Level 3D: Orientation, Integration, and Stokes

The Defect. The integral theorems of multivariable calculus arrive as three or four unrelated memorised identities: the fundamental theorem of line integrals, Green’s theorem, the divergence theorem, and the classical Stokes theorem, each with its own notation for the same boundary-versus- interior bookkeeping. They are one theorem. The language that reveals this is Level 3C’s exterior derivative plus the integration theory built here, and the proof of the unification is, at bottom, the one-variable fundamental theorem of calculus applied $n$ times.

The One-Line Reveal. Integrating an $n$-form over an oriented region means integrating its coefficient function in an oriented chart; $d$ is the unique operator for which the fundamental theorem of calculus survives the change of viewpoint; Stokes says the two copies of that theorem, one inside and one on the boundary, are the same statement in every coordinate system at once.

Scope. From Level 3C: the exterior derivative, $d^2 = 0$, graded Leibniz, and pullback commutation (Theorem 3.1). From Level 1B: determinants and their signs. Three inputs from analysis are declared, all standard and pointed at the Real Analysis track and any multivariable analysis text: Fubini’s theorem (multiple integrals compute as iterated integrals), the change-of-variables formula (replacement of variables in a multiple integral contributes the absolute value of the Jacobian determinant), and the existence of smooth partitions of unity on second-countable smooth manifolds subordinate to any open cover (a countable family $\{\rho_i\}$ of smooth functions, $0 \leq \rho_i \leq 1$, each with support inside one member of the cover, locally finitely many nonzero, with $\sum_i \rho_i = 1$; construction via the standard smooth bump $t \mapsto e^{-1/t}$, proved in Tu’s book). The one-variable fundamental theorem of calculus is carried from 3C. Everything else is proved.

Definition. Let $\mathbb{H}^n = \{x \in \mathbb{R}^n : x_n \geq 0\}$, the upper half-space, whose interior is $\{x_n > 0\}$ and whose boundary is $\{x_n = 0\}$. A smooth map on a subset of $\mathbb{H}^n$ means a map that extends to a smooth map on an open subset of $\mathbb{R}^n$ containing it. A manifold with boundary of dimension $n$ is a second countable Hausdorff space $M$ with an atlas of homeomorphisms $\varphi: U \to \varphi(U)$ from open sets of $M$ onto open subsets of $\mathbb{H}^n$, with smoothly compatible transitions: each transition $\psi \circ \varphi^{-1}$ extends to a smooth map on an open subset of $\mathbb{R}^n$, with smoothly extendable inverse, that carries the boundary hyperplane $\{x_n = 0\}$ into itself.

Remark (why the boundary clause is in the definition). For the boundary of $M$ to be a well-defined subset, independent of the chart used to look at it, transitions must respect the boundary hyperplane. For atlases arising from level sets and most geometric constructions this is automatic; proving it automatic for every extendable transition is the classical invariance of domain theorem (a homeomorphism between open subsets of $\mathbb{R}^n$ cannot send a boundary point of $\mathbb{H}^n$ to an interior point), which belongs to algebraic topology and is outside this course. Building the clause into the compatibility definition keeps every statement below proved. $\blacksquare$

Definition. The boundary $\partial M$ is the set of points $p$ whose charts map $p$ into $\{x_n = 0\}$; the clause above makes this independent of the chart (if one chart sends $p$ to the boundary hyperplane, the transition to any other chart keeps it there). $\partial M$ carries an $(n-1)$-dimensional manifold structure (without boundary): the restrictions of the charts to $\partial M$, landing in $\{x_n = 0\} \cong \mathbb{R}^{n-1}$, form an atlas with the restricted transitions. The interior is $M \setminus \partial M$. Smooth maps between manifolds with boundary, their differentials, and forms are defined exactly as in 3A, 3B and 3C: all chart representations are now smooth maps on subsets of half-spaces, i.e. extendable ones, and 3C Section 3’s pullback and commutation theorems apply verbatim, since the chain rule underlies them and extends to the boundary (both the representation and its extension restrict to the same map on the half-space part).

Quick check.

1. What is the half-space and where is its boundary?
2. What does smooth on a subset of the half-space mean?
3. Which compatibility clause keeps the boundary well defined?
4. What manifold is the boundary?
5. Which deep theorem would make the boundary clause automatic?

Answers: 1. The set of points with nonnegative last coordinate; the hyperplane where it is zero. 2. Extendable to a smooth map on an open subset of the full space. 3. Transitions carry the boundary hyperplane into itself. 4. An (n minus 1)-dimensional manifold with the restricted charts. 5. Invariance of domain (algebraic topology; outside this course).

Definition. An orientation of an $n$-dimensional vector space $V$ is a choice of one of the two connected components of its set of ordered bases, two bases $(v_1, \dots, v_n)$ and $(w_1, \dots, w_n)$ being equivalent when the change-of-basis matrix has positive determinant. By 1B (Theorem 4.8 and Corollary 4.9), this is an equivalence relation with exactly two classes, and a basis $(e_i)$ is declared positive once and for all, making $\det$ the sign detector: a basis $(w_i)$ is positive iff the matrix with columns $w_i$ (in the $(e_i)$ coordinates) has positive determinant. An oriented manifold is a manifold that admits an atlas in which every transition map has Jacobian determinant positive at every point of its domain; the tangent spaces are then oriented by declaring, in each chart of such an atlas, the transport of the standard basis to be positive. This is well defined: two charts of the atlas transport $T_pM$ to bases differing by the transition’s change-of-basis matrix, whose determinant is positive by hypothesis, so the two transports carry the same orientation (1B). A chart of such an atlas is itself called positively oriented.

Lemma 2.1 (transitions map outward vectors to outward vectors). Let $T$ be a transition of a manifold-with-boundary’s charts, defined (with its smooth extension, also written $T$) near $q \in \{x_n = 0\}$, with smooth extension $S = T^{-1}$ near $T(q)$. If $v \in \mathbb{R}^n$ has $v_n < 0$, then $(dT_q v)_n < 0$.

Proof. The compatibility clause gives: $T$ and $S$ carry the boundary hyperplane to the boundary hyperplane; and, being homeomorphisms of their domains, they carry interior points ($x_n > 0$) to interior points ($y_n > 0$) and exterior points ($x_n < 0$) to exterior points ($y_n < 0$): a bijection maps the complement of a set to the complement of its image, and the boundary hyperplane is carried onto the boundary hyperplane. Let $c(t) = q + tv$ for small $t \geq 0$, so $c_n(t) < 0$ for $t > 0$ and $c(t) \to q$. Then $T(c(t))$ is exterior for every small $t > 0$, so $T_n(c(t)) < 0$, and the one-sided derivative satisfies $(dT_q v)_n = \frac{d}{dt} T_n(c(t))\big|_{t=0} = \lim_{t \to 0^+} T_n(c(t))/t \leq 0$.

Next, $dT_q$ carries the boundary tangent hyperplane $H_0 = \{v_n = 0\}$ into $\{w_n = 0\}$: for $\xi$ with $\xi_n = 0$, the curve $q + t\xi$ stays in $\{x_n = 0\}$, so $T(q + t\xi)$ stays in $\{y_n = 0\}$, and differentiating at $t = 0$ gives $(dT_q \xi)_n = 0$. The differential $dT_q$ is invertible ($S \circ T = \operatorname{id}$, so $dS_{T(q)} \circ dT_q = I$ by the chain rule), and an invertible linear map carrying an $(n-1)$-dimensional subspace into an $(n-1)$-dimensional subspace carries it onto (the image is a subspace of dimension $n-1$ inside an $(n-1)$-dimensional subspace). Hence $v \notin H_0$ maps outside $\{w_n = 0\}$: $(dT_q v)_n \neq 0$. Combined with the inequality: $(dT_q v)_n < 0$. $\blacksquare$

Definition (boundary orientation). At $p \in \partial M$, a tangent vector $v \in T_pM$ is outward if in some (by Lemma 2.1, any) chart $d\varphi_p(v)$ has negative last coordinate; inward if positive. Since $d\varphi_p$ is an isomorphism and outward vectors exist (any vector mapping to a negative multiple of $e_n$), the outward vectors form an open half of $T_pM \setminus \{0\}$. A basis $(w_1, \dots, w_{n-1})$ of $T_p\partial M$ is positively oriented (in the boundary orientation) if, for any outward vector $n_p$, the basis $(n_p, w_1, \dots, w_{n-1})$ of $T_pM$ is positively oriented in the ambient orientation of $T_pM$.

Lemma 2.2 (the boundary orientation is well defined). The definition does not depend on the choice of outward vector $n_p$.

Proof. Let $n_p, n_p'$ be two outward vectors; write $n_p' = c\, n_p + u$ with $u$ in the boundary hyperplane $H = \{v : (d\varphi_p v)_n = 0\} = T_p\partial M$ (possible: both have negative last coordinate under $d\varphi_p$, so $d\varphi_p(n_p') - c\, d\varphi_p(n_p)$ has zero last coordinate for suitable $c$; concretely $c = (n_p')_n / (n_p)_n > 0$ after transport). The change-of-basis matrix from $(n_p, w_1, \dots, w_{n-1})$ to $(n_p', w_1, \dots, w_{n-1})$ is triangular in the chart coordinates, with diagonal entries $c$ (once) and $1$ ($n-1$ times): its determinant is $c > 0$. So the two ambient bases have the same orientation sign (determinants differ by a positive factor, 1B), and the boundary basis is positively oriented with respect to one outward vector iff with respect to the other. $\blacksquare$

Quick check.

1. What is an orientation of a vector space?
2. What makes an atlas oriented?
3. Define the outward vectors at a boundary point.
4. State the outward-first boundary orientation.
5. Why is it independent of the chosen outward vector?

Answers: 1. One of the two classes of bases under positive-determinant equivalence. 2. Every transition has positive Jacobian determinant. 3. Those mapping to negative last coordinate in one (hence any, Lemma 2.1) chart. 4. The outward vector followed by the boundary basis is positively oriented in the ambient tangent space. 5. The change-of-basis between two outward vectors is triangular with positive determinant (Lemma 2.2).

Definition (chart integral). Let $M$ be an oriented $n$-manifold with boundary, and let $\omega$ be a smooth $n$-form whose support is compact and contained in one chart domain $U$, with the chart $\varphi: U \to \varphi(U) \subseteq \mathbb{H}^n$ positively oriented (transition Jacobians positive, and this chart’s own orientation matching the ambient: $\varphi$ is positively oriented if its differential carries the standard basis of $\mathbb{R}^n$ to a positively oriented basis of $T_pM$; an atlas is oriented precisely when all its transitions are positive-Jacobian, so one positive chart makes all charts of that atlas positive). Writing $\omega_\varphi = a\, dx^1 \wedge \cdots \wedge dx^n$ in that chart (a top form has a single coefficient, by 3C’s Proposition 1.2), define

an ordinary multiple integral of a compactly supported continuous function (declared input). The integral over $\partial M$ of an $(n-1)$-form supported in a boundary chart is defined likewise, using the boundary orientation of Section 2 and the identification $\{x_n = 0\} \cong \mathbb{R}^{n-1}$.

Proposition 3.1 (the chart integral is well defined). If $\omega$ is supported in two positively oriented chart domains $U_1 \cap U_2 \supseteq \operatorname{supp}\omega$, the two chart integrals agree.

Proof. The two expressions are related by the transition $T = \psi \circ \varphi^{-1}$: the coefficient functions satisfy $b = (a \circ T^{-1}) \cdot \det(dT^{-1})$ (the top-form transformation rule, 3C Section 3: $(\psi\circ\varphi^{-1})^*$ of $a\,dx^1 \wedge \cdots \wedge dx^n$ is $(a \circ T^{-1}) \det(dT^{-1}) dy^1 \wedge \cdots \wedge dy^n$, the wedge of the pulled-back differentials producing exactly the determinant of the Jacobian, by the definition of the determinant as the alternating top form, 1B Section 4). Since both charts are positively oriented and the transition is orientation-compatible, $\det(dT^{-1}) > 0$ on the support’s image, and the change-of-variables formula (declared input) gives

using $|\det dT| = \det dT > 0$ where $T$ is defined and orientation preserving. $\blacksquare$

Definition (global integral). Let $\omega$ be a compactly supported smooth $n$-form on the oriented $M$. Choose a partition of unity $\{\rho_i\}$ (declared input) subordinate to positively oriented chart domains $\{U_i\}$, and define

each term being a chart integral by the definition above (the support of $\rho_i \omega$ is compact in $U_i$). The sum is finite over the support of $\omega$ (local finiteness of the partition).

Proposition 3.2 (independence of the partition). The value does not depend on the partition of unity chosen.

Proof. Let $\{\rho_i\}$ and $\{\sigma_j\}$ be two partitions. Then $\omega = \sum_i \rho_i \omega = \sum_j \sigma_j \omega$, and each $\rho_i \omega = \sum_j \rho_i \sigma_j \omega$ with $\rho_i \sigma_j \omega$ supported in $U_i \cap U_j$. Summing over $j$ and using Proposition 3.1 to convert each piece to the common chart over $U_i$:

the rearrangement being legitimate because locally only finitely many terms are nonzero (local finiteness), so all sums are finite over the support of $\omega$. $\blacksquare$

Quick check.

1. When is the integral of a form a chart integral?
2. What makes two chart integrals agree?
3. What is a partition of unity?
4. Why is the global integral partition-independent?
5. Which three analysis facts are declared for integration?

Answers: 1. The support is compact inside one positively oriented chart domain. 2. The change-of-variables formula with positive Jacobian (Proposition 3.1). 3. A locally finite smooth family between zero and one, supports in chart domains, summing to one. 4. Refine both partitions through the pairwise products and apply Proposition 3.1 (Proposition 3.2). 5. Fubini; the change-of-variables formula; existence of smooth partitions of unity.

This is the analytic heart of the whole enterprise: everything else is assembly.

Theorem 4.1 (Stokes on the half-space). Let $\omega$ be a smooth $(n-1)$-form on (an open set of) $\mathbb{H}^n$, with compact support. Then

where the left integral is over the half-space with its standard orientation and the right over the boundary hyperplane with the outward-first boundary orientation.

Proof. Expand $\omega = \sum_{I \text{ inc., } |I| = n-1} a_I\, dx^I$. By linearity of $d$ and of the integral, treat one summand, $a\, dx^I$. Then $d(a\, dx^I) = da \wedge dx^I = \sum_{j} \partial_j a\, dx^j \wedge dx^I$, and by Lemma 1.1 of 3C: if $j \in I$, the term is zero. The only surviving terms have $j \notin I$; since $|I| = n-1$, there is exactly one such $j$, namely $j = n$, and

where $\sigma$ is the sorting permutation of $(j_0) \concat I$ into increasing order (Lemma 1.1 of 3C). Hence

By Fubini (declared input), this multiple integral is the iterated integral integrating over $x_{j_0}$ first, where $j_0$ is the unique index in $\{1, \dots, n\} \setminus I$:

where $I_{j_0} = [0, \infty)$ if $j_0 = n$ and $(-\infty, \infty)$ otherwise, and the sign $\operatorname{sgn}(\sigma)$ is the insertion sign of $j_0$ into $I$: moving $j_0$ from the front to its sorted slot crosses exactly $\#\{i \in I : i < j_0\}$ elements, so $\operatorname{sgn}(\sigma) = (-1)^{\#\{i \in I: i < j_0\}}$. If $j_0 \neq n$, the inner integral runs over all of $\mathbb{R}$ and the fundamental theorem of calculus evaluates it at two endpoints, both killed by the compact support of $a$: the term contributes nothing. Only $j_0 = n$ survives, and there the insertion sign is $(-1)^{n-1}$ (no element of $I$ exceeds $n$), the interval is $[0, \infty)$, and the fundamental theorem gives

So

Now the right side of the theorem for this summand. The boundary restriction of $a\, dx^I$ to $\{x_n = 0\}$: for boundary tangent vectors $w_1, \dots, w_{n-1}$ (last coordinate zero), $dx^I(w_1, \dots, w_{n-1}) = \det\big(\delta\big)$ is the minor of their first $n-1$ coordinates, i.e. the $(n-1)$-dimensional volume sign detector $dx'$. Which multiple of the boundary’s oriented volume form is it? By the outward-first convention, $(w_1, \dots, w_{n-1})$ is positively oriented on the boundary iff $(-e_n, w_1, \dots, w_{n-1})$ is positively oriented in $\mathbb{R}^n$, i.e. iff $\det[-e_n \mid w] < 0$. Expanding that determinant along the first column: $\det[-e_n \mid w] = (-1) \cdot (-1)^{n+1} dx^I(w) = (-1)^n dx^I(w)$. So positivity on the boundary means $(-1)^n dx^I(w) > 0$: the boundary’s oriented volume form is $(-1)^n dx'$. Therefore

which equals the left side computed above. Summing over $I$ finishes the proof. $\blacksquare$

Sign audit (where mistakes classically hide). Three signs entered: the insertion sign $(-1)^{n-1}$ (3C, Lemma 1.1), the fundamental theorem’s minus sign ($\int \partial_n a = -a(\cdot, 0)$), and the boundary orientation’s $(-1)^n$. They combine to match exactly; a proof of Stokes whose signs “come out right” without this audit is not a proof.

Quick check.

1. State the local Stokes theorem.
2. Which summands of the exterior derivative die on the half-space, and why?
3. Name the three signs in the surviving term.
4. Why does the fundamental theorem of calculus appear in the proof?
5. How many terms survive for a single basis summand?

Answers: 1. The integral of the derivative over the half-space equals the integral of the form over its boundary (Theorem 4.1). 2. A missing index other than the last: the compactly supported derivative integrates to zero over the whole line. 3. The insertion sign; the fundamental theorem’s minus; the boundary orientation’s sign. 4. Fubini reduces the integral to the last-coordinate antiderivative evaluated at the boundary. 5. Exactly one: the term whose missing index is the last coordinate.

Theorem 5.1 (Stokes). Let $M$ be a compact oriented $n$-manifold with boundary, and let $\omega$ be a smooth $(n-1)$-form on $M$. Then

Proof. Choose a partition of unity $\{\rho_i\}$ subordinate to positively oriented charts $\{(U_i, \varphi_i)\}$, where the $U_i$ are chosen as in Section 1: each $\varphi_i(U_i)$ lies in $\mathbb{H}^n$, and for charts meeting $\partial M$ the boundary hyperplane part corresponds to $\partial M$ (charts not meeting the boundary can use $\mathbb{R}^n$-modelled domains, which are $\mathbb{H}^n$ with the boundary hyperplane avoided; we take them modelled in $\{x_n > 0\}$‘s opens, so no boundary face survives locally). Since $\omega = \sum_i \rho_i \omega$ and $d$ is linear,

Each $\rho_i \omega$ has compact support inside the chart domain $U_i$. The chart representation of $\int_M d(\rho_i\omega)$ is, by the definition of Section 3 and 3C’s Theorem 3.2 ($d$ read in charts is $d$), the integral over $\varphi_i(U_i) \subseteq \mathbb{H}^n$ of $d(\text{representation})$. Theorem 4.1 applies (the representation of $\rho_i\omega$ is a compactly supported $(n-1)$-form on $\mathbb{H}^n$):

For a chart $U_i$ not meeting $\partial M$, the support of $\rho_i \omega$ is compact inside the interior $\{x_n > 0\}$, so its trace on $\{x_n = 0\}$ is identically zero and the boundary integral vanishes. For a chart meeting $\partial M$, the trace on $\{x_n = 0\}$ is exactly the chart representation of $\rho_i \omega|_{\partial M}$, and the two orientation conventions agree point for point: Section 2’s outward-first boundary orientation is defined as the transport of the half-space convention of Theorem 4.1, so the integral of the trace with the boundary orientation is $\int_{\partial M} \rho_i \omega$ (Proposition 3.1’s mechanism, with the boundary chart identifications). Summing over $i$:

the last equality being Proposition 3.2’s independence argument run on $\partial M$ (locally finitely many nonzero terms). $\blacksquare$

Corollary 5.2 (no boundary). If $M$ is a compact oriented manifold without boundary (in particular any compact oriented manifold in the sense of 3A), then $\int_M d\omega = 0$ for every $(n-1)$-form $\omega$.

Proof. $\partial M = \emptyset$ and the boundary integral is zero by definition. $\blacksquare$

Quick check.

1. State the generalised Stokes theorem.
2. How does the partition of unity assemble its proof?
3. Why do interior charts contribute zero boundary terms?
4. What does Corollary 5.2 say?
5. Why is the boundary orientation convention essential to the assembly?

Answers: 1. The integral of the derivative over M equals the integral of the form over the boundary (Theorem 5.1). 2. The form is the sum of the cut-off pieces; apply the local statement to each and sum. 3. Their supports miss the boundary hyperplane, so the trace vanishes. 4. A compact oriented manifold without boundary has zero integral of every exact top form. 5. The local statement uses the half-space convention; Section 2 defines the boundary orientation as its transport, so the signs agree.

The fundamental theorem of calculus (one dimension). For $M = [a, b]$, oriented by increasing $t$, and a smooth function $f$: Stokes with $\omega = f$ (a 0-form) gives $\int_{[a,b]} df = \int_{\{b\}} f - \int_{\{a\}} f$; the right side is $f(b) - f(a)$ (the point orientation of $\{b\}$ is positive: outward is $+1$; of $\{a\}$ negative: outward is $-1$), and $df = f'\,dt$, so $\int_a^b f'(t)\, dt = f(b) - f(a)$. $\blacksquare$

Green’s theorem (two dimensions). For a compact oriented planar region $M$ with boundary, and $\omega = P\, dx + Q\, dy$: the graded Leibniz rule (Theorem 2.3 of 3C) gives $d\omega = dP \wedge dx + dQ \wedge dy = (\partial_y P)\, dy \wedge dx + (\partial_x Q)\, dx \wedge dy = (\partial_x Q - \partial_y P)\, dx \wedge dy$ (using $dy \wedge dx = -\, dx \wedge dy$). Stokes reads

the circulation of $(P, Q)$ around $\partial M$ equals the integral of the scalar curl over $M$. $\blacksquare$

The divergence theorem (three dimensions). For a compact oriented region $M \subseteq \mathbb{R}^3$ and the flux form $\omega = F_1\, dy \wedge dz - F_2\, dx \wedge dz + F_3\, dx \wedge dy$ (the signs being the 3C self-test’s insertion rule at work), the exterior derivative computes $d\omega = (\partial_x F_1 + \partial_y F_2 + \partial_z F_3)\, dx \wedge dy \wedge dz = (\operatorname{div} F)\, dV$, and Stokes reads $\int_M \operatorname{div} F\, dV = \int_{\partial M} \omega$: flux through the boundary equals the integral of the divergence inside. $\blacksquare$

The classical Stokes theorem (surfaces). For an oriented surface $S$ with boundary in $\mathbb{R}^3$ and $\omega = P\, dx + Q\, dy + R\, dz$: $d\omega = (\partial_y R - \partial_z Q)\, dy \wedge dz + (\partial_z P - \partial_x R)\, dz \wedge dx + (\partial_x Q - \partial_y P)\, dx \wedge dy$ (each term by 3C’s insertion rules), whose coefficient vector is $\operatorname{curl}(P, Q, R)$; Stokes reads $\int_S \operatorname{curl} \cdot \nu\, dA = \int_{\partial S} P\, dx + Q\, dy + R\, dz$ with $\nu$ the orientation normal. $\blacksquare$

One theorem, four costumes.

Quick check.

1. Which theorem is Stokes in one dimension?
2. What 2-form is the derivative of (P dx plus Q dy)?
3. What 3-form is the derivative of the flux form?
4. What does the classical Stokes theorem equate?
5. Which corollary covers the boundaryless case?

Answers: 1. The fundamental theorem of calculus. 2. The difference of the partials, times the area form. 3. The divergence, times the volume form. 4. The surface integral of the curl against the circulation of the boundary. 5. Corollary 5.2.

1. Prove that a transition of a manifold-with-boundary restricts to a transition of the boundary atlases, and conclude $\partial M$ is an $(n-1)$-manifold. (§1 – the boundary clause makes the restricted transitions land in $\{x_n = 0\} \cong \mathbb{R}^{n-1}$; smooth extendability survives restriction.)
2. Prove that the outward vectors at $p \in \partial M$ are exactly the vectors $v$ with $(d\varphi_p v)_n < 0$, independently of the chart, citing Lemma 2.1 in both directions. (§2 – the transition and its inverse.)
3. Prove that the boundary orientation of the boundary of the boundary is zero-dimensional and empty… instead: prove that $\partial(\partial M) = \emptyset$ in the sense that $\partial M$ is a manifold without boundary. (§1 – boundary charts of $\partial M$ land in $\{x_n = 0\} \cong \mathbb{R}^{n-1}$, whose points all have neighbourhoods in $\mathbb{R}^{n-1}$.)
4. Work out the sign of $dx^2 \wedge dx^1 \wedge dx^3$ as a multiple of $dx^1 \wedge dx^2 \wedge dx^3$, two ways: by Lemma 1.1 of 3C and by graded commutativity. (§2, 3C – one transposition; $(-1)^{1 \cdot 2} = +1$ for the two-past-one check confirms the method.)
5. Verify the divergence-form computation $d(F_1\, dy \wedge dz - F_2\, dx \wedge dz + F_3\, dx \wedge dy) = (\operatorname{div} F)\, dx \wedge dy \wedge dz$ term by term. (§6, 3C – each $dF_i$ lands in the slot the omitted differential leaves open, with the sign visible in the insertion rule.)
6. Prove that the integral over a point of a 0-form is the value of the form at the point if the point is positively oriented, and its negative otherwise. (§3, §6 – a point manifold with boundary has boundary the point itself, oriented outward-first.)
7. Prove that the chart integral of Section 3 extends to forms supported in finitely many disjoint chart domains, and prove additivity: $\int_M \omega + \eta = \int_M (\omega + \eta)$ for forms supported in disjoint chart domains. (§3 – linearity of the coefficient integrals; disjointness keeps the supports in separate charts so no double counting arises.)

Problems. The quick checks above tested recall. These chain several results and require full written arguments; hints are included where the jump is large.

8. Prove that the boundary of $[0, 1]^2$ consists of its four edges, and write each edge’s outward-first orientation explicitly. (Outward is $(0, -1)$ on the bottom, $(0, 1)$ on the top, $(-1, 0)$ on the left, $(1, 0)$ on the right; transport each through Lemma 2.1’s convention.)
9. Check that for $n = 2$ the boundary orientation of $\partial \mathbb{H}^2$ is $+dx$: take $w = e_x$ and verify $(-e_y, e_x)$ has positive determinant. (The determinant of the columns is $0 \cdot 0 - 1 \cdot (-1) = 1$.)
10. Compute $\int_{[0,1]^2} d(x y^2\, dy)$ directly and via Stokes, and confirm the two agree. (Direct: $d = y^2\, dx \wedge dy$, integral $1/3$. Stokes: only the right edge contributes, giving $\int_0^1 y^2 dy = 1/3$; the left edge carries $x = 0$.)

The entire page follows from one computation: the fundamental theorem of calculus applied to the coefficient of $dx^I$ in the $x_n$ direction. Orientation converts “signed volume” into a well-defined number, Fubini and the change-of-variables formula make that number independent of the chart, and the partition of unity then lets the same one-variable theorem run simultaneously in every chart, with the boundary collecting exactly one term per chart: that assembly is the generalised Stokes theorem, and the classical theorems are its coordinate expansions.

• The Core Course hub – Level 4 (measure and integration after Schilling, on the way to $L^p$ duality) and Level 5 (complex analysis through forms; de Rham cohomology with Bott and Tu; algebra with Aluffi) complete the course. Under construction.
• Level 3C: Differential Forms – the exterior derivative, graded Leibniz, and pullback commutation that this page’s proof assembles.
• Michael Spivak, Calculus on Manifolds, and Loring Tu, An Introduction to Manifolds (Springer) – the two classical treatments; Spivak proves Stokes over chains with the same local-to-global architecture, Tu over manifolds with boundary.