Zero in four dimensions

Euler’s formula for solids — corners minus edges plus faces equals two — has a natural extension to four dimensions, and the first thing anybody does with it is get it wrong. A solid in four dimensions has corners, edges, two-dimensional faces and three-dimensional cells, the solid pieces its boundary is made of, as a cube’s boundary is made of squares. The obvious guess is that the alternating sum of those four numbers is two again. It is zero.

The four-dimensional cube has 16 corners, 32 edges, 24 square faces and 8 cubical cells, and $16 - 32 + 24 - 8 = 0$. The same is true of every one of the six regular solids that four dimensions allows, including the two built of hundreds of cells, and of every convex solid in four dimensions whether regular or not. The answer is not a new constant for a new dimension. It is the old formula read correctly: the number it measures belongs to the boundary, and the boundary of a solid in four dimensions is a sphere of three dimensions, whose characteristic is zero.

Counting the six from their corners #

None of the numbers in the table was copied from a reference. Each polytope was given by the coordinates of its corners — the sixteen points $(\pm1, \pm1, \pm1, \pm1)$ for the tesseract, the eight points $\pm e_i$ for the 16-cell, the hundred and twenty points built from the golden ratio for the 600-cell — and the rest was found.

Edges are the pairs of corners at the smallest distance. For the three polytopes built entirely from tetrahedra — the 5-cell, the 16-cell and the 600-cell — every triangle in the edge graph is a face and every set of four mutually joined corners is a cell, so the faces and cells are found by searching the graph for triangles and for groups of four. On the 600-cell the search finds 720 edges, 1,200 triangles and 600 tetrahedra among 120 corners. The tesseract’s counts come out as 16, 32, 24 and 8, and they are the 16-cell’s counts reversed, which is what duality predicts: swapping a polytope for its dual turns corners into cells and edges into faces. The same reversal gives the 120-cell’s counts — 600 corners, 1,200 edges, 720 pentagonal faces and 120 dodecahedral cells — from the 600-cell’s. The 24-cell, which is its own dual, has 24 corners, 96 edges, 96 triangles and 24 octahedral cells, and it is symmetric in the same way.

Every row sums to zero. Five of the six rows are about solids nobody has seen, and their counts run from five to twelve hundred; a formula that fitted them all by accident would be remarkable. What they share is not their shapes but the shape of their boundaries.

The tesseract by hand #

The four-dimensional cube is the one where every count can be made without a computer, and the counting explains the zero.

Write each corner of the cube as a string of four digits, each $0$ or $1$ — the same strings as the rows of a truth table in four variables. An edge is a string with one digit replaced by a star, meaning free to be either: $01{*}0$ is the edge from $0100$ to $0110$. A square face has two stars, a cubical cell three, and the whole tesseract is ${*}{*}{*}{*}$. So a face of dimension $k$ is a choice of $k$ places for stars and a digit for each of the other $4 - k$. The squares, for instance, choose two of the four places to be free, in six ways, and fix the other two digits, in four ways, which is twenty-four. The cells fix a single digit: one of four places, set to $0$ or to $1$, which is eight — two opposite cubes in each of four directions, as an ordinary cube has two opposite squares in each of its three. In general

$$f_k = \binom{4}{k}\,2^{4-k}, \qquad f_0, f_1, f_2, f_3 = 16,\ 32,\ 24,\ 8.$$

The alternating sum now falls to one line of algebra. Including the whole tesseract as the single face of dimension four, the sum $\sum_k (-1)^k \binom{4}{k} 2^{4-k}$ is the binomial expansion of $(2 - 1)^4$, which is $1$. Dropping the whole tesseract, which contributes $(-1)^4 = +1$, leaves $1 - 1 = 0$. The same calculation for the ordinary cube gives $(2-1)^3 = 1$, and dropping the solid itself, which contributes $(-1)^3 = -1$, leaves $1 + 1 = 2$. The difference between two and zero is a single sign, the sign attached to the interior.

That way of reading the sum is worth making explicit, because it explains why the signs alternate at all. Think of each face as its interior only — a corner as a point, an edge without its ends, a square without its sides — so that the pieces fit together without overlapping, and give an open piece of dimension $k$ the weight $(-1)^k$. The alternating sum is the total weight, and it is additive: two pieces glued together have the weight of the first plus the weight of the second, with nothing to subtract, because nothing is counted twice. That additivity is what makes the number an invariant — cutting a face in two adds an edge and a face and changes nothing — and the sign of the interior is what separates a solid from its surface.

Two, zero, two, zero #

The calculation did not depend on the dimension being three or four, and the same algebra settles every dimension at once.

For the cube in $d$ dimensions, $(2 - 1)^d = 1$ gives a boundary sum of $1 - (-1)^d$. For the simplex, the solid whose corners are all joined to one another — the triangle, the tetrahedron, the 5-cell — a face of dimension $k$ is any $k + 1$ of the $d + 1$ corners, so $f_k = \binom{d+1}{k+1}$, and the binomial theorem for $(1 - 1)^{d+1} = 0$ gives the same $1 - (-1)^d$. For the cross-polytope, whose corners are $\pm e_i$ and whose faces choose at most one of each opposite pair, it is $(1 + 2 \cdot (-1))^d$ rearranged, and the answer is the same again. Three unrelated families and eight dimensions: the sum is $2$ when $d$ is odd and $0$ when it is even.

The one-dimensional row is a reminder of what the formula is about. A segment has two ends and nothing else in its boundary, and its alternating sum is two. The boundary of a segment is two points — a zero-dimensional sphere — and the pattern starts there. A polygon has as many corners as edges, and the sum is $0$; its boundary is a circle, which is the one-dimensional sphere. The solid of Euler’s own formula is the case $d = 3$, whose boundary is the ordinary sphere, and the two that everybody memorises is the characteristic of that sphere, not of the solid.

Building the boundary one cell at a time #

The reason the number is a property of the sphere can be watched directly, by assembling a boundary piece by piece and keeping the running sum.

Start with one face of a cube: a square, with $4 - 4 + 1 = 1$. Add a neighbouring face: two new corners, three new edges and one face, and the running sum is still $1$. Every face added while the result is still an open patch adds a piece glued along a path, and the sum does not move, because what has been built is always topologically a disc and every disc has characteristic $1$. The last face is different. It is glued along its entire boundary, adding no corners and no edges but one face, and the sum jumps to $2$.

For the tesseract the pieces are cubes, and the partial boundary is a solid ball of three dimensions at every stage but the last, with sum $1$. The last cube is glued along all six of its faces, adding only its own interior, which counts $(-1)^3 = -1$, and the sum falls to $0$. The same last step in any dimension adds $(-1)^{d-1}$, so the pattern $1 - (-1)^d$ is simply one plus the sign of the final piece.

Building a boundary this way, with every partial stage a ball, is called a shelling, and Heinz Bruggesser and Peter Mani proved in 1971 that every convex polytope in every dimension can be shelled. Their theorem turns this picture into a proof of the formula for all convex polytopes at once, and it is the direct descendant of the flattening proof for ordinary solids, in which faces are removed one at a time.

Where the formula was first written down #

The first person to state the formula in four dimensions and beyond was Ludwig Schläfli, a Swiss mathematician who between 1850 and 1852 wrote a long treatise on geometry in many dimensions. In it he found the six regular four-dimensional polytopes, showed that every higher dimension has only three, and gave the alternating sum for all of them. The Vienna and Berlin academies declined to publish it, partly for its length and partly because geometry in more than three dimensions was not yet taken seriously, and it appeared only in 1901, six years after his death. By then others had rediscovered most of it.

The general statement is named after Henri Poincaré, who stated it for polytopes in 1893 and returned to it in his papers on analysis situs. His first argument assumed things about how the cells fit together that were not yet proved — the same kind of silent hypothesis that the solid with a hole through it exposed in Euler’s own formula a century earlier. What repaired both was not a cleverer count but a better idea of what was being counted: the alternating sum stopped being a property of a pile of faces and became a property of the space those faces cover, independent of how it was cut up. The shelling proof of 1971 is the version that stays closest to the faces, and it came more than a century after Schläfli wrote the formula down.

Corners and edges decide the rest #

In three dimensions Euler’s formula is one equation in three unknowns, and it leaves a two-parameter family of possible counts. In four dimensions it is one equation in four unknowns, and for polytopes with all their cells tetrahedra there is a second equation that comes for free.

Every tetrahedron has four triangular faces, and every triangle of the boundary lies in exactly two tetrahedra, so $4 f_3 = 2 f_2$, or $f_2 = 2 f_3$. Put that into $f_0 - f_1 + f_2 - f_3 = 0$ and it becomes $f_0 - f_1 + f_3 = 0$. So

$$f_3 = f_1 - f_0, \qquad f_2 = 2(f_1 - f_0),$$

and the corners and edges determine everything else. These are the Dehn–Sommerville equations for four dimensions, found by Max Dehn in 1905 and extended by Duncan Sommerville to all dimensions in 1927. The 600-cell checks them: $720 - 120 = 600$ tetrahedra and $1{,}200$ triangles.

The cyclic polytopes on the lower rows are the extreme case. Put $n$ points on the curve $(t, t^2, t^3, t^4)$ and take their convex hull; every pair of them is joined by an edge, which is impossible in three dimensions for more than four points, and the facets are exactly the sets of four satisfying David Gale’s evenness condition — between any two points left out, an even number of chosen ones. The table counts those facets directly: 9 for six corners, 20 for eight, 35 for ten. Peter McMullen proved in 1970 that no four-dimensional polytope with $n$ corners can have more faces of any dimension than the cyclic one — the upper bound theorem — and the two equations above are why it suffices to maximise the edges.

What the drawing of the tesseract cannot show #

The tesseract figure is a picture of a picture. It flattens a solid of four dimensions onto a page of two, and the flattening costs nearly everything the solid has.

All thirty-two of the tesseract’s edges are equal in length, and in the drawing they come out in many different lengths. Its twenty-four square faces are congruent squares; most appear as trapezia or less regular quadrilaterals. Its eight cubical cells are identical cubes; in the drawing one is the large outer cube, one the small inner one, and six are the distorted frusta between them, and nothing on the page distinguishes the one drawn outside from the others as the arbitrary choice it is. The claim that every cell is a cube, and every face a square, is made by the coordinates and checked by the counting; the drawing contradicts it everywhere.

What the picture cannot show at all is the zero. The alternating sum is a statement about how the cells fit together into a closed three-dimensional sphere, and a two-dimensional shadow of that sphere is not a sphere of any kind — its edges cross, its faces overlap, and the pieces that are glued in four dimensions touch on the page only along lines. The count survives the drawing because it was never taken from it. And the larger polytopes are not drawn: a shadow of the 600-cell’s 720 edges is a ball of ink, and the table is the only honest way to present them. The sphere that a map of the world tries to flatten loses one point and all its distances; a tesseract flattened to a page loses two dimensions, and it is remarkable only that anything is left.

Still open: which lists of four numbers are the counts of a solid? #

In three dimensions the question of which counts are possible has a complete answer. Ernst Steinitz showed in 1906 that three numbers $V$, $E$ and $F$ are the corners, edges and faces of some convex solid exactly when $V - E + F = 2$, $V \le 2F - 4$ and $F \le 2V - 4$. Every such triple is realised, and nothing else is. The inequalities say that a solid cannot have many more corners than faces or the reverse, and the equation is Euler’s.

In four dimensions the corresponding question is open. The Euler–Poincaré equation leaves three free numbers, and the inequalities that cut out the possible quadruples $(f_0, f_1, f_2, f_3)$ are not known. It is not even known whether the fatness of a four-dimensional polytope — the ratio $(f_1 + f_2)/(f_0 + f_3)$, which measures how many edges and faces sit between the corners and the cells — can be arbitrarily large. Günter Ziegler asked this around 2002. The regular polytopes are not fat — the 24-cell, the fattest of them, has $(96 + 96)/(24 + 24) = 4$ — and examples considerably fatter have since been built, but no bound has been proved and none has been ruled out. The simplicial polytopes are well understood, because the Dehn–Sommerville equations reduce them to corners and edges and the g-theorem of Billera, Lee and Stanley describes the rest; the general case is where the formula stops helping.

The difficulty is partly that four dimensions has more room. In three dimensions any drawing of a planar graph with three edges at least at each corner can be realised as a convex solid — another theorem of Steinitz — so the counting question is really a question about graphs. In four dimensions the analogous statement fails: there are arrangements of cells that close up into a perfectly good three-dimensional sphere and that no convex polytope has as its boundary, and there is no simple test that tells the two kinds apart. The alternating sum is zero for both, and it is the only linear relation the counts are known to satisfy.

The number that belongs to the boundary #

The alternating sum of a solid’s faces is one of the oldest invariants in mathematics, and its behaviour in four dimensions is the clearest evidence of what it is. It does not change with the shape — five cells or six hundred, tetrahedra or dodecahedra, regular or lopsided — and it does not stay the same when the dimension changes. It alternates, and the alternation is exactly the characteristic of a sphere: two for the spheres of even dimension, zero for those of odd dimension.

That was already visible in the twelve pentagons and in the solid with a hole: what Euler’s number measures is the surface, not the lump of material inside it, and changing the surface changes the number. Henri Poincaré made that precise in 1895 by writing the characteristic as an alternating sum of the ranks of homology groups, which count holes of each dimension — a definition that no longer mentions corners at all and applies to spaces with no faces to count. The classification of surfaces uses it as one of two numbers that tell every closed surface apart. The corners, edges, faces and cells of the six polytopes are the shadow of that invariant on objects simple enough to count, and they sum to zero because the three-dimensional sphere has no more to say.