de Rham Cohomology, Periods, Duality, and Intersection

The exterior derivative turns real differential forms into a cochain complex, and its cohomology

$$H_{\mathrm{dR}}^k(M) = \frac{\ker(\mathrm d:\Omega^k(M)\to\Omega^{k+1}(M))} {\operatorname{im}(\mathrm d:\Omega^{k-1}(M)\to\Omega^k(M))}$$

measures closed forms that have no global potential. Stokes’ theorem makes integration descend to a pairing

$$H_{\mathrm{dR}}^k(M) \times H_k(M;\mathbb Z) \longrightarrow\mathbb R, \qquad ([\omega],[z])\longmapsto\int_z\omega.$$

The de Rham theorem identifies these smooth classes with real singular cohomology, so periods detect the entire real class. On a compact oriented manifold without boundary, Poincaré duality identifies complementary cohomology and homology data, while wedge integrals compute signed intersections of transverse cycles. Integral-period classes provide the additional input behind quantized flux. Real forms alone, however, cannot see torsion.

These are mathematical and kinematic statements. QFT dynamics, current algebras, Ward identities, the construction of bundles from transition functions, and the characteristic-class interpretation of curvature belong to later pages.

Required background. Differential Forms, Integration, Orientation, and Stokes Theorem. It supplies the wedge product, pullback, exterior derivative, integration, and the outward-vector-first boundary orientation used below.

Helpful background. Chains, Homology, Cohomology, and Exact Sequences supplies the cycle, boundary, cochain, and pairing language used to state the de Rham theorem.

de Rham setting and integration hypotheses

Unless a statement says otherwise, $M$ is a smooth Hausdorff, second-countable manifold, possibly disconnected and possibly with boundary, and forms are real-valued. Cycles have integer coefficients. Orientation and compactness are not needed to define de Rham cohomology or to state the de Rham theorem; they enter when a manifold or submanifold is integrated. No metric or Hodge star is used.

Closed locally, obstructed globally

Write

$$Z_{\mathrm{dR}}^k(M)=\ker\mathrm d, \qquad B_{\mathrm{dR}}^k(M)=\operatorname{im}\mathrm d.$$

The identity $\mathrm d^2=0$ gives $B_{\mathrm{dR}}^k\subseteq Z_{\mathrm{dR}}^k$, so the quotient defining $H_{\mathrm{dR}}^k$ exists. A form in $Z_{\mathrm{dR}}^k$ is closed; a form in $B_{\mathrm{dR}}^k$ is exact. At degree zero, there are no negative-degree forms, and

$$H_{\mathrm{dR}}^0(M) = \{f\in C^\infty(M):\mathrm df=0\}.$$

Thus $H_{\mathrm{dR}}^0(M)$ consists of locally constant functions and

$$H_{\mathrm{dR}}^0(M) \cong \prod_{C\in\pi_0(M)}\mathbb R.$$

In particular, it is $\mathbb R$ when $M$ is connected.

The Poincaré lemma gives the local answer in positive degree. If $U\subset\mathbb R^n$ is star-shaped and $k>0$, every closed $\omega\in\Omega^k(U)$ has a primitive $\eta\in\Omega^{k-1}(U)$:

$$\mathrm d\omega=0 \quad\Longrightarrow\quad \omega=\mathrm d\eta.$$

Every interior point of a smooth manifold has coordinate neighborhoods of this kind. At a boundary point, use a star-shaped half-ball in the half-space; the same radial homotopy proof applies. Consequently, closed forms are locally exact; a nonzero de Rham class is a global obstruction to joining the local primitives into one global primitive. A homotopy-operator proof and the resulting local-versus-global interpretation appear in Nakahara 2003, §§ 6.2–6.3, pp. 230–237.

The circle is the smallest useful example. On the unit circle $S^1\subset\mathbb R^2$, set

$$\eta = \left. \frac{x\,\mathrm dy-y\,\mathrm dx}{2\pi} \right|_{S^1}.$$

On an angular coordinate patch this is $\mathrm d\theta/(2\pi)$, but $\theta$ is not a globally defined real function. Since $S^1$ is one-dimensional, $\mathrm d\eta=0$, whereas the counterclockwise orientation gives

$$\int_{S^1}\eta=1.$$

If $\eta=\mathrm df$ for a global function, Stokes’ theorem on the closed cycle would give zero. Hence $\eta$ is closed but not exact and generates $H_{\mathrm{dR}}^1(S^1)\cong\mathbb R$.

Integration is a cochain map

Let $\sigma:\Delta^k\to M$ be a smooth singular $k$-simplex, with the standard simplex orientation and the alternating-face boundary convention from the preceding page. Define a real singular cochain by

$$I(\omega)(\sigma) = \int_{\Delta^k}\sigma^*\omega.$$

Extend this definition linearly to smooth singular chains. The coboundary has no extra sign:

$$(\delta c)(b)=c(\partial b).$$

For a smooth singular $(k+1)$-chain $b$, facewise Stokes gives

$$\begin{aligned} (\delta I(\omega))(b) &=I(\omega)(\partial b)\\ &=\int_{\partial b}\omega\\ &=\int_b\mathrm d\omega\\ &=I(\mathrm d\omega)(b). \end{aligned}$$

Therefore

$$\delta I=I\mathrm d.$$

Integration is a cochain map: it sends closed forms to cocycles and exact forms to coboundaries. Smooth singular chains compute the same homology as ordinary singular chains, so the induced cohomology map has the usual singular target. Calegari 2016, §§ 2.2–2.3, pp. 12–13, PDF gives this chain-level construction and sign convention.

Periods and the de Rham theorem

For a closed $k$-form $\omega$ and an integral $k$-cycle $z$, its period on $z$ is $\int_z\omega$. Both representative choices disappear by Stokes. If $\omega$ is replaced by $\omega+\mathrm d\lambda$, then

$$\int_z(\omega+\mathrm d\lambda)-\int_z\omega = \int_{\partial z}\lambda =0.$$

If $z$ is replaced by the homologous cycle $z+\partial b$, then

$$\int_{z+\partial b}\omega-\int_z\omega = \int_b\mathrm d\omega =0.$$

Thus the period depends only on $[\omega]$ and $[z]$. Both hypotheses are essential: an exact change can contribute on a chain with boundary, and a nonclosed form can distinguish homologous cycles.

de Rham theorem. For every smooth manifold $M$ and every $k\geq0$, integration induces a natural isomorphism

$$I_*: H_{\mathrm{dR}}^k(M) \xrightarrow{\ \cong\ } H^k(M;\mathbb R).$$

Neither compactness nor orientation is a hypothesis. The chain-map identity has just been proved. The assertion that the induced map is an isomorphism is cited here rather than proved in full. One proof first checks the statement on coordinate balls using the Poincaré lemma, then glues with partitions of unity and Mayer–Vietoris; another uses a smooth triangulation. The comparison between smooth and continuous singular chains completes the passage to ordinary singular cohomology. Calegari 2016, § 1.13 and §§ 2.4–2.5, pp. 11–17, PDF states and proves the boundaryless version. If $\partial M\neq\varnothing$, a collar pushes $M$ into its interior, making $\operatorname{Int}M\hookrightarrow M$ a homotopy equivalence; naturality and homotopy invariance of both cohomology theories then reduce the boundary case to the theorem for $\operatorname{Int}M$. Frankel 2012, § 13.4a, pp. 355–357 gives a complementary period-based formulation.

An immediate consequence is the period criterion:

$$[\omega]=0 \quad\Longleftrightarrow\quad \int_z\omega=0 \quad\text{for every integral }k\text{-cycle }z.$$

The forward implication is Stokes. For the reverse implication, the periods say that the singular cohomology class $I_*[\omega]$ evaluates to zero on homology. Over the field $\mathbb R$, the universal coefficient theorem identifies real cohomology with the linear dual of real homology, so the class, and hence $[\omega]$, vanishes. This is a global criterion: checking $\mathrm d\omega=0$ only establishes local exactness.

Integral periods and invisible torsion

The de Rham theorem is a theorem over $\mathbb R$. To retain an integral normalization, define the integral-period subgroup

$$H_{\mathrm{dR}}^k(M)_{\mathbb Z} = \operatorname{im} \left( H^k(M;\mathbb Z) \longrightarrow H^k(M;\mathbb R) \xrightarrow{\ I_*^{-1}\ } H_{\mathrm{dR}}^k(M) \right).$$

For a manifold of finite homotopy type, this is a lattice in the finite-dimensional real vector space. A closed form represents an integral class in this normalization exactly when

$$\int_z\omega\in\mathbb Z \qquad \text{for every integral }k\text{-cycle }z.$$

Indeed, the integral universal coefficient theorem maps $H^k(M;\mathbb Z)$ onto the integer-valued homomorphisms on $H_k(M;\mathbb Z)$. The group $\operatorname{Ext}_{\mathbb Z}^1(H_{k-1}(M;\mathbb Z),\mathbb Z)$ is the extension term in that theorem; for finitely generated homology it is torsion, and it maps to zero after changing coefficients to $\mathbb R$. Closedness alone does not impose the integral-period condition. If $\omega$ is closed, then $c\omega$ is closed for every $c\in\mathbb R$, and its periods vary continuously.

The same argument exposes what differential forms lose. Suppose an integral homology class $[z]$ has finite order $m$, so $mz=\partial b$ for some chain $b$. For every closed real form,

$$m\int_z\omega = \int_{\partial b}\omega = \int_b\mathrm d\omega =0,$$

and hence $\int_z\omega=0$. Real periods cannot detect torsion cycles. Likewise, every torsion class in $H^k(M;\mathbb Z)$ maps to zero in real and de Rham cohomology. For example,

$$H_1(\mathbb{RP}^2;\mathbb Z)\cong\mathbb Z_2, \qquad H_{\mathrm{dR}}^1(\mathbb{RP}^2)=0,$$

and the class in $H^2(\mathbb{RP}^2;\mathbb Z)\cong\mathbb Z_2$ also has zero real image. Integral lifts that differ by torsion therefore have the same de Rham class. The coefficient and universal-coefficient details are developed on the preceding page.

The torus fixes periods and wedge order

Take

$$T^2=\mathbb R^2/\mathbb Z^2, \qquad \operatorname{or}(T^2)=\mathrm dx\wedge\mathrm dy,$$

where $x$ and $y$ have period $1$. Let

$$\gamma_x(t)=(t,0), \qquad \gamma_y(t)=(0,t), \qquad 0\leq t\leq1,$$

with their increasing-coordinate orientations. The global one-forms $\mathrm dx$ and $\mathrm dy$ have period matrix

$$\begin{pmatrix} \displaystyle\int_{\gamma_x}\mathrm dx & \displaystyle\int_{\gamma_y}\mathrm dx\\[6pt] \displaystyle\int_{\gamma_x}\mathrm dy & \displaystyle\int_{\gamma_y}\mathrm dy \end{pmatrix} = \begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}.$$

They form the basis of $H_{\mathrm{dR}}^1(T^2)\cong\mathbb R^2$ dual to the two basic homology classes. Their wedge product satisfies

$$\int_{T^2}\mathrm dx\wedge\mathrm dy=1.$$

The wedge product descends to cohomology. If $\alpha$ and $\beta$ are closed, then $\alpha\wedge\beta$ is closed; replacing $\alpha$ by $\alpha+\mathrm d\mu$ changes the product by

$$\mathrm d\mu\wedge\beta = \mathrm d(\mu\wedge\beta),$$

and similarly for the second factor, with the graded Leibniz sign. Thus

$$[\alpha]\smile[\beta] = [\alpha\wedge\beta]$$

defines the graded-commutative de Rham cohomology ring. The torus calculation simultaneously displays two integral one-classes and their normalized top-degree product. Nakahara 2003, §§ 6.4.2–6.4.3, pp. 238–240 and Frankel 2012, § 13.4b, pp. 357–358 give parallel versions of this example.

Poincaré duality and its hypotheses

Now assume that $M^n$ is compact, oriented, and has no boundary. Then the wedge-integral pairing

$$H_{\mathrm{dR}}^k(M) \times H_{\mathrm{dR}}^{n-k}(M) \longrightarrow\mathbb R, \qquad ([\alpha],[\beta]) \longmapsto \int_M\alpha\wedge\beta$$

is nondegenerate; since the groups are finite-dimensional, it is a perfect pairing. This is the de Rham form of Poincaré duality.

The pairing is well defined for exactly the expected reasons. If $\alpha\mapsto\alpha+\mathrm d\mu$ and $\mathrm d\beta=0$, then

$$\int_M\mathrm d\mu\wedge\beta = \int_M\mathrm d(\mu\wedge\beta) = \int_{\partial M} i_{\partial}^*(\mu\wedge\beta) =0.$$

Changing $\beta$ by an exact form works similarly. Nondegeneracy is the theorem, not merely another application of Stokes. Through the de Rham isomorphism it is equivalent to the cap-product isomorphism

$$\begin{aligned} \mathcal D_M: H^k(M;\mathbb R) &\xrightarrow{\ \cong\ } H_{n-k}(M;\mathbb R),\\ \alpha &\longmapsto [M]\mathbin{\frown}\alpha , \end{aligned}$$

where $[M]$ is the oriented fundamental class. Hatcher 2002, Theorem 3.30, p. 241, and Proposition 3.38, pp. 249–250, PDF give the cap isomorphism and the nonsingular cup-product pairing. No metric is present: this is not Hodge duality, and Poincaré duality does not choose a unique form representative. Guillemin and Haine 2018, § 5.4, pp. 171–175, PDF proves the differential-form pairing, including its compact-support version.

The hypotheses have concrete replacements, not cosmetic exceptions.

• On an oriented noncompact manifold, one normally uses compact support on one factor. For an oriented boundaryless finite-type $M^n$, the pairing $H_{\mathrm{dR},c}^k(M)\times H_{\mathrm{dR}}^{n-k}(M)\to\mathbb R$ is nondegenerate. The primitive in the compact-support complex must itself have compact support.
• If $M$ has boundary, Stokes leaves a boundary term. The correct statement is Poincaré–Lefschetz duality, involving relative cohomology.
• If $M$ is nonorientable, the displayed untwisted integral pairing is not available. One uses the orientation local system or twisted coefficients.

Here the compact-support complex and its cohomology are

$$\begin{aligned} \Omega_c^k(M) &= \{\omega\in\Omega^k(M): \operatorname{supp}\omega\text{ is compact}\},\\ H_{\mathrm{dR},c}^k(M) &= \frac{\ker(\mathrm d:\Omega_c^k\to\Omega_c^{k+1})} {\operatorname{im}(\mathrm d:\Omega_c^{k-1}\to\Omega_c^k)}. \end{aligned}$$

The line makes the support issue visible. Choose $\rho\in C_c^\infty(\mathbb R)$ with $\int_{\mathbb R}\rho(x)\,\mathrm dx=1$. Then $\rho(x)\,\mathrm dx$ is ordinarily exact, with primitive

$$F(x)=\int_{-\infty}^x\rho(t)\,\mathrm dt,$$

but $F$ tends to $1$ as $x\to+\infty$ and is not compactly supported. No compactly supported primitive can exist, because the integral of its derivative would be zero. Hence $[\rho\,\mathrm dx]$ generates $H_{\mathrm{dR},c}^1(\mathbb R)\cong\mathbb R$, even though $H_{\mathrm{dR}}^1(\mathbb R)=0$.

Dual cycles and intersection numbers

Keep $M^n$ compact, oriented, and without boundary. Let $Y^{n-k}\subset M$ be a closed oriented embedded submanifold. We fix the normal-first convention: orient the normal bundle so that

$$(\text{positive normal basis},\text{positive tangent basis of }Y)$$

is a positive basis of $TM$. Its Poincaré-dual class $\operatorname{PD}(Y)\in H_{\mathrm{dR}}^k(M)$ is characterized by

$$\int_M\operatorname{PD}(Y)\wedge\eta = \int_Y i_Y^*\eta$$

for every closed $(n-k)$-form $\eta$. This equation determines a cohomology class, not one distinguished differential form. A smooth representative can be chosen as a Thom form supported in a tubular neighborhood of $Y$; a form literally concentrated on $Y$ would instead be a distributional current. Sources that place the test form before the dual form differ by $(-1)^{k(n-k)}$. Guillemin and Haine 2018, § 5.5, pp. 176–182, PDF develops Thom forms and their intersection pairing.

Let $A^a,B^b\subset M^n$ be closed oriented embedded submanifolds with $a+b=n$, and suppose they meet transversely. Their intersection is finite. At $p\in A\cap B$, define

$$\epsilon_p(A,B) = \begin{cases} +1, & (\text{positive basis of }T_pA, \text{positive basis of }T_pB) \text{ orients }T_pM,\\ -1,&\text{otherwise}. \end{cases}$$

The oriented intersection number is

$$A\cdot B = \sum_{p\in A\cap B}\epsilon_p(A,B) = \int_M \operatorname{PD}(A)\wedge\operatorname{PD}(B).$$

The wedge order is part of the convention. Since $\deg\operatorname{PD}(A)=b$ and $\deg\operatorname{PD}(B)=a$,

$$B\cdot A=(-1)^{ab}A\cdot B.$$

If representatives are not transverse, naive point counting is not an invariant; one first makes a transverse perturbation and then takes the signed count.

The torus fixes every sign. Let

$$A=\{y=0\}, \qquad B=\{x=0\},$$

oriented by $+\partial_x$ and $+\partial_y$, respectively. The ordered tangent pair $(\partial_x,\partial_y)$ is positive, so $A\cdot B=+1$. Under the normal-first convention,

$$\operatorname{PD}(A)=-\mathrm dy, \qquad \operatorname{PD}(B)=\mathrm dx.$$

Indeed,

$$\int_{T^2}(-\mathrm dy)\wedge\mathrm dx=1,$$

whereas

$$\int_{T^2}\mathrm dx\wedge(-\mathrm dy)=-1.$$

Thus $A\cdot B=1$ and $B\cdot A=-1$, exactly matching the local orientation definition.

Quantized flux and topological currents

The period pairing gives a controlled QFT-facing application before any dynamics is chosen. Let $\mathcal J$ be a closed $k$-form on a region of spacetime and let $\Sigma$ be a closed oriented integral $k$-cycle. Define

$$Q_\Sigma=\int_\Sigma\mathcal J.$$

If $\Sigma_1-\Sigma_0=\partial B$ inside the region, then

$$Q_{\Sigma_1}-Q_{\Sigma_0} = \int_B\mathrm d\mathcal J =0.$$

So closedness makes the charge depend only on the homology class of its support. The equation $\mathrm d\mathcal J=0$, the allowed supports, and the absence of contributions from physical boundaries or infinity are physical inputs; topology does not supply them.

For a compact $U(1)$ field, translate the site’s gauge convention by writing

$$D=\mathrm d-i\mathcal A, \qquad \mathcal A=gA,$$

so unit-charge holonomy is $\exp(i\oint\mathcal A)$. Its real curvature $\mathcal F$ is a global closed two-form. Compact $U(1)$ global data impose the additional condition

$$\left[ \frac{\mathcal F}{2\pi} \right]_{\mathrm{dR}} \in H_{\mathrm{dR}}^2(M)_{\mathbb Z},$$

and therefore

$$Q_m(\Sigma) = \frac{1}{2\pi} \int_\Sigma\mathcal F \in\mathbb Z.$$

Frankel 2012, § 17.4a, pp. 467–468 states this curvature integrality theorem in an anti-Hermitian convention, with $i\Theta/(2\pi)$ in place of the real form $\mathcal F/(2\pi)$ used here. Closedness gives invariance under homologous changes of $\Sigma$; compact $U(1)$ global data give integrality. These are distinct statements.

On $S^2$, oriented by $\sin\theta\,\mathrm d\theta\wedge\mathrm d\phi$, take

$$\mathcal F = \frac n2 \sin\theta\, \mathrm d\theta\wedge\mathrm d\phi, \qquad n\in\mathbb Z.$$

Then

$$\frac{1}{2\pi} \int_{S^2}\mathcal F =n.$$

For $n\neq0$, $\mathcal F$ cannot equal $\mathrm d\mathcal A$ for one globally defined potential, because every exact form has zero period on a closed cycle. Locally it still has potentials, in agreement with the Poincaré lemma. Frankel 2012, § 16.4e, pp. 444–446 and Nakahara 2003, § 10.5.2, pp. 400–401 give the monopole patch calculation and its flux normalization.

In a closed-form current convention, $\mathcal J_m=\mathcal F/(2\pi)$ is the magnetic topological current and its period is the flux charge. Gaiotto, Kapustin, Seiberg, and Willett 2015, § 3 and § 4.1, pp. 11–15 relates closed differential-form currents, charge supports, Poincaré duality, and intersection in higher-form symmetry. The developed Ward identities, operator algebra, defects, and backgrounds belong to Higher-Form Currents, Charges, Backgrounds, and Ward Identities. The bundle and characteristic-class explanation of integrality belongs to the next page.

Common pitfalls

Confusing closed with globally exact. The Poincaré lemma is local. The normalized angular form on $S^1$ is closed everywhere and locally exact, but its nonzero period rules out one global primitive.

Deriving quantization from closedness. Closed forms can be multiplied by arbitrary real constants. Integer periods express an integral-lattice input, not the equation $\mathrm d\omega=0$.

Expecting real forms to retain torsion. A closed real form has zero period on every torsion cycle. Differential cohomology, finite coefficients, or other discrete data are needed when torsion matters.

Using closed-manifold duality with missing hypotheses. Boundary, noncompactness, and nonorientability change the correct theorem. Use relative, compactly supported, or twisted variants as appropriate.

Calling Poincaré duality Hodge duality. Poincaré duality is topological and needs no metric. A Hodge star and harmonic representatives require a Riemannian metric and additional hypotheses.

Ignoring wedge order or transversality. Reversing complementary factors can change the intersection sign. Nontransverse geometric intersections cannot be counted naively.

Promoting a period to a physical solution. A cohomology class supplies a kinematic obstruction or pairing. It does not prove that a field configuration exists dynamically, is stable, defines a quantum sector, or satisfies a Ward identity.

Exercises

Each optional check is followed by a solution.

1. Retrieve the definitions and smallest obstruction

Define closed, exact, and $H_{\mathrm{dR}}^k(M)$. Why does the normalized angular form $\eta$ on $S^1$ fail to be exact?

Solution

A $k$-form is closed when $\mathrm d\omega=0$ and exact when $\omega=\mathrm d\lambda$ for some $(k-1)$-form. Since $\mathrm d^2=0$, exact forms are closed and

$$H_{\mathrm{dR}}^k(M) = \frac{\{\text{closed }k\text{-forms}\}} {\{\text{exact }k\text{-forms}\}}.$$

The circle form is closed and has period $\int_{S^1}\eta=1$. If it were exact, Stokes’ theorem on the closed cycle would make this integral zero. Therefore it represents a nonzero de Rham class.

2. Test the compact-support hypothesis

Let $\rho\in C_c^\infty(\mathbb R)$ satisfy $\int_{\mathbb R}\rho\,\mathrm dx=1$. Show that $\rho\,\mathrm dx$ is exact in ordinary de Rham cohomology but not in the compact-support complex.

Solution

The function

$$F(x)=\int_{-\infty}^x\rho(t)\,\mathrm dt$$

satisfies $\mathrm dF=\rho\,\mathrm dx$, so the form is ordinarily exact. But $F$ approaches $1$ at $+\infty$ and is not compactly supported. If a compactly supported $G$ satisfied $\mathrm dG=\rho\,\mathrm dx$, then

$$1 = \int_{\mathbb R}\rho\,\mathrm dx = G(+\infty)-G(-\infty) =0,$$

a contradiction. Thus the form is not exact in $\Omega_c^\bullet(\mathbb R)$.

3. Calculate the torus intersection sign

On the oriented torus, let $A=\{y=0\}$ have orientation $+\partial_x$ and $B=\{x=0\}$ have orientation $+\partial_y$. Using the normal-first convention, find $\operatorname{PD}(A)$ and $\operatorname{PD}(B)$ and compute both ordered intersection numbers.

Solution

For $A$, the positive normal must be $-\partial_y$ because $(-\partial_y,\partial_x)$ has the orientation $(\partial_x,\partial_y)$. Hence $\operatorname{PD}(A)=-\mathrm dy$. For $B$, the positive normal is $+\partial_x$, so $\operatorname{PD}(B)=\mathrm dx$. Therefore

$$A\cdot B = \int_{T^2}(-\mathrm dy)\wedge\mathrm dx =1,$$

while

$$B\cdot A = \int_{T^2}\mathrm dx\wedge(-\mathrm dy) =-1.$$

The signs agree with the ordered tangent bases at the unique intersection.

4. Transfer periods to the sphere flux

For

$$\mathcal F = \frac n2\sin\theta\, \mathrm d\theta\wedge\mathrm d\phi$$

on $S^2$, compute the normalized flux. Explain separately what closedness and integrality establish, and why $n\neq0$ forbids one global potential.

Solution

Using $0\leq\theta\leq\pi$ and $0\leq\phi<2\pi$,

$$\begin{aligned} \frac1{2\pi}\int_{S^2}\mathcal F &= \frac1{2\pi}\frac n2 \int_0^{2\pi}\mathrm d\phi \int_0^\pi\sin\theta\,\mathrm d\theta\\ &=n. \end{aligned}$$

Closedness makes the flux unchanged when the integration sphere is replaced by a homologous cycle in a region where $\mathrm d\mathcal F=0$ and no boundary term enters. The assertion $n\in\mathbb Z$ instead comes from the compact $U(1)$ integral-period condition. If one global $\mathcal A$ satisfied $\mathcal F=\mathrm d\mathcal A$, Stokes would force the period on $S^2$ to vanish, contradicting $n\neq0$.

Synthesis and continuations

The de Rham complex converts local differential data into global classes. Integration is a cochain map, so Stokes makes periods depend only on cohomology and homology classes; the de Rham theorem says those periods capture all real singular cohomology. Integral periods select extra global data and expose the limitation that real forms erase torsion. Under compact, oriented, boundaryless hypotheses, Poincaré duality turns complementary classes and cycles into a perfect pairing, and ordered wedge products compute signed transverse intersections.

For curvature polynomials, integral characteristic classes, and why their de Rham representatives do not depend on the connection, continue to Characteristic Classes and Chern–Weil Theory. For metric-dependent Hodge stars, Laplacians, and harmonic representatives, continue to Metrics, Volume Forms, Hodge Star, and Laplace Operators. For the developed physical interpretation of closed currents and charge supports, continue to Higher-Form Currents, Charges, Backgrounds, and Ward Identities.

References

• Danny Calegari, Notes on Differential Forms — Open PDF, course notes, University of Chicago, 2016, § 1.13 and §§ 2.2–2.5, pp. 11–17. These notes construct the integration cochain, prove Stokes’ chain-map identity, compare smooth and ordinary singular chains, and prove the de Rham theorem for manifolds without boundary.
• Theodore Frankel, The Geometry of Physics: An Introduction, third edition, Cambridge University Press, 2012, § 13.4, pp. 355–358, § 16.4e, pp. 444–446, and § 17.4a, pp. 467–468. Frankel supplies a period-based route, the torus calculation, and the monopole flux example.
• Davide Gaiotto, Anton Kapustin, Nathan Seiberg, and Brian Willett, “Generalized Global Symmetries”, Journal of High Energy Physics 02 (2015) 172, §§ 3 and 4.1, especially pp. 11–15. This section supplies the QFT application of closed-form currents, dual charge supports, and intersection pairings.
• Victor Guillemin and Peter J. Haine, Differential Forms — Open PDF, draft of March 28, 2018, §§ 5.1, 5.4, and 5.5, pp. 149–182. These sections develop de Rham cohomology, compactly supported Poincaré duality, Thom classes, and signed intersection numbers.
• Allen Hatcher, Algebraic Topology — Open PDF, Cambridge University Press, 2002, §§ 3.1 and 3.3. These results justify the integral-period, torsion, and Poincaré-duality statements.
• Mikio Nakahara, Geometry, Topology and Physics, second edition, Institute of Physics Publishing, 2003, §§ 6.2–6.4, pp. 230–240, and § 10.5.2, pp. 400–401. These sections support the local exactness, period, cohomology-ring, torus, and compact-flux calculations.