4.3 Differential Forms · Homology & Cohomology

Stokes’ theorem shares three important attributes with many fully evolved major theorems: 1. It is trivial. 2. It is trivial because the terms appearing in it have been properly defined. 3. It has significant consequences.

— Michael Spivak, Calculus on Manifolds (1965), p. 104

In the last two chapters you met a curious kind of calculus that lives on the edges of a graph. You put heights on the vertices, took differences along the edges, and discovered that some edge labellings could not come from any heights at all — the impossible staircase. That discrete calculus has an older, smooth sibling: the calculus of slopes, flows, areas and volumes that engineers and physicists have used for three centuries. This chapter rebuilds it from nothing, in a language called differential forms. In that language the calculus can feel the shape of the space it lives on, holes and all.

Here is a puzzle to carry with you. In 1854 the Swiss mathematician Jakob Amsler introduced the polar planimeter: two hinged arms, a pointer and a small measuring wheel (Foote et al. 2013). Set its pivot down beside a field on a map, trace the field’s outline once round with the pointer, and the wheel reports the field’s area. The instrument never visits the inside. How can it know what is in there?

The answer is a theorem about insides and boundaries, which comes in four famous versions: the fundamental theorem of calculus, Green’s theorem, Stokes’ theorem and the divergence theorem. In the language of forms all four collapse into one line, $\int_{\partial M} ω = \int_{M} d ω .$ You do not need to remember any calculus: we relearn slopes and areas with pictures first, and only then jump to forms.

A calculus refresher in pictures

We need three ideas from ordinary calculus: the derivative (a rate), the integral (a total), and the theorem that links them. If you already know them, skim this section for the pictures, because the pictures are what we will generalise.

Slopes and rates

Imagine walking along a hilly road. Let $F (x)$ be your height above sea level when you are at horizontal position $x .$ Between two nearby positions $x$ and $x + h,$ the road rises by $F (x + h) - F (x)$ over a horizontal distance $h,$ so its average steepness there is

\frac{rise}{run} = \frac{F ( x + h ) - F ( x )}{h} .

As $h$ shrinks towards $0,$ this ratio settles down (for the smooth roads we will consider) to a single number, the derivative $F^{'} (x),$ read “ $F$ prime of $x ”$ and also written $\frac{d F}{d x} .$ It is the slope of the graph at $x :$ positive going uphill, negative going downhill, zero at a summit. If $F (t)$ were your position along a straight road at time $t$ instead, $F^{'} (t)$ would be your velocity: the number on the speedometer, with a minus sign while you are reversing.

Try it on the parabola $F (x) = x^{2} .$ The rise over a run $h$ is $\frac{( x + h ) ^{2} - x ^{2}}{h} = \frac{2 x h + h ^{2}}{h} = 2 x + h,$ and as $h$ shrinks this settles to $F^{'} (x) = 2 x .$ At $x = 3$ the parabola climbs 6 units for every unit of run; at $x = 0$ it is momentarily flat. The same expansion gives $x^{3}$ the slope $3 x^{2},$ and in general $x^{n}$ the slope $n x^{n - 1} .$ Constant factors ride along ( $5 x^{2}$ has slope $10 x),$ and a sum has the sum of the slopes. That is nearly all the differentiating this chapter will ask of you.

Here is the reformulation that matters for this chapter. Turn the definition around: for a small step $Δ x$ (read “delta $x ”,$ a small change in $x),$ the change in height is approximately

Δ F \approx F^{'} (x) Δ x .

So the derivative is a little machine: feed it a small step, and it predicts how much you will climb. Notice that the prediction is linear in the step — twice the step, twice the climb; a step backwards, a negative climb. Keep this machine in mind. It is the seed from which differential forms grow.

Totals

The second idea runs the other way: from a rate to a total. If you know your speed at every moment, how far did you travel? At a steady 60 kilometres an hour for 2 hours the answer is 120 kilometres: speed times time, the area of a 2-by-60 rectangle under the graph of speed. When the speed varies, chop the time into many short pieces of length $Δ t;$ on each piece the speed hardly changes, so distance $\approx$ speed $\times$ $Δ t;$ add the pieces up. As the pieces get shorter, these sums approach a single number. For a function $f$ on an interval $[a, b],$ cut the interval into $n$ equal pieces of width $Δ x = (b - a) / n,$ pick a point $x_{i}$ in the $i$ -th piece, and let $n$ grow: the limit is the integral

\int_{a}^{b} f (x) d x = n \to \infty lim i = 1 \sum n f (x_{i}) Δ x .

The long S, $\int,$ is an old-fashioned letter S for “sum”; $a$ and $b$ are where the adding starts and stops; $d x$ remembers the little widths $Δ x .$ In a picture, each term $f (x_{i}) Δ x$ is the area of a thin rectangle, so the integral is the area under the graph of $f$ — with the convention that area below the axis counts as negative.

The fundamental theorem

Now put the two ideas together. Take $f = F^{'},$ the slope of our road, and add up the predicted rises $F^{'} (x_{i}) Δ x$ across the interval from $a$ to $b .$ Each predicted rise is close to the actual rise $F (x_{i + 1}) - F (x_{i})$ on its piece, and the actual rises add up in a wonderfully lazy way. With three pieces, $a = x_{0} < x_{1} < x_{2} < x_{3} = b :$

[F (x_{1}) - F (x_{0})] + [F (x_{2}) - F (x_{1})] + [F (x_{3}) - F (x_{2})] = F (x_{3}) - F (x_{0}) .

Every intermediate height appears once with a plus sign and once with a minus sign, and cancels. Only the two ends survive. Sums that collapse like this are called telescoping, after the way a pocket telescope folds up. Letting the pieces shrink gives the most important theorem of calculus.

Adding up little rises Drag a and b · change the number of pieces

Sum of predicted rises (signed area)

\sum_{i} F^{'} (x_{i}) Δ x = 0.770

Actual change in height

F (b) - F (a) = 0.732

Figure 4.3.1 Top: a road

F

and a gold broken line that starts at height

F (a)

and climbs, on each piece, by the rise that the slope predicts. Bottom: the slope

F^{'},

with rectangles whose signed areas are those same predicted rises (rose ones, below the axis, count as negative). As the pieces get thinner, the gold line hugs the road and the total of the rectangles approaches the actual change

F (b) - F (a),

the teal bar at

b .

Read the theorem as a sentence about insides and boundaries. The left side adds up a derivative over the inside of the interval $[a, b] .$ The right side looks only at the boundary of the interval — its two endpoints — counting the end $b$ with a plus sign and the start $a$ with a minus sign. You have seen exactly this signed boundary before: in §3.2 Chains and the Boundary Operator the boundary of an edge $[a, b]$ was $\partial [a, b] = b - a .$ Hold on to this sentence. The rest of the chapter is about making it true in every dimension:

Arrows everywhere: vector fields and work

Slopes in two directions

A landscape is a function of two variables: the height $f (x, y)$ at the point with east–west coordinate $x$ and north–south coordinate $y .$ Standing at a point, the slope depends on which way you face. Two directions are especially easy. Walking due east, $y$ stays fixed while $x$ changes, and the slope you feel is the partial derivative $\frac{\partial f}{\partial x},$ read “partial $f$ partial $x ”.$ It is computed exactly like an ordinary derivative while pretending $y$ is a constant. Walking due north gives $\frac{\partial f}{\partial y} .$ For example, if $f (x, y) = x^{2} y$ then $\frac{\partial f}{\partial x} = 2 x y$ (the $y$ rides along like the 5 in $5 x^{2})$ and $\frac{\partial f}{\partial y} = x^{2} .$ We will often shorten these to $f_{x}$ and $f_{y} .$

For a small step $(Δ x, Δ y)$ in any direction, the change in height is approximately the eastward part of the step times the eastward slope, plus the northward part times the northward slope:

Δ f \approx \frac{\partial f}{\partial x} Δ x + \frac{\partial f}{\partial y} Δ y .

Again a machine that eats a step and predicts a climb — and again linear in the step. The two slopes are often packaged as an arrow, the gradient $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ (the symbol $\nabla$ is read “nabla” or simply “grad”). The gradient points straight uphill, its length is the steepness, and it is perpendicular to the contour lines of the map, along which the height does not change.

Arrows everywhere

A vector field is a choice of arrow at every point: the wind on a weather map, the current in a river, the gravitational pull at each point of space. In the plane we write it as $F (x, y) = (P (x, y), Q (x, y)) :$ at the point $(x, y)$ the arrow goes $P$ units east and $Q$ units north. Bold letters like $F$ are a reminder that the value is an arrow, not a number.

The gradient of a landscape is one example. Others are the rotation $F = (- y, x),$ which spins the plane counterclockwise, and the source $F = (x, y),$ which streams out of the origin. Figure 4.3.2 shows them as they are best seen: alive. The particles ride along the arrows, and the streaky texture behind them traces the flow lines everywhere at once.

A gallery of vector fields Pick a field · switch the colouring · drag the paddle wheel

Arrows

The gradient of a landscape with two hills: every arrow points straight uphill.

F = \nabla f

curl F = 0

div F = \nabla^{2} f

Paddle wheel at

(0.80, - 0.90)

F = (- 0.18, 0.69)

curl = 0.00

div = 0.40

Figure 4.3.2 Six planar vector fields. The gold paddle wheel turns at the rate the fluid rotates around it; the colourings (curl, divergence, potential) are explained in the next section. Try the shear field: its flow lines are perfectly straight, yet the paddle wheel turns. And try vortex with the potential colouring: its level lines are rays, an object we will meet again.

Work along a path

Suppose $F$ is a force and you carry an object along a path $γ$ (the Greek letter gamma). On a short straight step $Δ r = (Δ x, Δ y),$ the work done by the force is the part of the force pointing along the step, times the length of the step. In coordinates this is the dot product

F \cdot Δ r = P Δ x + Q Δ y .

Pushing with the motion counts positively, against it negatively, sideways not at all. Chop the path into many small steps, add up the work of each, and let the steps shrink: the result is the line integral

\int_{γ} F \cdot d r = \int_{γ} P d x + Q d y .

A path has a direction of travel, and walking it backwards reverses every step and flips the sign of the answer. When $γ$ is a closed loop we write $\oint_{γ},$ with a little circle on the integral sign.

Now suppose the force is a gradient, $F = \nabla f .$ Then the work of each small step is approximately the change $Δ f$ of $f$ on that step, the changes telescope exactly as on the road, and

\int_{γ} \nabla f \cdot d r = f (end) - f (start) .

This is the fundamental theorem again, now along a curved path in the plane. It has two striking consequences. The work does not depend on the route, only on where you start and finish. And around any closed loop, where the start is the end, the work is zero. Fields with these properties are called conservative. Gravity near the ground is one, with $f = - m g y :$ the weight $m g$ times the height $y,$ with a minus sign because gravity pulls down. Haul a suitcase up three flights of stairs by any route you like, and gravity does the same (negative) work.

Swirl and spread: three famous theorems

The loop integral around a closed curve deserves a name: it is the circulation of $F$ around the curve, the total push you feel going round. Since a gradient has zero circulation around every loop, circulation is a test for “being a gradient”. But a loop is a global thing. Is there a local quantity, measured at single points, that adds up to the circulation?

Circulation and curl

Shrink the loop to a tiny square of side $h$ with its lower-left corner at $(x, y),$ and walk it counterclockwise. Along the bottom edge (going east) the work is about $P (x, y) h;$ along the top edge (going west) about $- P (x, y + h) h .$ Together they give $- (P (x, y + h) - P (x, y)) h \approx - \frac{\partial P}{\partial y} h^{2} .$ The right and left edges similarly give $\frac{\partial Q}{\partial x} h^{2} .$ So the circulation around the tiny square is approximately

(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) h^{2} = curl F \times area .

A physical picture: drop a tiny paddle wheel into the flow. It turns counterclockwise where the curl is positive, clockwise where it is negative, and the rate at which it turns is exactly half the curl. For the rotation $(- y, x)$ the curl is $1 - (- 1) = 2$ everywhere. For the shear $(y, 0)$ it is $0 - 1 = - 1 :$ the flow lines are straight, but the water above the wheel moves faster than the water below it, so the wheel turns clockwise. Switch on the curl colouring in Figure 4.3.2 and watch.

The vortex of Figure 4.3.2, $F = (\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}}),$ springs the opposite surprise. It is roughly how water swirls round a plughole, and its flow lines are circles, yet its curl is $0$ everywhere it is defined (everywhere except the origin). The water near the centre moves faster than the water farther out, by exactly enough to cancel the turning, so a paddle wheel carried round the plughole keeps facing the same way. Curl measures the spin of a tiny wheel, not the bending of the flow lines. We will meet this field again.

For a gradient, $P = f_{x}$ and $Q = f_{y},$ so $curl \nabla f = f_{y x} - f_{x y} :$ the difference of the two “mixed” second derivatives, taken in the two possible orders. Try it on $f = x^{2} y :$ differentiating $f_{x} = 2 x y$ by $y$ gives $2 x,$ and differentiating $f_{y} = x^{2}$ by $x$ gives $2 x$ as well. For every function with continuous second derivatives the two orders agree — a classical fact usually credited to Schwarz or Clairaut — so a gradient has zero curl everywhere. Zero curl is the local shadow of having zero circulation.

Flux and divergence

Circulation measures flow along a curve. The other natural question is how much flows across it. The flux of $F$ out of a closed curve is $\oint F \cdot n d s,$ where $n$ is the outward-pointing unit arrow perpendicular to the curve and $d s$ is the length of a small piece. Back to the tiny square. Fluid leaves through its east edge at a rate of about $P (x + h, y) h$ and enters through the west edge at about $P (x, y) h,$ a net outflow of about $\frac{\partial P}{\partial x} h^{2};$ the north and south edges compare $Q$ in the same way. So the outward flux of a tiny square is about

(\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}) h^{2} = div F \times area .

The divergence $div F = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$ is the outflow per unit area: positive at a source, where fluid appears; negative at a sink, where it disappears. The source $(x, y)$ has divergence $2$ everywhere; the rotation has divergence $0 .$

Green’s theorem

Here is the idea that turns local into global. Take a region $R$ in the plane and tile it with tiny squares. Add up the circulations of all the tiles. Each interior edge belongs to two neighbouring tiles, which walk it in opposite directions, so its contributions cancel. Only the edges on the outer boundary survive, and they make up the counterclockwise walk around the boundary of $R,$ written $\partial R .$ On the other hand, each tile’s circulation is about its curl times its area, and these add up to the total curl inside.

Why the inside talks to the boundary Step through, or press play

One small square

Figure 4.3.3 Tile a region with small squares, each walked counterclockwise. An edge shared by two tiles is walked once in each direction, so it cancels; only the outer boundary is left. This is the same cancellation that makes the boundary of a sum of triangles in §3.2 Chains and the Boundary Operator lose all its interior edges.

The tiling argument is the whole idea of the proof; making “about” precise is a matter of careful estimates. Rather than take it on trust, test it. In Figure 4.3.4 the computer evaluates the two sides independently — one by walking along your curve, the other by adding up over the region inside it — and you can drag the curve wherever you like.

Green’s theorem, live Drag the round handles to reshape · the diamond moves the loop

Loop:

Circulation around the loop

\oint_{\partial R} F \cdot d r = 0.650

Total curl inside

\iint_{R} curl F d A = 0.650

Two very different computations — one along the curve, one over the region inside — and they agree.

Figure 4.3.4 The background is coloured by the curl (rose counterclockwise, teal clockwise) or the divergence (gold source, blue sink). The number on the left is computed along the gold curve; the number on the right, over the shaded region inside it. Put the loop around one eddy, then around both: their swirls cancel. Make a figure eight: the two lobes are walked in opposite senses and count with opposite signs.

Example How the planimeter knows the area

Now the puzzle from the start of the chapter. Take the field $F = (0, x),$ so that $P = 0$ and $Q = x .$ Its curl is $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$ everywhere, so Green’s theorem says that for every region $R$

\oint_{\partial R} x d y = \iint_{R} 1 d A = the area of R .

A walk around the edge, adding up $x$ times each small northward step, measures the area inside. A planimeter does this kind of sum mechanically. Its wheel is mounted on the tracer arm with its axle along the arm, so it rolls when the arm moves sideways and merely skids when the arm slides lengthways: as the pointer travels, the wheel adds up the sideways part of each small step. That is the line integral of a field built from the geometry of the two arms, and a calculation with that geometry shows that its curl is the same at every point, $1/ L,$ where $L$ is the length of the tracer arm (Gatterdam 1981). By Green’s theorem, the wheel’s total roll is the area divided by $L .$ The instrument does not need to visit the inside, because Green’s theorem has already moved the inside onto the edge.

In three dimensions

In space a vector field has three components, $F = (P, Q, R) .$ A tiny loop can now face in any direction, so the curl becomes an arrow, $curl F = (R_{y} - Q_{z}, P_{z} - R_{x}, Q_{x} - P_{y}),$ whose component in a direction $n$ is the circulation per unit area around a tiny loop facing $n .$ Hold a paddle wheel in a flow in space and how fast it turns depends on where you point its axle; it turns fastest with the axle along the curl arrow. The divergence is $div F = P_{x} + Q_{y} + R_{z},$ the outflow per unit volume of a tiny box. The tiling argument, with tiny squares on a curved surface or tiny cubes in a solid, gives two more theorems:

\oint_{\partial S} F \cdot d r = \iint_{S} (curl F) \cdot n d A (Kelvin-Stokes), \iint_{\partial V} F \cdot n d A = ∭_{V} div F d V (Gauss) .

In the first, $S$ is a surface with boundary curve $\partial S;$ in the second, $V$ is a solid with boundary surface $\partial V,$ and $n$ points outward. The second one is the divergence theorem: what flows out through the skin is what was created inside.

The divergence theorem in space Drag the sphere to move it, its rim to resize · drag elsewhere to rotate

Add a wind

Flux out through the sphere

\iint_{S} F \cdot n d A = 0.846

Total divergence inside

∭_{B} div F d V = 0.846

Figure 4.3.5 The glowing blob is a source: the field’s divergence is concentrated there. Arrows on the glass sphere show the flow through its surface (gold out, teal in). Both totals are computed independently, by adding up over the sphere’s surface and over its solid inside. Drag the sphere off the blob and the flux drops to zero; switch on the wind and arrows point inward on one side and outward on the other, but the total does not change, because a uniform wind creates nothing.

Line up what we have so far:

Theorem	Inside	Boundary
Fundamental theorem	$\int_{a}^{b} F^{'} (x) d x$	$F (b) - F (a)$
Gradient theorem	$\int_{γ} \nabla f \cdot d r$	$f (end) - f (start)$
Green / Kelvin–Stokes	$\iint_{S} curl F \cdot n d A$	$\oint_{\partial S} F \cdot d r$
Gauss	$∭_{V} div F d V$	$\iint_{\partial V} F \cdot n d A$

Four theorems, one shape: a derivative integrated over the inside equals the original thing integrated over the boundary. Four different kinds of derivative (slope, gradient, curl, divergence), four kinds of integral (along intervals, curves, surfaces, solids). It is hard to believe this is a coincidence, and it is not. To see the single theorem behind all four, we need the right things to integrate.

Curl and divergence are also the language of fluids. A fluid that cannot be compressed, like water, has a velocity field with $div u = 0$ everywhere, and $curl u$ measures how it swirls. Newton’s law of motion for such a fluid, with friction included, is the system of Navier–Stokes equations, and one of the most famous problems in mathematics asks whether its solutions can break down.

Problem Navier–Stokes existence and smoothness

Solution claimed in 2026 Posed by Jean Leray, 1934; claimed by OpenAI, with an unreleased AI model, 2026

Do the Navier–Stokes equations of an incompressible fluid in three dimensions always have smooth solutions that last for all time, starting from any smooth initial velocity? Or can the velocity become infinite at some point after a finite time?

“Break down” means that the velocity, or its rate of change, becomes infinite at some point after a finite time, though everything started out smooth. Weather forecasts and aircraft design solve these equations numerically every day; the question is whether the mathematics behind them can fail.

How it was settled. On 8 September 2026 OpenAI announced a proof, checked in the Lean proof assistant, that a fluid initially at rest and pushed by a suitable smooth force can develop a singularity in finite time. That is one of the four alternatives in the Clay Mathematics Institute’s official statement of the problem. Charles Fefferman, who wrote that statement, said he was thrilled that the problem was solved, and credited the method of Diego Córdoba and Luis Martínez-Zoroa on which the construction builds. Others point out that the force must have special properties (it cannot be analytic, by work of Constantin, Ignatova and Vicol), and the Clay Institute still lists the problem as open.

What is known. The question has been one of the seven Millennium Prize Problems of the Clay Mathematics Institute since 2000. In two dimensions smooth solutions last forever. In three, Leray proved in 1934 that weaker solutions always exist, and Caffarelli, Kohn and Nirenberg showed in 1982 that their singularities, if there are any, are rare. Whether a smooth flow can blow up with no driving force at all is still open.

A way in. Write the velocity of the fluid as $u .$ Incompressibility says $div u = 0,$ which by the divergence theorem means that as much fluid flows into any ball as flows out. What can concentrate is the swirl, $curl u :$ in three dimensions a spinning tube of fluid can be stretched thinner and spin faster, which cannot happen in a flat, two-dimensional flow.

(Leray 1934; Fefferman 2000; Constantin, Ignatova & Vicol 2026) Status as of October 2026 All open problems

A little history A postscript and an exam question

William Thomson, Lord Kelvin 1824 – 1907

George Gabriel Stokes 1819 – 1903

The theorem we now call Stokes’ theorem first appeared in a postscript to a letter from William Thomson (later Lord Kelvin) to George Stokes, dated 2 July 1850; the two friends corresponded for more than fifty years. Thomson had entered Glasgow University at ten, and as a student there mastered Fourier’s book on heat in a fortnight (O’Connor & Robertson 2003). Stokes, a rector’s son from County Sligo, once told his wavering fiancée that without her he might “go to the grave a thinking machine” (O’Connor & Robertson 2003). He set Cambridge’s Smith’s Prize examination, and put Thomson’s theorem on the paper of February 1854 as question 8 (Stokes 1854; Spivak 1965, Preface). The 22-year-old James Clerk Maxwell, joint winner that year, stated it in his Treatise of 1873 with a footnote: “This theorem was given by Professor Stokes. Smith’s Prize Examination, 1854, question 8” (Maxwell 1873, art. 24). So the theorem carries the name of the examiner, not of its discoverer.

The single formula of this chapter came later. Vito Volterra in 1889 and Henri Poincaré in 1899 stated versions in any number of dimensions; Élie Cartan created the calculus of differential forms in 1899 (Cartan 1899), and in 1945 he wrote the theorem in the form $\int_{\partial M} ω = \int_{M} d ω$ (Katz 1979).

Covectors and 1-forms: stacks of sheets

Look again at the integrand of a line integral, $P d x + Q d y .$ What is it, really? Not quite the arrow $F :$ it is a recipe that turns a small step into a number. That recipe is the hero of this section.

Measuring devices

Picture a measuring device for arrows. You hand it an arrow, it hands you back a number, and it respects arrow arithmetic: the reading of a sum is the sum of the readings, and doubling the arrow doubles the reading. Such a device is called a covector. You met covectors in §1.5 Linear Algebra under another name: the covectors on a vector space $V$ are the elements of its dual space $V^{*},$ the linear functions from $V$ to the numbers.

In the plane there are two especially simple measuring devices. The covector $d x$ reads off how far east an arrow goes, and $d y$ reads off how far north:

d x (v) = v_{1}, d y (v) = v_{2} for v = (v_{1}, v_{2}) .

Every covector on the plane is a combination $α = a d x + b d y,$ with $α (v) = a v_{1} + b v_{2} .$ For instance, with $v = (3, 4) :$ $d x (v) = 3,$ $d y (v) = 4,$ and $(2 d x + d y) (v) = 2 \cdot 3 + 4 = 10 .$

How should you picture a covector? An arrow is the wrong picture. Draw instead the lines where the linear function $a x + b y$ takes the values $\dots, - 2, - 1, 0, 1, 2, \dots :$ a stack of evenly spaced parallel lines, with a marked direction in which the value increases. Then $α (v)$ is the number of lines the arrow $v$ crosses, counted positively in the marked direction and negatively against it. A large covector is a dense stack; $d x$ is the stack of vertical lines one unit apart. The picture was made famous by the physics textbook Gravitation by Charles Misner, Kip Thorne and John Wheeler, who draw a 1-form as a family of surfaces and its value on an arrow as the number of surfaces the arrow pierces (Misner et al. 1973).

A covector at every point

The most important example comes from a function. Our “step in, climb out” machine $Δ f \approx f_{x} Δ x + f_{y} Δ y$ is exactly a 1-form, the differential of $f :$

df = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y .

Apply this to the function $f (x, y) = x :$ its partial derivatives are $1$ and $0,$ so $d (x) = 1 d x + 0 d y .$ The basic covector $d x$ really is “ $d$ of $x ”,$ which justifies the notation.

The stack picture now becomes a map. A 1-form is a field of tiny stacks, one at each point. For $df$ the stacks fit together into the level curves $f = 0, \pm ε, \pm 2 ε, \dots$ of $f,$ for some small spacing $ε$ (epsilon): the contour lines of a hiking map. And $d x$ is the stack of vertical lines $x = const :$ “ $d x$ counts how many $x$ -sheets you cross”.

Integrating a 1-form

A 1-form eats small steps, so it can be added up along a path: chop the oriented curve $γ$ into small steps $Δ r_{i}$ starting at points $p_{i},$ add up $ω_{p_{i}} (Δ r_{i}),$ and refine. The limit is $\int_{γ} ω .$ For $ω = P d x + Q d y$ it is the same number as the work $\int_{γ} F \cdot d r$ of the field $F = (P, Q)$ — but notice that no arrows were needed. The 1-form is the thing you integrate along curves. In practice, if $γ (t) = (x (t), y (t))$ for $a \leq t \leq b,$ then

\int_{γ} ω = \int_{a}^{b} (P (γ (t)) x^{'} (t) + Q (γ (t)) y^{'} (t)) d t,

and the answer does not depend on how fast you travel along $γ,$ only on the route and its direction.

In the stack picture, $\int_{γ} ω$ is the net number of sheets the path crosses (forwards minus backwards), each sheet worth $ε .$ For $ω = df,$ crossing level curves from $f (A)$ to $f (B)$ gives the gradient theorem in its most memorable form:

\int_{γ} df = f (B) - f (A) (count the contour lines you climb).

A 1-form as a stack of sheets Drag the endpoints A, B and the middle of the path

The form and its function

ω = 2 d x + d y = d (2 x + y)

Sheets crossed 19 forward · 0 backward · net 19 (each sheet is worth ε = 0.4)

The line integral

\int_{γ} ω = f (B) - f (A) = 7.55 \approx 19 \times 0.4

Figure 4.3.6 Teal curves are the sheets of

ω = df

(the small arrows show which way across counts as

+ 1).

Green dots mark forward crossings, rose rings backward ones. Bend the path as much as you like: the net count — and so the integral — depends only on the endpoints. Drag

B

close to

A

to see a covector measuring a single short arrow.

The wedge product: oriented area and volume

A 1-form measures tiny arrows, so it can be integrated along curves. To integrate over a surface we need something that measures tiny parallelograms. Since a parallelogram is spanned by two arrows, we want a measuring device that eats two arrows and returns an area.

Oriented area

The parallelogram spanned by $u = (u_{1}, u_{2})$ and $v = (v_{1}, v_{2})$ has area $∣ u_{1} v_{2} - u_{2} v_{1} ∣ .$ Drop the absolute value and you get the signed area $u_{1} v_{2} - u_{2} v_{1},$ the determinant of the matrix with columns $u$ and $v$ from §1.5 Linear Algebra. Its sign records an orientation: it is positive when turning from $u$ to $v$ is a counterclockwise turn, and negative when it is clockwise. Three rules make the signed area tick: swapping $u$ and $v$ flips its sign; if $v$ is parallel to $u$ the area is $0;$ and it is linear in each arrow separately.

dx ∧ dy measures oriented area Drag the tips of u and v · swap them

(d x \land d y) (u, v) = u_{1} v_{2} - u_{2} v_{1} = (2) (1.75) - (0.5) (0.75) = 3.125

Figure 4.3.7 The signed area of the parallelogram spanned by

u

and

v .

When

v

lies counterclockwise from

u

the area is positive (gold); push

v

past the line of

u

and it turns negative (hatched rose). The dashed square is the unit of area.

Directly from the formula: $α \land β = - β \land α,$ and so $α \land α = 0;$ in particular $d x \land d x = 0$ and $d y \land d x = - d x \land d y .$ These two rules are all you need to compute. For example,

(a d x + b d y) \land (p d x + q d y) = a p 0 d x \land d x + a q d x \land d y + b p - d x \land d y d y \land d x + b q 0 d y \land d y = (a q - b p) d x \land d y .

Out pops the determinant $a q - b p$ of the coefficients. So $(d x + 2 d y) \land (3 d x - d y) = (1 \cdot (- 1) - 2 \cdot 3) d x \land d y = - 7 d x \land d y .$

2-forms, 3-forms, k-forms

A 2-form on the plane is a smoothly varying multiple of $d x \land d y,$ say $g (x, y) d x \land d y .$ It is integrated over a region $R$ by chopping $R$ into tiny parallelograms, measuring each, and adding: if $R$ carries the standard counterclockwise orientation, $\iint_{R} g d x \land d y$ is the ordinary area integral $\iint_{R} g d A,$ and reversing the orientation flips the sign.

In space, with coordinates $x, y, z,$ there is more room:

1-forms $P d x + Q d y + R d z$ measure arrows, and are integrated along curves.
2-forms $A d y \land d z + B d z \land d x + C d x \land d y$ measure parallelograms. Each basic piece measures a shadow: $d x \land d y$ gives the signed area of the parallelogram’s shadow on the floor (the $x y$ -plane), $d y \land d z$ its shadow on the $y z$ -wall, $d z \land d x$ on the $z x$ -wall. They are integrated over surfaces.
3-forms $f d x \land d y \land d z$ measure boxes: $d x \land d y \land d z$ gives the signed volume of the box spanned by three arrows, the $3 \times 3$ determinant. They are integrated over solids.

Counting the basic pieces gives $1, 3, 3, 1$ in degrees $0, 1, 2, 3 :$ a row of Pascal’s triangle. In $n$ dimensions a k-form measures tiny oriented $k$ -dimensional parallelepipeds, there are $(k n)$ basic ones, and there are no nonzero $k$ -forms with $k > n,$ because some $d x_{i}$ would have to appear twice and $d x_{i} \land d x_{i} = 0 .$ The space of $k$ -forms on $U$ is written $Ω^{k} (U) .$

The exterior derivative: where sheets end

We have one derivative already: $d$ turns a function $f$ into the 1-form $df,$ and $\int_{γ} df = f (B) - f (A)$ is the fundamental theorem in the new language. Green’s theorem asks for the next rung. Whatever “ $d$ of a 1-form” turns out to be, its integral over a region ought to equal the integral of the 1-form round the region’s edge. For $ω = P d x + Q d y$ that means it had better be the curl, $(Q_{x} - P_{y}) d x \land d y .$ One short rule delivers exactly this, and keeps working in every degree: differentiate the coefficient, and wedge the result onto what was already there.

Let us compute it for a general 1-form in the plane, $ω = P d x + Q d y .$ By the rule,

d ω = d P \land d x + d Q \land d y = (P_{x} d x + P_{y} d y) \land d x + (Q_{x} d x + Q_{y} d y) \land d y .

Two of the four products contain $d x \land d x$ or $d y \land d y$ and vanish, and $d y \land d x = - d x \land d y,$ leaving

d (P d x + Q d y) = (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x \land d y .

There is the curl, produced by a rule that never mentioned it. For example, $d (x^{2} y d x + x y d y) = (y - x^{2}) d x \land d y .$ Now Green’s theorem can be written without a single arrow:

\oint_{\partial R} ω = \iint_{R} d ω .

Where sheets end

What does $d ω$ look like? Take $ω = x d y,$ whose integral around a loop measured area in the planimeter example. Its covectors are multiples of $d y,$ so its sheets are horizontal lines; but at horizontal position $x$ the stack is $x$ times as dense as the stack of $d y,$ so the lines crowd together as you move right. Horizontal lines whose density grows towards the right cannot all run forever: new lines must begin. Mark each beginning with a dot. The dots turn out to be spread perfectly evenly, one dot for every $ε$ of area if each sheet is worth $ε,$ and that even sprinkling is the 2-form $d x \land d y = d (x d y) .$ By contrast the sheets of $d y,$ or the contour lines of any function, never begin or end, and indeed $d (d y) = 0$ and $d (df) = 0 .$

d marks where the sheets end Pick a form · drag the rectangle, or its corner to resize

ω = x d y

d (x d y) = d x \land d y

Net sheets piercing the boundary+9

Sheet-ends (dots) inside+9

The lines crowd together as you move away from the y-axis, so new lines must begin. Every beginning is a dot of dω — and the dots are spread evenly: dω = dx∧dy.

Figure 4.3.8 Each 1-form is drawn as co-oriented curves (ticks mark the side that counts

+ 1);

dots mark where curves begin — gold for positive, rose rings for negative — and together the dots are

d ω .

For every rectangle, the net number of curves piercing its counterclockwise boundary equals the signed number of dots inside: Stokes’ theorem as pure counting. The last-but-one form,

d θ = \frac{x d y - y d x}{x ^{2} + y ^{2}}

on the plane with the origin removed, has no dots at all, yet curves stream out of any rectangle around the origin.

Gradient, curl and divergence are one operator

In space, apply the same rule in each degree:

df = f_{x} d x + f_{y} d y + f_{z} d z,

d (P d x + Q d y + R d z) = (R_{y} - Q_{z}) d y \land d z + (P_{z} - R_{x}) d z \land d x + (Q_{x} - P_{y}) d x \land d y,

d (A d y \land d z + B d z \land d x + C d x \land d y) = (A_{x} + B_{y} + C_{z}) d x \land d y \land d z .

Read the coefficients: the first line is the gradient, the second the curl, the third the divergence. The three operators of vector calculus, each with its own formula to memorise, are one and the same $d,$ applied to forms of degree $0,$ $1$ and $2 .$

The de Rham ladder

Figure 4.3.9 Top row: forms on

R^{3},

with

d

raising the degree. Bottom row: the pieces they are integrated over, with the boundary operator

\partial

lowering the dimension. Integration pairs each column. Stokes’ theorem says that taking

d

on top and taking

\partial

underneath give the same number.

d of d is zero

In the language of vector calculus, $d \circ d = 0$ on functions says $curl (grad f) = 0,$ and on 1-forms says $div (curl F) = 0 .$ In the sheet picture it says something lovely: the ends of sheets have no ends. And it should ring a bell. In §3.2 Chains and the Boundary Operator you proved $\partial \circ \partial = 0,$ “the boundary of a boundary is empty”, and in §4.1 Cochains: Measuring Shapes its mirror image $δ \circ δ = 0 .$ The exterior derivative belongs to the same family. Whenever an operator squares to zero, its image sits inside its kernel, and the gap between them — the things killed by $d$ that are not produced by $d$ — is going to measure something.

One last rule, for computing: $d$ satisfies a product rule, $d (f ω) = df \land ω + f d ω,$ and more generally $d (α \land β) = d α \land β + (- 1)^{k} α \land d β$ when $α$ is a $k$ -form. You will not need it often in this book.

Pullbacks, manifolds and integration

Pulling forms back along maps

A disk is awkward to chop into little squares, but it is easy to describe in polar coordinates: every point of the disk of radius $R$ is $(r cos θ, r sin θ)$ for some distance $r$ between $0$ and $R$ and some angle $θ$ between $0$ and $2 π .$ So the map $φ (r, θ) = (r cos θ, r sin θ)$ takes a plain rectangle of $(r, θ)$ values and wraps it round onto the disk. Can we find the disk’s area, the integral of $d x \land d y,$ by working on the rectangle instead? To do that we must move the form from the disk back to the rectangle. The same need arises whenever we integrate over a curved surface, which is never a flat region but can be described by one.

In general, suppose $φ : U \to V$ (phi) is a smooth map between regions. A small arrow $v$ at a point $p$ of $U$ is carried by $φ$ to a small arrow $D φ_{p} (v)$ at $φ (p)$ — its image under the best linear approximation of $φ,$ the matrix of partial derivatives. A form $ω$ on $V$ can then measure arrows in $U :$ carry them across, and measure their images.

Pulling back area Drag the highlighted cell in either picture · arrow keys

area of the cell in the (r, θ) rectangle

Δ r Δ θ = 0.25 \times 0.393 = 0.098

area of its image in the plane

r ˉ Δ r Δ θ = 1.375 \times 0.098 = 0.135

Figure 4.3.10 A grid cell in the

(r, θ)

rectangle is carried by

φ

to a curved cell in the plane. Its area there is exactly

r ˉ Δ r Δ θ,

where

r ˉ

is the cell’s middle radius: cells farther out are stretched more. That stretching factor is why

φ^{*} (d x \land d y) = r d r \land d θ .

Notice the direction of travel. The map $φ$ carries points from $U$ to $V,$ but the pullback carries forms the other way, from $V$ back to $U .$ This is our fourth recurring idea, reversed arrows, again: preimages of sets in §1.1 Sets and Functions, transposes of matrices, cochains in §4.1 Cochains: Measuring Shapes — and now forms. Measuring devices always travel against the map.

Two facts make pullbacks indispensable. First, integrals are defined through them: for a curve $γ : [a, b] \to R^{2},$ the recipe for $\int_{γ} ω$ above is exactly $\int_{a}^{b} γ^{*} ω .$ Second, pullback respects everything we have built (Lee 2013, ch. 14):

φ^{*} (α \land β) = φ^{*} α \land φ^{*} β, φ^{*} (d ω) = d (φ^{*} ω) .

The second identity says that $d$ does not care which coordinates you compute it in. That is what allows us to leave flat space.

Forms on manifolds

Recall from §2.4 Manifolds and Surfaces that a smooth manifold $M$ of dimension $n$ looks, near each point, like a piece of $R^{n}$ — via charts that overlap smoothly — and that at each point $p$ it has a tangent space $T_{p} M$ of little arrows. A $k$ -form on $M$ is a choice, at every point, of a measuring device that eats $k$ tangent arrows, is linear in each, and flips sign when two arrows are swapped — varying smoothly. In each chart it looks exactly like the forms above; on overlaps, the two descriptions are related by pullback along the transition map. Because pullback commutes with $d,$ the exterior derivative computed in one chart agrees with the one computed in any other, so $d : Ω^{k} (M) \to Ω^{k + 1} (M)$ makes sense on the whole manifold.

To integrate an $n$ -form over a compact $n$ -dimensional manifold, chop the manifold into pieces that each lie in a chart, pull the form back to $R^{n}$ on each piece, integrate there, and add. (Doing the chopping smoothly uses a device called a partition of unity, which we will not need to see.) One requirement is essential: the manifold must be oriented, so that every chart agrees on which way round is positive. On a Möbius band there is no consistent choice, and 2-forms cannot be integrated over it.

More generally, a $k$ -form can be integrated over any oriented $k$ -dimensional piece of a manifold: a curve, a surface, a smooth triangle, or a formal sum of such pieces, $c = \sum a_{i} σ_{i},$ by $\int_{c} ω = \sum a_{i} \int_{σ_{i}} ω .$ Formal sums of pieces are exactly the chains of Part III. So integration gives a pairing

⟨ ω, c ⟩ = \int_{c} ω

between $k$ -forms and $k$ -chains — the same kind of pairing as between cochains and chains in §4.1 Cochains: Measuring Shapes.

Stokes’ theorem in one line

Everything is now in place to state the theorem that this chapter has been circling. An oriented region $M$ has a boundary $\partial M,$ and the boundary inherits an orientation by the rule “outward first”: at a boundary point, an arrow pointing out of $M,$ followed by the boundary’s own orientation, should give the orientation of $M .$ In the plane this means walking the boundary counterclockwise, with the region on your left; for an interval $[a, b]$ it means $b$ counts plus and $a$ counts minus.

Here is what it says in each dimension, with the old theorems as special cases:

$M$	$ω$	Stokes’ theorem becomes
an interval $[a, b]$	a function $F$	$F (b) - F (a) = \int_{a}^{b} F^{'} d x$
a curve $γ$ in space	a function $f$	$f (end) - f (start) = \int_{γ} df$
a region in the plane	$P d x + Q d y$	Green’s theorem
a surface in space	a 1-form	the Kelvin–Stokes theorem
a solid in space	a 2-form	the divergence theorem

Why is it true? You have already seen both reasons. The first is the tiling argument of Figure 4.3.3: chop $M$ into tiny cubes; for a single tiny cube the theorem is the fundamental theorem of calculus applied in each direction; and when the cubes are put back together, every interior face is counted twice with opposite orientations and cancels, leaving only $\partial M .$ The second is the sheet picture of Figure 4.3.8: $\int_{\partial M} ω$ counts the sheets of $ω$ entering and leaving through the boundary, each sheet that enters must leave again or end inside, and the ends inside are what $d ω$ counts.

Spivak’s verdict, in the epigraph of this chapter, is that the theorem is trivial — “because the terms appearing in it have been properly defined”. He goes on: “Since this entire chapter was little more than a series of definitions which made the statement and proof of Stokes’ theorem possible, the reader should be willing to grant the first two of these attributes to Stokes’ theorem” (Spivak 1965, p. 104). The difficulty was never in the theorem. It was in finding the right objects — forms, $d,$ oriented chains — for which it becomes a single honest line.

d and ∂ are mirror images

Write Stokes’ theorem with the pairing $⟨ ω, c ⟩ = \int_{c} ω :$

⟨ d ω, c ⟩ = ⟨ ω, \partial c ⟩ .

This is exactly the discrete Stokes theorem of §4.1 Cochains: Measuring Shapes, $⟨ δ φ, c ⟩ = ⟨ φ, \partial c ⟩ .$ There the coboundary $δ$ was defined to make it true, and turned out to be the transpose of $\partial;$ here the identity is a theorem about honest derivatives and integrals. So differential forms behave like cochains, with $d$ in the role of $δ .$ Two consequences follow in one line each, and they are the doorway to the next chapter.

If $ω = d η$ is exact and $z$ is a cycle ( $\partial z = 0),$ then $\int_{z} ω = \int_{\partial z} η = 0 .$ Exact forms integrate to zero over every cycle.
If $d ω = 0$ and $z = \partial b$ is a boundary, then $\int_{z} ω = \int_{b} d ω = 0 .$ Closed forms integrate to zero over every boundary.

So when a closed form is integrated over a cycle, the answer depends only on the cycle up to boundaries, and only on the form up to exact forms. Integrating closed forms over cycles is a way of pairing homology with something new. That something is de Rham cohomology.

Exercises

Exercise Green’s theorem on a square

Let $R = [0, 1] \times [0, 1]$ be the unit square with its counterclockwise boundary, and $ω = x d y .$ Compute $\oint_{\partial R} ω$ edge by edge, and compare with $\iint_{R} d ω .$ Then do the same for $ω = - y d x + x d y .$

Hint

Along a horizontal edge $d y = 0;$ along a vertical edge $x$ is constant.

Solution

For $ω = x d y :$ the bottom and top edges are horizontal, so $d y = 0$ and they contribute nothing. The left edge has $x = 0,$ so it contributes nothing either. The right edge has $x = 1$ and is walked upwards from $y = 0$ to $y = 1,$ contributing $\int_{0}^{1} 1 d y = 1 .$ On the other side, $d ω = d x \land d y,$ whose integral over the square is its area, $1 .$ They agree.

For $ω = - y d x + x d y :$ $d ω = 2 d x \land d y,$ so the area integral is $2 .$ Along the edges: bottom ( $y = 0,$ going right) gives $0;$ right ( $x = 1,$ going up) gives $\int_{0}^{1} d y = 1;$ top ( $y = 1,$ going left, so $x$ runs from $1$ to $0)$ gives $\int_{1}^{0} - 1 d x = 1;$ left ( $x = 0)$ gives $0 .$ The total is $2 .$

Exercise A closed form that refuses to be exact

On the plane with the origin removed, let $ω = \frac{- y d x + x d y}{x ^{2} + y ^{2}} .$ Show that $d ω = 0,$ and that $\oint ω = 2 π$ around the unit circle. Explain why $ω$ cannot equal $df$ for any smooth function $f$ on the punctured plane.

Hint

With $P = - y / (x^{2} + y^{2})$ and $Q = x / (x^{2} + y^{2}),$ compute $Q_{x}$ and $P_{y}$ with the quotient rule. For the circle use $x = cos t,$ $y = sin t .$

Solution

By the quotient rule, $Q_{x} = \frac{( x ^{2} + y ^{2} ) - x \cdot 2 x}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}}$ and $P_{y} = \frac{- ( x ^{2} + y ^{2} ) + y \cdot 2 y}{( x ^{2} + y ^{2} ) ^{2}} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}},$ so $d ω = (Q_{x} - P_{y}) d x \land d y = 0 .$ On the unit circle, $x^{2} + y^{2} = 1,$ $d x = - sin t d t$ and $d y = cos t d t,$ so $ω = (sin^{2} t + cos^{2} t) d t = d t,$ and $\oint ω = 2 π .$ If $ω$ were $df,$ the integral around any closed loop would be $f (end) - f (start) = 0 .$ This is the form $d θ$ of Figure 4.3.8 and the vortex of Figure 4.3.2, and it is the hero of §4.4 de Rham Cohomology.

Summary

Recap

The derivative is a rate; the integral is a total; the fundamental theorem says the total of a derivative over an interval is the signed boundary values: $\int_{a}^{b} F^{'} = F (b) - F (a) .$
Vector fields assign arrows to points. Their line integrals measure work; curl measures circulation per area, divergence outflow per area. Green’s, Stokes’ and the divergence theorem all say “derivative inside = original on the boundary”, and all come from cancelling interior edges.
A covector is a linear measuring device for arrows, pictured as a stack of lines; a 1-form is a covector at every point, pictured as local stacks of sheets. Line integrals count sheets crossed, and $\int_{γ} df = f (B) - f (A) .$
The wedge product measures oriented area: $(d x \land d y) (u, v) = u_{1} v_{2} - u_{2} v_{1} .$ $k$ -forms measure oriented $k$ -dimensional pieces and are integrated over them.
The exterior derivative $d$ is gradient, curl and divergence in one; pictorially it marks where sheets end. It satisfies $d \circ d = 0,$ the cousin of $\partial \circ \partial = 0 .$
Forms pull back along maps (arrows reverse), and pullback commutes with $d,$ so forms and $d$ live on manifolds.
Stokes’ theorem $\int_{\partial M} ω = \int_{M} d ω$ unifies the classical theorems and says that $d$ is the mirror image of $\partial$ under integration.

Differential Forms

A calculus refresher in pictures

Slopes and rates

Totals

The fundamental theorem

Arrows everywhere: vector fields and work

Slopes in two directions

Arrows everywhere

Work along a path

Swirl and spread: three famous theorems

Circulation and curl

Flux and divergence

Green’s theorem

In three dimensions

Covectors and 1-forms: stacks of sheets

Measuring devices

A covector at every point

Integrating a 1-form

The wedge product: oriented area and volume

Oriented area

2-forms, 3-forms, k-forms

The exterior derivative: where sheets end

Where sheets end

Gradient, curl and divergence are one operator

d of d is zero

Pullbacks, manifolds and integration

Pulling forms back along maps

Forms on manifolds

Stokes’ theorem in one line

d and ∂ are mirror images

Exercises

Summary

Further reading

References