Differential Forms

The space of $k$-covectors at $p$ is $\Lambda^k(T_p^* M)$, the space of alternating $k$-linear maps $T_p M \times \cdots \times T_p M \to \mathbb{R}$.

Definition. A differential $k$-form on $M$ is a smooth section of $\Lambda^k T^* M$, i.e., a smooth map $\omega : M \to \Lambda^k T^* M$ assigning to each $p$ an alternating $k$-linear functional $\omega_p$ on $T_p M$.

Examples.

• A $0$-form is a smooth function $f \in C^\infty(M)$.
• A $1$-form is a section of $T^* M$ (a covector field). In local coordinates, $\omega = \sum \omega_i\, dx^i$.
• A $2$-form on $M$ has the local expression $\omega = \sum_{i < j} \omega_{ij}\, dx^i \wedge dx^j$.

Definition. If $F : M \to N$ is a smooth map, the pullback $F^* : \Omega^k(N) \to \Omega^k(M)$ is defined by:

for $v_i \in T_p M$.

Proposition 4.1. The pullback satisfies:

1. $F^*(\alpha \wedge \beta) = F^*\alpha \wedge F^*\beta$
2. $F^*(d\omega) = d(F^*\omega)$
3. $(G \circ F)^* = F^* \circ G^*$

Definition. The interior product (contraction) of a vector field $X$ with a $k$-form $\omega$ is the $(k-1)$-form $\iota_X \omega$ defined by:

Proposition 4.2 (Cartan’s Magic Formula). The Lie derivative $\mathcal{L}_X$ of a differential form satisfies:

The exterior derivative is the operator $d : \Omega^k(M) \to \Omega^{k+1}(M)$ defined by:

• For $f \in \Omega^0(M)$: $df = \sum_{i=1}^n \frac{\partial f}{\partial x^i}\, dx^i$ (the total differential).
• For general $k$-forms: defined by requiring $d(df) = 0$ and the product rule $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\deg(\alpha)} \alpha \wedge d\beta$.

Proposition 4.3. $d \circ d = 0$ ($d^2 = 0$).

Theorem 4.4 (Poincare Lemma). If $M$ is a star-shaped open subset of $\mathbb{R}^n$ (or more generally, a contractible manifold), then every closed $k$-form is exact: if $d\omega = 0$, then $\omega = d\eta$ for some $(k - 1)$-form $\eta$.

The wedge product $\wedge : \Omega^k(M) \times \Omega^\ell(M) \to \Omega^{k+\ell}(M)$ is the bilinear, associative, anti-commutative operation:

Theorem 4.5 (Stokes’ Theorem). Let $M$ be an oriented $n$-dimensional manifold with boundary $\partial M$ (with the induced orientation). If $\omega$ is a compactly supported $(n-1)$-form on $M$, then:

Special Cases:

• Fundamental Theorem of Calculus ($n = 1$): $\int_a^b f'(x)\, dx = f(b) - f(a)$.
• Green’s Theorem ($n = 2$): $\oint_{\partial D} P\, dx + Q\, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\, dy$.
• Classical Stokes’ Theorem ($n = 3$): $\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$.
• Divergence Theorem ($n = 3$): $\oint_{\partial V} \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F})\, dV$.

Three proofs. The first is the identity the whole subject stands on; $d^2 = 0$ - and it deserves the computation rather than the phrase “one line to verify”: the line hides why it is true, which is that mixed partial derivatives are symmetric while the wedge product is antisymmetric, and a symmetric object contracted with an antisymmetric one vanishes. The second is naturality of $d$. The third is Stokes’ theorem, reduced to the fundamental theorem of calculus on a cube.

Proposition 4.3 ($d^2 = 0$), with the computation. For any $k$-form $\omega$, $d(d\omega) = 0$.

Proof. By linearity it suffices to check a monomial $\omega = f\,dx^{i_1} \wedge \cdots \wedge dx^{i_k}$. By the definition of $d$,

Applying $d$ again, with the product rule on each term $\frac{\partial f}{\partial x^j}\,dx^j \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_k}$:

Now pair the terms $m = j$ and $m \neq j$. When $m = j$ the wedge $dx^j \wedge dx^j = 0$ vanishes outright. When $m \neq j$, swap the names of the summation indices $m \leftrightarrow j$ in the double sum: the second-derivative coefficient is unchanged; Schwarz’s theorem, $\partial^2 f/\partial x^m \partial x^j = \partial^2 f/\partial x^j \partial x^m$; while the wedge factor picks up one transposition, hence a factor $-1$:

so the sum equals minus itself and must be zero. $\blacksquare$

The computation says exactly where the identity comes from: $d^2 = 0$ holds because $d$ builds second derivatives and the wedge product is antisymmetric. In the other order; symmetric coefficients, antisymmetric basis; every term cancels its transposed partner. Any putative “boundary of a boundary” argument must reduce to this cancellation, and any function $d'$ with symmetric second derivatives and an antisymmetric wedge would satisfy it equally, which is why the definition of $d$ is pinned down by requiring $d^2 = 0$, normalisation, and the product rule together.

Proposition (naturality). For any smooth $f : M \to N$ and any form $\omega$ on $N$: $d(f^*\omega) = f^*(d\omega)$.

Proof. Both sides are linear in $\omega$, so reduce to a monomial $\omega = g\,dy^{i_1} \wedge \cdots \wedge dy^{i_k}$ with $g$ a function on $N$. Then $f^*\omega = (g \circ f)\,d(f^*y^{i_1}) \wedge \cdots$, and applying $d$ with the product rule:

The $d^2$ terms vanish by Proposition 4.3. And $d(g \circ f) = f^*(dg)$ by the chain rule, so

Naturality is what makes cohomology functorial; the proof on the de Rham page uses exactly this; and it is why the pullback is the right map between form spaces: it commutes with the operator the cohomology is built from.

Theorem 4.5 (Stokes). For an oriented $n$-manifold $M$ with boundary and a compactly supported $(n-1)$-form $\omega$: $\int_M d\omega = \int_{\partial M}\omega$, with $\partial M$ carrying the induced (boundary-first) orientation.

Proof. Choose an oriented atlas and a subordinate partition of unity $\{\psi_i\}$ with each $\psi_i$ supported in a chart; both sides of the identity are linear in $\omega$, so it suffices to prove the case $\omega = \psi_i\,\omega_0$ for one chart, i.e. to work on $\mathbb{R}^n$ or on the half-space $\mathbb{H}^n = \{x^n \geq 0\}$; the two model cases.

The $\mathbb{R}^n$ case, no boundary. By linearity and the support, reduce to a monomial missing one differential. If $\omega$ contains all $n$ of the $dx^i$, then $d\omega$ is proportional to $dx^1\wedge\cdots\wedge dx^n$ times derivatives of the compactly supported coefficient, whose integral over $\mathbb{R}^n$ vanishes by Fubini and the fundamental theorem of calculus applied one variable at a time; the integral of a pure derivative in $x^i$ over the whole line is zero. So both sides are zero and agree. If $\omega$ is missing $dx^i$, write $\omega = a(x)\,dx^1 \wedge \cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^n$; then

the sign $(-1)^{i-1}$ counting how many transpositions move $dx^i$ to the front. Integrating $x^i$ first and applying the fundamental theorem of calculus in that variable:

since $\partial\mathbb{R}^n = \emptyset$.

The half-space case. $\mathbb{H}^n = \{x^n \geq 0\}$ has boundary the hyperplane $x^n = 0$. If the monomial $\omega$ is missing $dx^n$; the boundary direction; write $\omega = a\,dx^1\wedge\cdots\wedge dx^{n-1}$; then $d\omega = (-1)^{n-1}\frac{\partial a}{\partial x^n}dx^1 \wedge \cdots \wedge dx^n$, and Fubini + the fundamental theorem in $x^n$ give

while $\int_{\mathbb{H}^n}\omega$-monomials containing $dx^n$ integrate to zero by the $\mathbb{R}^n$ case. The boundary orientation of $\mathbb{H}^n$ is decreasing $x^n$; outward normal first; which supplies exactly the factor $-1$: $\int_{\partial\mathbb{H}^n}\omega = -\int_{x^n=0}\omega$ in the coordinates as written. So $\int_{\mathbb{H}^n}d\omega = \int_{\partial\mathbb{H}^n}\omega$, and summing over the partition of unity gives the theorem on general $M$. $\blacksquare$

The sign $(-1)^{n-1}$ is where every treatment hides a small lie: the boundary orientation is “outward normal first”, and getting that convention backwards flips the sign of every Stokes formula in the subject. The reduction to cubes is what lets the proof be elementary; the general theorem follows by the partition of unity, which is local by construction.

Integration of an $n$-form over an $n$-dimensional oriented manifold is defined by pulling back to $\mathbb{R}^n$ and integrating in coordinates. If $\omega = f\,dx^1 \wedge \cdots \wedge dx^n$ on a coordinate chart $U$ with parametrisation $\phi : U \to \mathbb{R}^n$:

Problem. Compute $\int_{S^2} \omega$ where $\omega = x\,dy \wedge dz + y\,dz \wedge dx + z\,dx \wedge dy$ on the unit sphere $S^2 \subset \mathbb{R}^3$.

An orientation on an $n$-dimensional manifold is a nowhere-vanishing $n$-form. A manifold is orientable if such a form exists. The Möbius strip is the classic example of a non-orientable manifold: any attempt to define a global $n$-form results in a sign change along the closed loop.

Integration of an $n$-form over an oriented manifold is independent of the choice of atlas (as long as the charts are orientation-preserving). The integral changes sign if the orientation is reversed.

Mistake 1: Forgetting the sign rule in the wedge product The wedge product is anti-commutative: $\alpha \wedge \beta = (-1)^{k\ell} \beta \wedge \alpha$ for a $k$-form and an $\ell$-form. Students often write $\alpha \wedge \beta = \beta \wedge \alpha$ or forget the sign entirely. For 1-forms, $dx \wedge dy = -dy \wedge dx$, so $dx \wedge dx = 0$.

Mistake 2: Applying the exterior derivative incorrectly to products The product rule for the exterior derivative is $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\deg(\alpha)} \alpha \wedge d\beta$. The sign factor $(-1)^{\deg(\alpha)}$ is frequently omitted. Forgetting it leads to incorrect computations of $d$ of higher-degree forms.

Mistake 3: Assuming $d^2 = 0$ means every closed form is exact on any manifold The Poincare lemma guarantees that closed forms are exact only on contractible (or star-shaped) domains. On manifolds with nontrivial topology like $S^1$ or $S^2$, there exist closed forms that are not exact. For example, the angular form $d\theta$ on $S^1$ is closed but not exact.

Differential forms are the natural objects to integrate on manifolds. A 0-form is a function, a 1-form is something you integrate along a curve (like work done by a force), a 2-form is something you integrate over a surface (like flux), and a 3-form is something you integrate over a volume. The exterior derivative $d$ increases the degree by one and encodes differentiation: $df$ is the gradient, $d$ of a 1-form is the curl, and $d$ of a 2-form is the divergence. The fundamental identity $d^2 = 0$ unifies the vector calculus identities $\nabla \times \nabla f = 0$ and $\nabla \cdot (\nabla \times \mathbf{F}) = 0$. Stokes’ theorem generalises the fundamental theorem of calculus to manifolds.

Differential forms unify the classical vector calculus operators in $\mathbb{R}^3$ via the identifications:

• $0$-forms $\leftrightarrow$ scalar functions
• $1$-forms $\leftrightarrow$ vector fields (via $F_1 dx + F_2 dy + F_3 dz \leftrightarrow \mathbf{F}$)
• $2$-forms $\leftrightarrow$ vector fields (via $F_1 dy\wedge dz + F_2 dz\wedge dx + F_3 dx\wedge dy \leftrightarrow \mathbf{F}$)
• $3$-forms $\leftrightarrow$ scalar functions (via $f\, dx\wedge dy\wedge dz \leftrightarrow f$)

Under these identifications:

• $df \leftrightarrow \nabla f$ (gradient)
• $d\omega \leftrightarrow \nabla \times \mathbf{F}$ (curl) for a $1$-form $\omega$
• $d\omega \leftrightarrow \nabla \cdot \mathbf{F}$ (divergence) for a $2$-form $\omega$

The identity $d^2 = 0$ becomes $\nabla \times (\nabla f) = 0$ and $\nabla \cdot (\nabla \times \mathbf{F}) = 0$.

• Smooth Manifolds: Differential forms are defined on smooth manifolds and require the smooth structure for their construction.

• Riemannian Geometry: The Hodge star operator on differential forms uses the Riemannian metric to relate $k$-forms to $(n-k)$-forms.

• Curvature: The curvature tensor can be expressed using differential forms and the exterior derivative.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

• Applications
• Curvature
• De Rham Cohomology
• Elliptic Regularity