why integrating only alternating forms?

Question

Hello I was reviewing some concepts of differential forms. I cannot recall why only multilinear alternating forms can be integrated on manyfolds and not general multilinear forms... Why is the hypothesis of being alternating so important? In an intuitive way I'd like to understand it... Is it for having a covariant concept? Or for making some riemann sums convergent?

Answer 1

When defining an integral over manifolds, you want it to resemble integration in local coordinates. Apart from the possibility of a vector valued integral (which I have not seen yet), this means you can only integrate sections of some line bundle $L$ (i.e. a vector bundle of rank 1).

So it is not true, that you can integrate arbitrary differential forms. Furthermore the integral in $ℝ^n$ transforms under diffeomorphisms $ϕ$ with $|\det(Dϕ)|$ (compare also Why do differential forms and integrands have different transformation behaviours under diffeomorphisms?), so the coordinate representation of $L$ should transform accordingly. This leads to the notion of the density bundle $L = |Λ|^1(M)$, where integration is well-defined.

As for form bundles $Λ^k(M)$, you need (at least implicitly) an isomorphism from $Λ^k(M)$ to $L$, which then defines integration on $Λ^k(M)$. This leaves only $0$-forms (i.e. the trivial line bundle) and $\dim M$-forms (i.e. volume forms). To define an isomorphism $Λ^n(M) → L$ the manifold has to be orientable and you need to choose an orientation. For $C^∞(M) → L$ multiplication by a certain density $μ ∈ Γ(L)$ suffices.

So alternating forms do not have much to do with it. They are just a way to construct $Λ^n(M)$, which is isomorphic to $|Λ|^1(M)$ if the manifold is orientable.

Answer 2

Let $\alpha$ be a covariant $k$ -tensor on a finite-dimensional $n$ vector space $V .$ The following are equivalent:

1. $\alpha$ is alternating.
2. $\alpha\left(v_{1}, \ldots, v_{k}\right)=0$ whenever the $k$ -tuple $\left(v_{1}, \ldots, v_{k}\right)$ is linearly dependent.
3. $\alpha$ gives the value zero whenever two of its arguments are equal: $$ \alpha\left(v_{1}, \ldots, w, \ldots, w, \ldots, v_{k}\right)=0 $$

The determinant can be defined as an $n$ alternating form such that $\det(I_n)=1$. This hints that alternating forms behave well with volume. As you want lower dimensional subspace to have zero volume in space.

For a $C^{\infty}$ function $f(x)$ on $\mathbb{R}^n$ with compact support, its Riemann integral \begin{equation}\label{eq:Riemann_integral} \int_{\mathbb{R}^{n}} f(x) d x_{1} \cdots d x_{n}=\lim _{\left|\sigma_{j}\right| \rightarrow 0} \sum_{j} f\left(x_{j}\right)\left|\sigma_{j}\right| \end{equation} is defined. Here the $\sigma_{j}$ are $n$ -dimensional small cubes which altogether cover supp $f$, $x_{j}$ is a point on $\sigma_{j},$ and $\left|\sigma_{j}\right|$ is the volume of $\sigma_{j}$. we have that translating $\sigma_i$ to the origin gives $|\sigma_i|=|\det \sigma_i|$. However, if we change the viewpoint slightly and integration of $n$-forms (putting in the symbols $\wedge$ also), and set $$ \omega=f(x) d x_{1} \wedge \cdots \wedge d x_{n} $$ Then this is nothing but an $n$ -form on $\mathbb{R}^{n}$. We may consider that the integral $\int_{\mathbf{R}^{n}} \omega$ of $\omega$ on $\mathbb{R}^{n}$ is defined by the riemann integral However, two problems arise. One is that if we change the order of $d x_{1}$ and $d x_{2},$ for example, in the formula of $\omega,$ the sign changes, while in the formula for the riemann integral the integral value does not change. This is where the covariant concept comes to play.

In another set of coordinates we have $$\omega=f(x(y)) \operatorname{det}\left(\frac{\partial x_{i}}{\partial y_{j}}\right) d y_{1} \wedge \cdots \wedge d y_{n}$$ By the formula of variable change, we have $$ \int_{\mathbf{R}^{n}} f(x) d x_{1} \cdots d x_{n}=\int_{\mathbf{R}^{n}} f(x(y))\left|\operatorname{det}\left(\frac{\partial x_{i}}{\partial y_{j}}\right)\right| d y_{1} \cdots d y_{n} $$ This means that if we take coordinates $y_1, \dots y_n$ that have the same orientation the integral will have the same value. The change in order of $dx_1$ and $dx_2$ was given by $(x_1,x_2,x_3 \dots,x_n) \mapsto (x_2,x_1,x_3\dots,x_n)$ an orientation reversing diffeomorphism thus we solved the problem of sign ambiguity of the riemann integral once we choose an orientation.

Bodo has a nice answer and this answer tries to focus on the isomorphism to the density bundle he wrote about and connect it to the concepts you talked about. An added bonus of this discussion is that you may understand why alternating forms of lower degrees are also interesting as you can integrate them on cycles the content of the celebrated de rham theorem De Rham's theorem.

Answer 3

$\def\d{\mathrm{d}} \def\np{\star} \newcommand\pder[2][]{\frac{\partial #1}{\partial #2}}$Here we motivate this fact with a small example.

Suppose we wish to integrate something of the form $\d x^1\np \d x^2$, where $\np$ is some, not necessarily alternating, product. (The superscripts here are simply labels.) We assume that the product is distributive and acts nicely with respect to scalar multiplication. Under coordinate transformation we find \begin{align*} \d x^1\np \d x^2 &= \left(\pder[x^1]{y^1}\d y^1+\pder[x^1]{y^2}\d y^2\right) \np \left(\pder[x^2]{y^1}\d y^1+\pder[x^2]{y^2}\d y^2\right) \\ &=\pder[x^1]{y^1}\pder[x^2]{y^1}\d y^1\np\d y^1 + \pder[x^1]{y^1}\pder[x^2]{y^2}\d y^1\np\d y^2 \\ & \quad + \pder[x^1]{y^2}\pder[x^2]{y^1}\d y^2\np\d y^1 + \pder[x^1]{y^2}\pder[x^2]{y^2}\d y^2\np\d y^2. \end{align*} To agree with our standard notion of integration, we must have $$\d x^1\star\d x^2 = \det\pder[x^i]{y^j} \d y^1\np \d y^2.$$ We immediately find \begin{align*} \d y^1\np\d y^1 &= \d y^2\np\d y^2 = 0 \\ \d y^2\np\d y^1 &= -\d y^1\np\d y^2, \end{align*} that is, the product $\np$ is antisymmetric.