The room a projective space needs, read off Pascal's triangle

The projective plane cannot be built in three-dimensional space without passing through itself. The bottle that needs a fourth dimension gave the reason for any one-sided closed surface — a closed surface sitting in space separates an inside from an outside, and so has two sides — and every section of the band must vanish restated it in terms of bundles: the surface’s own twist and the twist of the directions sticking out of it must cancel, and for the projective plane they cannot.

That essay ended on the question in every dimension. The $n$-dimensional projective space — the lines through the origin of a space one dimension larger — is a closed shape of dimension $n$; what is the smallest room it fits in? The bundle argument, made into arithmetic, gives a lower bound, and the arithmetic turns out to be one of the most familiar objects in mathematics: Pascal’s triangle, with every entry replaced by whether it is odd.

A ledger of twists #

The band essays followed one sign. Carry a frame round a loop and ask whether it comes back reversed; the answer is a sign, $+1$ or $-1$, and a shape is orientable when no loop reverses. Written additively, as $0$ or $1$ counted mod 2, that sign is the first entry of a longer record.

The record is called the Stiefel–Whitney class, after Eduard Stiefel and Hassler Whitney, who found it independently in the 1930s. Here it will be called a ledger, because that is how it behaves. For the projective spaces it is a polynomial in one symbol $a$, with every coefficient $0$ or $1$ and every power above $a^n$ thrown away — on the $n$-dimensional projective space, $a^{n+1} = 0$. The coefficient of $a$ is the sign the band essays followed: it is $1$ exactly when some loop turns the frame over. The later coefficients are the same kind of sign one level up each time: the coefficient of $a^k$ is a mod-2 obstruction, met when one tries to choose enough independent directions continuously over the $k$-dimensional pieces of the shape and finds that the choice cannot be completed.

Two rules make the ledger usable, and both have already been seen in their smallest case.

A bundle with no twist at all has ledger $1$. That is what an untwisted bundle is.

The ledger of two bundles laid side by side is the product of their ledgers. This is Whitney’s product formula, and its smallest instance is the figure in which two Möbius bands add up to a product. Over the circle the Möbius band has ledger $1 + a$, and the circle is one-dimensional, so $a^2 = 0$ there. Two bands side by side have ledger $(1 + a)^2 = 1 + 2a + a^2$, and mod 2 the middle term vanishes and the last is above the circle’s dimension. What is left is $1$: the sum is untwisted, as the frame turning steadily through a half-turn showed.

The projective space’s own ledger #

The bundle the band essays kept returning to was the tautological one: over each line through the origin, put the line itself. On the circle of lines in the plane it is a Möbius band, and over the $n$-dimensional projective space it is the same thing in a larger setting. Its ledger is $1 + a$; indeed $a$ is defined as the twist of that bundle.

The directions tangent to the projective space at a line $L$ are the ways of tilting $L$, and a tilt of $L$ is a linear map from $L$ to the directions perpendicular to it. Add to these the one map from $L$ to itself and the result is all linear maps from $L$ into the whole space of dimension $n + 1$, which is $n + 1$ copies of the tautological line. Adding the one extra map adds a trivial line, which multiplies the ledger by $1$. So the tangent ledger of the $n$-dimensional projective space is

$$(1 + a)^{n+1}, \quad \text{with powers above } a^n \text{ dropped.}$$

Its coefficients are the binomial coefficients $\binom{n+1}{k}$, read mod 2 — row $n + 1$ of Pascal’s triangle with every entry replaced by its parity, cut off after $n + 1$ entries. For the projective plane, $n = 2$, that is row 3, $1, 3, 3, 1$, all odd, cut to $1 + a + a^2$. The coefficient of $a$ is $1$: the plane is one-sided, as a disc sewn to a Möbius band must be.

What the room demands #

Now put the shape in a room of dimension $n + k$, as an immersion — allowed to cross itself, but smooth and without creases, so that at every point it has a tangent space of dimension $n$ and the room supplies $k$ more directions sticking out. Those $k$ directions form the normal bundle. Laid side by side, tangent and normal give every direction of the room at that point, and the room is untwisted. So by the two rules,

$$\text{tangent ledger} \times \text{normal ledger} = 1.$$

The normal ledger is forced: it must be the inverse of the tangent ledger. And a bundle with $k$ directions has nothing in its ledger past $a^k$, since there is nothing in $k$ directions for a higher obstruction to obstruct. So the last term of the inverse is a lower bound on $k$.

For the projective plane the inverse of $1 + a + a^2$ is found one coefficient at a time: the constant term is $1$; the coefficient of $a$ must cancel the tangent’s $a$, so it is $1$; the coefficient of $a^2$ must cancel $a^2 + a \cdot a$, which is $2a^2 = 0$ mod 2, so it is $0$. The normal ledger is $1 + a$, its last term is $a^1$, and so the room must supply at least one extra direction. An immersion needs three dimensions — which Werner Boy’s surface of 1901 achieves, so for immersions the answer for the plane is exactly three.

An embedding asks more. When the shape does not cross itself, the top term of the normal ledger must vanish as well — in codimension one this is exactly the inside-and-outside argument, since a closed surface in space without crossings has an outward direction everywhere and a normal line with a never-vanishing choice is untwisted. The normal ledger $1 + a$ has its top term in degree one, so in a room with one extra direction the plane cannot embed, and it needs four. The argument from which side of the line is inside and the argument from the ledger are the same argument; the ledger’s version has the advantage of working in every dimension.

Four dimensions need seven #

Take the four-dimensional projective space. Its tangent ledger is row 5 of the triangle, $1, 5, 10, 10, 5, 1$, which mod 2 is $1, 1, 0, 0, 1, 1$; cut after $a^4$, it is $1 + a + a^4$.

The inverse comes out one coefficient at a time again. Each new coefficient is whatever cancels the terms already produced, and here every one up to $a^3$ comes out $1$. Only at $a^4$ does the tangent’s own $a^4$ arrive in time to make the sum even. So the normal ledger is $1 + a + a^2 + a^3$, its last term is $a^3$, and the room must supply three extra directions: the four-dimensional projective space cannot be immersed in fewer than seven dimensions, nor embedded in fewer than eight.

That is a surprising amount of room. A four-dimensional shape is being told that a room of six dimensions — half as much again as its own — is not enough to hold it even when crossings are allowed. Nothing about the shape looks large; the ledger, a few coefficients mod 2, is the whole of the reason.

Why the triangle #

The inverse looked like a coefficient-by-coefficient chore, and in both cases it came out as something recognisable — $1 + a$ is row 1 of the triangle, and $1 + a + a^2 + a^3$ is row 3. That is not luck, and it is the reason the triangle is the right picture.

Mod 2, squaring is additive: $(x + y)^2 = x^2 + 2xy + y^2 = x^2 + y^2$, since the middle term is even. Applied repeatedly, $(1 + a)^{2^r} = 1 + a^{2^r}$. Once $2^r$ exceeds the dimension $n$, the second term is thrown away and $(1 + a)^{2^r} = 1$. So if $2^r$ is the first power of two above $n$, then

$$(1 + a)^{n+1} \cdot (1 + a)^{2^r - n - 1} = (1 + a)^{2^r} = 1,$$

and the inverse of row $n + 1$ is simply row $2^r - n - 1$. Every ledger drawn here is computed both ways — by elimination, coefficient by coefficient, and as a row of the triangle — and the two are checked to agree. For the plane, $2^r = 4$ and the inverse is row $4 - 3 = 1$; for the four-dimensional space, $2^r = 8$ and it is row $8 - 5 = 3$.

Which entries of a row are odd is itself a matter of binary digits: by Lucas’s theorem, $\binom{m}{k}$ is odd exactly when every binary 1 of $k$ sits where $m$ also has a 1. That is why the triangle read mod 2 is the pattern of nested triangles in Pascal mod two, and why the rows of the form $2^j - 1$, which are all 1s in binary, are odd all the way across.

Every power of two meets Whitney #

Put $n = 2^j$. The next power of two is $2n$, and the normal ledger is row $2n - n - 1 = n - 1$. That row is all 1s in binary, so every entry is odd, and its last term is $a^{n-1}$. The room must supply $n - 1$ extra directions: the $n$-dimensional projective space needs a room of at least $2n - 1$ dimensions to immerse, and $2n$ to embed.

The other side of that bound is a theorem about every shape. Whitney proved in 1944 that every smooth closed shape of dimension $n$ can be immersed in $2n - 1$ dimensions, for $n$ above one, and embedded in $2n$. No shape needs more room than that. So at every power of two the projective space needs exactly the most room any shape of its dimension can need. The bounds meet, the answer is known, and both halves of the proof are short — one a general construction, the other a row of the triangle.

These projective spaces are the worst-behaved shapes there are, as far as room goes. The point is not that they are large or complicated. It is that their ledgers carry a run of odd coefficients as long as the dimension allows, and the room has to pay for every one.

A staircase, and what it cannot see #

Between powers of two the ledger says less. If $2^r \le n < 2^{r+1}$, the normal ledger is row $2^{r+1} - n - 1$, whose last odd entry is at its very end, so the bound on the room is $n + (2^{r+1} - n - 1) = 2^{r+1} - 1$ — the same number all the way across the block. The fourth through the seventh projective spaces all get the bound seven; the eighth through the fifteenth all get fifteen.

So the lower bound is a staircase and the upper bounds are lines, and the gap between them opens steadily through each block. The figure draws the better of two upper bounds: Whitney’s $2n - 1$, and the bound Ralph Cohen proved in 1985 for every closed shape of dimension $n$, namely $2n$ minus the number of 1s in the binary expansion of $n$. William Massey had shown in 1960 that no shape’s normal ledger can run past that degree, so Cohen’s theorem says the ledger is the only obstruction that applies to every shape at once. For particular shapes there can be more room to spare.

At the bottom of the blocks the staircase sits below the truth. The projective spaces of dimensions 1, 3 and 7 have ledgers that are exactly $1$ — rows 2, 4 and 8 of the triangle are odd only at their ends, and the end beyond $a^n$ is dropped — so the bound says nothing. In fact these three spaces are as untwisted as a shape can be: their tangent directions can be chosen independently all over, an echo of the multiplication of the complex numbers, the quaternions and the octonions. A closed shape can never be immersed in a room of its own dimension, so each needs at least $n + 1$, and a theorem of Morris Hirsch says a shape this untwisted gets exactly that. Three dimensions need four, seven need eight, and the ledger’s bound of $n$ is one short.

The sixth asks for one direction and gets a large gap #

The middle of a block is where the ledger says least, and the sixth projective space is the first clear case.

Its tangent ledger is row 7, all odd, so it is as twisted as the ledger can record — every coefficient up to $a^6$ is $1$. And yet its inverse is tiny. Row 7 and row 1 multiply to row 8, which is $1 + a^8$, and $a^8$ is far past the dimension. So the normal ledger is $1 + a$, one extra direction is forced, and the bound on the room is seven.

A shape whose own ledger is full needs a room the ledger thinks is small; a shape whose own ledger is nearly empty, like the eighth projective space with $1 + a + a^8$, needs the most room possible. What the room must pay for is not how twisted the shape is but how far the inverse runs, and inverting mod 2 turns long rows into short ones and short rows into long ones. That reversal is the content of the staircase.

For the sixth projective space the truth sits between the ledger’s seven and Whitney’s eleven, and settling it takes invariants finer than any sign — the refinements of K-theory, and tables computed term by term. None of that is visible in the triangle, and from dimension six onward most of the problem lives there.

What remains open #

The exact smallest room for the $n$-dimensional projective space is known at every power of two, for the three untwisted cases, and for a scattering of other dimensions where some construction happened to meet some lower bound. For general $n$ it is not known. Donald Davis has kept tables of the best bounds for decades, and the gaps in them have narrowed without closing; the question first posed in Whitney’s day is still a working problem. The same is true without crossings: the smallest room in which the projective space embeds is open in just the same way, and the ledger’s bound for it is the staircase raised by one step.

What the triangle does settle is where the difficulty starts. Every power of two is decided, and the staircase marks out, block by block, how much of the remaining gap a mod-2 count can see. Beyond that the obstructions are no longer signs and no longer fit in Pascal’s triangle — which is the point at which the one idea these essays have followed, a twist recorded as a sign, has been carried as far as it goes.

A sign, and then a polynomial of signs #

The band essays started with half a twist in a strip of paper and a single sign. Here the sign became the first coefficient of a polynomial; the product rule turned laying bundles side by side into multiplication; and the room’s lack of twist became the equation that the normal ledger is the inverse of the tangent one.

Nothing in the argument needs a picture of the shape. The projective plane’s failure to sit in three-dimensional space without crossing itself — the fact that made one-sided surfaces strange — is the statement that $1 + a + a^2$ times $1 + a$ is $1$ mod 2 and $1 + a$ has a term in degree one. The four-dimensional space’s need for seven is the statement that row 3 of Pascal’s triangle is odd all the way across. What looked like a fact about the room has been turned into a fact about the shape’s own arithmetic, which was the band essays’ habit from the first paper strip on.