A hole is a cycle that bounds nothing

The cycles and the cuts took a graph’s edges and points and wrote the matrix that sends each edge to its end minus its start. The kernel of that map — the combinations of edges whose ends all cancel — was the space of cycles, and its dimension, edges minus points plus pieces, counted the graph’s independent loops. That essay closed by pointing one dimension up: the same matrix is the first of a sequence, and the next one takes triangles to their edges.

Adding triangles changes what a loop is worth. Three points joined in a ring have one independent cycle, the walk round the ring. Fill the ring in with a triangle and the walk is still a cycle — its ends still cancel, nothing about the edges has changed — but it is now also the edge of something. It goes round a region, and a loop that goes round a region is not going round a hole.

That is the whole idea, and the arithmetic follows from it in one line. A hole is a cycle that is not a boundary, so the number of independent holes is the number of independent cycles minus the number of independent boundaries. Both of those are ranks and nullities of matrices, and the rest of this essay is what that subtraction computes.

The matrix that takes a triangle to its edge #

Each edge is given a direction, from its smaller label to its larger, and each triangle an order of its corners, smallest first. The boundary of the edge from $a$ to $b$ is $b - a$: arrive with a plus, leave with a minus. The boundary of the triangle with corners $a < b < c$ is its three edges taken round it,

$$\partial [a, b, c] = [b, c] - [a, c] + [a, b],$$

where the middle edge carries a minus because walking $a \to b \to c \to a$ traverses it from $c$ to $a$, against its direction.

Written as matrices, $\partial_1$ has a row for every point and a column for every edge, with a $-1$ and a $+1$ in each column — the transpose of the incidence matrix from the cycles and the cuts — and $\partial_2$ has a row for every edge and a column for every triangle, with three entries of $\pm 1$ in each column. Nothing about either is chosen: the directions are a bookkeeping convention, and reversing an edge flips the sign of one row, which changes no rank.

The one fact everything rests on is that the boundary of a boundary is nothing. Apply $\partial_1$ to the boundary of a triangle: $(c - b) - (c - a) + (b - a) = 0$, each corner arriving once and leaving once. So $\partial_1 \partial_2 = 0$ as a matrix product, which every figure here checks entry by entry before it draws. And that says every boundary is a cycle — the image of $\partial_2$ lies inside the kernel of $\partial_1$ — which is exactly what makes the subtraction meaningful. A boundary is a special cycle, and the holes are the cycles left when the special ones are divided out.

Counting holes is subtraction #

With the two matrices in hand, the count is rank–nullity applied at each end of the edges.

The cycles are the kernel of $\partial_1$, so there are $E - r_1$ of them, where $r_1$ is the rank of $\partial_1$ and $E$ the number of edges. The boundaries are the image of $\partial_2$, so there are $r_2$ of them. The independent holes are the first number minus the second:

$$b_1 = (E - r_1) - r_2.$$

The same pattern at the other two levels gives the other two counts. At the points, everything is a cycle — a point has no boundary — and the boundaries are the image of $\partial_1$, so $b_0 = V - r_1$, which is the number of connected pieces, as the incidence matrix already showed. At the triangles, nothing arrives from above, so every cycle survives: $b_2 = F - r_2$, the combinations of triangles whose boundaries cancel completely. Those are the closed surfaces, and $b_2$ counts the independent enclosed voids.

The two loops at a point show why the word is independent and not different. There are three obvious loops — each triangle, and the figure-eight walk round both — but the figure-eight is the sum of the other two as a combination of edges, and a count of dimensions does not count it again. That is the discipline the cycles and the cuts learned for graphs, and it carries over unchanged: the numbers $b_0, b_1, b_2$ are dimensions of quotient spaces, and they count independent alternatives, never lists.

Seven small spaces #

The subtraction can be carried out on any space built from points, edges and triangles, and the interesting ones are the closed surfaces, where every edge lies on exactly two triangles.

The surface of a tetrahedron is a sphere made of four triangles. Its $\partial_2$ has rank three, not four, because the four triangles’ boundaries add to nothing when each is taken with the right sign — every edge appears in two faces, once each way — and that dependency is the sphere itself: one combination of triangles that is a cycle and bounds no solid, since there is no solid. So $b_2 = 1$ and $b_1 = 6 - 3 - 3 = 0$. A sphere has one void and no hole: every loop drawn on it is the edge of a patch.

The torus built from nine points has 27 edges and 18 triangles. The ranks come out as eight and seventeen, and the holes as $1$, $2$, $1$: one piece, one enclosed void, and two independent loops that bound nothing — the one round the tube and the one through it. Those are the two loops everybody draws on a torus, and here they are counted without being drawn, as the dimension of a quotient of two spaces of edge-combinations.

The nine points of the torus are the grid $\{0,1,2\}^2$ with its opposite sides glued, and the Klein bottle is the same grid with one pair glued after a flip. The two surfaces have the same number of every kind of cell. They differ in one rank — seventeen for the torus and eighteen for the Klein bottle — and that one rank is the difference between having an inside and not having one.

Why the alternating sum is special #

Every row of the table satisfies the same equation,

$$V - E + F = b_0 - b_1 + b_2,$$

and the reason is visible once each space of cells is drawn split into its parts.

The triangles split into those whose boundaries are independent — $r_2$ of them — and the cycles, $b_2$. The edges split into the $r_1$ that $\partial_1$ sends onward independently, the $r_2$ that arrive as boundaries of triangles, and the $b_1$ holes. The points split into the $r_1$ boundaries of edges and the $b_0$ pieces. Take the alternating sum and each rank appears once with a plus and once with a minus:

$$F - E + V = (r_2 + b_2) - (r_1 + r_2 + b_1) + (r_1 + b_0) = b_2 - b_1 + b_0.$$

That is why the corner count — Euler’s $V - E + F$ — refuses to change however a surface is cut up. Each subdivision changes $V$, $E$ and $F$, and it changes the ranks, and the ranks were never in the alternating sum to begin with. What is in it is the holes, and the holes are a property of the surface rather than of the triangles drawn on it.

That last clause is the part the figures cannot supply. That $b_0, b_1, b_2$ are the same for every triangulation of a surface is a theorem — proved by comparing any two triangulations through a common subdivision — and nothing here checks more than one triangulation of each. The cancellation, which is checked, shows that $V - E + F$ is exactly as invariant as the holes are; the invariance of the holes is borrowed.

The same move one level up #

The hollow and filled triangles are the smallest case of a pattern that repeats at every level, and the next case is already in the table.

The surface of a tetrahedron has one enclosed void: the combination of all four triangles, each taken with the orientation that makes it point outward, has boundary zero, and nothing has it as a boundary because there is nothing inside. Now fill it. Add one solid tetrahedron $[a, b, c, d]$ and a third matrix, $\partial_3$, which sends it to its four faces with alternating signs,

$$\partial_3 [a, b, c, d] = [b, c, d] - [a, c, d] + [a, b, d] - [a, b, c].$$

That combination of faces is exactly the void. The new matrix has rank one, the void is now the boundary of the solid, and $b_2$ drops from one to nought — the same subtraction that took the filled triangle’s loop away, one dimension higher. The boundary of the boundary is still nothing: every edge of the tetrahedron lies on two of its faces, entering once with each sign.

The alternating sum takes one more term and loses nothing. The solid tetrahedron has four points, six edges, four triangles and one solid, and $4 - 6 + 4 - 1 = 1$, the Euler characteristic of a filled ball and of a single point, which is what it can be shrunk to. The hollow version has $4 - 6 + 4 = 2$, the characteristic of the sphere. Filling a void subtracts one from the alternating sum, exactly as filling a loop does, and it does so for the same reason: a new rank appears at two neighbouring levels at once, once leaving and once arriving, and the holes are what change.

Nothing about this depends on the pieces being triangles and tetrahedra rather than squares and cubes, or on the dimension stopping at three. At every level there is a matrix, the composite of two consecutive ones is zero, and the holes are kernel modulo image. The sequence of matrices is called a chain complex, and everything drawn here is its first two links.

A loop that is half a boundary #

The projective plane in the table has holes $1$, $0$, $0$ — one piece and nothing else — which looks like a disc. It is not a disc. It is the closed one-sided surface, a disc sewn to a Möbius band, and something about it has been lost in the count.

The lost thing is a loop. Walk once round the Möbius band’s core: that loop is a cycle, and it bounds nothing — no combination of triangles has it as its boundary. But walk round it twice and the result is a boundary, of the whole surface’s triangles taken with suitable signs. Over the rational numbers, a cycle whose double is a boundary is itself a boundary — it is half of one — so it contributes no hole. The rationals can divide by two, and dividing by two erased the loop.

The same computation with coefficients that are only $0$ and $1$, where $1 + 1 \equiv 0$, cannot divide by two, and the result changes at two levels at once. The ten triangles’ boundaries now have a dependency — the whole surface, every triangle taken once, has boundary zero because every edge is used twice and two of anything is nothing — so $\partial_2$ drops to rank nine, the top level acquires a cycle, and the edges acquire a hole.

Both counts give the same alternating sum, $1$, as they must: the equation above holds over any field. But the individual holes depend on the arithmetic, and the dependence is exactly the one-sidedness. On the torus and the sphere every triangle can be oriented consistently, the whole surface is a cycle over any numbers, and the two counts agree. On the projective plane and the Klein bottle no consistent orientation exists — that is what one-sided means — and the whole surface is a cycle only when signs stop mattering.

The holes over both kinds of number #

Side by side the pattern is plain. On five of the seven spaces the two counts are identical. On the Klein bottle and the projective plane the mod-two count is larger by one in each of the two top dimensions, and the extra pair appears and cancels in the alternating sum.

This is the same phenomenon the cycles and the cuts met and was rescued from. A graph’s incidence matrix has the same rank over every field, because every square piece of it has determinant $0$ or $\pm 1$ and no such number can vanish modulo a prime without vanishing outright. The triangle-to-edge matrix of a surface has no such protection: every ten-by-ten piece of the projective plane’s $\partial_2$ has an even determinant, and some of them are not zero — invisible over the rationals, where a non-zero minor is all a rank needs, and fatal modulo two. A rank drops modulo a prime exactly when every one of its largest non-vanishing minors is divisible by that prime, and the minors of a boundary matrix record the twists of the space.

The complete account works over the integers, where no field has to be chosen, and it records the loop as what it is: a cycle of order two, which does not vanish and whose double does. The integer count of the projective plane’s loops is the group with two elements, and the Klein bottle’s is the integers alongside a group of order two. Both field counts can be read off that one integer answer, and neither of them can recover it alone.

What the linear algebra supplies and what it borrows #

It is worth being exact about the division of labour, because the pleasure of this construction is how much it gets for nothing and the danger is claiming too much.

It supplies the numbers mechanically. Given a space built from triangles, the holes are two ranks and three subtractions, and the ranks are computed by elimination with no insight required. That is why computers can count holes in spaces nobody can draw — in data sets of points in hundreds of dimensions, for instance, where the method is applied to a triangulation built from the points and the holes that persist across scales are reported as features of the data.

It supplies the Euler characteristic’s meaning. The alternating sum of cells is the alternating sum of holes because the ranks cancel, and that is a one-line argument once the matrices exist. The cancellation is the same one that makes a graph’s cycles and cuts add to its edges, taken one step further.

It does not supply invariance. That two triangulations of the same surface give the same holes is a theorem about spaces, not about matrices, and it is where the topology actually lives. Nor does it say what a hole looks like: a basis of cycles for the torus is two loops, and choosing which two is a choice the dimension count is silent about, exactly as a spanning tree had to be chosen to name a graph’s cycles.

Seven spaces of at most nine points #

Every space drawn here has at most nine points, and the torus and Klein bottle appear only as rows of a table and bars in a chart. No figure shows a loop on the torus or the one-sided loop on the projective plane; the holes are counted as dimensions and never exhibited as curves. A figure that drew them would be a different figure, and it would need a choice of basis that the count does not make.

One triangulation of each surface is used, so the figures cannot show that the counts are independent of the triangulation, and the text says where that fact comes from. Two fields are used, the rationals and the field of two elements; the integer answer, with its element of order two, is described and not computed. And the spaces stop at dimension two: the whole construction continues to tetrahedra and beyond, with a matrix at every level and the same cancellation in the alternating sum, and nothing here goes past triangles.

Still open: what the holes cannot hear #

Homology is a coarse invariant, and its coarseness has a famous measure. Henri Poincaré conjectured in 1900 that a closed three-dimensional space with the holes of a three-sphere must be a three-sphere, and then found in 1904 a counterexample of his own: the Poincaré homology sphere, a space with exactly the holes of the three-sphere — one piece, nothing in dimensions one and two, one enclosed void — which is not a sphere, because loops in it cannot all be shrunk to points. He corrected the conjecture to require that every loop can be shrunk, and Grigori Perelman proved the corrected version in 2003.

The question survives one dimension up in its hardest form. Is every smooth four-dimensional space that can be continuously deformed into a four-sphere, with loops and holes and every higher invariant matching, smoothly the four-sphere itself? That is the smooth four-dimensional Poincaré conjecture, and it is open. In dimensions one to three the answer is yes, and from five upwards the question was turned into a computation in the 1960s and answered wherever that computation has been done; in dimension four the holes, the loops and every algebraic invariant anyone has defined agree between the four-sphere and any candidate, while whether a genuinely different smooth structure exists is not known.

Holes are the part of a cycle that survives #

The pattern worth keeping is the one the hollow and filled triangles show at their smallest.

A space built from pieces carries a sequence of matrices, each taking a piece to its edge, and the edge of an edge is always nothing. Everything the cycles and the cuts found for graphs was the first step of that sequence; adding triangles adds a second matrix, and the new information is exactly what the second matrix’s image removes from the first matrix’s kernel. Holes are kernel modulo image, and each word in that phrase is a piece of linear algebra that was already familiar.

The counts it produces are the numbers a topologist uses to tell spaces apart, the Euler characteristic is their alternating sum, and the arithmetic chosen for the coefficients can see a twist that another arithmetic divides away. None of that needed a picture of a torus. It needed rank and nullity, applied twice, and the observation that a boundary is always a cycle.