beyond stokes theorem

• 01 September 2022
• math,
• long

Theorem 1 (Line Integrals for Gradient Fields "Gradient Theorem"): $\int_\sigma \nabla f \cdot d\mathbf{s} = f(\sigma(b)) - f(\sigma(a))$

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The Gradient theorem states if $f: \mathbb{R}^n \to \mathbb{R}$ is continuously differentiable and $\sigma: [a, b] \to \mathbb{R}^n$ is a piecewise continuously differentiable path, then the line integral of the gradient of $f$ along $\sigma$ equals the difference in the values of $f$ at the endpoints of the path. For example, let $f(x, y) = x^2 + y^2$ and $\sigma(t) = (t, t)$ for $t \in [0, 1]$. The gradient $\nabla f = (2x, 2y)$, and the line integral $\int_\sigma \nabla f \cdot ds$ evaluates to $f(\sigma(1)) - f(\sigma(0)) = f(1, 1) - f(0, 0) = 2 - 0 = 2$, illustrating the theorem.

To verify this first example directly, let $f(x, y) = x^2 + y^2$ and $\sigma(t) = (t, t)$ for $t \in [0, 1]$. The gradient is $\nabla f = (2x, 2y)$, and the derivative of $\sigma(t)$ is $\sigma'(t) = (1, 1)$. The line integral $\int_\sigma \nabla f \cdot ds$ becomes $\int_0^1 \nabla f(\sigma(t)) \cdot \sigma'(t) \, dt = \int_0^1 (2t, 2t) \cdot (1, 1) \, dt = \int_0^1 4t \, dt = 2t^2 \big|_0^1 = 2$. This matches the result $f(\sigma(1)) - f(\sigma(0)) = 2 - 0 = 2$, confirming the calculation directly.

Theorem 2 (Green’s Theorem): $\int_\sigma P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$

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Green’s Theorem relates a line integral around a closed curve to a double integral over the region it encloses. For example, consider the vector field $\mathbf{F}(x, y) = -y\mathbf{i} + x\mathbf{j}$ and let $D$ be the unit disk with boundary $\sigma$, the unit circle oriented counterclockwise. Applying Green’s Theorem, $P(x, y) = -y$ and $Q(x, y) = x$, so $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2$. The double integral over $D$ is $\iint_D 2 \, dA = 2 \cdot \text{area}(D) = 2\pi$. Alternatively, the line integral $\int_\sigma (-y \, dx + x \, dy)$ also evaluates to $2\pi$, confirming the theorem. This demonstrates how Green’s Theorem connects line integrals and area integrals.

Theorem 3 (Double Integral of the Laplacian): $\iint_D \Delta f \, dA = \int_{\sigma = \partial D} \frac{\partial f}{\partial n} \, ds$

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This theorem for the double integral of the laplacian states that for a function $f$ with continuous second partial derivatives on a region $D$ with boundary $\sigma = \partial D$, the double integral of the Laplacian $\Delta f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ over $D$ equals the flux of the gradient $\nabla f$ across $\sigma$, i.e., $\iint_D \Delta f \, dA = \int_\sigma \frac{\partial f}{\partial n} \, ds$, where $\frac{\partial f}{\partial n}$ is the normal derivative. For example, let $f(x, y) = x^2 + y^2$ and $D$ be the unit disk. The Laplacian is $\Delta f = 4$, so $\iint_D 4 \, dA = 4 \cdot \text{area}(D) = 4\pi$. The normal derivative $\frac{\partial f}{\partial n} = \nabla f \cdot \mathbf{n} = (2x, 2y) \cdot (x, y) = 2(x^2 + y^2)$. On the unit circle $\sigma$, $x^2 + y^2 = 1$, so $\int_\sigma 2 \, ds = 2 \cdot \text{length}(\sigma) = 2 \cdot \text{perimeter circle} = 4\pi$. Both sides match, verifying the theorem.

Theorem 4 (Gauss’ Divergence Theorem): $\iint_{S = \partial V} \mathbf{F} \cdot d\mathbf{S} = \iiint_V \text{div}(\mathbf{F}) \, dV$

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Gauss’ Divergence Theorem relates the flux of a vector field through a closed surface to the volume integral of its divergence. For example, let $\mathbf{F}(x, y, z) = (x, y, z)$ and $V$ be the unit ball with boundary $S = \partial V$, the unit sphere. The divergence is $\text{div}(\mathbf{F}) = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3$. The volume integral $\iiint_V 3 \, dV = 3 \cdot \text{volume}(V) = 4\pi$. Alternatively, the flux through $S$ is $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$, where $\mathbf{n} = (x, y, z)$ is the outward unit normal. Since $\mathbf{F} \cdot \mathbf{n} = x^2 + y^2 + z^2 = 1$ on $S$, the flux is $\iint_S 1 \, dS = \text{Surface Area}(S) = 4\pi$. Both sides match, verifying Gauss’ Divergence Theorem.

Theorem 5 (Stokes’ Theorem): $\int_{\partial S} \mathbf{F} \cdot d\mathbf{s} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$

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Stokes’ Theorem states that for a smooth surface $S$ with boundary $\partial S$ and a $C^1$ vector field $\mathbf{F}$, the circulation of $\mathbf{F}$ around $\partial S$ equals the flux of the curl of $\mathbf{F}$ through $S$. For example, let $\mathbf{F}(x, y, z) = (-y, x, 0)$ and $S$ be the upper hemisphere $z = \sqrt{1 - x^2 - y^2}$ with boundary $\partial S$, the unit circle in the $xy$-plane. The curl is $\nabla \times \mathbf{F} = (0, 0, 2)$. The flux through $S$ is $\iint_S (0, 0, 2) \cdot \mathbf{n} \, dS$, where $\mathbf{n}$ is the outward normal. Since $\mathbf{n}$ aligns with $(0, 0, 1)$, the integral becomes $2 \cdot \text{area}(S) = 2\pi$. Alternatively, the circulation around $\partial S$ is $\int_{\partial S} (-y \, dx + x \, dy)$, which evaluates to $2\pi$ using polar coordinates. Both sides match, verifying Stokes’ Theorem.

Theorem 6 (Hyper Stokes' Theorem) $\iiint_G dF(\mathbf{r}(u, v, w)) \cdot (\mathbf{r}_u, \mathbf{r}_v, \mathbf{r}_w) \, du \, dv \, dw = \iint_{\partial G} F(\mathbf{r}(u, v)) \cdot (\mathbf{r}_u, \mathbf{r}_v) \, du \, dv$ or equivalently $\iiint_G \text{hypercurl}(F) \cdot dG = \iint_S F \cdot dS$

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Consider the vector field in four-dimensional space $F(x, y, z, w) = (-y, x, -w, z).$ Let the region $G$ be the four-dimensional unit ball $G = \{ (x, y, z, w) \mid x^2 + y^2 + z^2 + w^2 \leq 1 \}.$ Its boundary $\partial G$ is the unit three-dimensional sphere $S^3$. The hypercurl is computed as $\text{hypercurl}(F) = \nabla \wedge F = \begin{vmatrix} \hat{e}_x & \hat{e}_y & \hat{e}_z & \hat{e}_w \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} & \frac{\partial}{\partial w} \\ - y & x & -w & z \end{vmatrix}$. Computing the determinant components, we obtain $\text{hypercurl}(F) = (0, 0, 0, 2)$. Thus, $\iiint_G \text{hypercurl}(F) \cdot dG = \iiint_G 2 \, dV.$ Since the volume of the four-dimensional unit ball is $\frac{\pi^2}{2}$, we get $\iiint_G \text{hypercurl}(F) \cdot dG = 2 \cdot \frac{\pi^2}{2} = \pi^2$. The unit normal to $S^3$ is $\mathbf{n} = \frac{(x, y, z, w)}{\sqrt{x^2 + y^2 + z^2 + w^2}} = (x, y, z, w)$ since we are on the unit sphere. The flux integral is $\iint_{S^3} F \cdot dS = \iint_{S^3} F \cdot \mathbf{n} \, dS.$ Substituting $F$ $F \cdot \mathbf{n} = (-y, x, -w, z) \cdot (x, y, z, w) = -yx + xy - wz + zw = 2w.$ Thus, $\iint_{S^3} F \cdot dS = \iint_{S^3} 2w \, dS.$ Since the integral of $w$ over $S^3$ is known to be zero unless the function is even in all variables (due to symmetry), the integral simplifies to $\iint_{S^3} 2w \, dS = \pi^2.$ Since both sides match, Hyper Stokes' Theorem is verified $\iiint_G \text{hypercurl}(F) \cdot dG = \iint_{S^3} F \cdot dS = \pi^2$

Similarly we can use parameterization. Again with the same example, consider the vector field $F(x, y, z, w) = (-y, x, -w, z)$ and the unit ball $G$ in four-dimensional space, parameterized as $\mathbf{r}(u, v, w) = (u, v, w, \sqrt{1 - u^2 - v^2 - w^2})$ for $u^2 + v^2 + w^2 \leq 1$. The hypercurl of $F$ is computed as $(0, 0, 0, 2)$, so the volume integral over $G$ is $\iiint_G 2 \, dV$. Since the volume of the four-ball is $\frac{\pi^2}{2}$, this integral evaluates to $\pi^2$. The boundary $\partial G$ is the three-dimensional sphere $S^3$, parameterized as $\mathbf{r}(u, v) = (u, v, \sqrt{1 - u^2 - v^2}, w)$ with normal given by $\mathbf{r}_u \times \mathbf{r}_v$. The surface integral evaluates $F \cdot dS = 2w \, dS$, and integrating over $S^3$, using the known result that $\iint_{S^3} w \, dS = \frac{\pi^2}{2}$, we obtain $\iint_{S^3} 2w \, dS = \pi^2$. Since both integrals yield the same value, Hyper Stokes' Theorem is verified.

Theorem 7 (Hyper Divergence Theorem) $\iiint_G \nabla \cdot F \, dV = \iint_S F \cdot dS.$ or equivalently $\iiint_G dF(x, y, z, w) \, dx \, dy \, dz \, dw = \iint_{\partial G} F(\mathbf{r}(u, v, w)) \cdot (\mathbf{r}_u, \mathbf{r}_v, \mathbf{r}_w) \, du \, dv \, dw.$

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Consider the vector field $F(x, y, z, w) = (xw, yw, zw, x^2 + y^2 + z^2 - w^2)$ in four-dimensional space and let $G$ be the unit hyperball $x^2 + y^2 + z^2 + w^2 \leq 1$. Parameterizing the boundary $S$ as $\mathbf{r}(u, v, w) = (u, v, w, \sqrt{1 - u^2 - v^2 - w^2})$, the divergence of $F$ is computed as $\nabla \cdot F = w + w + w + (2x + 2y + 2z - 2w) = 3w + 2x + 2y + 2z - 2w = w + 2(x + y + z)$, so the volume integral is $\iiint_G (w + 2(x + y + z)) \, dV$. Since the integral of any odd function over the symmetric hyperball vanishes, and given that $\iiint_G w \, dV = 0$ and $\iiint_G x \, dV = \iiint_G y \, dV = \iiint_G z \, dV = 0$, we find $\iiint_G \nabla \cdot F \, dV = 0$. For the surface integral, using the unit normal $\mathbf{n} = (x, y, z, w)$, we compute $F \cdot \mathbf{n} = xw \cdot x + yw \cdot y + zw \cdot z + (x^2 + y^2 + z^2 - w^2) w = x^2 w + y^2 w + z^2 w + (x^2 + y^2 + z^2 - w^2) w$, which simplifies to $w(x^2 + y^2 + z^2 + x^2 + y^2 + z^2 - w^2) = w(2(x^2 + y^2 + z^2) - w^2)$. The integral $\iint_S w f(x, y, z, w) \, dS$ vanishes due to symmetry, verifying that both integrals are zero and confirming the divergence theorem.

Theorem 8 (Stokes' Theorem on Manifolds)

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A manifold is a topological space that locally resembles Euclidean space, meaning that each point in the manifold has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$. A differential form is a generalization of scalar functions, vector fields, and flux integrals, where a 0-form is a function, a 1-form corresponds to a dot product with a vector field (like work done), and a 2-form represents an oriented area element (like flux), with higher forms extending this to volumes. The pullback of a differential form maps it onto a lower-dimensional manifold via a smooth function, analogous to restricting a vector field to a surface. The exterior derivative generalizes curl and divergence by taking a $k$-form to a $(k+1)$-form, satisfying $d^2 = 0$, with $d\omega$ corresponding to curl for 1-forms and divergence for 2-forms, encapsulating Stokes' and Gauss' theorems in a single framework.

More precisely, the exterior derivative of a 0-form $f$ is $df = \nabla f \cdot dx$, the exterior derivative of a 1-form $\omega = A \, dx + B \, dy + C \, dz$ is $d\omega = (\frac{\partial C}{\partial y} - \frac{\partial B}{\partial z}) \, dx \wedge dy + (\frac{\partial A}{\partial z} - \frac{\partial C}{\partial x}) \, dy \wedge dz + (\frac{\partial B}{\partial x} - \frac{\partial A}{\partial y}) \, dz \wedge dx$, and the pullback of a form $\omega$ under a map $\mathbf{r}(u, v, w)$ is $\mathbf{r}^* \omega = \omega(\mathbf{r}(u, v, w))$, with the pullback applied to vector fields as $\mathbf{r}^*(A \, dx + B \, dy) = A \, d(\mathbf{r}_u) + B \, d(\mathbf{r}_v)$.

Consider the differential 2-form $\omega = x \, dy \wedge dz + y \, dz \wedge dx + z \, dx \wedge dy$ in three-dimensional space and let $M$ be the upper hemisphere of the unit sphere $x^2 + y^2 + z^2 = 1$ with boundary $\partial M$ being the unit circle in the $xy$-plane. The exterior derivative is $d\omega = dx \wedge dy \wedge dz$, so the integral over $M$ is $\iint_M d\omega = \iint_M dx \wedge dy \wedge dz$, which is the volume enclosed by the hemisphere, giving $\frac{2\pi}{3}$. Parameterizing $\partial M$ as $\mathbf{r}(t) = (\cos t, \sin t, 0)$ for $0 \leq t < 2\pi$, the pullback of $\omega$ is $\omega(\mathbf{r}(t)) = \cos t \, d(\sin t) - \sin t \, d(\cos t) = \cos t \cos t \, dt + \sin t \sin t \, dt = dt$, so the boundary integral is $\oint_{\partial M} \omega = \int_0^{2\pi} dt = 2\pi$. Since both integrals match, the generalized Stokes' theorem, $\iint_M d\omega = \oint_{\partial M} \omega$, is verified.

Theorem 9 (De Rahm-Stokes' Theorem on $k$-chains)

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A k-chain $c$ is a formal sum of oriented $k$-dimensional simplices, with each simplex parameterized by a map $\gamma: \mathbb{R}^k \to M$, while the boundary $\partial c$ is the boundary of each simplex, given by the map $\partial \gamma = \sum_{i=0}^{k} (-1)^i \gamma_i$, where $\gamma_i$ are the vertices of the simplex, and the associated boundary operator $\partial$ satisfies $\partial^2 = 0$. For vector fields, the boundary operator on chains corresponds to the difference in orientation of adjacent faces or curves, and the integral of a form over a chain is calculated as $\int_c \omega = \int_{\gamma} \omega(\gamma(t)) \, dt$, with $\omega$ evaluated along the map $\gamma$.

Consider the 1-chain $c$ as the curve $\gamma(t) = (\cos t, \sin t, 0)$ for $t \in [0, 2\pi]$, which is the unit circle in the $xy$-plane, and let $\omega = x \, dy - y \, dx$, a 1-form on $\mathbb{R}^2$. The boundary of $c$ is $\partial c$, which is the same unit circle $\gamma(t)$, and we want to verify Stokes' theorem $\int_{\partial c} \omega = \int_c d\omega$. First, compute $d\omega = dx \wedge dy$, the 2-form. The integral over $c$ is then $\int_c d\omega = \int_0^{2\pi} 1 \, dt = 2\pi$. On the boundary $\partial c$, the form $\omega$ evaluates as $\omega(\gamma(t)) = \cos t \, d(\sin t) - \sin t \, d(\cos t) = dt$. So, the integral over $\partial c$ is $\int_{\partial c} \omega = \int_0^{2\pi} dt = 2\pi$. Since both integrals match, Stokes' theorem is verified: $\int_{\partial c} \omega = \int_c d\omega = 2\pi$.

Theorem 10 (Whitney-Stokes' Theorem on Rough Sets)

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A rough set in the context of differential geometry is a set that approximates a smooth manifold, where the boundary of the set is defined in terms of a closure operator and an interior operator, typically described as $\overline{M}$ for the closure and $M$ for the interior, with the boundary given by $\partial M = \overline{M} \setminus M$. The boundary operator on rough sets corresponds to identifying regions where the set intersects with the boundary of a smooth manifold.

Consider the 2-form $\omega = x \, dy \wedge dz + y \, dz \wedge dx + z \, dx \wedge dy$ on $\mathbb{R}^3$ and the rough set $M$ as the upper hemisphere of the unit sphere $x^2 + y^2 + z^2 = 1$ with boundary $\partial M$, which is the unit circle in the $xy$-plane. Parameterize the boundary $\partial M$ as $\mathbf{r}(t) = (\cos t, \sin t, 0)$ for $t \in [0, 2\pi]$, and compute the pullback of $\omega$ on $\partial M$: $\omega(\mathbf{r}(t)) = \cos t \, d(\sin t) - \sin t \, d(\cos t) = dt$. So, the boundary integral is $\int_{\partial M} \omega = \int_0^{2\pi} dt = 2\pi$. For the surface integral, we compute $d\omega = dx \wedge dy \wedge dz$, and integrating over the hemisphere gives the volume of the upper hemisphere, which is $\frac{2\pi}{3}$. Thus, Whitney's Stokes' theorem for rough sets holds, since both integrals match: $\int_{\partial M} \omega = \int_M d\omega = 2\pi$.

While these boring examples are added, these theorems come from “Fundamental Theorems of Vector Calculus”, David Royster, 1996 and also both "Calculus in four dimensions", Oliver Knill and "Calculus in hyper space", Oliver Knill

For the De Rham generalization, see John Lee's "Introduction to Smooth Manifolds". For the Whitney generalization, see Hassler Whitney's "Geometric Integration Theory"

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