4.1 Cochains: Measuring Shapes · Homology & Cohomology

Visual paradoxes like the Penrose staircase present a fundamental tension: locally coherent geometric relationships that cannot be realized globally.

— Robert Ghrist and Zoe Cooperband, Obstructions to Reality: Torsors & Visual Paradox (2025)

On the roof of M. C. Escher’s Ascending and Descending, monks walk a staircase on which every step goes up, yet one lap of the courtyard brings them back to where they started. Inspect any flight and it is an honest flight of stairs. So where is the mistake? Not on any step. It lives in the loop, in the courtyard the stairs go round — and finding mistakes of that kind, which are invisible everywhere you look and fatal as a whole, is what cohomology is for.

Part III was about places. A chain was an inventory of pieces of a shape — these edges, those triangles, each with a multiplicity — and homology asked which inventories go around holes. Now we turn the telescope around and measure the pieces instead: write a number on every vertex, or on every edge, the way a hiker writes elevations on a trail map or an engineer writes voltages on a circuit diagram. Then we ask one innocent question: when do numbers on the edges come from numbers on the vertices? The answer, “exactly when every loop adds up to zero”, sounds like accounting. It unmasks the staircase, explains why a ring of gears can jam and finds free money on the currency markets. And it leads straight back to the holes: homology counted them, and cohomology tells you what each one forbids.

Places and measurements

Here is a trail map of a small mountain. The dots are junctions, the lines are trails. In the language of §2.5 Simplicial Complexes this is a simplicial complex of dimension one — a graph. Next to each junction is its elevation above sea level, in metres.

A hiker does not really care about elevations; she cares about climbs. How much does the trail from the Spring to the Ridge go up? That is the elevation at the Ridge minus the elevation at the Spring: $1780 - 1460 = 320$ metres. So the elevations on the junctions automatically produce a number on every trail. Play with this in the figure: drag a junction up or down and watch the numbers on the trails next to it change.

Heights make climbs Drag a junction up or down · tap a trail to flip its arrow

Lake 1530 m

Terrain view

Climbs around the gold loop:

150 + 180 - 70 - 260 = 0

Figure 4.1.1 Heights on the junctions (gold) produce climbs on the trails (teal): each trail carries the height at its arrow’s head minus the height at its tail. Raising one junction changes only the trails that touch it. However you set the heights, the climbs around every closed loop add up to

0

— try the three loops with “Next loop”, and try the tilted terrain view.

Let us write this down carefully, because every symbol in it will stay with us for the rest of the book.

The arrow on each edge matters. Walk the trail from the Spring to the Ridge and you climb $320$ metres; walk it the other way and you “climb” $- 320$ metres. A number on an edge only makes sense once we have agreed which way the edge points. In this book every edge of a simplicial complex points from its lower-numbered vertex to its higher-numbered one (the convention of §3.2 Chains and the Boundary Operator), and in the figures the arrows are drawn. Tap a trail in Figure 4.1.1 to flip its arrow: the number on it changes sign, and nothing else changes. This is the same orientation bookkeeping that gave chains their signs.

A second model: voltages

Exactly the same arithmetic describes an electrical circuit. The vertices are the connection points, $f (v)$ is the voltage (the electric potential) at $v,$ and $(δ f) (e)$ is the voltage across the component sitting on the edge $e$ — the thing a voltmeter clipped to its two ends would read. John Baez puts it in the language we are building: “It’s good to think of $V$ as a 1-cochain, which assigns to each edge the voltage across that edge” (Baez 2010). The word cochain is coming in a moment; for now, read it as “a number on every edge”.

Walking a path: the climbs telescope

Suppose you walk from the Trailhead to the Summit by way of the Spring and the Ridge. Adding up the climbs along the way gives

+ 260 (1460 - 1200) + + 320 (1780 - 1460) + + 140 (1920 - 1780) = 1920 - 1200 = 720.

Every intermediate height appears once with a plus sign and once with a minus sign, and cancels. Only the two ends survive. A sum that collapses like this is called telescoping, after the old folding telescope whose segments slide inside each other. In general, if a walk visits the vertices $v_{0}, v_{1}, \dots, v_{k}$ in order, then

i = 1 \sum k (f (v_{i}) - f (v_{i - 1})) = f (v_{k}) - f (v_{0}) .

(If the walk uses an edge against its arrow, the climb counts with a minus sign, which is exactly what the formula needs.) This is the discrete fundamental theorem of calculus: adding up the changes along a route gives the total change, and the total change depends only on where you start and where you finish — not on the route. The other route in the figure, Trailhead → Meadow → Lake → Hut → Summit, gives $150 + 180 + 80 + 310 = 720$ as well.

And here is the consequence that will drive the whole chapter. If the walk is a closed loop, so that it ends where it began, then $v_{k} = v_{0}$ and the total is $f (v_{0}) - f (v_{0}) = 0 .$ The climbs around any closed loop add up to zero: you cannot walk in a circle and end up higher than you started. In circuit language this is Kirchhoff’s voltage law: the voltages around any closed loop of a circuit sum to zero.

The gradient puzzle

Now run the story backwards. A hiker has kept a notebook: for every trail she walked, she wrote down how much it climbed. She never wrote down a single elevation. Can we reconstruct the elevations from her notebook?

In symbols: we are handed a number $ψ (e)$ on every edge $e$ (the letter $ψ$ is psi), and we ask whether there is a potential $f$ with $δ f = ψ .$ If there is, we call $ψ$ a gradient. Sometimes the answer is yes and sometimes no, and the difference between the two is the first glimpse of cohomology. Try it yourself before reading on.

Find the potential Tap a vertex, set its height · or integrate from it

Every step is +1, so the loop sum is 4: an impossible staircase.

0 of 4 edges satisfied · loops to check: E − V + c = 4 − 4 + 1 = 1

Figure 4.1.2 The gold numbers on the edges are given; your job is to put heights on the vertices so that every edge equals head minus tail. An edge turns green when it is satisfied and rose when it is not. “Integrate” does it systematically: start at one vertex, walk a spanning tree (teal), then test every leftover edge. The test fails exactly when the leftover edge closes a loop whose numbers do not add up to

0,

and that whole loop turns rose.

On a tree, it always works

If the graph is a tree — connected, with no loops at all — the puzzle always has a solution, and here is a recipe. Pick any vertex $r$ (the root) and give it height $0 .$ Every other vertex $v$ is joined to $r$ by exactly one path, because a tree has no loops to provide a second route. Walk that path and add up the numbers on its edges (with a minus sign for edges walked against their arrows); call the result $f (v) .$ Then every edge is satisfied: if an edge goes from $u$ to $v,$ the path to $v$ is the path to $u$ followed by that edge, so $f (v) = f (u) + ψ (e) .$ This recipe, like much of this chapter, comes from the four pages Allen Hatcher calls “The Idea of Cohomology” (Hatcher 2002, pp. 186–189). There the root is a base vertex $v_{0},$ and each value is fixed “by induction along the unique path from $v_{0}$ to $v$ in the tree.” (Hatcher writes $φ$ for the potential where we write $f .)$

On a loop, the numbers must add up to zero

The square puzzle “A loop that does not” in Figure 4.1.2 has $+ 1$ on each of its four edges, all pointing the same way around. Say the bottom-left vertex has height $h .$ Following the arrows, the next vertex must be at $h + 1,$ the next at $h + 2,$ the next at $h + 3$ — and the last edge leads back to the start, which would then have to be at $h + 4 .$ But it is at $h .$ Since $h \neq = h + 4,$ there is no solution, whatever $h$ you choose.

The culprit is the loop. We showed in the last section that any gradient sums to $0$ around every loop; here the sum is $1 + 1 + 1 + 1 = 4 .$ So a non-zero loop sum is a certificate of impossibility: one number that proves no potential can exist. The surprise is that the converse holds too: if every loop sums to zero, a potential exists.

How many loops must we check?

A graph can have a great many loops, but we do not have to test them all. Recall from §3.1 Cycles and Boundaries how a spanning tree organises the loops of a graph: the edges outside the tree are the troublemakers, and each one closes exactly one fundamental loop with the tree. Every loop of the graph is a combination of fundamental loops, and loop sums add up the same way, so it is enough to check the fundamental ones. That is exactly what “Integrate” does in Figure 4.1.2: it builds heights along a spanning tree (the tree edges are automatically satisfied) and then tests each leftover edge. The mismatch on a leftover edge, $ψ (e) - (f (head) - f (tail)),$ is precisely the sum of $ψ$ around that edge’s fundamental loop.

For a graph with $V$ vertices, $E$ edges and $c$ connected pieces, a spanning forest has $V - c$ edges, so the number of leftover edges — the number of independent checks — is

E - V + c .

You have seen this number before: it is the first Betti number $b_{1}$ of the graph, the number of independent holes, from §3.1 Cycles and Boundaries. The trail map of Figure 4.1.1 has $9 - 7 + 1 = 3$ of them, which is why “Next loop” cycles through three loops.

What never changes: $H^{0}$

When the puzzle has a solution, how many solutions are there? If $δ f = ψ$ and $δ g = ψ,$ then subtracting, $δ (f - g) = 0 .$ So the question is really: which potentials $h$ have $δ h = 0$ — no climb on any edge at all?

$δ h = 0$ means $h (v) - h (u) = 0$ across every edge, so neighbours have equal heights. Walking along edges, the height never changes, so $h$ is constant on each connected piece of the graph. But different pieces are free to sit at different heights: nothing connects them, so nothing forces them to agree. Such a function is called locally constant.

Islands Drag the vertices · flatten each island

Tap a vertex to raise or lower it Terrain view

δf = 0 on every edge — each island is flat, but the 3 islands sit at their own heights.

Figure 4.1.3 A graph in three pieces. An edge glows rose whenever its two ends have different heights.

δ f = 0

everywhere exactly when each island is flat — but the three islands can sit at three independent heights. The functions with

δ f = 0

form a space of dimension

3,

the number of islands.

The space of locally constant functions has one dimension for each connected piece: choose one height per island. It is the first of the cohomology groups, and it gets the name

H^{0} (G) = {h : V \to R ∣ δ h = 0} = {locally constant functions on G},

and its dimension is the number of connected pieces:

dim H^{0} (G) = c = b_{0} .

So the solutions of the gradient puzzle, when they exist, are unique up to adding a locally constant function — exactly like the “ $+ C ”$ you add to an antiderivative in calculus, one constant per island. Hatcher makes the same comparison: solving $δ φ = ψ$ is “rather like the calculus problem of finding a function having a specified derivative, with the difference operator $δ$ playing the role of differentiation.”

Local tests, global obstructions

So far our shapes have been graphs. Now fill in some triangles. Picture the trail map again, but where three trails enclose a small meadow you can walk straight across, we fill the triangle in: it is now a 2-dimensional simplex of our complex. What does a filled triangle say about measurements on the edges?

If the edge numbers come from heights, then walking once around the triangle must return you to your starting height. For a triangle with vertices $a, b, c$ and edges pointing $a \to b,$ $b \to c$ and $a \to c,$ walking $a \to b \to c \to a$ uses the last edge against its arrow, so the test reads

ψ (ab) + ψ (b c) - ψ (a c) = 0.

The left-hand side is called the circulation of $ψ$ around the triangle — or, borrowing a word from vector calculus, its curl. For a gradient $ψ = δ f$ the curl of every filled triangle is $(f (b) - f (a)) + (f (c) - f (b)) - (f (c) - f (a)) = 0 :$ every term cancels. That is the discrete version of a fact you may remember from calculus, “the curl of a gradient is zero”.

The local test Tap a vertex or an edge · use ±1

Height of C 6

ψ (A B) + ψ (B C) - ψ (A C) = 2 + 3 - 5 = 0

Figure 4.1.4 One filled triangle. When the edge numbers are differences of heights, the circulation in the middle is always

0,

however you change the heights. When you choose the edge numbers yourself, the circulation can be anything — and a non-zero circulation proves, on the spot, that no heights can produce these numbers.

The curl test is local: it looks at only three edges at a time. That makes it cheap, and it is a genuine necessary condition. The question is whether it is also sufficient: if a labelling passes the curl test on every filled triangle, must it be a gradient?

When there are no holes, local is enough

Suppose the complex is a filled-in disk, triangulated. Take any loop in it. As we saw in §3.1 Cycles and Boundaries, a loop in a disk is a boundary: it is the boundary of the union of the triangles it encloses. Add up the curls of all those triangles. Every edge inside the region is shared by two of them and is walked once in each direction, so it cancels; only the edges on the loop survive. So the loop sum equals the sum of the curls inside it. If every curl is $0,$ every loop sum is $0,$ and by the gradient test $ψ$ is a gradient. In a region without holes, passing every local test is the same as being a gradient.

When there is a hole, it is not

Now make a hole. The complex in the next figure is a triangulated annulus — a ring — with an inner triangle $a_{0} a_{1} a_{2},$ an outer triangle $b_{0} b_{1} b_{2},$ and six filled triangles between them. The labelling $ψ$ has a $1$ on the three edges $a_{0} \to a_{1},$ $a_{0} \to b_{1}$ and $b_{0} \to b_{1},$ and $0$ everywhere else: a little “fence” of ones running from the hole out to the edge.

Closed but not exact Tap an edge (±1) or switch to bumps · try “Smooth it out”

Selected edge:

Local tests: 6 of 6 triangles pass a₀ → a₁ → a₂ → a₀:

1 + 0 + 0 = 1

Figure 4.1.5 Every one of the six triangle tests passes (the green ticks), yet the loop around the hole sums to

1,

so these numbers are not differences of heights. Add a bump at any vertex — that is, add a gradient — and the edge numbers change, but no triangle test and no loop sum changes. “Smooth it out” spreads the obstruction as evenly as possible:

\frac{1}{3}

on every edge around the hole,

\pm \frac{1}{6}

on the spokes, and still exactly

1

once around.

Check the triangle $a_{0} a_{1} b_{1}$ by hand: its edges carry $ψ (a_{0} a_{1}) = 1,$ $ψ (a_{1} b_{1}) = 0$ and $ψ (a_{0} b_{1}) = 1,$ and walking around it gives $1 + 0 - 1 = 0 .$ The triangle $a_{0} b_{0} b_{1}$ gives $0 + 1 - 1 = 0,$ and the other four triangles see only zeros. Every local test passes. But walk around the hole along the inner triangle: $ψ (a_{0} a_{1}) + ψ (a_{1} a_{2}) + ψ (a_{2} a_{0}) = 1 + 0 + 0 = 1 \neq = 0 .$ So $ψ$ is not a gradient. The loop around the hole is a cycle that is not a boundary — the very definition of a hole in §3.1 Cycles and Boundaries — and that is why no collection of triangle tests can reach it.

Hatcher describes this situation with two phrases worth memorising (Hatcher 2002, p. 188). The curl of $ψ$ — in his notation $δ ψ,$ which we will justify shortly — is a local obstruction:

We can think of $δ ψ$ as a local obstruction to solving $ψ = δ φ$ since it depends only on the values of $ψ$ within individual 2 simplices of $X .$ … This class in $H^{1} (X; G)$ is thus the global obstruction to solving $ψ = δ φ .$

The “class in $H^{1} ”$ is what is left of $ψ$ once we stop caring about bumps. Adding a bump at a vertex — that is, adding the gradient $δ g$ of a function $g$ that is $1$ at that vertex and $0$ elsewhere — changes the edge numbers but cannot change any loop sum, because the loop sum of a gradient is $0 .$ For the annulus, the only thing that survives is a single number, the loop sum around the hole.

Intuition The same story in calculus

A vector field in the plane is a little arrow at every point. It is a gradient (conservative) if it is the slope field of some height function, and it is curl-free if it has no infinitesimal swirl anywhere. Every gradient is curl-free. In a region without holes the converse holds; in the plane with a point removed it fails — the classic example swirls around the missing point, has curl $0$ everywhere, and yet its integral around the puncture is $2 π .$ Hatcher again: “The local obstruction here is the vanishing of the curl of the vector field, and the global obstruction is the vanishing of all line integrals around closed loops in the domain of the vector field.” Our annulus is the triangulated, whole-number version of that vortex; the smooth version is the star of §4.4 de Rham Cohomology.

Impossible staircases and other paradoxes

The pattern we have found — data that is perfectly consistent everywhere you look closely, yet impossible as a whole — turns up all over the place once you know its shape. Here are four instances: one from art, one from geometry, one from engineering and one from finance.

The staircase that climbs forever

The Penrose staircase is a flight of stairs that goes around the four sides of a square courtyard, climbing at every step, and arrives back where it started. Look at any short stretch and it is an ordinary staircase; look at the whole and it is impossible. The geneticist Lionel Penrose and his son, the mathematician Roger Penrose, published it in 1958 in a short note on “impossible objects” (Penrose & Penrose 1958), and two years later M. C. Escher put it on the roof of Ascending and Descending, the lithograph this chapter opened with.

In our language the staircase is an edge labelling. Make a graph with one vertex per step and one edge from each step to the next; the edges form a single loop. “Each step goes up by one” labels every edge $+ 1 .$ If the staircase could exist, there would be a height function on the steps with these differences — but the loop sum is the number of steps, not $0,$ so by the gradient test no height function exists. The staircase is the “loop that does not close” from Figure 4.1.2, drawn by an artist.

Possible after all — from one direction Drag to turn · use the button to find the trick

↑ 0

steps climbed — every step is +1

one lap = 20 steps, loop sum = +20

Colour by height Walk

Figure 4.1.6 These are real three-dimensional objects, seen through a camera without perspective. From one special direction the top of the staircase hides exactly in front of its bottom, so the walker seems to climb for ever; turn the object and the trick is revealed — the real staircase is a spiral with a cliff. The colours, which follow the true heights, must jump somewhere: that is where the cliff is. The tribar is built the same way.

The real object in Figure 4.1.6 does have a height function: step $i$ sits at height $i .$ Its differences are $+ 1$ on every edge except the one that closes the loop, where it drops by $N - 1$ — the cliff, hidden behind the camera angle. Compare that with the impossible labelling $ψ$ that is $+ 1$ everywhere: the two differ only on the cliff edge, by exactly $N,$ the loop sum. You may put the cliff wherever you like (that is adding a bump, changing $f),$ but you cannot get rid of it, because the loop sum $N$ does not change. The cliff is the obstruction, made of stone — or of plywood: between 1955 and 1959 the Penroses built small models of exactly this kind, meant to be photographed from the one angle at which they look impossible, and London’s Science Museum keeps them (Penrose & Penrose 1955–59).

The tribar

The Penrose triangle, or tribar, is three square beams joined at right angles into a triangle. Each corner is a perfectly good right-angled joint; the three together are impossible. (The Swedish artist Oscar Reutersvärd had drawn an impossible triangle made of cubes as early as 1934; Lionel and Roger Penrose published theirs in 1958.) Decades later Roger Penrose wrote a short paper, On the cohomology of impossible figures, explaining the tribar in exactly our terms (Penrose 1992). Cut the drawing into three overlapping pieces, each of which can be built. A drawing fixes each piece only up to its distance from your eye, so on each overlap the two pieces must be rescaled to match, by some positive factor. The figure can be built exactly when these factors come from one scale per piece — when they are “differences” — and that happens exactly when their product around the triangle is $1 .$ For the tribar it is not. This is our gradient test again, with one change: the numbers on the edges are multiplied around the loop instead of added, because scale factors combine by multiplication. We will meet Penrose’s argument properly in §4.7 Sheaves and Čech Cohomology.

A little history The paradox and the theory

M. C. Escher 1898 – 1972

Lionel Penrose 1898 – 1972

One was a Dutch printmaker who had failed mathematics at school; the other an English geneticist, a Quaker who had served with the Friends’ Ambulance Unit in France in the First World War. Escher and Lionel Penrose were born six days apart, in June 1898. The influence ran both ways. Roger Penrose drew his triangle in 1954, after seeing Escher’s work; father and son sent Escher a copy of their 1958 paper, and he turned the staircase into Ascending and Descending (1960) and the tribar into the endless aqueduct of Waterfall (1961) (Penrose & Penrose 1955–59). It took more than thirty years for Roger Penrose to say in print that the impossibility was a cohomology class. Today applied topologists build whole galleries of such figures, on cylinders, Möbius bands and Klein bottles; Ghrist and Cooperband’s phrase for the common principle is “the obstruction to globalizing locally consistent geometric relationships” (Ghrist & Cooperband 2025).

A ring of gears

Arrange $n$ identical gears in a ring, each meshing with its two neighbours. When two gears mesh, they turn in opposite directions. Lewis Ghrist and Robert Ghrist describe what happens: “If $n$ is even, alternating spins around the loop satisfies all opposition constraints, and the system spins freely. If $n$ is odd, no such alternation exists – the final gear must simultaneously agree and oppose its neighbor. The system locks: a mechanical paradox that we will identify as a nontrivial $H^{1}$ class” (Ghrist & Ghrist 2026, §2.2).

Let us identify it. Record the spin of each gear as an element of $Z /2 = {0, 1}$ — the integers mod 2, also written $F_{2}$ — with $0$ for clockwise and $1$ for anticlockwise. Each mesh says “the spins differ”, which in $Z /2$ reads $s (v) - s (u) = 1 .$ So the meshes put the label $1$ on every edge of the ring, and the gears can turn exactly when this labelling is a gradient: when it comes from spins on the gears. Around the ring the labels add up to $n,$ and in $Z /2$ that is $0$ exactly when $n$ is even.

A ring of gears Change the number of gears · follow the spins

6 gears: the spins alternate all the way round — the ring turns.

Number of gears 6

Show the ℤ/2 labels

Loop sum in ℤ/2:

6 meshes 1 + 1 + \dots + 1 = 6 \equiv 0 (mod 2)

Figure 4.1.7 Every mesh carries the label

1

Z /2 :

“turn the other way”. With an even number of gears the labels add up to

0

around the ring and the spins alternate consistently; with an odd number they add up to

1,

the last mesh contradicts the first gear, and the ring locks. Locally every pair of gears is fine.

Nothing in the gradient test used the fact that our numbers were real numbers; all we needed was to be able to add and subtract. So measurements may take values in any abelian group: heights in $R,$ counts in $Z,$ spins in $Z /2,$ scale factors in the positive real numbers under multiplication. The choice is called the coefficients, and changing it changes what the obstruction can see. Over $Z /2,$ a loop of length $n$ is impossible only when $n$ is odd.

Free money

Our last example is about currencies. Suppose dollars, euros, pounds and yen trade only around a square: dollars for euros, euros for pounds, pounds for yen, yen for dollars. Each edge carries an exchange rate $r,$ meaning one unit of the first currency buys $r$ units of the second. Start with \$1000 and go once around. If you come back with more than \$1000, you have found an arbitrage: free money, from nothing but the rates.

An arbitrage loop Slide the pound–yen rate · open a euro–yen market

Yen per pound (£ → ¥) 202 Show logarithms Open a € ↔ ¥ market

Once around: $1000×0.9​€900×0.85​£765×202​¥154,530×1/153​$1,010.00
Profit: +$10.00 free money from nothing — an arbitrage loop
Log loop sum: ln0.9+ln0.85+ln202+ln1531​=ln200202​=0.0100

Figure 4.1.8 Exchange rates multiply around a loop; their logarithms add. At £1 = ¥200 the loop is fair and the log loop sum is

0;

at any other rate you can make money by going around. Open a direct euro–yen market and the square gets two filled triangles: now the arbitrage shows up inside a single triangle, where any trader checking three currencies at a time would spot it.

Rates multiply around a loop, but logarithms turn multiplication into addition ( $ln (x y) = ln x + ln y),$ so label each edge with $ψ (e) = ln r (e)$ instead. A market is fair if every currency has a “value” $V$ (in some common unit) and every rate is the ratio of values, $r (u \to v) = V (u) / V (v) .$ Taking logarithms, $ψ (u \to v) = ln V (u) - ln V (v) :$ the log-rates are exactly the differences of the potential $- ln V .$ So fair market = gradient, and the gradient test says: a market is fair if and only if no loop of trades makes money. In Figure 4.1.8 the log loop sum is $ln (r /200),$ zero only at the fair rate $r = 200 .$

A trader who checks every three-currency triangle is running the local curl test. If all six pairs of currencies trade, every triangle is filled, every loop of trades is a boundary, and the local test is enough: no triangular arbitrage means no arbitrage at all (the last exercise of this chapter asks you to prove it). But in our square market there are no triangles to check, so a perfectly real arbitrage hides from every local inspection. Arbitrage can only hide in the holes.

The coboundary operator

It is time to give our objects their proper names. Nothing new will happen mathematically; we only need to notice that the two operations of this chapter — “differences of heights” and “circulation around a triangle” — are one operation, and that it is the mirror image of the boundary operator of Part III.

Now look again at our two operations, side by side with the boundary formulas of §3.2 Chains and the Boundary Operator:

\partial [u, v] = v - u, (δ f) ([u, v]) = f (v) - f (u) = f (\partial [u, v]),

\partial [a, b, c] = [b, c] - [a, c] + [a, b], (δ ψ) ([a, b, c]) = ψ ([b, c]) - ψ ([a, c]) + ψ ([a, b]) = ψ (\partial [a, b, c]) .

In both cases the new cochain is computed by evaluating the old cochain on the boundary. That is the definition in every degree.

The discrete Stokes theorem

Since $(δ φ) (σ) = φ (\partial σ)$ for each simplex, adding up over a chain gives the same identity for every chain $c :$

⟨ δ φ, c ⟩ = ⟨ φ, \partial c ⟩ .

In words: measuring the change of $φ$ over a region is the same as measuring $φ$ on the region’s edge. In degree $0$ this is telescoping: $⟨ δ f, path ⟩ = f (end) - f (start),$ because the boundary of a path is its end minus its start. In degree $1$ it is the argument we used for the disk: the sum of the curls over a region equals the loop sum around its boundary. In calculus the same statement is Stokes’ theorem, and there it takes real work. Here it costs nothing, because we defined the coboundary to make it true.

The coboundary of a coboundary is zero

We saw that the curl of a gradient vanishes. Here is the one-line reason, valid in every degree:

(δ δ φ) (σ) = (δ φ) (\partial σ) = φ (\partial\partial σ) = φ (0) = 0,

using the fundamental fact $\partial\partial = 0$ from §3.2 Chains and the Boundary Operator. So $δ \circ δ = 0 :$ “the curl of a gradient is zero” is the shadow, in the world of measurements, of “the boundary of a boundary is zero” in the world of places.

A matrix picture: $δ$ is a transpose

Take the hollow triangle with vertices $0, 1, 2$ and edges $[0, 1], [0, 2], [1, 2] .$ The boundary matrix $\partial_{1}$ from §3.2 Chains and the Boundary Operator has a row for each vertex and a column for each edge, and the column of an edge lists its boundary. The coboundary $δ_{0}$ has a row for each edge and a column for each vertex, and the row of an edge lists how $δ f$ on that edge depends on the heights:

\partial_{1} = - 1 10 - 1 01 0 - 1 1, δ_{0} = - 1 - 1 0 10 - 1 011 = \partial_{1}^{T} .

The second matrix is the first one flipped across its diagonal — its transpose, from §1.5 Linear Algebra. That is not a coincidence of this example: the entry of $\partial$ in row $v,$ column $e$ is the coefficient of $v$ in $\partial e,$ and the entry of $δ$ in row $e,$ column $v$ is the change in $δ f$ on $e$ when $f$ is $1$ at $v$ and $0$ elsewhere — which is the same coefficient. In every degree, $δ_{k} = \partial_{k + 1}^{T} .$ Keenan Crane, comparing the two theories in his notes on discrete geometry, sums it up: “the only difference is a matrix transpose!” (Crane 2025, p. 138) We will make full use of this in §4.2 Cohomology Groups.

The mirror

Figure 4.1.9 Chains (places) and cochains (measurements) sit in the same three columns. The boundary maps

\partial

lower the dimension; the coboundary maps

δ

raise it, and each

δ

is the transpose of the

\partial

next to it. The pairing joins each column, and Stokes’ formula

⟨ δ φ, c ⟩ = ⟨ φ, \partial c ⟩

ties the two rows together.

The new vocabulary

Everything we have discovered now has a name, built by putting “co-” in front of a word from homology. (The prefix means “dual”; the names are due to Hassler Whitney in the late 1930s, as we will see in the next chapter (Whitney 1938).)

Homology: places	Cohomology: measurements	In this chapter
chain $c \in C_{k}$	cochain $φ \in C^{k}$	heights, edge labellings
boundary operator $\partial$ (lowers degree)	coboundary operator $δ$ (raises degree)	gradient, curl
cycle: $\partial c = 0$	cocycle: $δ φ = 0$	passes every local test
boundary: $c = \partial b$	coboundary: $φ = δ g$	comes from a potential
$\partial\partial = 0$	$δ δ = 0$	curl of a gradient is zero
$H_{k}$ = cycles / boundaries	$H^{k}$ = cocycles / coboundaries	the global obstructions

The last row is a preview. In the language of §1.4 Abelian Groups and Formal Sums, the cocycles form a group, the coboundaries form a subgroup inside it, and the cohomology group $H^{k} (K)$ is the quotient: cocycles, with two of them counted as the same when they differ by a coboundary. That is exactly the move we made when we decided that “adding a bump” should not count as a change. For the annulus, the cocycles modulo coboundaries form a single copy of $R$ (or $Z,$ with whole-number labels), and the number that identifies a class is its loop sum around the hole. Lek-Heng Lim calls the general recipe — two matrices $A$ and $B$ with $A B = 0,$ and the quotient $ker A / im B$ — “cohomology on a bumper sticker” (Lim 2020, §2.1). The precise definitions, and many computations, are the subject of §4.2 Cohomology Groups.

Gradient, curl and harmonic: the Hodge decomposition

Real data is never exactly a gradient. Suppose sports teams play each other and each match produces a number on an edge — how many points the winner won by. If there were a “true strength” for each team, the margins would be differences of strengths, a gradient. In practice the data is noisy and inconsistent: A beats B, B beats C, and C beats A. What is the best we can do?

The answer sounds too good to be true. Every edge labelling $X$ (call it a flow when you think of the numbers as amounts flowing along the arrows) splits in exactly one way into three pieces,

X = gradient δ f + swirls around filled triangles X_{curl} + circulation around the holes X_{harm},

and the three pieces are perpendicular to each other, in the sense that $\sum_{e} a (e) b (e) = 0$ for any two of them. This is the discrete Hodge decomposition, named after W. V. D. Hodge, who developed its smooth version, for differential forms, in the 1930s.

Taking a flow apart Choose a flow · tap an edge in the first panel and use ±1

Your flow

Tap an edge, then use ±1.

= Gradient

heights (a ranking): 3 > 2 = 4 > 1

+ Curl

swirl around the filled triangle: 1 per edge

+ Harmonic

loop sum around the hole: 8

Selected edge: Energy (sum of squares): 35 = 8 + 3 + 24

Figure 4.1.10 The upper triangle

1, 2, 3

is filled; the lower triangle

1, 3, 4

is a hole. Any flow (gold) is the sum of a gradient (teal: differences of the heights on the vertices), a curl part (violet: a pure swirl around the filled triangle) and a harmonic part (rose: a balanced circulation around the hole). The energies add like Pythagoras: for the example,

35 = 8 + 3 + 24 .

Look at the example in Figure 4.1.10, the flow $X = (3, 3, 3, 2, - 2)$ on the edges $1 \to 2, 2 \to 3, 1 \to 3, 3 \to 4, 1 \to 4 .$

The gradient part $δ f = (1, 1, 2, - 1, 1)$ comes from the heights $f = (0, 1, 2, 1)$ on vertices $1, 2, 3, 4 .$ It is the gradient closest to $X$ — the one that minimises the total squared error, the “least squares” fit. Read as a ranking, it says $3 > 2 = 4 > 1 .$ Finding $f$ means solving $δ^{T} δ f = δ^{T} X,$ one equation per vertex. The matrix $L = δ^{T} δ$ is the graph Laplacian: each vertex’s number of edges on the diagonal, and $- 1$ in row $u,$ column $v$ whenever an edge joins $u$ and $v .$ Here $δ^{T} X = (- 4, 0, 4, 0),$ and the heights above solve the equations.
The curl part $(1, 1, - 1, 0, 0)$ is one unit of swirl around the filled triangle: $+ 1$ along $1 \to 2$ and $2 \to 3,$ and $- 1$ along $1 \to 3,$ so it goes once around $1 \to 2 \to 3 \to 1 .$ It lives entirely on triangles that are filled, and it is what the local curl test detects.
The harmonic part $(1, 1, 2, 3, - 3)$ is what is left. It passes every local test (its curl on the filled triangle is $1 + 1 - 2 = 0),$ and it is balanced: at every vertex, what flows in equals what flows out. Yet its sum around the hole is $2 + 3 + 3 = 8 .$ It is the global obstruction, standing on its own.

Check the energies: $3^{2} + 3^{2} + 3^{2} + 2^{2} + 2^{2} = 35,$ while the parts have $8,$ $3$ and $24,$ and $8 + 3 + 24 = 35 .$ Because the pieces are perpendicular, their squared lengths add, just as for the sides of a right-angled triangle.

This is not a toy. In a 2011 paper Xiaoye Jiang, Lek-Heng Lim, Yuan Yao and Yinyu Ye proposed HodgeRank, which ranks items from pairwise comparisons — teams from match results, films from viewers’ preferences — in exactly this way: the gradient part is the ranking, the curl part records local three-way inconsistencies, and the harmonic part records inconsistencies that are, in their words, “locally acyclic but globally cyclic” (Jiang et al. 2011). Those are the ones no amount of local checking will find.

Exercises

Exercise Bumps never change loop sums

On the annulus of Figure 4.1.5, let $g$ be the potential that is $1$ at $a_{1}$ and $0$ at the other five vertices. The vertex $a_{1}$ lies on four edges. Write down $δ g$ on each of them. Add $δ g$ to $ψ$ and recompute the sum around the inner loop $a_{0} \to a_{1} \to a_{2} \to a_{0} .$ Then explain, without computing, why adding $δ g$ can never change the sum of a labelling around any loop, on any complex.

Solution

The four edges at $a_{1}$ are $a_{0} \to a_{1},$ where $a_{1}$ is the head, so $δ g = 1 - 0 = + 1;$ and $a_{1} \to a_{2},$ the rung $a_{1} \to b_{1}$ and the diagonal $a_{1} \to b_{2},$ where $a_{1}$ is the tail, so $δ g = 0 - 1 = - 1 .$ On every other edge $δ g = 0 .$ The new inner loop sum is $(1 + 1) + (0 - 1) + (0 + 0) = 1,$ unchanged. In general, the sum of $δ g$ around a closed loop telescopes to $g (v_{0}) - g (v_{0}) = 0,$ so the loop sums of $ψ + δ g$ equal those of $ψ .$ In the language of the coboundary section: $⟨ δ g, z ⟩ = ⟨ g, \partial z ⟩ = 0$ for every cycle $z .$

Exercise Which gear trains can turn?

Gears are placed at the vertices of a graph, and two gears mesh when they are joined by an edge. Which of these can turn? (a) Four gears at the corners of a square, each meshing with its two neighbours. (b) The same four gears, with one diagonal pair also meshing. (c) A $3 \times 3$ grid of gears, each meshing with the gears directly left, right, above and below it. Find a general rule.

Hint

Label every edge $1 \in Z /2$ and look for loops with an odd number of edges.

Solution

The train turns exactly when the labelling “ $1$ on every edge” is a gradient over $Z /2,$ i.e. when every loop has an even number of edges. (a) Turns: the only loop has length $4 .$ (b) Locks: the diagonal creates two loops of length $3 .$ (c) Turns: colour the grid like a chessboard and let the dark squares turn clockwise; every mesh joins a dark and a light square, so every loop is even. General rule: a gear train can turn if and only if its graph has no loop of odd length — such graphs are called bipartite.

Exercise No hidden arbitrage in a complete market

Four currencies, and every one of the six pairs trades, so every three-currency triangle is filled. Suppose no triangle offers an arbitrage (each triangle’s rates multiply to $1).$ Prove that no loop of trades at all offers an arbitrage. Then explain why the square market of Figure 4.1.8, with only four trading pairs, does not have this property.

Hint

Work with log-rates, so “no triangular arbitrage” says the labelling is closed. Try to build the potential directly: fix the dollar at $0$ and use the three edges at the dollar.

Solution

Let $ψ = ln r .$ Set $f ($) = 0$ and, for each other currency $x,$ $f (x) = ψ ($ \to x)$ (using the direct dollar edge, with a minus sign if the arrow points into the dollar). Now take any edge $x \to y$ between two non-dollar currencies. The triangle $$, x, y$ is filled and its curl is $0 :$ $ψ ($ \to x) + ψ (x \to y) - ψ ($ \to y) = 0,$ so $ψ (x \to y) = f (y) - f (x) .$ Edges at the dollar are satisfied by construction. So $ψ = δ f$ is a gradient and, by the gradient test, every loop sum vanishes: no arbitrage anywhere. (Geometrically: the complex with all four triangles filled is a hollow tetrahedron, whose loops are all boundaries.) In the square market there are no filled triangles at all, so there is nothing for the curl test to check, and the loop around the square is a cycle that bounds nothing: its log-sum can be anything.

Summary

Recap

Chains are places; cochains are measurements: numbers on the $k$ -simplices, extended to chains by adding up. The pairing $⟨ φ, c ⟩$ “integrates” a measurement over a place.
Heights $f$ on vertices give climbs $(δ f) (u \to v) = f (v) - f (u)$ on edges. Climbs telescope, so around any closed loop they sum to $0$ (Kirchhoff’s voltage law).
The gradient test: an edge labelling is a gradient if and only if its sum around every loop is $0 .$ It suffices to check $E - V + c = b_{1}$ fundamental loops: one independent obstruction per hole.
$H^{0}$ = functions with $δ f = 0$ = locally constant functions; its dimension is the number of components.
On filled triangles the curl $ψ (ab) + ψ (b c) - ψ (a c)$ is a local test. Without holes it is enough; around a hole (the annulus) a labelling can be closed but not exact, and the loop sum is the global obstruction.
Impossible staircases, the tribar, odd rings of gears ( $Z /2$ coefficients) and hidden arbitrage are all closed-but-not-exact labellings.
The coboundary $(δ φ) (σ) = φ (\partial σ)$ is the gradient in degree $0$ and the curl in degree $1;$ $⟨ δ φ, c ⟩ = ⟨ φ, \partial c ⟩$ (discrete Stokes), $δ δ = 0,$ and $δ = \partial^{T} .$
Cocycles ( $δ φ = 0)$ modulo coboundaries ( $φ = δ g)$ give cohomology. Every flow splits as gradient + curl + harmonic; the harmonic part is the obstruction, spread out evenly.

Cochains: Measuring Shapes

Places and measurements

A second model: voltages

Walking a path: the climbs telescope

The gradient puzzle

On a tree, it always works

On a loop, the numbers must add up to zero

How many loops must we check?

What never changes: $H^{0}$

Local tests, global obstructions

When there are no holes, local is enough

When there is a hole, it is not

Impossible staircases and other paradoxes

The staircase that climbs forever

The tribar

A ring of gears

Free money

The coboundary operator

The discrete Stokes theorem

The coboundary of a coboundary is zero

A matrix picture: $δ$ is a transpose

The new vocabulary

Gradient, curl and harmonic: the Hodge decomposition

Exercises

Summary

Further reading

References

Places and measurements

A second model: voltages

Walking a path: the climbs telescope

The gradient puzzle

On a tree, it always works

On a loop, the numbers must add up to zero

How many loops must we check?

What never changes: H0

Local tests, global obstructions

When there are no holes, local is enough

When there is a hole, it is not

Impossible staircases and other paradoxes

The staircase that climbs forever

The tribar

A ring of gears

Free money

The coboundary operator

The discrete Stokes theorem

The coboundary of a coboundary is zero

A matrix picture: δ is a transpose

The new vocabulary

Gradient, curl and harmonic: the Hodge decomposition

Exercises

Summary

Further reading

References

What never changes: $H^{0}$

A matrix picture: $δ$ is a transpose