. like, a differential $n$-form on an $n$-dimensional manifold, when expressed in a chart looks like $\omega=F\mathrm{d}x^1\wedge...\wedge\mathrm{d}x^n$, where $F$ is a scalar function, plus I'd appreciate a smiliar explanation to what I outlined above for difforms).

2) How are they related to measures?

I mean, as an example, I thought about the following: As far as I am aware, every smooth non-orientable manifold is locally orientable, so I guess if I take a differential form defined in some chart, $F\mathrm{d}x^1\wedge...\wedge\mathrm{d}x^n$ (so that the basis $n$-form is nonvanishing) and then define a measure as $$ \mu_x(\mathcal{D})=\left|\int_\mathcal{D}\mathrm{d}x^1\wedge...\wedge\mathrm{d}x^n\right| ,$$ then this case the integral $$ \int_\mathcal{D}Fd\mu_x $$ makes sense in the chart in a way that is independent of coordinates, because if I change the coordinates, I first transform the differental form, then construct another measure $\mu_y$, then extend this via a partition of unity, but this isn't really explicit in the sense that I don't think $$F\ d\mu_x $$ is a well defined object without the integral symbol, and I am not even sure what I'm doing here is even correct.

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The basic concept is that of a density of weight $s$ (or just an $s$-density) for any $s\in \mathbb R$ on an $n$-dimensional real vector space $V$, which is a map $\mu\colon V\times\dots\times V\to \mathbb R$ that satisfies the following identity for every linear map $A\colon V\to V$: $$ \mu(AX_1,\dots,AX_n) = |\det A|^s\mu(X_1,\dots,X_n). $$ (If $s<0$, you have to require this just for nonsingular $A$.) Compare this to an $n$-form $\omega$, which satisfies $$ \omega(AX_1,\dots,AX_n) = (\det A)\omega(X_1,\dots,X_n). $$ Any $n$-form $\omega$ determines an $s$-density $|\omega|^s$ by $$ |\omega|^s(X_1,\dots,X_n) = |\omega(X_1,\dots,X_n)|^s. $$

On a smooth $n$-manifold $M$, for each $s$ there is a smooth real line bundle $\Omega^s M$, whose fiber at each point $p\in M$ is the $1$-dimensional vector space of densities of weight $s$ on $T_pM$. Any choice of coordinates $(x^i)$ yields a local frame $|dx^1\wedge \dots dx^n|^s$ for $\Omega^s$, so a section of $\Omega^s M$ (also called a density of weight $s$ on $M$) can be written locally as $f\,|dx^1\wedge \dots dx^n|^s$ for some function $f$. The functions $f$ and $\widetilde f$ associated with different coordinates transform by $\widetilde f = |J|^s f$, where $J$ is the Jacobian determinant of the coordinate transformation: $J=\det (\partial \widetilde x^i/\partial x^j)$.

A density of weight $1$ can be integrated over $M$ in a coordinate-independent fashion, since the transformation law for $1$-densities exactly matches the change of variables formula for integrals. Thus $1$-densities are also sometimes called volume densities or just densities.

A tensor density is just a section of a tensor product bundle of the form $T^{(k,l)}M\otimes \Omega^s M$. The transition functions for such a section are the transition functions for an ordinary $(k,l)$-tensor multiplied by $|J|^s$. To distinguish them from tensor densities, ordinary densities are sometimes called scalar densities.

Note that the counting of weights is sometimes done differently in conformal geometry. A density of conformal weight $r$ is typically defined to be a density of weight $r/n$. This is so that the scaling of of densities matches that of lengths.

This is all explained pretty well in this Wikipedia article.

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  • $\begingroup$ Thank you! I was aware of the wiki page, but it was only enough to give me a general overview, I found many descriptions there to be either too vague, too abstract (I don't mind that, but when I first encounter something, I tend to only understand it when I see it in a form that is directly calculatable) or went too deep into bundles which I am not familiar with. I have two questions if you don't mind. $\endgroup$ Mar 29 '15 at 18:21
  • $\begingroup$ First is, when you mean tensor product of bundles, is that construction the same as taking a tensor product space at $p$, then taking its tensor product with the vector space of densities, then taking the disjunct union over $M$? Secondly, I assume tensor densities are not integrable, right? Just scalar densities? If so, what sort of applications exist for them? I know vector-valued differential forms are used in physics, but I'd assume their tensorial and skew-symmetric nature is why they're used. Is there any obvious application of tensor densities? $\endgroup$ Mar 29 '15 at 18:24
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    $\begingroup$ About tensor products of bundles: yes, if $E$ and $F$ are vector bundles, then $E\otimes F$ is the bundle whose fiber at each point $p\in M$ is $E_p \otimes F_p$. $\endgroup$
    – Jack Lee
    Mar 29 '15 at 20:31
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    $\begingroup$ @Uldreth: About tensor densities: Right, they're not integrable all by themselves. But you can pair them with other tensor densities and get something integrable. For example, two sections of $\Omega^{1/2}M$ can be multiplied together to get a volume density, which can be integrated; this is a way of creating an intrinsic $L^2$ inner product without a metric. Also, a section of $TM\otimes \Omega^{1/2}M$ can be paired with a section of $T^*M\otimes \Omega^{1/2}M$ to get a volume density, which can be integrated. $\endgroup$
    – Jack Lee
    Mar 29 '15 at 20:35
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    $\begingroup$ @Uldreth: Another important application of tensor densities is in conformal geometry, where certain curvature tensors can be viewed invariantly as tensor densities without having to choose a particular conformal representative for the metric. $\endgroup$
    – Jack Lee
    Mar 29 '15 at 20:36

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