The n-Category Café

Idea:

Consider a scaled copy $A$ of the first shell in the second shell that contains $2e_0$, treated as twice the identity in the octonions. Note that any point on the second shell not in $A$ is adjacent to some point in $A$.

Consider a point on the second shell that is adjacent to $2e_0$. This yields a unit octonion via scaling; call it $s$. There are 64 possible values of $s$, since each $7$-orthoplex in the $E_8$ polytope is adjacent to $64$ other $7$-orthoplexes.

The two sets

$$s A = \{s a | a \in A\}$$

and

$$A s = \{a s | a \in A\}$$

are copies of the first shell in the second shell. So we get a total of 128 pairs of disjoint copies of the first shell in the second shell where one of the copies is $A$.

Thus, since each pair contains two shells, we get $2160\cdot 128/2 = 17280$ disjoint pairs by this procedure.

Does this match the descriptions given by the computer calculations?

Nice work.

This reminds me of the spectrum of states in the heterotic string theories at the first excited level. To compute the degeneracies, one needs to know how many points of any given length exist on the lattice. This information is encoded in the theta function of the lattice:

$\Theta_{\Gamma}(\tau)=\displaystyle\sum_{w\in\Gamma}e^{\pi i \tau |w|^2}$

which for an even lattice becomes

$\Theta_{\Gamma}(\tau)=\displaystyle\sum_{n=0}^{\infty}d_n e^{2\pi i n \tau}$

where $d_n$ is the number of lattice sites $w$ with $w\cdot w=2n$. The degeneracies correspond to the coefficients of the various frequencies in the theta series expansion.

The theta function of an even-self dual lattice in $d$ dimensions is a modular form of weight $\frac{d}{2}$, and by the uniqueness theorem there is just one modular form of weight four (up to normalization, where $d=8$), the theta function for the $E_8$ lattice:

$\Theta_{\Gamma_8}=1+240e^{2\pi i \tau}+9\cdot 240 e^{4\pi i \tau}+\ldots$

The number of points in each shell you mentioned can be read off from the coefficients, and in turn, give the degeneracies of the heterotic string states on a single copy of $E_8$. Moving up, there is a unique modular form of weight eight:

$\Theta_{\Gamma_16}=\Theta_{\Gamma_8\times\Gamma_8}=(\Theta_{\Gamma_8})^2=1+480\displaystyle\sum_{m=1}^{\infty}\sigma_7(m)e^{2\pi i m \tau}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=1+480e^{2\pi i\tau}+129\cdot 480 e^{4\pi i \tau}+\ldots$

The degeneracy of bosonic left-moving modes at the $N$th mass level compactified on $\Gamma_8\times\Gamma_8$ or $\Gamma_16$ is given by

$\displaystyle\sum_{N=-1}^{\infty}d_L(N)x^N=\frac{1}{x}(1+480\displaystyle\sum_{m=1}^{\infty}\sigma_7(m)x^m)\displaystyle\prod_{n=1}^{\infty}(1-x^n)^{-24}$

where the last factor is the partition function for 24 bosonic dimensions. Considering both the bosonic left and fermionic right-moving modes, we get the number of heterotic string states at the $N$th mass level

$d(N)=d_R(N)d_L(N)$

arising from tensoring the left and right-moving modes. At $N=0$ we have

$d(0)=16\cdot (480+24)=$

which counts the $16\cdot 8$ states of the supergravity multiplet and the $16\cdot 496$ states of the super Yang-Mills multiplet in the adjoint of $E_8\times E_8$ or $spin(32)/\mathbb{Z}_2$.

If we work in Wilson’s particular version of $E_8$, there is a way to systematically obtain 64 different large $E_8$ lattices that are disjoint from one particular large $E_8$ lattice.

Wilson’s version of $E_8$ consists of all 8-tuples that are either integers or half-integers, and the sum of the coordinates is even in the integer case but odd in the half-integer case. He calls this set $L$.

Suppose we define $L_1$ to be twice the usual Kirmse integers, for some choice of Fano plane structure. If we define:

$S=\{(-\frac{3}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4},\pm\frac{1}{4}) \}$

where there are an odd number of coordinates $-\frac{1}{4}$, then $S$ has 64 elements that are all unit vectors. The 64 lattices:

$$\{L_1 s| s\in S\}$$

turn out to be distinct sublattices of $L$ that are all disjoint from $L_1$. Here we are using octonionic right multiplication based on the same choice of Fano plane structure as we used to define $L_1$.

It’s possible to map all of this to any other example with appropriate rotations, whatever choice we make for $E_8$ and $L_1$. But it’s still not very satisfying! It would be much nicer to find a combinatorial/geometric enumeration of the 64 disjoint lattices that doesn’t rely on octonion multiplication.

Here’s a human-checkable proof that for every large $\mathrm{E}_8$ lattice $L_1$, there are 64 others whose sets of roots are disjoint from it. Because we know that there are 270 different large $\mathrm{E}_8$ lattices, it follows that there are $270 \times 64 = 17280$ ordered pairs of lattices like this, and these are precisely the pairs we can use to construct Leech lattices as sublattices of $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$.

The key to the proof turns out to be the answer to a nice geometrical/combinatorial puzzle! Suppose I have a 7-cube, and then I delete every second vertex to produce a demi-7-cube. To make this concrete, suppose the vertices of my demi-7-cube are the 64 vectors of the form:

$$(\pm 1, \pm 1, \pm 1, \pm 1, \pm 1, \pm 1, \pm 1)$$

for any odd number of minus signs.

Now suppose I want to enumerate all the regular 7-simplexes whose 8 vertices are subsets of these 64 points. I claim there are 240 such 7-simplexes, and I first found them by an exhaustive computer search, but there’s actually an easy way to generate and count them.

In a regular 7-simplex, all the vertices have an angle between them of $\cos^{-1}{(-\frac{1}{7})}$, which for vectors of length $\sqrt{7}$ corresponds to dot products of $-1$. That in turn, for the particular set of vectors we have here, means they must agree in sign in three positions, and disagree in sign elsewhere.

How can we generate subsets of 8 vectors with this property? We use a Fano plane structure!

In a Fano plane, (a) any two distinct lines have a point in common, and (b) the complements of any two distinct lines have two points in common. We will label the points in this diagram with the numbers from 1 to 7, and then use each of the seven lines as follows.

Starting from any vector $v_0$ in our set of vertices of the demi-7-cube, we construct a new vector by changing the signs of those coordinates whose numbers do not lie on the chosen line of the Fano plane. Doing this for all seven lines gives us seven new vectors $\{v_1,\dots,v_7\}$ that agree with $v_0$ at three positions and disagree elsewhere.

What about the relationship between $v_i$ and $v_j$ when $i\neq 0$ and $j\neq 0$? Since $v_i$ and $v_j$ are based on distinct lines in the Fano plane, they will have one sign in common because of the Fano plane point their lines had in common (with their signs both agreeing with $v_0$ at the corresponding coordinate), and another two signs in common because of the two Fano plane points that their complements had in common (with their signs both disagreeing with $v_0$ at the corresponding coordinates). What’s more, these three positions will again correspond to a line in the Fano plane. So, $v_i$ and $v_j$ will always agree at three positions (corresponding to colinear points in the Fano plane) and disagree elsewhere. It follows that their dot product will be $-1$.

Because there are 30 different Fano plane structures we can put on any 7-element set, we can generate 30 different 7-simplexes from the same starting vector $v_0$. But what happens if we change $v_0$? If we fix a Fano plane structure and choose any triple of coordinates that lie on a line according to that structure, the 8 different choices of sign we can make for those 3 coordinates of $v_0$ will lead to 8 different 7-simplexes. Doing this for each of the 30 choices of Fano plane structures will produce 240 distinct 7-simplexes.

OK, now here’s a simple twist we can apply to this construction. Suppose we pick one Fano plane structure out of the 30, and then impose a somewhat odd-sounding condition on our choice of 7-simplexes: if any of the seven lines of the chosen Fano plane structure appear in one of the 30 structures, we declare that structure to be “inadmissible”. How many 7-simplexes do we have, then?

It’s not hard to check that we are left with only 8 of the original 30 Fano plane structures, and so the number of “admissible” 7-simplexes is 64, rather than 240.

We’ve reached the number 64 … but what does any of this have to do with $\mathrm{E}_8$ lattices?

In a lemma I proved earlier, I showed that when a pair of large $\mathrm{E}_8$ lattices $L_1$ and $L_2$ have no roots in common, each root of $L_1$ (a vector in shell 2 of the normal $\mathrm{E}_8$ lattice) will have roots of $L_2$ as exactly 8 of its nearest neighbours.

Now, the only way 8 roots of $L_2$ can be equidistant from a vector in shell 2 at the correct distance is if they all belong to one of the 17,280 7-simplexes of the root polytope of $L_2$, which is a scaled-up, rotated version of the usual $\mathrm{E}_8$ root polytope. So for every root of $L_1$, 8 roots of $L_2$ surround it in a 7-simplex.

Let’s focus on a particular root of $L_1$, say $r_1$. Any vector in shell 2 has 64 nearest neighbours, so 8 of the 64 nearest neighbours of $r_1$ will be the 8 roots of $L_2$ that form a 7-simplex around it. What’s more, those 64 nearest neighbours in shell 2 are arranged in a demi-7-cube, because they correspond to the centres of the 7-orthoplex faces of the ordinary $\mathrm{E}_8$ root polytope, and only half of the 128 6-faces of any given 7-orthoplex are shared by neighbouring 7-orthoplexes (with the others being shared by 7-simplexes). In an earlier post in this series, John discussed all this geometry in detail, and dubbed the two kinds of 6-faces of the 7-orthoplexes “black” and “white”.

So, we know there are 240 ways of finding a 7-simplex among the vertices of a demi-7-cube, and we might try to use any one of them to fix the exact position of the 7-simplex of roots of $L_2$ around $r_1$, our chosen root of $L_1$. However, some of those 240 ways will actually cause things to go awry elsewhere, forcing some roots of $L_1$ and $L_2$ to coincide. How does this happen, and how can we avoid it?

If we pick any root in an $\mathrm{E}_8$ polytope (at any scale), its nearest neighbours will be 60 degrees away, and there will be 56 of them. In the case of our construction of a large $\mathrm{E}_8$ lattice making use of a particular Fano plane structure, if we pick $2 e_0$ as our root, then the 56 roots that lie 60 degrees away are:

$$e_0 \pm e_i \pm e_j \pm e_k$$

where $i,j,k$ all lie on one of the 7 lines in our Fano plane.

However, a less-well-known fact about the $\mathrm{E}_8$ root polytope is that 60 degrees away from the centre of any of its 17,280 7-simplex faces are 28 vertices of the polytope, i.e. 28 roots. To see this, consider the vector:

$$s = (5,1,1,1,1,1,1,1)$$

I claim this is a multiple of a 7-simplex centre in the root polytope of the “standard” $\mathrm{E}_8$. It has a dot product of 6 with the 8 roots:

$$e_0 + e_i\qquad i=1,\dots,7$$ $$\sum_{i=0}^7 \frac{1}{2} e_i$$

which implies an angle of $\cos^{-1}{\frac{3}{4}}$ with all of them (the same as the angle between nearest neighbours in shell 2). But at an angle of 60 degrees, corresponding to a dot product of 4, are the 28 roots:

$$e_0 - e_i\qquad i=1,\dots,7$$ $$\frac{1}{2} e_0 + \sum_{i=1}^7 \pm\frac{1}{2} e_i\qquad \text{with exactly two -ve signs}$$

So, the danger is that some of the 56 roots of $L_1$ that lie 60 degrees away from $r_1$ in the $L_1$ root polytope will end up coinciding with some of the 28 roots of $L_2$ that lie 60 degrees away from the centre of the 7-simplex of the $L_2$ root polytope that coincides with $r_1$.

How can we avoid this? It turns out that the 28 roots that lie 60 degrees away from the centre of any 7-simplex in the root polytope all lie in “directions” away from the centre that match up exactly with the midpoints of the 28 edges of the 7-simplex! Here I mean “directions” in the tangent space sense: if you project everything onto a 7-sphere, and draw a geodesic from the centre of the 7-simplex to each of those 28 roots 60 degrees away, those geodesics will bisect the 28 edges of the 7-simplex. Another way to put this is to note that the vectors for these three things are coplanar. For example, if we pick two vertices of the 7-simplex in our example above:

$$v_1 = (1,1,0,0,0,0,0,0)$$ $$v_2 = (1,0,1,0,0,0,0,0)$$

then the midpoint of the edge between them is a multiple of:

$$m_{12} = (2,1,1,0,0,0,0,0)$$

One of our 28 roots 60 degrees away is:

$$\rho = \left(\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$$

And we have:

$$s = 2(m_{12} + \rho)$$

showing that the three vectors are coplanar.

OK, that gives us a map between the 28 roots in the $L_2$ polytope and the edges of the 7-simplex. But what about the 56 roots in the $L_1$ polytope?

As we previously noted, in the case where we construct the large $\mathrm{E}_8$ lattice from a choice of Fano plane structure, if we pick $2 e_0$ as our reference root $r_1$, then the 56 roots that lie 60 degrees away from it are:

$$e_0 \pm e_i \pm e_j \pm e_k$$

where $i,j,k$ all lie on one of the 7 lines in our Fano plane. If we stick with this example, the demi-7-cube from whose vertices the 7-simplex of the $L_2$ root polytope needs to be selected consists of the points:

$$\left(\frac{3}{2}, \pm\frac{1}{2}, \pm\frac{1}{2}, \pm\frac{1}{2}, \pm\frac{1}{2}, \pm\frac{1}{2}, \pm\frac{1}{2}, \pm\frac{1}{2}\right)$$

where there is an odd number of minus signs. These are the 64 shell 2 vectors that are the nearest neighbours of $2 e_0$. If we drop the first coordinate and double the rest, we get exactly the demi-7-cube we were working with initially.

To avoid giving the $L_2$ root polytope’s 7-simplex any edges whose midpoints lie in the same direction as one of the 56 roots of $L_1$, all we need to do is carry out our trick for choosing 7-simplexes, with all the lines from one particular Fano plane structure ruled inadmissible! To see why this works, recall that when we discussed the dot product between any pair of vectors in one of these 7-simplexes, we noted that the coordinates where their signs agreed would always correspond to a line in the Fano plane. This means that if we sum any such pair of vectors to find the midpoint of the edge between them, the non-zero coordinates of that sum will again correspond to a line in the Fano plane. By avoiding all the lines of the particular Fano plane structure that was used to create the large $\mathrm{E}_8$ lattice $L_1$, we avoid all the 7-simplexes that would place some of the 28 roots of $L_2$ in the same positions as some of the 56 roots of $L_1$.

If we declare the 7 lines from a particular Fano plane structure inadmissible, as well as losing that one structure out of the 30, we lose 21 others, leaving just 8. So when we multiply that count by the 8 possible sign choices in the starting vector, we have 64 valid choices for the way the 7-simplex from the $L_2$ root polytope fits in the demi-7-cube of nearest neighbours in shell 2 of the $L_1$ root $r_1$.

Now, although we’ve shown that these 64 choices avoid one possible source of collisions between the roots of the two lattices, we haven’t yet proved that nothing else can go wrong. There are further sets of roots at 90 degrees and 120 degrees from both the vertices and the 7-simplex centres of the $\mathrm{E}_8$ root polytope. All the roots at 120 degrees are just the opposites of those at 60 degrees, so if the first kind fail to collide, so will the second.

What about the roots at 90 degrees, orthogonal to either a vertex or a 7-simplex centre? In the context of a large $\mathrm{E}_8$ lattice based on a choice of Fano plane structure, there are $7 \times 16 = 112$ roots orthogonal to $2 e_0$ of the form:

$$\pm e_a \pm e_b \pm e_c \pm e_d$$

where $a, b, c, d$ are 4 numbers that do not lie on a line in the Fano plane, along with 14 more roots:

$$\pm 2 e_i\qquad i\in{1,\dots,7}$$

When we look at roots orthogonal to a 7-simplex centre in the ordinary $\mathrm{E}_8$ root polytope, there are 56. If we return to our favourite 7-simplex centre:

$$s = (5,1,1,1,1,1,1,1)$$

this will be orthogonal to 42 roots with 0 in the first coordinate and $\pm 1$ in any other two positions, with opposite signs, such as $(0,-1,1,0,0,0,0,0)$, as well as 14 roots with $\pm\frac{1}{2}$ for all coordinates, with the first and any one of the seven other coordinates agreeing with each other and the rest taking the opposite sign.

This time, these 56 roots correspond directly to vectors that point along each edge of the 7-simplex! For example, if we again take the two vertices of the 7-simplex to be:

$$v_1 = (1,1,0,0,0,0,0,0)$$ $$v_2 = (1,0,1,0,0,0,0,0)$$

then their difference is:

$$e_{12} = (0,1,-1,0,0,0,0,0)$$

which is one of the 56 roots orthogonal to the simplex centre.

Corresponding to this, if we take a difference between any two vertices of a 7-simplex inscribed in our standard demi-7-cube according to a Fano plane structure, that vector will have non-zero coordinates only for the four positions that lie off some line in the Fano plane. And if we declare these 7 complements-of-lines inadmissible, and rule out any Fano plane structure that contains any of them, we end up with the same 8 as we chose with our previous collision-avoidance strategy.

That rules out any clash with the roots orthogonal to $2 e_0$ that depend on the Fano plane structure used to construct the large $\mathrm{E}_8$ lattice $L_1$, but what about the other 14 roots:

$$\pm 2 e_i\qquad i\in{1,\dots,7}$$

How can we be sure that these won’t collide with any roots of $L_2$? The answer is that all the roots of $L_2$ can be identified with the “Fano-type” orthogonal roots for some choice of Fano plane structure, and these are always distinct from these “non-Fano” cases.

Greg’s remarks on demicubes and simplices helped me make progress on a little puzzle I’d given up on.

In 3 dimensions, a ‘demicube’, formed by taking every other vertex of a cube, is a tetrahedron. So, we can partition the vertices of a cube into two tetrahedra in a unique way. The puzzle was: how does this pattern generalize in higher dimensions?

Here Greg took a demicube in 7 dimensions, and found a lot of 7-simplices among its vertices. If the vertices of the cube are

$$(\pm 1, \pm 1, \pm 1, \pm 1,\pm 1, \pm 1,\pm 1 )$$

we find that two of these can be vertices of a 7-simplex centered at the origin if their signs agree in 3 out of the 7 places, and differ in the remaining 4.

Why? Because then their dot product is $3 - 4 = -1$, while their lengths are $\sqrt{7}$, so the angle between them is $\arccos(-\frac{1}{7})$.

In general, we get the vertices of a regular $n$-simplex by taking $n$ vectors of equal length with angles $\arccos(-\frac{1}{n})$ between them, so this is just what we want.

How much does this generalize to other dimensions? In any odd dimension $n = 2k+1$ we can hope to find $n+1$ vectors of the form

$$(\pm 1, \pm 1, \dots, \pm 1, \pm 1 )$$

such that each pair agrees in $k$ places and differs in the remaining $k+1$. If we can succeed, the angle between each pair of these vectors will be $\arccos(-\frac{1}{n})$, so we’ll get the vertices of a regular $n$-simplex!

Puzzle 1: can we always succeed?

I don’t know the answer to this one.

Greg succeeded for $n = 7$, and in this case the vertices of the 7-simplex necessarily lie in a demicube.

Puzzle 2: in which dimensions $n$ will the vertices of a regular simplex, chosen among those of a cube, lie in demicube?

Answer: we need the dimension $n$ to be one less than a multiple of 4, since then $n = 2k+1$ where $k$ is odd and $k+1$ is even, since then each pair of vectors differs in an even number of places.

Puzzle 3: in which dimensions can we partition the vertices of an $n$-demicube into the vertices of some number of regular $n$-simplices?

I don’t know the answer to this one. But since the demicube has $2^{n-1}$ vertices and the simplex has $n+1$, we need $2^{n-1}$ to be divisible by $n+1$. So, there’s only hope when $n$ is one less than a power of two.

It works when $n = 3$, since then $2^{n-1}/(n+1) = 4 / 4 = 1$, and a 3-demicube just is a regular 3-simplex.

It works when $n = 7$, since then we get $2^{n-1}/(n+1) = 64/8 = 8$, and Greg showed we can partition the vertices of a 7-demicube into 8 regular 7-simplices.

So, the first unsolved case comes when $n = 15$. Then $2^{n-1}/(n+1) = 2048$.

Puzzle 4: Can we partition the vertices of a $15$-demicube into the vertices of 2048 regular $15$-simplices?

It would be interesting if this failed: 3d space is the imaginary quaternions, and 7d space is the imaginary octonions, but 15d space is the imaginary sedenions, which are not a division algebra. It would be nice if the partition arose by taking a given simplex and rotating it by multiplying with elements of the given normed division algebra that lie in a subgroup (in the 3d case) or sub-loop (in the 7d case).