The correct interpretation for the integral of a differential $k$-form in terms of typical vector-calculus multiple integrals

differential-formsdifferential-geometryintegration

Preliminary context:

Prof. Kennan Crane from CMU posted a course on discrete differential geometry. I've been able to follow so far, until Lecture 7. Please let me give some context on how he defines the stuff before posing my question. Basically, a vector is defined in terms of its components as a linear combination of the basis vectors denoted with $\left\{e_i=\frac{\partial}{\partial x_i}\right\}_{i=1}^n$. Moreover $k$-vectors are defined as linear combinations of $e_{i_1}\wedge\dots\wedge e_{i_k}$ for some indices $i_1,\dots,i_k$ (see Lecture 3). Then, it is explained that forms are basically a generalization of covectors (in euclidean space, the dual of vectors which can be thought as row vectors vs column vectors), which are linear transformations from vectors to scalars. Then, covectors (or 1-forms) are given as linear combinations of the basis 1-forms $\{dx_i\}_{i=1}^n$, and $k$-forms are linear combinations of $dx_{i_1}\wedge\dots\wedge dx_{i_k}$. Until this point, the symbols $dx_i$ and $\frac{\partial}{\partial x_i}$ are not related to derivatives or differentials, but are just basis vectors/forms. Prof. Crane explains that $k$-forms are meant to be applied to $k$-vectors in order to obtain a scalar measure. For example the area of the parallelepiped $u\wedge v$ for two vectors $u,v$ can be measured by a 2-form $\alpha\wedge\beta$ for two 1-forms $\alpha,\beta$, just scaled by the "size" of $\alpha,\beta$. In particular the concrete relation between forms and vectors basis is given by
$$
dx_i\left(\frac{\partial}{\partial x_j}\right) = \delta_{ij}
$$
where $\delta_{ij}$ is the Kronecker delta. Moreover, the application $(\alpha\wedge \beta)(u, v)=\alpha(u)\beta(v)-\alpha(v)\beta(u)$ which can be written as a determinant (and generalized for the $k$-forms).

Now comes my questions all regarding Lecture 7.

  1. At 6:23, the motivation behind the definition of the integral of a 2-form $\omega$ is that it is the summation of $\omega_{p_i}$ (evaluate $\omega$ at the point $p_i$) applied to a pair of vectors $u_i=u({p_i}),v_i=v(p_i)$ (for some vector fields $u,v$) along with the whole region $\Omega$:
    $$
    \sum_{i}\omega_{p_i}(u_i,v_i)\to \int_{\Omega}\omega
    $$     However, in 8:31 we have the integral of a form $(x+xy)dx\wedge dy$ over $\Omega$ which is the unit square. Then, somehow it follows that
    $$
    \int_{\Omega} (x+xy)dx\wedge dy = \int_0^1\int_0^1 (x+xy)dxdy
    $$
    where I don't see to what vectors $u,v$ are we applying the form $dx\wedge dy$. Mainly, I'm lost at the motivation behind the interpretation of the 2-form $dx\wedge dy$ as a scalar $dxdy$ inside the integral, given that $dx$ and $dy$ were supposed to be just basis forms (covectors). The following may be too hand wavy, but I'm trying to get intuition about this: My thoughts are that we make a parallelepiped spanned by the vectors $u=(dx)\frac{\partial}{\partial x}$ and $v=(dy)\frac{\partial}{\partial y}$ where here $dx$ and $dy$ are small displacements in the directions $\frac{\partial}{\partial x},\frac{\partial}{\partial y}$. This parallelepiped has area $dxdy$, and we can measure such area by the form $dx\wedge dy$. Hence, we use $dx\wedge dy$ to measure this parallelepiped obtaining $dx\wedge dy(u,v)=dxdy$. However, this reasoning seems artificial to me (since we didnt't take advantage of the notation dx,dy for the basis covectors, instead we used other displacements $dxdy$ different from the basis covectors). It would be better to have defined $dx(\frac{\partial}{\partial x})$="$dx$" (where the second $dx$ is the displacement and not the covector) instead of $dx(\frac{\partial}{\partial x})=1$ as we did before (move the displacements $dx$ and $dy$ to the definition of the covectors $dx$ and $dy$ instead of the parallelepiped $u\wedge v$).

  2. At 14:00, one finds the integral of a 1-form $\alpha=dy$ over the circular curve $S^1$ (parametrized as $\gamma:[0,2\pi)\to\mathbb{R}^2, \gamma(s) = (\cos(s),\sin(s))$), which somehow results in:
    $$
    \int_{S^1}\alpha = \int_0^{2\pi} \alpha_\gamma(T(s))ds
    $$
    where $T(s)$ is the tangent vector to $\gamma(s)$. Here we are clearly applying the form $\alpha$ to a vector, different from my previous doubt. Howeve, different than before, instead of using $dy$ as the differential in the integral, we introduced an additional $ds$ where I don't know where it came from.

I already know how to compute this integrals, and I know that their expressions make sense from the typical vector calculus perspective. But I want to understand what is the systematic way of translating the integral of $k$-forms to a multiple-integral. My confusion comes from the fact that it appears that very different "rules" were used to translate $\int_{\Omega} (x+xy)dx\wedge dy = \int_0^1\int_0^1 (x+xy)dxdy$ as in the case $\int_{S^1}\alpha = \int_0^{2\pi} \alpha_\gamma(T(s))ds$.

Best Answer

Let's say $\varphi \colon U \rightarrow \mathbb{R}^n$ is an embedding where $U \subseteq \mathbb{R}^k$ is an open subset so that $\varphi(U)$ is a $k$-dimensional submanifold of $\mathbb{R}^n$ and let $\omega$ be a $k$-form on $\mathbb{R}^n$. Write $\varphi = \varphi(x^1, \dots, x^k)$. The integral of $\omega$ over $\varphi(U)$ is then $$ \int_{\varphi(U)} \omega = \int_{U} \omega_{\varphi(x^1, \dots, x^k)} \left( \frac{\partial \varphi}{\partial x^1}, \dots, \frac{\partial \varphi}{\partial x^k} \right) \, dx^1 \dots dx^k. $$

Here, the left hand side is an integral of a different $k$-form on a submanifold while the right hand side is the Riemann/Lebesgue integral of a function over an open set (so there are no wedges and the notation $dx^1 \dots dx^k$ is just symbolic). More explicitly, at any point $(x^1, \dots, x^k)$ in the domain $U$ of your parametrization, the partial derivatives $\frac{\partial \varphi}{\partial x^1}, \dots, \frac{\partial \varphi}{\partial x^k}$ give you a special ($\varphi$-dependent) basis for the tangent space to $\varphi(U)$ at $\varphi(x^1,\dots,x^k)$. At each point you plug this basis into the $k$-form (in the specific order!) and get a number. This process gives you a function which you then integrate over $U$ (in the sense of Riemann/Lebesgue integral).

Given this description, let's see how your two examples fit in:

  1. When $\Omega \subseteq \mathbb{R}^2$ is an open subset and $\omega$ is a two-form on $\mathbb{R^2}$, we take $\varphi$ to be the inclusion. Let's write $\omega = f(x,y) \, dx \wedge dy$. Then $$ \int_{\varphi(\Omega)} \omega = \int_{\Omega} \omega = \int_{\Omega} \left( f(x,y) \, dx \wedge dy \right) \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) \, dx dy = \int_{\Omega} f(x,y) dxdy$$ as $\left( dx \wedge dy \right) \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) \equiv 1$. Thus, the direct effect is just "removing the wedge symbol" and treating it as a regular integral of a function.
  2. For the case of the circle, take $U = (0,2\pi)$ and $\varphi(s) = \alpha(s) = (\cos(s), \sin(s))$. Then $$ \int_{\varphi(U)} \alpha = \int_{S^1 \setminus \{ \textrm{pt} \}} \alpha = \int_{0}^{2\pi} \alpha_{\varphi(s)} \left( \alpha'(s) \right) \, ds.$$ In the particular case $\alpha = dy$, we have $$ \alpha_{\varphi(s)} \left( \alpha'(s) \right) = dy \left( -\sin(s) \frac{\partial}{\partial x} + \cos(s) \frac{\partial}{\partial y} \right) = \cos(s) $$ so you are left with the integral $$ \int_{0}^{2\pi} \cos(s) \, ds. $$ Strictly speaking, this calculates the integral of $\alpha$ over $S^1 \setminus \{ \textrm{pt} \}$ and not over $S^1$ but a point has measure zero in $S^1$ and so can be ignored in the calculation under appropriate assumptions.
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