One operator $d$ takes $k$-forms to $(k+1)$-forms, and in $\mathbb{R}^3$ it is gradient, curl and divergence depending on what you feed it. It satisfies $d^2=0$, which encodes both "curl of a gradient is zero" and "divergence of a curl is zero" in a single line. The gap between $d\omega=0$ and $\omega=d\eta$ is where topology enters the course.
Three unrelated-looking differential operators, three unrelated-looking integral theorems, and a pile of vector identities all turn out to be one operator and one theorem. The unification is not cosmetic: it is what lets the whole apparatus move to manifolds, where grad, curl and div do not individually make sense.
A conservative force field is an exact $1$-form; the potential is the $0$-form it comes from. In thermodynamics, heat is a $1$-form that is not exact and entropy is what an integrating factor makes exact. Maxwell's $dF=0$ is three of the four equations at once.
Poincaré proved the lemma in the 1880s while studying differential equations. That closed-minus-exact measures holes became de Rham cohomology, which computes topological invariants by pure calculus — a bridge between analysis and topology that is still one of the central tools in geometry.
For a $0$-form $f$, $df=\sum_j\dfrac{\partial f}{\partial x_j}dx_j$. For a general $k$-form $\omega=\sum_Ia_I\,dx_I$, where $I$ runs over increasing index tuples,
Differentiate each coefficient and wedge the result on the left. That is the entire computational content.
These four properties determine $d$ uniquely, which is why every text's definition agrees despite differing wedge conventions.
For a $0$-form, $d(df)=\sum_{i,j}\dfrac{\partial^2f}{\partial x_i\partial x_j} dx_i\wedge dx_j$. The second partials are symmetric in $i,j$ by Clairaut, while $dx_i\wedge dx_j$ is antisymmetric, so the terms cancel in pairs and the diagonal terms vanish outright.
The general case follows from the Leibniz rule by induction. Note the hypothesis: $\omega$ must be $C^2$, so Clairaut applies. Smoothness is not decoration here.
$g^{*}(d\omega)=d(g^{*}\omega)$ for smooth $g$.
On $0$-forms this is the chain rule; the general case follows from the Leibniz rule and the fact that pullback respects wedges. This identity is what makes $d$ a genuinely geometric operator: it does not depend on the coordinates you compute it in, which is what will let Unit 13 define it on a manifold.
So $d^2=0$ read on $0$-forms says $\nabla\times\nabla f=0$, and read on $1$-forms says $\nabla\cdot(\nabla\times F)=0$. Two identities usually verified by grinding through components are one theorem.
Let $\omega=xy\,dx+yz\,dy+zx\,dz$. Then
Reading off components gives the curl $(-y,-z,-x)$, and it agrees with the determinant formula — but with no formula to memorise.
$\omega$ is closed if $d\omega=0$, and exact if $\omega=d\eta$ for some form $\eta$. Since $d^2=0$, every exact form is closed. The converse is a question about the shape of the domain.
On $\mathbb{R}^2$ minus the origin, let
Locally $\omega=d\theta$, where $\theta$ is the polar angle. Globally there is no single-valued $\theta$ on the punctured plane, and the failure is caused entirely by the puncture.
On a star-shaped open set — one containing the segment from some fixed point to each of its points — every closed form is exact.
The proof constructs an explicit homotopy operator $K$ with $dK+Kd=\mathrm{id}$ on forms of positive degree, by integrating the form along the radial segments. Applied to a closed $\omega$, this reads $\omega=d(K\omega)$.
Every open set is locally star-shaped, so closed always implies locally exact. The punctured plane shows the global statement needs the hypothesis.
• Dropping the sign in the graded Leibniz rule. The
$(-1)^k$ depends on the degree of the left factor.
• Concluding exact from closed without checking the domain. The angle
form is the counterexample to keep in mind.
• Applying $d^2=0$ to a form whose coefficients are not $C^2$. The proof
needs Clairaut.
• Wedging $da_I$ on the right instead of the left. It changes the sign in
odd degree.
• Believing the vector calculus dictionary works outside $\mathbb{R}^3$.
Only the form language survives.
Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.
Find $d\omega$ for $\omega=x^2y\,dx+e^z\,dy$ in $\mathbb{R}^3$.
Verify the angle form is closed.
Is $\omega=(2xy+z)\,dx+x^2\,dy+x\,dz$ exact on $\mathbb{R}^3$? If so find $f$.
For a $1$-form $\alpha$ and $2$-form $\beta$, expand $d(\alpha\wedge\beta)$.
Compute $d$ of $\omega=x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy$.
Give a star-shaped set that is not convex, and one that is not star-shaped.
Check $g^{*}(df)=d(g^{*}f)$ directly for $f(x,y)=xy$ and $g(u,v)=(u+v,u-v)$.
Show $\omega=y\,dx-x\,dy$ is not closed, but $\omega/y^2$ is exact on $y>0$.
Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:
Spivak defines $d$ and proves the Poincaré lemma with the explicit homotopy operator; Munkres gives more detail on the vector calculus dictionary. Check which proof the syllabus expects.
Clairaut gives $d^2=0$, which makes "exact implies closed" free. The Poincaré lemma gives the converse on star-shaped sets. Unit 14 pairs $d$ with the boundary operator, and the pairing is Stokes' theorem.
Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.
Which vector identities does $d^2=0$ contain?
State the implication that always holds and the one that does not.
Name a closed non-exact form and its domain.
Why does $g^{*}d=dg^{*}$ matter for manifolds?