Writing Integrals using Differential Forms

Consider some smooth curve $C \subset \mathbb{R^n}$ and $\gamma:[a,b] \subset\mathbb{R}\rightarrow C$ a parametrisation of $C$ and a continuous vector field $K:\mathbb{R^n} \rightarrow \mathbb{R^n}$. Let $\omega = K_{1}dx^{1}+...+K_{n} dx^{n}$ where $K_{1},...,K_{n}$ are the components of $K$ with respect to the standard basis of $\mathbb{R^n}$. Now the following holds: $$\int_{c}\vec{K}\cdot\vec{ds} = \int_{c}\vec{K}\cdot\hat{n}\space ds:=\int_{a}^{b}\langle K(\gamma(t)),\dot{\gamma}(t)\rangle \space dt =\int_{a}^{b}\sum_{i=1}^{n}K_{i}(\gamma(t))\space\dot{\gamma_{i}}(t)\space dt$$ where $\langle.,.\rangle$ is the standard inner product.

One can also write the same integral using a differential form:

$$\int_{\gamma}\omega:=\int_{a}^{b}\gamma^{*}\omega=\int_{a}^{b}\omega(\gamma(t))\space \dot{\gamma}(t)\space dt =\int_{a}^{b}\sum_{i=1}^{n}K_{i}(\gamma(t))\space\dot{\gamma_{i}}(t)\space dt$$

Similarly let $S \subset \mathbb{R^3}$ be a smooth surface (2-dim submanifold) and $\phi:U\subset\mathbb{R^2}\rightarrow S$ a parametrisation of $S$. $\space F:\mathbb{R^3}\rightarrow\mathbb{R^3}$ a continuous vectorfield. Let $\eta = F_{1}\space dx\wedge dy -F_{2}\space dx \wedge dz +F_{3}\space dy \wedge dz$ where $F_{1},F_{2},F_{3}$ are the components of F with respect to the standard basis of $\mathbb{R^3}$. Now the following holds:

$$\int_{S}\vec{F}\cdot \vec{dA} = \int_{S}\vec{F}\cdot\hat{n}\space dA :=\int_{U}\langle F,\frac{\partial\phi}{\partial u}\times\frac{\partial\phi}{\partial v}\rangle\space d\mu(u,v)$$ And the same Integral using the differential form: $$\int_{S}\eta:=\int_{U}\phi^{*}\eta= \int_{U}\langle F,\frac{\partial\phi}{\partial u}\times\frac{\partial\phi}{\partial v}\rangle\space d\mu(u,v)$$

My Question is: How do I express the following integrals using differential forms?
Let $\space f:\mathbb{R^n}\rightarrow\mathbb{R}$ and $g:\mathbb{R^3}\rightarrow \mathbb{R}$ be continuous functions.

$$\int_{c}f\space ds := \int_{a}^{b}f(\gamma(t))\space\lVert\dot{\gamma}(t)\rVert\space dt$$

$$\int_{S}g\space dA := \int_{U} g(\phi(u,v)) \space\lVert\frac{\partial\phi}{\partial u}\times\frac{\partial\phi}{\partial v}\rVert\space d\mu(u,v)$$

Help is greatly appreciated.
Vincent Pfenninger

Solution 1:

For an oriented m-dimensional Riemannian manifold $(M,g)$ there is a unique m-form $\omega$ such that $\omega_{p}(e_{1},\ldots,e_{m})=1$ for $\lbrace e_{i} \rbrace_{i=1}^{m}\subset T_{p}M$ a g-orthonormal basis ordered according to the orientation. For a function $f$ on $M$ which is sufficiently nice one defines $\int\limits_{M} f:= \int\limits_{M}f \cdot \omega$. For a chart $\phi: U \to \phi(U)=:O$ it is easy to check that $\phi^{*}\omega=\sqrt{\det(g(x))} dx^{1}\wedge \cdots \wedge dx^{m}$ where $g$ is the matrix of $\phi^{*}g = \sum\limits_{i,j}g_{ij}(x) dx^{i} \otimes dx^{j}$. Thus $$ \begin{aligned}\int\limits_{O}f &= \int\limits_{O}f \cdot \omega = \int\limits_{U} \phi^{*}(f \cdot \omega) \\ &=\int\limits_{U}f(\phi(x))\sqrt{\det(g(x))} dx^{1}\wedge \cdots \wedge dx^{m} \equiv \int\limits_{U}f(\phi(x))\sqrt{\det(g(x))} dx. \end{aligned}$$

You are dealing with submanifolds $M$ of Euclidean space $(\mathbb{R}^{n},\delta)$, for which you naturally use the Riemannian metric induced by restricting the Euclidean metric to your submanifold, $g=\delta\big\vert_{M}$. As $\delta=\sum\limits_{i=1}^{n}dy^{i} \otimes dy^{i}$ for the usual coordinates you have for a chart $\det(g(x))=\det((D\phi(x))^{t}D\phi(x))$ where $D\phi(x)$ is the matrix of the differential of $\phi$, leaving us with $$ \int\limits_{O} f = \int\limits_{U}f(\phi(x))\sqrt{\det((D\phi(x))^{t}D\phi(x))} dx.$$ For curves $\gamma: I \to \mathbb{R}^{n}$ it obviously reduces to $\det(g(t))=\Vert \dot{\gamma}(t) \Vert_{2}^{2}$, giving the formula you wrote down. For a surface in $\mathbb{R}^{3}$ it also gives the formula you want, but I leave it to you to check that.

Remark. You probably are not familiar with certain notions I have used here, since no one actually introduces these notions in Analysis I/II in the same generality. You might want to look those things up if you are really interested into certain details or wait for the course in Differential Geometry.

Solution 2:

Most of the literature about differential forms is about exterior differential forms, and the differential forms that you are integrating in your question are not exterior differential forms. Nevertheless, they are perfectly sensible things.

Note that $\vec{\mathrm{d}s} = \hat{n} \,\mathrm{d}s = (\mathrm{d}x_1, \mathrm{d}x_2, \ldots, \mathrm{d}x_n)$, which is why $\omega = \vec{K} \cdot \mathrm{d}\vec{s} = (K_1, K_2, \ldots, K_n) \cdot (\mathrm{d}x_1, \mathrm{d}x_2, \ldots, \mathrm{d}x_n) = K_1 \mathrm{d}x_1 + K_2 \mathrm{d}x_2 + \cdots + K_n \mathrm{d}x_n$. So it is no mystery what $\mathrm{d}s$ is; it is $\mathrm{d}s = \|\vec{\mathrm{d}s}\| = \|(x_1, x_2, \ldots, x_n)\| = \sqrt{\mathrm{d}x_1^2 + \mathrm{d}x_2^2 + \cdots + \mathrm{d}x_n^2}$. (I switched your superscripts to subscripts to make this last expression easier to write. Since the Euclidean metric plays an essential role here, we wouldn't be able to maintain the distinction anyway.) Consequently $f \,\mathrm{d}s = f \sqrt{\mathrm{d}x_1^2 + \mathrm{d}x_2^2 + \cdots + \mathrm{d}x_n^2}$, and that is the differential form that you are integrating.

Similarly, $\vec{\mathrm{d}A} = \hat{n} \,\mathrm{d}A = (\mathrm{d}y \wedge \mathrm{d}z, -\mathrm{d}x \wedge \mathrm{d}z, \mathrm{d}x \wedge \mathrm{d}y)$, which is why $\eta = \vec{F} \cdot \vec{\mathrm{d}A} = (F_1, F_2, F_3) \cdot (\mathrm{d}y \wedge \mathrm{d}z, -\mathrm{d}x \wedge \mathrm{d}z, \mathrm{d}x \wedge \mathrm{d}y) = F_1 \mathrm{d}y \wedge \mathrm{d}z - F_2 \mathrm{d}x \wedge \mathrm{d}z + F_3 \mathrm{d}x \wedge \mathrm{d}y$, so $\mathrm{d}A = \|\vec{\mathrm{d}A}\| = \|(\mathrm{d}y \wedge \mathrm{d}z, -\mathrm{d}x \wedge \mathrm{d}z, \mathrm{d}x \wedge \mathrm{d}y)\| = \sqrt{(\mathrm{d}y \wedge \mathrm{d}z)^2 + (\mathrm{d}x \wedge \mathrm{d}z)^2 + (\mathrm{d}x \wedge \mathrm{d}y)^2}$. Consequently, $g \,\mathrm{d}A = g \sqrt{(\mathrm{d}y \wedge \mathrm{d}z)^2 + (\mathrm{d}x \wedge \mathrm{d}z)^2 + (\mathrm{d}x \wedge \mathrm{d}y)^2}$ is the other differential form that you're integrating.

Now, you might ask, so what are these things, that I am calling differential forms, even though they are not exterior differential forms? How I answer this depends on how you would answer what exterior differential forms are. One common answer to this last question is that they are fields of multilinear alternating functions of tangent vectors, and if that's how you think of differential forms, then all that you have to do is to drop the word ‘multilinear’. (You can also drop the word ‘alternating’, but it's not necessary to do that today.) For example, while $\mathrm{d}x_1$ takes a tangent vector and returns its first component in the standard basis, and the vector-valued $1$-form $\vec{\mathrm{d}s} = (\mathrm{d}x_1, \mathrm{d}x_2, \ldots, \mathrm{d}x_n)$ takes a tangent vector and returns, well, the exact same vector (since we're in $\mathbb{R}^n$), so $\mathrm{d}s = \sqrt{\mathrm{d}x_1^2 + \mathrm{d}x_2^2 + \cdots + \mathrm{d}x_n^2}$ takes a tangent vector and returns its length. Similarly, while $\mathrm{d}y \wedge \mathrm{d}z$ takes two vectors $\vec{v}$ and $\vec{w}$ and returns $v_2 w_3 - v_3 w_2$ (possibly divided by $2$, depending on what conventions you adopt), and the vector-valued $2$-form $\vec{\mathrm{d}A} = (\mathrm{d}y \wedge \mathrm{d}z, -\mathrm{d}x \wedge \mathrm{d}z, \mathrm{d}x \wedge \mathrm{d}y)$ takes two vectors and returns their cross product (possibly divided by $2$), so $\mathrm{d}A = \sqrt{(\mathrm{d}y \wedge \mathrm{d}z)^2 + (\mathrm{d}x \wedge \mathrm{d}z)^2 + (\mathrm{d}x \wedge \mathrm{d}y)^2}$ takes two vectors and returns the area of the parallelogram (or triangle if you've been dividing by $2$) spanned by them.

These are not linear operations; they are not even homogeneous, although they are absolute-homogenous; that is, if you multiply an input vector by a scalar $c$, then the result is multiplied by $|c|$ rather than by $c$. The latter of the two is also not antisymmetric, but symmetric instead; that is, if you swap the two vectors, then the result is not replaced with its opposite, but instead is unchanged. It is still alternating, however; that is, if the two input vectors are equal, then the result is zero. (A multilinear alternating function must be antisymmetric, but linearity is necessary here.)

I have called such things absolute differential forms, and you can read about them on the nLab. However, they were earlier studied by Gelfand & Gindikin, and you can read about that starting from a MathOverflow answer. Actually, you can consider differential forms in far more generality than that; if you can write a function of a vector as a formula in terms of its components, then by writing $\mathrm{d}x_i$ for the $i$th component in the formula, you get a differential form, and this can be extended to functions of several vectors with some care. However, if these forms are not at least alternating and positive-homogeneous (meaning that if you multiply an input vector by a positive scalar $c$, then the result is multiplied by $c$, a property that both homogenous forms and absolute-homogeneous forms have), then their integrals tend to be uninteresting, that's all.