## December 11, 2014

### Integral Octonions (Part 11)

#### Posted by John Baez

Take a bunch of equal-sized solid balls in 8 dimensions. Pick one… and then get as many others to touch it as you can.

You can get 240 balls to touch it — no more. And unlike in 3 dimensions, there’s no ‘wiggle room’: you’re forced into a specific arrangement of balls, apart from your ability to rotate the whole configuration.

You can continue packing balls so that each ball touches 240 others — and unlike in 3 dimensions, there’s no choice about how to do it: their centers are forced to lie at a lattice of points called the E8 lattice.

If we pick a point in this lattice, it has 240 nearest neighbors. Let’s call these the first shell. It has 2160 second-nearest neighbors. Let’s call these the second shell.

And here’s what fascinates me now…

You can take the first shell, rotate it, and expand it so that the resulting 240 points form a subset of the second shell!

In fact, there are 270 different subsets of this type. And if you pick two of them that happen to be disjoint, you can use them to create a copy of the Leech lattice inside $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$ — that is, the direct sum of three copies of the $\mathrm{E}_8$ lattice!

There are exactly 17,280 ways to pick two disjoint subsets of this type. And each way gives a distinct embedding of the Leech lattice in $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$.

We’re seeing some big numbers here! Some seem harder to understand than others. Though it’s not easy to prove that the $\mathrm{E}_8$ lattice gives the largest number of balls that can touch a given central one in 8 dimensions, it’s easy to check by hand that it has 240 points in the first shell, and 2160 in the second shell. This is old, well-known stuff.

Greg Egan discovered the other two numbers using computer calculations. By now we understand one of them fairly well: we found a simple construction that gives 270 subsets of the second shell that are rotated, rescaled versions of the first shell. The construction is nice, because it involves the Fano plane: the projective plane over the field with 2 elements! So, I’ll explain that today.

The number 17,280 remains mysterious.

### Why 270?

I want to show you that there are at least 270 subsets of the second shell of the $\mathrm{E}_8$ lattice that are rotated, rescaled copies of the first shell. In fact that’s the exact number — but I only know that because I trust Egan’s computing skills. We did, however, figure out a very pretty proof that there are at least 270. Here are the key steps:

1. There’s a rotational symmetry of the $\mathrm{E}_8$ lattice mapping any point of the first shell to any point to any other point, and similarly for the second shell.
2. There are 240 points in the first shell.
3. There are 2160 points in the second shell.
4. There are 30 rotated, expanded copies of the first shell containing your favorite point in the second shell.

Item 1 lets us exploit symmetry. Since your favorite point in the second shell is just like any other point, we can construct 30 × 2160 rotated, expanded copies of the first shell in the second shell. But these copies are not all distinct: if we counted them by naive multiplication, we would be overcounting by a factor of 240, since each has 240 points. So, the correct count is

$30 \times 2160 / 240 = 30 \times 9 = 270$

as desired.

Why are items 1-4 true?

In Part 4 and Part 5 of these series we saw that the points in the first shell are the vertices of a highly symmetrical polytope, the E8 root polytope. We saw that this polytope has 240 vertices, 2160 7-orthoplex faces, and 17,280 7-simplex faces. Remember, an n-orthoplex is the $n$-dimensional generalization of an octahedron:

while an n-simplex is the $n$-dimensional generalization of a tetrahedron:

General facts about Coxeter groups, reviewed in Part 5, imply that the symmetries of the $\mathrm{E}_8$ root polytope act transitively on its vertices, its 7-orthoplex faces and its 7-simplex faces. We also saw that each 7-orthoplex face has a vector in the second shell as its outward-pointing normal. Thus there are 2160 points in the second shell, and the symmetries act transitively on these too. This takes care of items 1-3.

So, the only really new fact is item 4: there are 30 rotated, expanded copies of the first shell containing a chosen point in the second shell! This is what I want to show. More precisely, I’ll show that there are at least 30. So far I only know there are at most 30 thanks to Egan’s computer search. We can probably dream up a proof, but we haven’t gotten around to it.

### Why 30?

Remember the standard description of the $\mathrm{E}_8$ lattice. Define a half-integer to be an integer plus $\frac{1}{2}$. The $\mathrm{E}_8$ lattice consists of all 8-tuples of real numbers

$(x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7)$

such that

• the $x_i$ are either all integers or all half-integers, and
• the sum of all the $x_i$ is even.

The shortest nonzero vectors in this lattice — the vectors in the first shell — have length $\sqrt{2}$, like this:

$(1,1,0,0,0,0,0,0)$

The second shortest nonzero vectors — the vectors in the second shell — have length $2$, like this:

$(1,1,1,1,0,0,0,0)$

So, we have the opportunity to find an expanded copy of $\mathrm{E}_8$ lattice, dilated by a factor of $\sqrt{2}$, inside the $\mathrm{E}_8$ lattice.

Our goal is to find 30 such copies containing our favorite point in the second shell. Let’s take this as our favorite:

$(2,0,0,0,0,0,0,0)$

And here’s the cool part: there’s one such copy for each way to give the set

$\{1,2,3,4,5,6,7\}$

the structure of a projective plane!

Up to isomorphism there’s just one projective plane of order 7: the projective plane over the field of 2 elements, called the Fano plane. It has 7 points and 7 lines, 3 points on each line and 3 lines through each point:

It’s a projective plane in the axiomatic sense: each pair of distinct points lies on a unique line, and each pair of distinct lines intersects in a unique point.

So, how many projective plane structures are there on the set

$\{1,2,3,4,5,6,7\} ?$

We get one from any way of labelling the diagram above by numbers from 1 to 7. There are $7!$ of these labellings. However, not every one gives a distinct Fano plane structure, since the Fano plane has symmetries. Its symmetry group is famous: it’s the second smallest nonabelian simple group, and it has 168 elements. So, we get

$\frac{7!}{168} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{24 \cdot 7} = 30$

different Fano plane structures on the set

$\{1,2,3,4,5,6,7\}$

How can we use one of these Fano plane structures to find an expanded copy of the $\mathrm{E}_8$ lattice inside the $\mathrm{E}_8$ lattice?

Let $e_i$ be the $i$th basis vector of $\mathbb{R}^8$, where $i = 0,1,2,3,4,5,6,7$. We form the lattice consisting of all integer linear combinations of:

• the vectors $2e_i$,
• the vectors $\pm e_0 \pm e_i \pm e_j \pm e_k$ where $i,j,k$ lie on some line in the Fano plane,
• the vectors $\pm e_p \pm e_q \pm e_r \pm e_s$ where $p,q,r,s$ all lie off some line in the Fano plane.

Pondering the discussion of ‘Kirmse integers’ in Part 6, you’ll see this lattice is a rotated copy of the $\mathrm{E}_8$ lattice, rescaled by a factor of $\sqrt{2}$. However, you can also immediately see that it’s a sublattice of the $\mathrm{E}_8$ lattice as defined above!

With a bit more work, you can see that each different projective plane structure on $\{1,2,3,4,5,6,7\}$ gives a different lattice of this sort. So, we get 30 of them.

Since each such lattice consists of linear combinations of its shortest vectors, we get 30 distinct subsets of the second shell that are rotated, rescaled copies of the first shell. Moreover, all these contain our favorite point in the second shell,

$2e_0 = (2,0,0,0,0,0,0,0)$

### Why 17,280?

So, I’ve shown how to construct at least 270 subsets of the second shell of the $\mathrm{E}_8$ lattice that are rotated, rescaled copies of the first shell. And in fact, that’s all there are.

I said any way to pick two of these 270 subsets that are disjoint gives a copy of the Leech lattice inside $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$. Egan explained how this works in Part 9.

But why are there 17,280 ways to choose two disjoint subsets of this sort? We don’t have a proof yet; his computer calculations just reveal that it’s true. Since

$270 \times 64 = 3^3 \times 10 \times 4^3 = 12^3 \times 10 = 17,280$

it would suffice to check that for each of our 270 subsets we can find 64 others that are disjoint from it. And indeed, his computer calculations verify this, and have told us a lot about what these 64 are like. But we haven’t yet turned this into a human-readable proof that there are 64.

Another thing he checked that each of these 17,280 disjoint pairs gives a distinct sublattice of $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$. So, this procedure gives 17,280 distinct copies of the Leech lattice sitting in here. But again, this deserves a nice proof.

By the way: it’s not as if we’re short of Leech lattices sitting inside $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$. In Part 10 we saw, without any computer assistance, that there are at least 244,035,421. However, the ones obtained by the construction we’re discussing now are especially nice: for example, we’ve seen they give Jordan subrings of the exceptional Jordan algebra. So I think these are worth understanding in more detail.

And besides, they’re leading us into fun questions about the geometry of the $\mathrm{E}_8$ lattice! Any excuse for thinking about $\mathrm{E}_8$ is a good thing as far as I’m concerned. It’s endlessly rewarding in its beauty: today we’ve seen how the Fano plane is hiding inside it, in a rather abstract way. Of course, we already knew the Fano plane was involved in defining octonion multiplication — but nothing today required octonion multiplication. To define the product of octonions we need to orient the lines of the Fano plane, breaking its symmetry group — but today we didn’t need to do this!

Posted at December 11, 2014 3:35 AM UTC

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### Re: Integral Octonions (Part 11)

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?

Posted by: Layra on December 11, 2014 8:56 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Hi, Layra! It’s great to see you here again!

My first guess was similar to yours, and it didn’t work. I’m just waking up, so it will take me a while to figure out if your guess is the same as mine. But let me explain my guess: even though it’s wrong, explaining it will give me excuse to present some interesting clues.

While working on this problem, I indeed spent most of my time thinking of a scaled copy of the first shell in the second shell as a particularly nice choice of 7-orthoplex faces of the $\mathrm{E}_8$ root polytope! In any such choice, we must pick 240 orthoplexes. If we choose some orthoplex, we cannot choose any of its nearest neighbors. But we must choose exactly 56 of its second nearest neighbors.

The hard part was figuring out the rules that govern these 56 second nearest neighbors. It was helpful to look at an example. If our orthoplex corresponds to this point in the second shell:

$(2, 0,0,0,0,0,0,0)$

then it has 64 nearest neighbors:

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

and 280 second nearest neighbors:

$(1, [\pm 1, \pm 1, \pm 1, 0, 0, 0 , 0])$

Here the square brackets mean that we can permute the last 7 components any way we want, as well as independently adjusting the signs. That gives $2^3 \cdot \binom{7}{3} = 280$ choices.

So, what are the allowed ways to choose 56 of these second nearest neighbors? Greg had already realized there were 30 allowed ways. He had also realized that the symmetry group of the Fano plane played a key role in this problem. So, it was not too hard to guess that the allowed choices of

$[\pm 1, \pm 1, \pm 1, 0, 0, 0 , 0]$

come from picking a Fano plane structure on $\{1,2,3,4,5,6,7\}$ and then choosing the $\pm 1$s to lie on any line. There are 30 Fano plane structures, and each gives $2^3 \cdot 7 = 56$ choices of this sort.

We soon realized that this story could be massively simplified, and I presented the simplified account in my blog article here! But the more complicated story is very cute, especially if we think about it in a coordinate-independent way. The 7 coordinates in the vector above correspond to the 7 axes of the originally chosen orthoplex. So, we are putting a Fano plane structure on its 7 axes, and using that to make 56 choices of second nearest neighbors.

Now let’s return to the question at hand. We’ve chosen a recaled copy of the first shell in the second shell, and we want to know why there are exactly 64 ways to choose a second, disjoint copy.

Since each orthoplex has 64 nearest neighbors, my first guess was that each second, disjoint copy contains exactly one of these 64 nearest neighbors.

But Greg quickly saw this was false. In fact, it contains exactly 8 of these 64 nearest neighbors!

When I wake up, I’ll think about whether your guess differs from mine, and whether this last clue tends to confirm or disconfirm your guess.

Posted by: John Baez on December 11, 2014 12:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I’m afraid Layra’s idea doesn’t work out, because although the sets produced this way by octonion multiplication are isometric to $A$, they won’t generally be subsets of the $E_8$ lattice. There are some specific choices of $s$ and $A$ that make $s A$ or $A s$ subsets of the lattice, but there is no single choice of $A$ that makes every pair $(s A, A s)$ for every $s$ a pair of subsets of the lattice.

John wrote:

Another thing he checked is that each of these 17,280 disjoint pairs gives a distinct sublattice of $\mathrm{E}_8 \oplus \mathrm{E}_8 \oplus \mathrm{E}_8$. So, this procedure gives 17,280 distinct copies of the Leech lattice sitting in here. But again, this deserves a nice proof.

That, at least, turns out to be easy. If you take a Leech lattice constructed according to the generalised Wilson method, you can recover the two lattices used in its construction. If the elements of the Leech lattice are $(a,b,c)$, then one lattice consists of all pairwise sums $a+b, a+c, b+c$, and the other lattice consists of all triple sums $a+b+c$. This means it’s impossible for two different pairs of lattices to generate the same Leech lattice.

Posted by: Greg Egan on December 11, 2014 1:39 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Thanks! You’d told me that once before—I’d forgotten.

Posted by: John Baez on December 11, 2014 2:48 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

It might be worth posting a proof of something that previously I’d only noted by means of computer calculations.

Lemma: If a pair of large $E_8$ lattices $(L_1, L_2)$ have no roots in common, then each root of $L_1$ will have exactly 8 roots of $L_2$ among its 64 nearest neighbours in shell 2, and vice versa. [By a “large $E_8$ lattice” I mean a copy of $E_8$ rescaled by $\sqrt{2}$ and rotated to be a sublattice of $E_8$. Such a lattice has its shortest vectors, or roots, in shell 2 of the original $E_8$ lattice.]

Proof: Shell 2 of the $E_8$ lattice contains 2160 vectors. Every vector in shell 2 has 64 nearest neighbours, separated from it by an angle of $\cos^{-1}(3/4)$.

Colour the 240 roots of $L_1$ red, then colour the remaining vectors in shell 2 blue. None of the red vectors will be nearest neighbours to each other, because that would contradict the minimum angular separation of $\cos^{-1}(1/2)$ between $E_8$ roots. So there are:

$2160 - 240 = 1920$

blue vectors, and they must supply all the nearest neighbours to the 240 red vectors. That means they must each be the nearest neighbour of:

$(240 \times 64) / 1920 = 8$

different red vectors.

Now focus on one blue vector that is a root of $L_2$. Among its own 64 nearest neighbours, precisely 8 of them will be red, i.e. in $L_1$. But we could run the same argument with $L_1$ and $L_2$ swapped, so each kind of vector will have 8 nearest neighbours of the other kind.

Posted by: Greg Egan on December 11, 2014 1:52 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

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 $spin(32)/\mathbb{Z}_2$.

Posted by: Metatron on December 11, 2014 6:40 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Thanks! I really want to bring lattices and theta functions into the discussion at some point. I think this is where categories and higher categories — the central topic of this blog — can become relevant. And as you note, the theta functions show up automatically when we use these lattices for string theory compactifications! Since string theory is categorified particle physics, that’s another clue that lattices and higher categories should be studied together.

Posted by: John Baez on December 12, 2014 6:56 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Yes, the patterns so far uncovered, show that these theta functions, as well as mock theta functions describe degeneracies of string and brane configurations.

Take a look at:

Quantum Black Holes, Wall Crossing, and Mock Modular Forms Atish Dabholkar, Sameer Murthy, Don Zagier

which places Ramanujan’s mysterious mock theta functions in the realm of counting BPS black hole degeneracies.

The larger picture, does indeed involve the category of effective geometric motives, as the “dream”, so to speak, is to construct the compactification geometry from the modular data alone. For the Calabi-Yau case, see:

Emergent spacetime from modular motives Rolf Schimmrigk

This stuff takes us directly into the generalization of the Taniyama-Shimura conjecture (which states that every elliptic curve is really a modular form in disguise) into the broader Langland’s program. In other words, instead of looking at worldsheets, we can also look at worldvolumes and study Shimura varieties - the higher dimensional analogues of modular curves.

Other forms of Moonshine, in the guise of umbral moonshine and Mathieu moonshine, have also been discovered in the context of stringy physics:

Mathieu Moonshine and N=2 String Compactifications Miranda C. N. Cheng, Xi Dong, John F. R. Duncan, Jeffrey A. Harvey, Shamit Kachru, Timm Wrase

In light of all this, the theta function for the $K_27$ lattice should encode some useful information.

Posted by: Metatron on December 13, 2014 6:35 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Before going up to 27-dimensions, it’s worthwhile to look at 24. For d=24, let $c_1=|L|$ be the number of roots in our 24 dimensional unimodular lattice $L$. Then the theta series is given as:

$\theta_L=1+c_1q+(196560-c_1)q^2+\ldots$

For the Leech we have no roots, so $c_1=0$ and we have:

$\theta_L=1+196560q^2+\ldots$

where 196,560 is the number of vectors of norm four.

In 27-dimensions, Borcherds gave a lattice with no roots and theta function:

$\theta_L=1+1640q^3+119574q^4+1497600q^5+\ldots$

As stated before, this lattice has an automorphism group with 27-dimensional representation given by the exceptional Jordan algebra.

These q-expansions are related to L-series by replacing $q^n \mapsto n^{-s}$. And the L-series give L-functions that can give information about the relevant Kac-Moody algebras and the corresponding arithmetic geometry for the 27-dimensional Motivic physics that is lurking behind this.

Posted by: Metatron on February 8, 2015 1:38 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

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.

Posted by: Greg Egan on December 11, 2014 10:56 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Oops, in that comment I talk about “disjoint lattices” when I meant to say “lattices with disjoint sets of roots”. The lattices themselves are definitely not disjoint; their intersection is equal to twice $E_8$.

Posted by: Greg Egan on December 11, 2014 11:01 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

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.

Posted by: Greg Egan on December 13, 2014 3:01 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

The mention of the $30$ ways to put a Fano Plane structure on $7$ points made me sit up, as there is a well-known construction involving these. Here is a little bit of extra context for one small part of your proof.

The $30$ ways to put a Fano Plane on $7$ points are all interchanged under the symmetric group $S_7$, of course, but fall into two orbits of $15$ elements under $A_7$. Call the elements of one orbit points, and elements of the other orbit planes. One can also draw $\left(\array{7\\3}\right)=35$ unordered triples from the $7$ points. Call these lines.

A point lies on a line precisely if the triple corresponding to the line forms a line in the Fano Plane corresponding to the point. Dually, a plane contains a line precisely if the triple corresponding to the line forms a line in the Fano Plane corresponding to the plane. A point therefore lies in a plane precisely when the corresponding Fano Planes share a line.

Then the resulting structure is just $PG(3, 2)$ (i.e. the projective $3$-space over the field with $2$ elements), with the usual interpretation of points, lines and planes.

So suppose that you pick one of the $30$ Fano structures. Relative to this structure, when is a structure admissible? Suppose your chosen structure is a plane. If another structure is also a plane, then they share a line precisely when intersect in the corresponding line. But any two planes of $PG(3, 2)$ intersect in some line, so no plane is admissible.

Suppose another structure is a point. Then it shares a line with your chosen structure precisely when the point lies on the plane that is your chosen structure.

So the admissible structures are just the $8$ points not lying on the chosen plane.

There should be more stuff like this later in your proof. Still reading and thinking …

Posted by: Tim Silverman on December 13, 2014 11:41 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Thanks, Tim! That’s very enlightening. I’m convinced now that Fano plane structures index pretty much every aspect of this problem, so understanding this “higher level” incidence geometry should make everything much clearer.

Posted by: Greg Egan on December 13, 2014 11:57 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Tim—that’s cool! I’m still digesting Greg’s proof, but where can I read about that “well-known” construction involving the 30 Fano plane structures on 7 points? As one becomes increasingly erudite, ones standard for what counts as “well-known” diverges increasingly from that of the man on the street.

Posted by: John Baez on December 14, 2014 12:25 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Hi John!

I learnt about this from Peter Cameron’s unpublished (but widely referenced!) Notes on Classical Groups, section 2.5 Small fields. I’ve also seen a proof by Conway (but I don’t think it was actually in a paper by Conway) which was more group-y, to the point that I couldn’t really follow it properly (not comfortable enough manipulating group extensions, I think).

It’s part of a cascade of (more- or less-) exceptional things involving small fields and small groups of Lie type, with the $E_8$ lattice at the pinnacle. The work you and Greg Egan are doing looks as though it ought to fit right into this, but the viewpoint is unfamiliar to me so I’m trying to manipulate it into something I understand better.

Posted by: Tim Silverman on December 14, 2014 10:33 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Is there an analogous construction for thinking about the ways to make a finite affine plane out of the points $\{1,2,3,\ldots,n^2\}$?

Posted by: Blake Stacey on December 14, 2014 6:08 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Thanks, Tim! I took the liberty of linkifying your reference to Notes on Classical Groups, so more people would get ahold of it. Looks good!

It’s part of a cascade of (more- or less-) exceptional things involving small fields and small groups of Lie type, with the $\mathrm{E}_8$ lattice at the pinnacle.

Great! This reminds me of a chart you gave me, which has the Weyl group of $\mathrm{E}_8$, or maybe its commutant, at the top.

Greg and I took a helicopter to this pinnacle and are treating it as a base camp for an assault on Mt. Leech, so it’s not surprising that we may need to backtrack and think harder about how $\mathrm{E}_8$ relates to various smaller structures.

The construction we’re looking at, that builds the Leech lattice inside 3 copies of the $\mathrm{E}_8$ lattice, is sometimes called the Turyn construction. This more properly refers to a way to build the Golay code inside 3 copies of the extended Hamming code of length 8, but they’re almost the same idea. Greg found a very helpful paper on this:

You might like it, because it involves a lot of $\mathbb{F}_2$ stuff.

Posted by: John Baez on December 15, 2014 12:24 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Thanks. I’m still slowly working to understand that chart properly. The Weyl group of $E_8$ is isomorphic to $GO_8^+(2)$, the general orthogonal group of plus type in $8$ dimensions over $\mathbb{F}_2$. But I was interested in the simple orthogonal group, so have to go to something smaller.

I’m sure a lot of what you and Greg Egan are doing can be reinterpreted in terms of this cascade, but I’m trying to understand how much. One problem is that I haven’t quite worked up to the point you two are standing in, i.e. I don’t know enough about $E_8$. Yet.

Thanks for the reference!

Posted by: Tim Silverman on December 16, 2014 12:13 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

What’s the “the general orthogonal group of plus type?” I call the group of all linear transformations of a vector space $V$ that preserve a quadratic form $Q$ on $V$ the orthogonal group $\mathrm{O}(V,Q)$, and the subgroup consisting of transformations with determinant 1 the special orthogonal group $SO(V,Q)$. I know about the general linear and special linear transformation groups, but I’ve never hard of a ‘general orthogonal group’, and I don’t know what ‘plus type’ means.

Presumably some things get trickier in characteristic 2, where you can’t pass effortlessly between quadratic forms and symmetric bilinear forms using the polarization identity. And I’m still terrified to contemplate a world where symmetric bilinear forms are the same as antisymmetric ones, though presumably that’s good in lots of ways.

Posted by: John Baez on December 16, 2014 2:13 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

OK, the general orthogonal group is what you call the orthogonal group.

Plus and minus types are analogues of signature for finite fields. In odd dimensions, there’s only one type of quadratic form, but in dimension $2n$ there two types. With one type, a $2n$-dimensional space is tne is a sum of $n$ hyperbolic planes, and therefore contains $n$-dimensional isotropic subspaces—this is the plus type. With the other, it’s a sum of $n-1$ hyperbolic planes and a $2$-dimensional “anisotropic” space (one with no non-zero null vectors), and therefore has no isotropic subspaces of dimension greater than $n-1$; this is the minus type.

In any characteristic, you get a unique symmetric bilinear form from a quadratic form, but in characteristic $2$ you can’t go the other way. Rather, there are multiple quadratic forms for a given bilinear form. Over $\mathbb{F}_2$, there is one for each vector in the dual space. You can add a dual vector to a quadratic form to get another quadratic form (basically because in $\mathbb{F}_2$, every element is its own square, so quadratic things are secretly linear).

On top of this, in characteristic $2$, symmetric bilinear forms are also alternating and therefore give symplectic structures. (So for a quadratic form $Q$ and a corresponding bilinear form $B$, we always have $B(v,v)=0$ but not always $Q(v)=0$, unlike in odd characteristic.)

Moreover, the collection of $2n$-dimensional quadratic forms corresponding to a given $2n$-dimensional symmetric bilinear form turns out, when put together, to be equivalent to an orthogonal form in $2n+1$ dimensions, so symplectic in dimension $2n$ is orthogonal in dimension $2n+1$ (effectively, in characteristic $2$, the $B$ and $C$ type Dynkin diagrams give the same groups). (That’s on top of, and not isomorphic to, the Klein correspondence between $B_2$ and $C_2$, so there’s an outer automorphism of the symplectic groups in $4$ dimensions in characteristic $2$, obtained by composing the Klein correspondence from, e.g., orthogonal to symplectic, and the other kind of isomorphism in the other direction.)

Posted by: Tim Silverman on December 16, 2014 9:41 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

OK, here’s a bit more. It’s a bit sketchy because I haven’t properly thought it through, but I think it’s on the right track.

Take the coordinates of the points of the demi-$7$-cube. Replace the $-1$s with $0$s. Add an extra coordinate fixed at $0$. We now have a $6$ dimensional subspace of an $8$ dimensional space over $\mathbb{F}_2$. All points in the subspace have an even number of $0$s and an even number of $1$s, and their last coordinate is fixed at $0$.

Stick the following quadratic form (over $\mathbb{F}_2$) on it: take half the number of $1$s. If this is even, the quadratic form is $0$, otherwise $1$.

Let $v_0$ be all $0$s. Then the points which “have a dot product of $-1$” with it in the original formulation all have four $1$s and four $0$s, hence a quadratic form has the value $0$. In fact, they are precisely those points.

Likewise, two points in this set “have a dot product of $-1$” precisely if they differ in $4$ points, which equivalent to saying that their sum is in the same set.

Now if, as I hope and expect, the quadratic form is of plus type, then we can invoke the Klein correspondence. The isotropic points (i.e. the points, apart from $v_0$, with quadratic form $0$) lie on the Klein quadric, and correspond to lines in $PG(3, 2)$. Isotropic lines (lying entirely in the Klein quadric) correspond to plane pencils of $PG(3, 2)$. As we saw above, these consist of points with mutual dot products of $-1$ in the original demicube.

Isotropic planes (lying entirely in the Klein quadric) fall into two classes, corresponding respectively to the points and planes of $PG(3, 2)$. So there are $30$ of them, and since we are working over $\mathbb{F}_2$, they are Fano planes. And these are precisely groups of $7$ points which all have mutual dot products of $-1$ in the world of the demicube.

The $6$-dimensional vector space over $\mathbb{F}_2$ with a quadratic form of plus type can be derived from the $E_8$ lattice mod $2$ with its Euclidean form, but it isn’t usually done in this way, so I need to think a bit more about translating this connection.

I hope I haven’t garbled this or slipped up in a serious way here …

Posted by: Tim Silverman on December 14, 2014 2:14 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Let me just quickly jot down some notes before I forget them.

The requirement that two points in the demi-$7$-cube have a dot product of $-1$ corresponds to their differing in $4$ coordinates, which in the $6$ dimensional vector space over $\mathbb{F}_2$ corresponds to their having a sum in which $4$ coordinates are $1$ and $4$ coordinates are $0$, which means it lies in the Klein Quadric and therefore corresponds to a line in the projective space $PG(3, 2)$ (aka $\mathbb{F}_2P^3$).

So our basic requirement is to have a set of eight vectors such that for any two of them, their sum lies in the Klein Quadric. Projectively, over $\mathbb{F}_2$, every line has $3$ points, and we are requiring a set of $8$ points such that for any two of them, the line between them has its $3$rd point on the Klein Quadric.

1) If one of those points is the vector space $0$, our requirement amounts to

a) Each of the other $7$ points lies on the Klein Quadric.

b) Each of the pairs of points drawn from the $7$ lies on a line lying entirely in the Klein Quadric, i.e. an isotropic line, i.e. any two points on that line are mutually orthogonal. So any two points in the set of $7$ are mutually orthogonal. Back down in $PG(3, 2)$, this means the corresponding lines intersect. The only way we can have a set of $7$ intersecting lines in $PG(3, 2)$ is if they either

i) all lie in the same plane, or

ii) all pass through the same point.

These both correspond to the $7$ points of an isotropic plane up in the Klein Quadric—there are two families of isotropic planes corresponding to points and lines respectively.

Now suppose that none of the points is the vector $0$. It cannot be the case that all $8$ points lie in the Klein Quadric, because then the requirement that any two points sum to a line in the Klein Quadric would mean that the lines of $PG(3, 2)$ corresponding to any two points intersect, and you can’t have eight mutually intersecting lines in $PG(3, 2)$. So at least one of the points lies off the Klein Quadric, and therefore corresponds to a symplectic form in $PG(3, 2)$.

Call this point $s$.

We want an octad containing $s$. Take a second point. There are two possibilities.

1) The second point itself lies in the QK, i.e. corresponds to a line. Let’s call this second point $l$. In this case, the line corresponding to $l$ is non-isotropic under the symplectic form corresponding to $s$. Also, $s$ and $l$ are non-orthogonal. I.e. if the bilinear form corresponding to our quadratic form is $B$, then $B(s, l)=1$.

What is the maximum possible number of points in the octad that can correspond to lines? Well, since any two such points must sum to a point in the KQ, they must lie on isotropic lines, i.e. the corresponding lines of $PG(3,2)$ must all mutually intersect. But those lines must also all be non-isotropic under (the symplectic form corresponding to) $s$. The maximum number of such lines is $4$—either the $4$ non-isotropic lines lying in a single plane, or the $4$ non-isotropic lines passing through a single point.

2) Alternatively consider the case where the second point does not lie in the QK, i.e. it corresponds to another symplectic structure. In this case, we’ll call the second point $s_1$.

But $s+s_1$ corresponds to a line $l_1$, and in this case $l_1$ must by isotropic in both $s$ and $s_1$. Also, the line $\left\{s,s_1,s_2\right\}$ is isotropic, i.e. $s$, $s_1$ and $l_1$ are all mutually orthogonal.

What is the maximum possible number of points in the octad that can correspond to symplectic forms?

Well, suppose you have two such points $s_1$ and $s_2$, with $s+s_1=l_1$ and $s+s_2=l_2$. $s_1$ and $s_2$ sum to a point in the KQ, so they themselves must be mutually orthogonal.

Then

$\array{\arrayopts{\collayout{left}}\\ B(l_1, l_2)&=B(s+s_1,s+s_2)\\ &=B(s,s)+B(s,s_1)+B(s,s_2)+B(s_1,s_2)\\ &=0+0+0+0\\ &=0 }$

The zeros come from the mutual orthogonality implied by the sums lying in the KQ.

This means that $l_1$ and $l_2$ must intersect. So the number of symplectic forms is limited to the number of mutually intersecting lines of $PG(3,2)$ which are all isotropic under the symplectic form corresponding to $s$. This number is $3$, and the $3$ lines must be both coplanar and concurrent.

Apart from $s$, there are $7$ points in the octad, some of which correspond to lines and some to symplectic forms. From above, no more than $4$ are lines. Also, no more than $3$ are symplectic forms, hence no fewer than $4$ are lines. So exactly $4$ are lines and $3$ are symplectic forms.

OK, let $l$ be a point of the octad that corresponds to a line, and let $s_1$ be one that corresponds to a symplectic form.

We have $s+s_1=l_1$ where $l_1$ corresponds to a line.

Now, any two points in the octad must sum to a point in the KQ, so $s_1$ and $l$ must sum to such a point, which means (since one lies in the KQ and other off it) that they are not mutually orthogonal, i.e. $B(s_1, l)=1$.

Now

$\array{\arrayopts{\collayout{left}}\\ B(l_1, l)&=B(s+s_1,l)\\ &=B(s,l)+B(s_1,l)\\ &=1+1\\ &=0 }$

So $l_1$ and $l$ also intersect.

So, we have $4$ points of the octad which correspond to lines, and, apart from $s$ itself, we have $3$ points corresponding to symplectic forms. All the lines mutually intersect. And adding the three symplectic forms to $s$ gives three more lines which intersect each other and all of the first set of lines. So we have a set of seven mutually intersecting lines, so they must all lie in a plane or pass through a point.

Let’s suppose they lie on a plane. Then we have the following description:

Pick a symplectic form $s$. This will make one of the points in the plane the pole of the plane, i.e. orthogonal to all the points of the plane (including itself). Call that point $p$. The three isotropic points of the plane will pass through $p$. Add $s$ to each of the isotropic lines to get three more symplectic forms, $\left\{s_1, s_2, s_3\right\}$.

Keep the $4$ lines of the plane which don’t pass through $p$. Keep the form $s$ and the forms $\left\{s_1, s_2, s_3\right\}$. This gives the eight points of the octad.

Now, $s_1$, $s_2$ and $s_3$ also make $p$ into the pole of the plane. We reason as follows: the three isotropic lines are precisely $l_1=s+s_1$, $l_2=s+s_2$ and $l_3s+s_3$. The three lines are concurrent and coplanar, so the sum of any two of them is the third. Suppose that we had originally picked $s_1$ instead of $s$. Then we have $s_1+l_1=s_1+s+s_1=s$, $s_1+l_2=s+l_1+l_2=s+l_3=s_3$ and $s_1+l_3=s+l_1+l_3=s+l_2=s_2$.

So

a) $s_1$ sums with each $l_i$ to give another symplectic form, implying that the line is isotropic under $s_1$

b) We get the same set of symplectic forms as we did starting with $s$.

In fact, given a plane and point on the plane, there are exactly $4$ symplectic forms which make that point the pole of the plane, and those are the symplectic forms we have been getting.

So we have the following construction:

We pick a plane. We pick a point on the plane. We take the four lines that do not pass through the point, and the four symplectic forms that make the point the pole of the plane. By the Klein correspondence, these correspond to $8$ vectors in $6$ dimensional space, and these form an octad.

Actually, we can do something slightly neater. Let $L$ be one of the non-isotropic lines of the plane—one of the points not passing through the pole. Pick one of the symplectic forms, $S$. Add $S$ to $L$. Then we get another line, the polar of $L$—i.e. the line, skew to $L$, all of whose points are orthogonal to all the points of $L$. This polar line will pass through $p$ (but obviously won’t lie in the plane). By sticking with $S$ and changing $L$, we get all the four non-isotropic lines passing through $p$ (viz. precisely those lines through $p$ which do not lie in the plane).

So we have the following construction:

Take a plane $\pi$ and a point on the plane, $p$. Take the $4$ lines of $\pi$ that do not lie on $p$ and the four lines through $p$ that do not lie in $\pi$. Take one line of the first set and sum it with the four lines of the second set to get $4$ symplectic forms. Take the four lines in $\pi$ together with the four symplectic forms, and that gives you an octad. Or, dually, take the four lines through $p$ together with the four symplectic forms, and that gives you another octad. Since there are $15$ planes, with with $7$ points, each giving $2$ octads, we get $15\times7\times2=210$ octads this way.

Let’s go back up to $6$ dimensions. A point on a plane of $PG(3, 2)$ corresponds to two isotropic planes of the KQ which intersect in a line. Pick such a pair of planes. Delete the lines. That leaves you a set of four points, $\Pi$, from one plane, and a set of four points, $P$, from the other plane. Pick a point of $\Pi$. Draw the line through it to each of the $4$ points of $P$. That gives you $4$ lines. Each of those lines will have a 3rd point on it, giving you another set of $4$ points, $S$. Picking a different point in $\Pi$ gives you the same set $S$, as does starting, dually, with a point of $\P$ instead. Then we get two octads, $\Pi\coprod S$ and $\P\coprod S$.

This also works if the two planes we start with are actually the same, and the “line of interection” is just an arbitrary line in that plane (the same line in both “copies” of the plane, so that $\Pi$ and $P$ are the same set). Then we replicate the case where we started off with $v_0=0$. So we get all $240$ octads by this construction.

OK, this is kind of a weird construction, so there’s something deeper underneath.

Posted by: Tim Silverman on March 26, 2015 4:07 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Oh, yeah, should have seen this. The points of $S$ lie in the third plane through the line of intersection between the two isotropic planes.

Posted by: Tim Silverman on March 26, 2015 4:18 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

And since each isotropic line uniquely determines the pair of isotropic planes it lies in, this is basically two octads per isotropic line plus one per isotropic plane. Weird.

Posted by: Tim Silverman on March 26, 2015 4:31 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Pick an isotropic plane. Then the isotropic plane that you draw your four (or seven) points from cannot intersect that plane. Down in $PG(3,2)$, you can only (e.g.) use lines from planes that don’t contain a particular point.

Posted by: Tim Silverman on March 26, 2015 4:52 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Just a quick question: do you really mean to be talking about Klein’s quadric instead of Klein’s quartic?

If so, that’s a sort of wonderful coincidence, because I know how Klein’s quartic is related to the Fano plane, which we’ve seen showing up in the process of building up the Leech lattice from $\mathrm{E}_8$.

(It’s amusing that the tiny letter switch from ‘quadric’ to ‘quartic’ takes us from degree 2 to degree 4. Are these words etymologically related or not? I.e., is ‘quarter’ related to ‘quadrant’? I wouldn’t be surprised if they were. Somehow 2 is related to the process of squaring, but a square has $2^2 = 4$ corners.)

Posted by: John Baez on March 26, 2015 5:08 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Yes I do!

If so, that’s a sort of wonderful coincidence, because I know how Klein’s quartic is related to the Fano plane

Hmm, interesting. Here’s something probably relevant in some way:

Generically, a plane quartic has $28$ bitangents. These form a configuration of lines whose automorphism group is $Sp(6, 2)$. Now, suppose we pick two non-intersecting isotropic projective planes in a 5-dimensional projective space with a symplectic structure. Suppose we try to restrict to symmetries of the space that fix each of the planes (as planes, not pointwise). If we had no symplectic structure, the corresponding subgroup of $PSL(6, 2)$ would be $PSL(3,2)\times PSL(3,2)$. But maintaining the symplectic structure forces the planes to move in sync, so the subgroup of $Sp(6, 2)$ is just $PSL(3, 2)$. In fact, the points of the two planes behave like the points and lines, respectively, of a single Fano plane. And this $PSL(3, 2)$ is the automorphism group of the Klein quartic.

But this is also related to the Weyl group of $E_7$. In fact, $W(E_7)\simeq2\times Sp(6,2)$ (where $2$ is the group of order $2$). (I think the $56$ contact points of the bitangents behave like the $56$ roots of the minuscule representation of $E_7$—these are the $56$ roots at an angle of $60^\circ$ to a given root.) I think this fact ought to give us a slicker way to prove some of this stuff, but I can’t yet see how.

Are these words etymologically related or not?

Yes, they all come via various routes from Latin quattuor meaning “four”: “quadric” from quadra meaning “square” and “quartic” from quartus meaning “quarter” (which is also the origin of the word “quarter” itself).

Posted by: Tim Silverman on March 26, 2015 11:11 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I don’t understand your description of the $8$ choices for $v_0$. The way I read it, you pick a triple of coordinates, and there are $2^3=8$ choices of sign for those three coordinates, and each choice gives an alternative $v_0$. But this would violate the requirement that there always be an odd number of minus signs: there should only be $4$ choices of the signs for the $3$ coordinates. So I’ve misunderstood something.

Posted by: Tim Silverman on December 14, 2014 10:55 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Sorry, I didn’t explain this properly!

Having chosen a triple of coordinates and one of $2^3=8$ choices of sign, you then need to ensure the correct parity of $v_0$ by making use of the remaining four coordinates. So if your choice of signs for the triple had an odd number of minus signs, you could set the other four coordinates of $v_0$ to 1, but if your choice for the triple had an even number of minus signs, you need to balance this by, say, setting the first of the other four coordinates to $-1$ and the rest to 1.

These two possible choices, $(1,1,1,1)$ or $(-1,1,1,1)$, for the complement of the triple are arbitrary: there’s no reason why you couldn’t choose any other pair of 4-tuples of even and odd parity, so long as you stick to the same pair. So let’s define $e$ and $o$ to be any choice of 4-tuples of even and odd parity, respectively.

If you fix a Fano plane structure and a choice of one line, $\mathcal{l}_0$, from that structure to serve as your triple, then within each 7-simplex the other four coordinates will range over the 8 possible 4-tuples of a given parity. Recall that, starting with $v_0$, you then negate the complement of each line. When you do this for the line you used to define your triple you will get $e\to-e$ or $o\to -o$, and in the other six cases you will negate each of the six pairs of the four coordinates. These eight distinct possibilities cover all 4-tuples of the same parity.

Now, given any vertex $v$ of the demi-7-cube, I claim that I can find it in one of the eight 7-simplexes we get for each choice of Fano plane structure. Let $t$ be the coordinates of $v$ from the triple defined by $\mathcal{l}_0$, and $f$ be the other four coordinates. If $f=e$ or $f=o$, then $v$ will simply be $v_0$ in the 7-simplex generated by $v$. Otherwise, we can convert $f$ to $e$ or $o$ by negating either 4 or 2 coordinates. If we need to negate 4 coordinates, $v$ will appear in the 7-simplex generated by $(t,-f)$, as the point that arises from the complement of our chosen line $\mathcal{l}_0$.

If we need to negate 2 coordinates of $f$, then there will be a unique line $\mathcal{l}_1$ whose complement and that of $\mathcal{l}_0$ intersect in the corresponding 2 points. If we negate the 2 coordinates of $t$ that also lie in the complement of $\mathcal{l}_1$, obtaining a new triple $t_1$, then $v$ will appear in the 7-simplex generated by $(t_1,e)$ or $(t_1,o)$, as the point that arises from the complement of $\mathcal{l}_1$.

Since we can find all 64 vertices of the demi-7-cube among the eight 7-simplexes, it follows that a choice of:

• a Fano plane structure
• one line $\mathcal{l}_0$ within that plane
• 4-tuples $e$ and $o$ of even and odd parity

gives a partition of the 64 vertices into 8 disjoint 7-simplexes.

Posted by: Greg Egan on December 15, 2014 3:14 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Oh, that’s perfect! Many thanks.

Posted by: Tim Silverman on December 15, 2014 8:58 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

As soon as we choose a Fano plane structure, that gives us an equivalence relation between vertices of the demi-7-cube. If $f$ is the set of triples of colinear points that defines the Fano plane structure on $\{1,\dots,7\}$, and $v$ and $w$ are two vertices of the demi-7-cube, we define:

$\delta(v,w) = \{i | v_i=w_i\}$

and then define the equivalence relation:

$v\sim w \; \text{iff} \; v=w \; \text{or} \; \delta(v,w) \in f$

We’ve made this reflexive by definition, and it’s obviously symmetric, but what about transitivity? Suppose $v\sim w$ and $w\sim x$; we need to show that $v\sim x$.

If either equivalence is actually an equality, we’re done.

If $\delta(v,w) = \delta(w,x)$, then both $v$ and $x$ have opposite signs to $w$ for the complement of $\delta(v,w)$, as well as agreeing with $w$ for $\delta(v,w)$, so $v=x$.

Finally, suppose $\delta(v,w) \neq \delta(w,x)$. Since $\delta(v,w) \in f$ and $\delta(w,x) \in f$, they are two distinct lines which share a single point $a$, but it’s also the case that their complements share two points, $b$ and $c$, with $\{a,b,c\}\in f$. So we have $\delta(v,x)=\{a,b,c\}\in f$, so $v\sim x$.

The 64 vertices of the demi-7-cube will be partitioned into equivalence classes by this relation. Because of the geometry, it’s obvious that no equivalence class can contain more than 8 vertices, but how can we rule out any smaller equivalence classes? We just pick one vertex in any equivalence class and treat it as $v_0$ in the original construction, generating $v_i$ with $i\in\{1,\dots,7\}$ by altering the signs of $v_0$ corresponding to the complement of each of the seven lines in $f$. That guarantees that there are at least 8 vertices in each equivalence class, so they must all have exactly the same size, 8, and there must always be 8 of them, to include all 64 vertices.

Posted by: Greg Egan on December 15, 2014 10:36 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

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).

Posted by: John Baez on December 15, 2014 4:31 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I don’t have the time now to sort out whether the resemblance is only superficial, but this comment reminded me of the Hadamard conjecture.

Posted by: Mark Meckes on December 15, 2014 4:36 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Over on G+, Philip Gibbs seems to be claiming that Puzzle 3 and thus Puzzle 4 has an affirmative solution based on the punctured Hadamard code. I don’t see how it works, but it seems promising, and perhaps worth solving for the light it might shed on the case $n = 7$.

Puzzle 3 was about fitting simplexes into demicubes, but we can restate it this way:

Puzzle 3${}'$: for which $n$ can we partition the vertices of an $n$-cube into the vertices of some number of regular $n$-simplexes?

This can only happen when $2^n$ is divisible by $n+1$, so $n$ has to be one less than a power of 2. When $n = 7$, Greg showed the answer is yes.

(If and only if $n+1$ is a multiple of 4, each simplex will lie in a demicube, so an affirmative answer to Puzzle 3${}'$ gives an affirmative answer to Puzzle 3. But never mind that now: let’s focus on partitioning the vertices of a cube into simplexes.)

The punctured Hadamard code is a way to take $k$ bits and encode them as $2^k - 1$ bits in an extremely robust way. We think of these bits as elements of the 2-element field $\mathbb{F}_2$, and define an inner product on $\mathbb{F}_2^k$ in the obvious way:

$\langle x, y \rangle = \sum_{i = 1}^k x_i y_i$

Then, we encode $x \in \mathbb{F}_2^k$ by writing down the list of its inner products with all vectors $y \in \mathbb{F}_2^k$ except $y = 0$. Taking the inner product with $0$ gives no information, so we can skip that; this is the reason for the word ‘punctured’.

This list of inner products is a list of $n = 2^k - 1$ bits. In other words, it’s a point $\mathbb{F}_2^n$. Let’s call this point $pHad(x)$:

$pHad(x) = (\langle x, y\rangle)_{y \in \mathbb{F}_2^k - \{0\}}$

So, we’ve encoded a string of $k$ bits, $x$, into a string of $n$ bits, $pHad(x)$.

All this looks promising, but how does it give a way to fit an $n$-simplex into an $n$-cube? I can see $\mathbb{F}_2^n$ as an $n$-cube. I need to choose $n+1$ points in here, forming the corners of an $n$-simplex. Does the punctured Hadamard code help me do it?

Maybe so. Since $n+1 = 2^k$, the obvious choice is to take all the points $x \in \mathbb{F}_2^k$ and encode them, getting $n+1$ points $pHad(x) \in \mathbb{F}_2^n$.

Puzzle 5: Do these form the vertices of a regular $n$-simplex?

I now feel sure they must. We just need to check that whenever $x \ne x'$, the points $pHad(x)$ and $pHad(x')$ differ in $2^{k-1}$ coordinates. Anyone want to do that?

Next, assuming this works, can we partition the vertices of an $n$-cube into the vertices of some number of regular $n$-simplexes?

I think we just take the $n$-simplex above and form all its translates in the vector space $\mathbb{F}_2^n$!

Posted by: John Baez on December 16, 2014 5:49 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

John wrote:

Puzzle 5: Do the points $pHad(x) \in \mathbb{F}_2^n$ form the vertices of a regular $n$-simplex when $n = 2^k - 1$?

We just need to check that whenever $x \ne x'$, the points $pHad(x)$ and $pHad(x')$ differ in $2^{k-1}$ coordinates. Anyone want to do that?

Okay, I’ll do it. We need to check that if $x,x' \in \mathbb{F}_2^k$ are different, they have different inner products with exactly $2^{k-1}$ vectors $y \in \mathbb{F}_2^k$.

It suffices to work with $v = x - x'$ and show that the inner product $\langle v, y \rangle$ is nonzero for exactly $2^{k-1}$ vectors $y \in \mathbb{F}_2^k$. That is, exactly half the vectors.

Let $S \subseteq \{1,\dots, k\}$ be the subset of $i$ such that $v_i \ne 0$. The inner product $\langle v, y \rangle$ is nonzero iff $y_i = 1$ for an odd number of choices $i \in S$.

However, no matter what $S$ is, there are the same number of vectors $y \in \mathbb{F}_2^k$ with $y_i = 1$ for an odd number of $i \in S$ as there are with $y_i = 1$ for an even number of $i \in S$.

So indeed, $\langle v, y \rangle$ is nonzero for exactly half the vectors $y \in \mathbb{F}_2^k$

So the answer to Puzzle 5 is yes. The points $pHad(x)$ pick out the corners of regular $n$-simplex among the corners of the $n$-cube whenever $n = 2^k - 1$.

Even better, we can partition the corners of the $n$-cube into congruent copies of this $n$-simplex! The reason is that the points $pHad(x)$ form a linear subspace of $\mathbb{F}_2^n$. The translates of this subspace, i.e. its cosets, will give congruent $n$-simplexes that form a partition of the $n$-cube.

Moreover, we’ve already seen this kind of partition can only exist when $n = 2^k - 1$.

So, we’ve completely settled Puzzle 3${}'$ and thus Puzzles 3 and 4, with a lot of help from Hadamard and Philip Gibbs:

Puzzle 3${}'$: for which $n$ can we partition the vertices of an $n$-cube into the vertices of some number of regular $n$-simplexes?

Answer: we can do this if and only if $n = 2^k - 1$ where $k \ge 0$.

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

Answer: we can do this if and only if $n = 2^k - 1$ where $k \ge 2$.

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

Posted by: John Baez on December 16, 2014 7:36 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

That’s a nice proof!

I don’t think this has any bearing on it, but the Wikipedia article on the Hadamard code defines pHad as:

$\mathrm{pHad}(x) = \left(\langle x,y \rangle\right)_{y\in\{1\}\times\{0,1\}^{k-1}}$

rather than:

$\mathrm{pHad}(x) = \left(\langle x, y\rangle\right)_{y \in \mathbb{F}_2^k - \{0\}}$

That is, the punctured Hadamard code encodes $x$ as the $2^{k-1}$ bits that are the inner products of $x$ with all bit strings of length $k$ whose first bit is 1, rather than as the $2^k-1$ bits that are the inner products of $x$ with all bit strings of length $k$ other than 0.

So, unless I’m confused, you’ve defined something different, but that different thing does exactly the job you want, so it doesn’t really matter if it isn’t the punctured Hadamard code.

Actually, if you use the punctured Hadamard code as Wikipedia defines it, what you get are $2n$ points from the $2^k$ points in $\mathbb{F}_2^k$ rather than $n+1$ (that is, $n=\frac{1}{2}\times 2^k$ rather than $n=2^k - 1$), and they are the vertices of an $n$-orthoplex.

Posted by: Greg Egan on December 16, 2014 10:22 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Hah, that’s funny — I think this is a case of knowing what you want and not letting what you read stand in your way! I completely ignored the difference between $\{1\} \times \{0,1\}^{k-1}$ and $\mathbb{F}_2^k - \{0\}$. I got fewer code words, which I suppose is bad in some way, but they form a simplex, which is what I was after!

Embedding orthoplexes in cubes interesting; I guess the $k = 4$ case of that construction gives the ‘well-known’ fact that the 4-demicube is a 4-orthoplex.

Posted by: John Baez on December 17, 2014 1:16 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

For puzzle 1 I am surprised to find solutions in 11 dimensions using a brute force search as follows

00000000000
11111100000
11100011100
10011011010
01010110110
00101101110
00110111001
10001110101
01011001101
01101010011
11000101011
10110000111

I used 0 and 1 instead of -1 and +1

To find an equilateral triangle on the vertices of a hypercube you need n to be one less than a multiple of four so this was the next case to try.

This is just one of many solutions that are found very quickly. It would be good to know how many inequivalent solutions there are under permutations of rows and columns. Notice that you get another solution from the columns. There must be some known combinatorial pattern in use here but what is it?

Posted by: Philip Gibbs on December 17, 2014 8:33 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Thanks! I meant to post a little summary comment saying that I’d solved all my puzzles except:

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 $k$. If we 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 only proved we can succeed in dimensions that are one less than a power of 2 — in which case we can do much more, actually partitioning the vertices of the cube into simplexes of this type.

So you succeeded in dimension 11. Cool!

Did you try dimensions 5 and 9? Only for dimensions that are one less than a multiple of four will we get a simplex that lies in a demicube, but that’s not really the issue in Puzzle 1.

Posted by: John Baez on December 17, 2014 8:51 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Even if you don’t limit yourself to a demicube you cannot form an equilateral triangle in 4k+1 dimensions on the vertices of the cube.

Just try to find one in 5 dimensions and you will quickly see why you can’t. The first two points can be taken as 00000 and 00111 without loss of generality, but then it is easy to see that there is no third.

By the way a strange fact is that if you do find an equilateral triangle in any number of dimensions you can always add one more point to make it a regular tetrahedon. Out of revenge I will make it a puzzle to find the very simple construction for such a point.

I am looking for solutions in 19 dimensions but brute force is too long.

Posted by: Philip Gibbs on December 17, 2014 9:15 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

As Mark Meckes already suggested, I think Puzzle 1 is essentially equivalent to the Hadamard conjecture.

The most important open question in the theory of Hadamard matrices is that of existence. The Hadamard conjecture proposes that a Hadamard matrix of order $4k$ exists for every positive integer $k$.

If the Hadamard conjecture is true, then if you give me a $4k\times 4k$ Hadamard matrix (that is, a matrix with orthogonal rows whose entries are all $\pm 1$) , I can negate any row that starts with $-1$ without changing the orthogonality. Then if I drop the first entry from each row, I’m left with $4k$ vectors in $\mathbb{R}^{4k-1}$ whose dot products with each other are all $-1$ (because the dot products started as 0 and we’ve omitted a coordinate where all the vectors agreed).

Posted by: Greg Egan on December 18, 2014 3:10 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

The two Paley constructions give very nice ways to get Hadamard matrices, and hence simplexes, in certain cases. The Wikipedia article is very clear, so I won’t try to repeat its description here. But for, say, $n=19$, on a computer it becomes trivial to get a simplex this way:

$\left( \begin{array}{ccccccccccccccccccc} - & - & - & - & - & - & - & - & - & - & - & - & - & - & - & - & - & - & - \\ + & - & + & + & - & - & - & - & + & - & + & - & + & + & + & + & - & - & + \\ + & + & - & + & + & - & - & - & - & + & - & + & - & + & + & + & + & - & - \\ - & + & + & - & + & + & - & - & - & - & + & - & + & - & + & + & + & + & - \\ - & - & + & + & - & + & + & - & - & - & - & + & - & + & - & + & + & + & + \\ + & - & - & + & + & - & + & + & - & - & - & - & + & - & + & - & + & + & + \\ + & + & - & - & + & + & - & + & + & - & - & - & - & + & - & + & - & + & + \\ + & + & + & - & - & + & + & - & + & + & - & - & - & - & + & - & + & - & + \\ + & + & + & + & - & - & + & + & - & + & + & - & - & - & - & + & - & + & - \\ - & + & + & + & + & - & - & + & + & - & + & + & - & - & - & - & + & - & + \\ + & - & + & + & + & + & - & - & + & + & - & + & + & - & - & - & - & + & - \\ - & + & - & + & + & + & + & - & - & + & + & - & + & + & - & - & - & - & + \\ + & - & + & - & + & + & + & + & - & - & + & + & - & + & + & - & - & - & - \\ - & + & - & + & - & + & + & + & + & - & - & + & + & - & + & + & - & - & - \\ - & - & + & - & + & - & + & + & + & + & - & - & + & + & - & + & + & - & - \\ - & - & - & + & - & + & - & + & + & + & + & - & - & + & + & - & + & + & - \\ - & - & - & - & + & - & + & - & + & + & + & + & - & - & + & + & - & + & + \\ + & - & - & - & - & + & - & + & - & + & + & + & + & - & - & + & + & - & + \\ + & + & - & - & - & - & + & - & + & - & + & + & + & + & - & - & + & + & - \\ - & + & + & - & - & - & - & + & - & + & - & + & + & + & + & - & - & + & + \end{array} \right)$

Posted by: Greg Egan on December 18, 2014 4:24 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I’ve been thinking about ways to count the numbers of these simplexes in general, but when there isn’t some association with a known structure (like the Fano plane that we can exploit for the 7-dimensional case), so far I can only see how to do this algorithmically rather than with a closed formula.

An $n$-simplex whose symmetries are easy to understand is the one that sits in $\mathbb{R}^{n+1}$, where you take the vertices to be some constant multiple $\alpha$ of the vectors $\{e_i\}$ of the standard basis for $\mathbb{R}^{n+1}$.

It’s not hard to connect this to the kind of simplex we’ve been talking about, whose vertices belong to an $n$-cube (or demicube, but I think the demicube part is the least of our worries). Start with an $(n+1)\times (n+1)$ Hadamard matrix, and negate columns as necessary to end up with a matrix whose first row is $(1,1,\dots,1)$. Call the rows of the Hadamard matrix $\{h_j\}$, and construct the $(n+1)$-cube whose vertices are:

$\pm h_1 \pm h_2 \dots \pm h_{n+1}$

If our simplex vertices are $\{\alpha e_i\}$, and we express them in terms of the basis $\{h_j\}$, we have:

$\begin{array}{rcl} \alpha e_i & = &\sum_{j}{ \beta_i^j h_j}\\ \alpha e_i \cdot h_k & = &\beta_i^k h_k \cdot h_k\\ \alpha h_k^i& = &(n+1) \beta_i^k\\ \end{array}$

So if we set $\alpha = n+1$, all the $\beta_i^k$ will just be entries of the Hadamard matrix, i.e. $\pm 1$, so the vertices of the simplex will be vertices of the $(n+1)$-cube whose axes are the rows of the Hadamard matrix. What’s more, since we’ve chosen:

$h_1 = (1,1,\dots,1)$

we will have $\beta_i^1=1$ for all $i$, so the vertices of the simplex will be vertices of one particular $n$-cube face of the $(n+1)$-cube.

To count the number of simplexes, we can count the number of symmetries of this simplex that are also symmetries of the $n$-cube; suppose this is $Y$. The number of simplexes will then equal the total order of the group of symmetries of the $n$-cube divided by $Y$, i.e.

$\text{number of simplexes}\;=\;\frac{2^n n!}{Y}$

or half this for a demicube.

The hard part, of course, is computing $Y$. So far the only way I can see to do this in general is to check each of the $(n+1)!$ permutations of the standard basis $\{e_i\}$, and see whether or not it is a symmetry of the $n$-cube. That means checking whether each permutation of the $n+1$ coordinates maps the set $\{h_2,h_3,\dots,h_{n+1},-h_2,-h_3,\dots,-h_{n+1}\}$ into itself (where we don’t have to worry about $h_1$ because it is obviously fixed by all these permutations).

This isn’t too bad (by computer) for $n=7$, where $(n+1)!=8!=40320$, but obviously it becomes intractable quite quickly. The only thing I can say is that it’s always better than checking the symmetry group of the $n$-cube, with $2^n n!$ elements.

What we’re really trying to find is the order of the intersection of the two groups – the symmetry group of the $n$-simplex and that of the $n$-cube – and there might be some sneaky way to figure that out directly, perhaps by looking closely at the subgroups of both.

Posted by: Greg Egan on December 18, 2014 11:14 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Greg wrote:

As Mark Meckes already suggested, I think Puzzle 1 is essentially equivalent to the Hadamard conjecture.

Wow! It seems Hadamard is the go-to guy for questions involving bit strings and simplexes. Which is odd, because I mainly know of his work on “number theory, complex function theory, differential geometry and partial differential equations.”

You showed one direction of the equivalence, but the other seems to work just as well: given a bunch of bit strings with dot product $-1$ we can stick a $1$ in front of each and get a bunch of orthogonal bit strings. So, if we can always find the vertices of a regular $n$-simplex among the vertices of an $n$-cube when $n = 4k - 1$, the Hadamard conjecture is true!

I guess I can rest content and declare all my puzzles dealt with, unless someone wants to win everlasting fame by proving the Hadamard conjecture. (I wonder how hard this conjecture is considered.)

Posted by: John Baez on December 18, 2014 5:51 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

The Hadamard conjecture looks pretty daunting to me, given that people have been working on it since the 1860s!

One intriguing aspect of this subject is that the sets of $k$-simplexes for each $k\in\{0,1,\dots,n\}$ with vertices in an $n$-cube, all separated by an angle of $\arccos(-\frac{1}{n})$, must have quite a rich structure: each $k$-simplex will belong to higher dimensional simplexes up to some maximum dimension, $m$, but even for dimensions where an $n$-simplex exists, many lower-dimensional simplexes will have $m$ less than $n$.

And even for $k=n$, things eventually get complicated. The Wikipedia article says there are unique Hadamard matrices of order 1, 2, 4, 8, and 12 up to equivalence (where the notion of equivalence they use corresponds to isometries of an $n$-cube), but after that:

There are 5 inequivalent matrices of order 16, 3 of order 20, 60 of order 24, and 487 of order 28. Millions of inequivalent matrices are known for orders 32, 36, and 40.

So from $n=15$ and up, even the $n$-simplexes will start to exist in multiple orbits of the symmetry group of the $n$-cube. This means my suggestion that we could count them by looking at the stabiliser subgroup of a single example wouldn’t work; that would only tell us the size of one orbit.

Posted by: Greg Egan on December 19, 2014 5:21 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

To prove the conjecture you only need one example for each $n$ that is a multiple of 4, so in one sense you are making the problem sound harder than it is.

On the other hand, one strategy would be to explore all the solutions for low $n$ and find those that can not be built from Paley constructions. Then you would try to understand those solutions through their symmetries and invariants to see if they are unique irregular cases or part of some more general series that can be used for higher $n$.

This seems to be a problem where computers have already been utilised so it will not be easy to go beyond what has already been done. Perhaps it has already been found that the irregular solutions do not follow any useful pattern, in which case it is going to be very hard. However, it is always possible that more powerful computers may be enough to reveal hidden patterns.

Posted by: Philip Gibbs on December 19, 2014 11:08 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

This seems to be a problem where computers have already been utilised so it will not be easy to go beyond what has already been done.

Sloane’s A Library of Hadamard Matrices gives a good sense of where the cutting edge has reached.

Posted by: Greg Egan on December 20, 2014 11:11 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Greg wrote:

One intriguing aspect of this subject is that the sets of $k$-simplexes for each $k\in\{0,1,\dots,n\}$ with vertices in an $n$-cube, all separated by an angle of $\arccos(-\frac{1}{n})$, must have quite a rich structure: each $k$-simplex will belong to higher dimensional simplexes up to some maximum dimension, $m$, but even for dimensions where an $n$-simplex exists, many lower-dimensional simplexes will have $m$ less than $m$.

For a topologist or category theorist, a natural impulse would be to create the simplicial complex built from all these simplices. A simplicial complex is just a set $S$ of vertices together with a collection $K$ of subsets of $S$, called simplexes, with the property that

1) every singleton is in $K$ and

2) $\sigma \in K, \tau \subseteq \sigma \implies \tau \in K$.

1) says that every vertex is a simplex, and 2) says that every face of a simplex is a simplex.

In our example $S$ is the set of vertices of the $n$-cube and $K$ consists of subsets of vertices that are all separated by an angle of $\arccos(-\frac{1}{n})$. These are, quite literally, regular simplexes!

There’s a way to turn any simplicial complex into a space, called its geometric realization, simply by treating all the simplexes as actual chunks of space and gluing them together in the obvious way.

The advantage is that simplicial complexes are lurking in many interesting problems, and turning them into spaces lets us apply topology to these problems. You can also turn any simplicial complex into a commutative ring, its Stanley–Reisner ring, which has one generator $x_i$ for each vertex and one relation $x_{i_1} \cdots x_{i_n} = 0$ for each collection of vertices $\{i_1, \dots, i_n\}$ that’s not a simplex.

I have no idea how these tricks would help solve Hadamard’s conjecture, but they at least raise new questions, like: what can we say about the topology of the geometric realization of the simplicial complex we’re talking about? And sometimes new questions are the key to solving old ones.

Hadamard’s conjecture is that the simplicial complex associated to the $n$-cube is $n$-dimensional whenever $n = 4k - 1$. But it looks like there’s a lot more to wonder about than merely its dimension!

However, I don’t have the energy to pursue this; I want to get back to integral octonions, the Leech lattice, etc.

Posted by: John Baez on December 19, 2014 6:35 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I had not appreciated the role of the Hadamard matrces in this story before but now I see that they can help answer some questions about the symmetries of the exceptional structures.

When you add an extra column in the Paley construction to extend the field with $p$ elements to something with $p+1$ elements something deeper is going on. In fact you are using the projective plane over the finite field. This plane has $p+1$ elements because of the addition of a line with infinite gradient. This gives the plane and the Hadamard matrix the symmetry $PSL(2,p)$

But for the 12 by 12 Hadamard matrix something special happens. It can be constructed in more than one way using either the fields $\mathbb{F}_11$ or $\mathbb{F}_5$, so its symmetry group must include both $PSL(2,11)$ and $PSL(2,5)$ In fact the full symmetry group is known to be the Mathieu group $M_12$, so are the two PSL groups sufficient to generate the Mathieu group? When you double up to the 24 by 24 matrix the same thing happens again and $PSL(2,23)$ is added into the mix. The full group is $M_24$.

I think there is a lot more to be learnt here especially if you can see how to use the simplexes to construct the Leech lattice. How do its symmetries come into play?

Another thing that may be relevant is the interplay between the field $\mathbb{F}_2$ and $\mathbb{F}_p$ when using the quadratic residues. Does quadratic reciprocity play a role too? Is there something more general?

I don’t think you should be put off by the fact that this problem is old. Sometimes just the fact that we have better computers and better algorithms can take us further than very clever people were able to go before. That is what happened with Lebesgue’s covering problem right? One time I even used computers to advance a problem posed by Diophantus and worked on by Euler. It was the problem to find sets of rational numbers such that the product of any two is one less than a square. Diophantus has found sets of three and four rationals, while Euler increased it to five. Then with a computer search using brute force enhanced by some analysis I found the first sets of six. That also led to interesting discoveries about elliptic curves and hyperdeterminants. Perhaps Hadamard’s conjecture is also ripe for a new computer attack.

Posted by: Philip Gibbs on December 19, 2014 8:15 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

John and Philip: lots of nice ideas here! The only thing I can think to add at this point is that the Hadamard conjecture is also easily restated as a maximum clique problem in graph theory, and the information encoded in the simplicial complex should also be encoded in the graph whose vertices are vertices of the $n$-cube and whose edges join vertices with dot product $-1$.

Posted by: Greg Egan on December 19, 2014 1:07 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Greg wrote:

So from $n=15$ and up, even the $n$-simplexes will start to exist in multiple orbits of the symmetry group of the $n$-cube.

Is that really true? It seems amazing. I’m wondering if maybe inequivalent Hadamard matrices could still give $n$-simplexes that are related by symmetries of the cube. But quite possibly I’m suffering from a deficit of geometrical imagination which kicks in around 15 dimensions.

Posted by: John Baez on December 19, 2014 3:26 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

John wrote:

Is that really true? It seems amazing.

It amazed me too, so maybe I’m mistaken! Let me try to check it carefully (assuming Wikipedia’s claim about inequivalent Hadamard matrices existing is true).

The article says:

Two Hadamard matrices are considered equivalent if one can be obtained from the other by negating rows or columns, or by interchanging rows or columns. […] There are 5 inequivalent matrices of order 16.

I don’t think any of the uses of “or” here are meant exclusively. That is, I think they’re saying two matrices are equivalent if you can obtain one from the other by performing any number of the following operations, in any order you like:

1. Swap any two rows.
2. Swap any two columns.
3. Negate any row.
4. Negate any column.

The map I suggested to produce a simplex-in-an-$n$-cube from an $(n+1)\times (n+1)$ Hadamard matrix uses the following steps:

1. If any rows of the Hadamard matrix start with $-1$, negate those rows.
2. Drop the first entry of each row.
3. Treat the resulting $n+1$ vectors in $\mathbb{R}^n$ as the vertices of an $n$-simplex, which are also vertices of the $n$-cube $\{-1,1\}^n$.

The full symmetries of the $n$-cube, the Coxeter group $BC_n$, consist of the $2^n n!$ linear operators that can be produced by performing any of the following operations, in any order you like:

1. Swap any two coordinates.
2. Negate any coordinate.

Now, suppose we start with two inequivalent Hadamard matrices, $H_1$ and $H_2$. We negate any of their rows that start with $-1$, giving us $H_{1}'$ and $H_{2}'$, which belong to the same two equivalence classes as the original matrices, and so are still inequivalent.

We then drop the first entry of each row, to obtain two simplexes, $S_1$ and $S_2$.

Suppose $S_1$ and $S_2$ belonged to the same orbit of $BC_n$. There is no special ordering of the vertices of a simplex that allows us to form a unique matrix from the vertices, so what we have is that any matrix whose rows are the vertices of $S_1$ can be obtained from any matrix whose rows are the vertices of $S_2$ by some sequence of the following operations:

1. Swapping two columns.
2. Negating a column.
3. Swapping two rows.

That in turn implies that we can obtain $H_{1}'$ from $H_{2}'$ by a sequence of such operations (since the presence of an extra column of all 1s doesn’t change the fact that we can bring the rest of the matrices into agreement) … which contradicts our assumption that $H_1$ and $H_2$ are inequivalent.

I’m still amazed, but it seems to be true!

Posted by: Greg Egan on December 19, 2014 11:01 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Thanks for presenting the argument in detail, so people can find a hole if there is one. I cannot.

I’m still amazed. I guess I’d have to try to disprove this claim to really see where my intuition goes bad. I imagine picking vertices of a simplex one at a time, and showing that whenever there’s more than one choice, we can use a symmetry of the $n$-cube to map one choice to another, while still preserving the partial simplex we’d already built. It’s hard to imagine there’s room for truly different choices.

But I’d have to try to carry out this argument in more detail to see what’s wrong with it. And if it first fails in 15 dimensions, it probably means I’d see a hole but have trouble being sure there was no way to fix it.

Posted by: John Baez on December 20, 2014 12:10 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

For what it’s worth, it helps my own intuition a little to contemplate something a bit simpler that happens even in lower dimensions.

Consider the equilateral triangle $E$ whose vertices are:

$\left( \begin{array}{ccccccc} -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & -1 & -1 & 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 & -1 & 1 & 1 \end{array} \right)$

There are 9 other vertices of the same 7-demicube that have dot products of $-1$ with all three vertices of $E$:

$\left( \begin{array}{ccccccc} -1 & 1 & 1 & 1 & 1 & -1 & -1 \\ 1 & -1 & 1 & -1 & 1 & -1 & 1 \\ 1 & -1 & 1 & -1 & 1 & 1 & -1 \\ 1 & -1 & 1 & 1 & -1 & -1 & 1 \\ 1 & -1 & 1 & 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 & 1 & -1 & 1 \\ 1 & 1 & -1 & -1 & 1 & 1 & -1 \\ 1 & 1 & -1 & 1 & -1 & -1 & 1 \\ 1 & 1 & -1 & 1 & -1 & 1 & -1 \end{array} \right)$

The Gram matrix of this set of 9 vectors is:

$\left( \begin{array}{ccccccccc} 7 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\ -1 & 7 & 3 & 3 & -1 & 3 & -1 & -1 & -5 \\ -1 & 3 & 7 & -1 & 3 & -1 & 3 & -5 & -1 \\ -1 & 3 & -1 & 7 & 3 & -1 & -5 & 3 & -1 \\ -1 & -1 & 3 & 3 & 7 & -5 & -1 & -1 & 3 \\ -1 & 3 & -1 & -1 & -5 & 7 & 3 & 3 & -1 \\ -1 & -1 & 3 & -5 & -1 & 3 & 7 & -1 & 3 \\ -1 & -1 & -5 & 3 & -1 & 3 & -1 & 7 & 3 \\ -1 & -5 & -1 & -1 & 3 & -1 & 3 & 3 & 7 \end{array} \right)$

So, some of the 9 vectors have dot products of $-1$ with 8 vertices of the 7-demicube, while others have dot products of $-1$ with only 4 vertices of the 7-demicube. If we append a vector of the first kind to $E$ to get a tetrahedron $T_1$, and a vector of the second kind to $E$ to get a tetrahedron $T_2$, then the image of $T_1$ under any symmetry of the 7-demicube will have a set of 8 vertices of the 7-demicube with dot product $-1$ with all vertices of the tetrahedron, but for $T_2$ that set will contain only 4 vertices. So $T_1$ and $T_2$ must be in different orbits.

For $n=7$ it turns out to be impossible to “paint yourself into a corner”: you never end up without any choices for adding one more vertex, and once you reach a 7-simplex the differences along the way all wash out: all 240 of the 7-simplexes in a 7-demicube lie on one orbit. But the fact that you can be in different orbits for some lower-dimensional $k$-simplexes makes it a bit less surprising to me that for higher values of $n$ that difference can persist all the way up to $k=n$.

Posted by: Greg Egan on December 20, 2014 2:14 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I wrote:

So, some of the 9 vectors have dot products of $−1$ with 8 vertices of the 7-demicube, while others have dot products of $−1$ with only 4 vertices of the 7-demicube.

I meant to say “other vectors in the set of 9” rather than “vertices of the 7-demicube”. All these vectors have dot products of $-1$ with 35 vertices of the 7-demicube, but in relation to the triangle $E$, and this set of 9 vectors derived from $E$, that symmetry is broken.

Posted by: Greg Egan on December 20, 2014 2:42 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I’ve found a systematic construction for all 270 large $\mathrm{E}_8$ lattices, which simply repeats the pattern we used to construct the first 30 cases.

Given an orthonormal basis $\{e_0,\dots,e_7\}$, and a Fano plane structure $f$ consisting of triples of points taken from $\{1,\dots,7\}$, we can form integer-linear combinations of points of three kinds:

$\begin{array}{cl} 2 e_i & i \in \{0,\dots,7\} \\ \pm e_0 \pm e_i \pm e_j \pm e_k & \{i,j,k\} \in f \\ \pm e_p \pm e_q \pm e_r \pm e_s & \{p,q,r,s\} \in \tilde{f} \end{array}$

where $\tilde{f}$ is the set of 4-tuples that are complements of lines in the Fano plane.

We already did this for the case where $\{e_0,\dots,e_7\}$ is the standard basis of $\mathbb{R}^8$, giving us 30 lattices for the 30 Fano plane structures, but there is a (non-unique) set of 8 more bases we can use that will give us the remaining 240 lattices.

One basis consists of the rows of this matrix:

$\left( \begin{array}{cccccccc} \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} \end{array}\right)$

The other seven all look like this:

$\left( \begin{array}{cccccccc} \frac{3}{4} & -\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{3}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} \end{array} \right)$

To get the seven different matrices, we keep the first row/column as one that has a single minus sign, and then choose each of the other seven possibilities for the second row/column with that structure.

As we range over the 9 bases and 30 Fano plane structures, we get each of the 270 large $\mathrm{E}_8$ lattices.

So far I only know this because of computer calculations, but it shouldn’t be too hard to make rigorous. It really just comes down to showing that all these lattices are sublattices of the standard $\mathrm{E}_8$, and that we don’t get any double counting of the same lattice from different choices of basis and Fano plane structure.

Posted by: Greg Egan on December 16, 2014 12:28 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Here’s a nicer way to describe those eight bases I mentioned earlier: they are just the standard basis reflected in eight different vectors, namely the eight rows of this matrix:

$\left( \begin{array}{cccccccc} -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \end{array}\right)$

And these eight vectors are just the vertices of a 7-simplex in the standard $\mathrm{E}_8$ root polytope!

So, we can obtain all 270 large $\mathrm{E}_8$ lattices by starting with the 30 we get from the usual construction with the standard basis, and then reflecting each of those 30 lattices in each of the 8 vertices of this 7-simplex. Since reflection in a root is always an automorphism of the underlying lattice, that makes it manifest that the lattices we obtain this way are sublattices of the standard $\mathrm{E}_8$ lattice.

There are actually 64 other 7-simplexes (and their opposites) that would do the job just as well. At first I thought they might be the 64 7-simplexes that are incident on the 7-orthoplex whose centre is $e_0$, but that 7-orthoplex has purely integer coordinates and these 7-simplexes all have half-integer coordinates. I’ll need to ponder the geometry a bit longer to think of a nice way to describe them.

Posted by: Greg Egan on December 16, 2014 4:40 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Here’s a much easier proof of the fact that, given a large $\mathrm{E}_8$ lattice $L_1$, there are 64 other large $\mathrm{E}_8$ lattices whose root sets are disjoint from that of $L_1$.

We can construct 270 large $\mathrm{E}_8$ lattices by starting with 30 whose root vectors are:

$\begin{array}{cl} 2 e_i & i \in \{0,\dots,7\} \\ \pm e_0 \pm e_i \pm e_j \pm e_k & \{i,j,k\} \in f \\ \pm e_p \pm e_q \pm e_r \pm e_s & \{p,q,r,s\} \in \tilde{f} \end{array}$

where $f$ is one of 30 Fano plane structures consisting of triples of points taken from $\{1,\dots,7\}$, which define the lines in the plane, and $\tilde{f}$ is the set of 4-tuples that are complements of those lines. Note that all the coordinates of all the roots of all 30 lattices are integers.

We can then reflect any of these 30 lattices in any of these 8 vectors:

$\begin{array}{lccccccccr} (& -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2} & )\\ (& \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & \frac{1}{2} & )\\ (& \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & \frac{1}{2}, & -\frac{1}{2} & )\\ (& \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2} & )\\ (& \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2} & )\\ (& \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2} & )\\ (& \frac{1}{2}, & -\frac{1}{2}, & \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2} & )\\ (& \frac{1}{2}, & \frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2}, & -\frac{1}{2} & ) \end{array}$

These vectors are the vertices of a 7-simplex in the root polytope of the standard $\mathrm{E}_8$ lattice, so their Gram matrix of mutual dot products is:

$\left( \begin{array}{cccccccc} 2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 2 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 2 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 2 \end{array} \right)$

We will also be relying on the fact that every vector $v$ in this set of mirror vectors has coordinates drawn solely from $\{-\frac{1}{2},\frac{1}{2}\}$.

We will define $R(v)$ to be the reflection in the mirror vector $v$:

$R(v) = I - \left(\frac{2}{v\cdot v}\right) v\otimes v = I - v\otimes v$

where the last step follows because $v\cdot v=2$ for all our mirror vectors.

I claim that these reflections produce another 240 distinct lattices, for a total of 270. It’s clear that these reflections do produce lattices that are sublattices of $\mathrm{E}_8$, since the mirror vectors are all roots of $\mathrm{E}_8$. But how do we know that there is no double counting of the same lattice more than once among the 270?

Two lattices will be identical if and only if they have identical root sets. Clearly none of the original 30 lattices have exactly the same root sets as each other, since most of the roots depend on the choice of Fano plane structure. It follows that if we use the same mirror vector to reflect two of these original lattices, we must again have two different lattices.

It’s also clear that none of the reflected lattices can have the same root sets as any of the original 30 lattices, because the original 30 lattices all have solely integer coordinates, but every reflected lattice will contain, as the reflection of $2 e_0$ in a mirror vector $v$, a vector of the form:

$r = 2 R(v) e_0 = 2 (e_0 - v^0 v)$

with

$r^0 = 2 (1 - (v^0)^2)= 2 \left(1 - \frac{1}{4}\right) = \frac{3}{2}$

since every coordinate of every mirror vector $v$ is $\pm \frac{1}{2}$.

The final possibility we need to rule out is that two different lattices among the original 30, when reflected in two different mirror vectors, might end up the same. That is:

$R(v_1) L_1 = R(v_2) L_2$

Since reflections are their own inverse, this is equivalent to:

$\begin{array}{rcl} L_1 & =& R(v_1) R(v_2) L_2 \\ & = & (I - v_1\otimes v_1) (I - v_2\otimes v_2) L_2 \\ & = & (I - v_1\otimes v_1 - v_2\otimes v_2 + (v_1 \cdot v_2)(v_1\otimes v_2)) L_2 \\ & = & (I - v_1\otimes v_1 - v_2\otimes v_2 + (v_1\otimes v_2)) L_2 \end{array}$

The last step follows because $v_i \cdot v_j = 1$ for all $i\neq j$.

The first coordinate of the image of $2 e_0$ under this map will be:

$\begin{array}{rcl} 2(1 - (v_1^0)^2 - (v_2^0)^2 + v_1^0 v_2^0) & = & 2\left(1 -\frac{1}{4} - \frac{1}{4} \pm \frac{1}{4}\right)\\ & = & 1 \pm \frac{1}{2} \end{array}$

So the image of $2 e_0$ can’t belong to $L_1$ after all.

Now, suppose we start with two lattices among the original 30, say $L_1$ and $L_2$, based on Fano plane structures $f_1$ and $f_2$ that have no lines in common. For a given choice of $f_1$, there will be 8 choices for $f_2$ that satisfy that condition.

We then reflect $L_2$ in any mirror vector $v$ and call the reflected lattice $L_3$. Is it possible that $L_3$ has any roots in common with $L_1$? If that were the case, there would be a root $r_1$ of $L_1$ that was a reflection in $v$ of some root $r_2$ in $L_2$. When two vectors are each other’s reflection, either their difference is a non-zero multiple of the mirror vector $v$, or they are equal and they are orthogonal to the mirror vector.

Let’s address the case of equal vectors first. It’s not hard to see that when $f_1$ and $f_2$ have no lines in common, the only roots shared by $L_1$ and $L_2$ will be vectors of the form $2 e_i$. But such vectors are not orthogonal to any of our mirror vectors.

The mirror vectors all have eight non-zero coordinates. So the remaining question is, can a difference between a root $r_1$ of $L_1$ and a root $r_2$ of $L_2$ have eight non-zero coordinates? Recall that the possible forms for the roots are:

$\begin{array}{cl} 2 e_i & i \in \{0,\dots,7\} \\ \pm e_0 \pm e_i \pm e_j \pm e_k & \{i,j,k\} \in f \\ \pm e_p \pm e_q \pm e_r \pm e_s & \{p,q,r,s\} \in \tilde{f} \end{array}$

• If both are of the form $2 e_i$, the difference will have at most 2 non-zero coordinates.

• If one is of the form $2 e_i$ and the other is of either form with 4 non-zero coordinates, the difference will have at most 5 non-zero coordinates.

• If both are based on lines in their respective Fano plane structures, the two sets of 4 non-zero coordinates in each will overlap, with both including $e_0$, so the difference will have at most 7 non-zero coordinates.

• If both are based on complements of lines in their respective Fano plane structures, neither set of 4 non-zero coordinates will include $e_0$.

• Finally, suppose one root is based on a line and the other is based on the complement of a line. Since we have chosen the Fano plane structures $f_1$ and $f_2$ to have no lines in common, the union of the non-zero coordinates must fall short of covering all seven points of the plane. This makes it impossible for the difference of the roots to have eight non-zero coordinates.

So, for a given $f_1$ we have 8 choices for $f_2$ and 8 choices for the mirror vector $v$, yielding a total of 64 reflected lattices $L_3$ with no roots in common with $L_1$.

Posted by: Greg Egan on December 16, 2014 11:04 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

I think I’ve worked out a lot of the relationship between large $E_8$ and geometry over $\mathbb{F}_2$, which makes some of the stuff above clearer.

First, let me recapitulate some of the underlying finite geometry, particularly over $\mathbb{F}_2$, and set up some notation.

Projective and polar spaces

I’ll work with projective spaces over $\mathbb{F}_q$ and try not to suddenly start jumping back and forth between projective spaces and the underlying vector spaces as is my wont, at least not unless it really makes things clearer.

So we have an $n$-dimensional projective space over $\mathbb{F}_q$. We’ll denote this by $PG(n,q)$.

The full symmetry group of $PG(n,q)$ is $GL_{n+1}(q)$, and from that we get subgroups and quotients $SL_{n+1}(q)$ (with unit determinant), $PGL_{n+1}(q)$ (quotient by the centre) and $PSL_{n+1}(q)$ (both). Over $\mathbb{F}_2$, the determinant is always $1$ (since that’s the only non-zero scalar) and the centre is trivial, so these groups are all the same.

In projective spaces over $\mathbb{F}_2$, there are $3$ points on every line, so we can “add” two any points and get the third point on the line through them. (This is just a projection of the underlying vector space addition.)

In odd characteristic, we get two other families of Lie type by preserving two types of non-degenerate bilinear form: symmetric and skew-symmetric, corresponding to orthogonal and symplectic structures respectively. (Non-degenerate Hermitian forms, defined over $\mathbb{F}_{q^2}$, also exist and behave similarly.)

Denote the form by $B(x,y)$. Points $x$ for which $B(x, x)=0$ are isotropic. For a symplectic structure all points are isotropic. A form $B$ such that $B(x,x)=0\forall x$ is called alternating, and in odd characteristic, but not characteristic $2$, skew-symmetric and alternating forms are the same thing.

A line spanned by two isotropic points, $x$ and $y$, such that $B(x,y)=1$ is a hyperbolic line. Any space with a non-degenerate bilinear (or Hermitian) form can be decomposed as the orthogonal sum of hyperbolic lines (i.e. as a vector space, decomposed as an orthogonal sum of hyperbolic planes), possibly together with an anisotropic space containing no isotropic points at all. There are no non-empty symplectic anisotropic spaces, so all symplectic spaces are odd-dimensional (projectively—the corresponding vector spaces are even-dimensional).

There are anisotropic orthogonal points and lines (over any finite field including in even characteristic), but all the orthogonal spaces we consider here will be a sum of hyperbolic lines—we say they are of plus type. (The odd-dimensional projective spaces with a residual anisotropic line are of minus type.)

A quadratic form $Q(x)$ is defined by the conditions

i) $Q(x+y)=Q(x)+Q(y)+B(x,y)$, where $B$ is a symmetric bilinear form.

ii) $Q(\lambda x)=\lambda^2Q(x)$ for any scalar $\lambda$.

There are some non-degeneracy conditions I won’t go into.

Obviously, a quadratic form implies a particular symmetric bilinear form, by $B(x,y)=Q(x+y)-Q(x)-Q(y)$. In odd characteristic, we can go the other way: $Q(x)=\frac{1}{2}B(x,x)$.

We denote the group preserving an orthogonal structure of plus type on an $n$-dimensional projective space over $\mathbb{F}_q$ by $GO_{n+1}^+(q)$, by analogy with $GL_{n+1}(q)$. Similarly we have $SO_{n+1}^+(q)$, $PGO_{n+1}^+(q)$ and $PSO_{n+1}^+(q)$. However, whereas $PSL_n(q)$ is simple apart from $2$ exceptions, we usually have an index $2$ subgroup of $SO_{n+1}^+(q)$, called $\Omega_{n+1}^+(q)$, and a corresponding index $2$ subgroup of $PSO_{n+1}^+(q)$, called $P\Omega_{n+1}^+(q)$, and it is the latter that is simple. (There is an infinite family of exceptions, where $PSO_{n+1}^+(q)$ is simple.)

Symplectic structures are easier—the determinant is automatically $1$, so we just have $Sp_{n+1}(q)$ and $PSp_{n+1}(q)$, with the latter being simple except for $3$ exceptions.

Just as a point with $B(x,x)=0$ is an isotropic point, so any subspace with $B$ identically $0$ on it is an isotropic subspace.

And just as the linear groups act on incidence geometries given by the (“classical”) projective spaces, so the symplectic and orthogonal act on polar spaces, whose points, lines, planes, etc, are just the isotropic points, isotropic lines, isotropic planes, etc given by the bilinear (or Hermitian) form. We denote an orthogonal polar space of plus type on an $n$-dimensional projective space over $\mathbb{F}_q$ by $Q_n^+(q)$.

In characteristic $2$, a lot of this goes wrong, but in a way that can be fixed and mostly turns out the same.

1) Symmetric and skew-symmetric form are the same thing! There are still distinct orthogonal and symplectic structures and groups, but we can’t use this as the distinction.

2) Alternating and skew-symmetric forms are not the same thing! Alternating forms are all skew-symmetric (aka symmetric) but not vice versa. A symplectic structure is given by an alternating form—and of course this definition works in odd characteristic too.

3) Symmetric bilinear forms are no longer in bijection with quadratic forms: every quadratic form gives a unique symmetric (aka skew-symmetric, and indeed alternating) bilinear form, but an alternating form is compatible with multiple quadratic forms. We use non-degenerate quadratic forms to define orthogonal structures, rather than symmetric bilinear forms—which of course works in odd characteristic too. (Note also from the above that in characteristic $2$ an orthogonal structure has an associated symplectic structure, which it shares with other orthogonal structures.)

We now have both isotropic subspaces on which the bilinear form is identically $0$, and singular subspaces on which the quadratic form is identically $0$, with the latter being a subset of the former. It is the singular spaces which go to make up the polar space for the orthogonal structure.

To cover both cases, we’ll refer to these isotropic/singular projective spaces inside the polar spaces as flats.

Everything else is still the same—decomposition into hyperbolic lines and an anisotropic space, plus and minus types, $\Omega_{n+1}^+(q)$ inside $SO_{n+1}^+(q)$, polar spaces, etc.

Over $\mathbb{F}_2$, we have that $GO_{n+1}^+(q)$, $SO_{n+1}^+(q)$, $PGO_{n+1}^+(q)$ and $PSO_{n+1}^+(q)$ are all the same group, as are $\Omega_{n+1}^+(q)$ and $P\Omega_{n+1}^+(q)$.

The vector space dimension of the maximal flats in a polar space is the polar rank of the space, one of its most important invariants—it’s the number of hyperbolic lines in its orthogonal decomposition.

$Q_{2m-1}^+(q)$ has rank $m$. The maximal flats fall into two classes. In odd characteristic, the classes are preserved by $SO_{2m}^+(q)$ but interchanged by the elements of $GO_{2m}^+(q)$ with determinant $-1$. In even characteristic, the classes are preserved by $\Omega_{2m}^+(q)$, but interchanged by elements of $GO_{2m}^+(q)$.

Finally, I’ll refer to the value of the quadratic form at a point, $Q(x)$, as the norm of $x$, even though in Euclidean space we’d call it “half the norm-squared”.

Here are some useful facts about $Q_{2m-1}^+(q)$:

1a. The number of points is $\displaystyle\frac{\left(q^m-1\right)\left(q^{m-1}+1\right)}{q-1}$.

1b. The number of maximal flats is $\prod_0^{m-1}\left(1+q^m\right)$.

1c. Two maximal flats of different types must intersect in a flat of odd codimension; two maximal flats of the same type must intersect in a flat of even codimension.

Here two more general facts.

1d. Pick a projective space $\Pi$ of dimension $n$. Pick a point $p$ in it. The space whose points are lines through $p$, whose lines are planes through $p$, etc, with incidence inherited from $\Pi$, form a projective space of dimension $n-1$.

1e. Pick a polar space $\Sigma$ of rank $m$. Pick a point $p$ in it. The space whose points are lines (i.e. $1$-flats) through $p$, whose lines are planes (i.e. $2$-flats) through $p$, etc, with incidence inherited from $\Sigma$, form a polar space of the same type, of rank $m-1$.

The Klein correspondence at breakneck speed

The bivectors of a $4$-dimensional vector space constitute a $6$-dimensional vector space. Apart from the zero bivector, these fall into two types: degenerate ones which can be decomposed as the wedge product of two vectors and therefore correspond to planes (or, projectively, lines); and non-degenerate ones, which, by, wedging with vectors on each side give rise to symplectic forms. Wedging two bivectors gives an element of the $1$-dimensional space of $4$-vectors, and, picking a basis, the single component of this wedge product gives a non-degenerate symmetric bilinear form on the $6$-dimensional vector space of bivectors, and hence, in odd characteristic, an orthogonal space, which turns out to be of plus type. It also turns out that this can be carried over to characteristic $2$ as well, and gives a correspondence between $PG(3,q)$ and $Q_5^+(q)$, and isomorphisms between their symmetry groups. It is precisely the degenerate bivectors that are the ones of norm $0$, and we get the following correspondence:

$\array{\arrayopts{\collayout{left}\collines{dashed}\rowlines{solid dashed}\frame{solid}} \mathbf{Q_5^+(q)}&\mathbf{PG(3,q)}\\ \text{point}&\text{line}\\ \text{orthogonal points}&\text{intersecting lines}\\ \text{line}&\text{plane pencil}\\ \text{plane}_1&\text{point}\\ \text{plane}_2&\text{plane} }$

Here, “plane pencil” is all the lines that both go through a particular point and lie in a particular plane: effectively a point on a plane. The two type of plane in $Q_5^+(q)$ are two families of maximal flats, and they correspond, in $PG(3,q)$, to “all the lines through a particular point” and “all the lines in a particular plane”.

From fact 1c above, in $Q_5^+(q)$ we have that two maximal flats of of different type must either intersect in a line or not intersect at all, corresponding to the fact in $PG(3,q)$ that a point and a plane either coincide or don’t; while two maximal flats of the same type must intersect in a point, corresponding to the fact in $PG(3,q)$ that any two points lie in a line, and any two planes intersect in a line.

In $Q_7^+(q)$, you may observe from facts 1a and 1b that the following three things are equal in number: points; maximal flats of one type; maximal flats of the other type. This is because these three things are cycled by the triality symmetry.

Counting things over $\mathbb{F}_2$

Over $\mathbb{F}_2$, we have the following things:

2a. $PG(3,2)$ has $15$ planes, each containing $7$ points and $7$ lines. It has (dually) $15$ points, each contained in $7$ lines and $7$ planes. It has $35$ lines, each containing $3$ points and contained in $3$ planes.

2b. $Q_5^+(2)$ has $35$ points, corresponding to the $35$ lines of $PG(3,2)$, and $30$ planes, corresponding to the $15$ points and $15$ planes of $PG(3, 2)$. There’s lots and lots of other interesting stuff, but we will ignore it.

2c. $Q_7^+(2)$ has $135$ points and $270$ $3$-spaces, i.e. two families of maximal flats containing $135$ elements each. A projective $7$-space has $255$ points, so if we give it an orthogonal structure of plus type, it will have $255-135=120$ points of norm $1$.

$E_8$ mod $2$

Now we move onto the second part.

We’ll coordinatise the $E_8$ lattice so that the coordinates of its points are of the following types:

a) All integer, summing to an even number
b) All integer+$\frac{1}{2}$, summing to an odd number.

Then the roots are of the following types

a) All permutations of $\left(\pm1,\pm1,0,0,0,0,0,0\right)$
b) All points like $\left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)$ with an odd number of minus signs.

We now quotient $E_8$ by $2E_8$. The elements of the quotient can by represented by the following:

a) All coordinates are $1$ or $0$, an even number of each.
b) All coordinates are $\pm\frac{1}{2}$ with either $1$ or $3$ minus signs.
c) Take an element of type b and put a star after it. The meaning of this is: you can replace any coordinate $\frac{1}{2}$ and replace it with $-\frac{3}{2}$, or any coordinate $-\frac{1}{2}$ and replace it with $\frac{3}{2}$, to get an $E_8$ lattice element representing this element of $E_8/2E_8$.

This is an $8$-dimensional vector space over $\mathbb{F}_2$.

Now we put the following quadratic form on this space: $Q(x)$ is half the Euclidean norm-squared, mod $2$. This gives rise to the following bilinear form: the Euclidean dot product mod $2$. This turns out to be a perfectly good non-degenerate quadratic form of plus type over $\mathbb{F}_2$.

There are $120$ elements of norm $1$, and these correspond to roots of $E_8$, with $2$ roots per element (related by switching the sign of all coordinates).

a) Elements of shape $\left(1,1,0,0,0,0,0,0\right)$ are already roots in this form.

b) Elements of shape $\left(0,0,1,1,1,1,1,1\right)$ correspond to the roots obtained by taking the complement (replacing all $1$s by $0$ and vice versa) and then changing the sign of one of the $1$s.

c) Elements in which all coordinates are $\pm\frac{1}{2}$ with either $1$ or $3$ minus signs are already roots, and by switching all the signs we get the half-integer roots with $5$ or $7$ minus signs.

There are $135$ non-zero elements of norm $0$, and these all correspond to lattice points in shell $2$, with $16$ lattice points per element of the vector space.

a) There are $70$ elements of shape $\left(1,1,1,1,0,0,0,0\right)$. We get $8$ lattice points by changing an even number of signs (including $0$). We get another $8$ lattice points by taking the complement and then changing an odd number of signs.

b) There is $1$ element of shape $\left(1,1,1,1,1,1,1,1\right)$. This corresponds to the $16$ lattice points of shape $\left(\pm2,0,0,0,0,0,0,0\right)$.

c) There are $64$ elements like $\left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)^*$, with $1$ or $3$ minus signs. We get $8$ actual lattice points by replacing $\pm\frac{1}{2}$ by $\mp\frac{3}{2}$ in one coordinate, and another $8$ by changing the signs of all coordinates.

This accounts for all $16\cdot135=2160$ points in shell $2$.

Isotropic

$\array{\arrayopts{\collayout{left}\rowlines{solid}\collines{solid}\frame{solid}} \mathbf{shape}&\mathbf{number}\\ \left(1,1,1,1,1,1,1,1\right)&1\\ \left(1,1,1,1,0,0,0,0\right)&70\\ \left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)^*&64\\ \mathbf{total}&\mathbf{135} }$

Anisotropic

$\array{\arrayopts{\collayout{left}\rowlines{solid}\collines{solid}\frame{solid}} \mathbf{shape}&\mathbf{number}\\ \left(1,1,1,1,1,1,0,0\right)&28\\ \left(1,1,0,0,0,0,0,0\right)&28\\ \left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)&64\\ \mathbf{total}&\mathbf{120} }$

Since the quadratic form in $\mathbb{F}_2$ comes from the quadratic form in Euclidean space, it is preserved by the Weyl group $W(E_8)$. In fact the homomorphism $W(E_8)\rightarrow GO_8^+(2)$ is onto, although (contrary to what I said in an earlier comment) it is a double cover—the element of $W(E_8)$ that reverses the sign of all coordinates is a (in fact, the) non-trivial element element of the kernel.

Large $E_8$ lattices

Pick a Fano plane structure on a set of seven points.

Here is a large $E_8$ containing $\left(2,0,0,0,0,0,0,0\right)$:

(where $1\le i,j,k,p,q,r,s\le7$)

$\pm2e_i$

$\pm e_0\pm e_i\pm e_j\pm e_k$ where $i$, $j$, $k$ lie on a line in the Fano plane

$\pm e_p\pm e_q\pm e_r\pm e_s$ where $p$, $q$, $r$ , $s$ lie off a line in the Fano plane.

Reduced to $E_8$ mod $2$, these come to

i) $\left(1,1,1,1,1,1,1,1\right)$

ii) $e_0+e_i+e_j+e_k$ where $i$, $j$, $k$ lie on a line in the Fano plane. E.g. $\left(1,1,1,0,1,0,0,0\right)$.

iii) $e_p+e_q+e_r+e_s$ where $p$, $q$, $r$, $s$ lie off a line in the Fano plane. E.g. $\left(0,0,0,1,0,1,1,1\right)$.

Each of these corresponds to $16$ elements of the large $E_8$ roots.

Some notes on these points:

1) They’re all isotropic, since they have a multiple of $4$ non-zero entries.

2) They’re mutually orthogonal.

a) Elements of types ii and iii are all orthogonal to $\left(1,1,1,1,1,1,1,1\right)$ because they have an even number of ones (like all all-integer elements).

b) Two elements of type ii overlap in two places: $e_0$ and the point of the Fano plane that they share.

c) If an element $x$ of type ii and an element $y$ of type iii are mutual complements, obviously they have no overlap. Otherwise, the complement of $y$ is an element of type ii, so $x$ overlaps with it in exactly two places; hence $x$ overlaps with $y$ itself in the other two non-zero places of $x$.

d) From $c$, given two elements of type iii, one will overlap with the complement of the other in two places, hence (by the argument of c) will overlap with the other element itself in two places.

3) Adjoining the zero vector, they give a set closed under addition.

The rule for addition of all-integer elements is reasonably straightforward: if they are orthogonal, then treat the $1$s and $0$s as bits and add mod $2$. If they aren’t orthogonal, then do the same, then take the complement of the answer.

a) Adding $\left(1,1,1,1,1,1,1,1\right)$ to any of the others just gives the complement, which is a member of the set.

b) Adding two elements of type ii, we set to $0$ the $e_0$ component and the component corresponding to the point of intersection in the Fano plane, leaving the $4$ components where they don’t overlap, which are just the complement of the third line of the Fano plane through their point of intersection, and is hence a member of the set.

c) Each element of type iii is the sum of the element of type i and an element of type ii, hence is covered implicitly by cases a and b.

4) There are $15$ elements of the set.

a) There is $\left(1,1,1,1,1,1,1,1\right)$.

b) There are $7$ corresponding to lines of the Fano plane.

c) There are $7$ corresponding to the complements of lines of the Fano plane.

From the above, these $15$ elements form a maximal flat of $Q_7^+(2)$. (That is, $15$ points projectively, forming a projective $3$-space in a projective $7$-space.)

That a large $E_8$ lattice projects to a flat is straightforward:

First, as a lattice it’s closed under addition over $\mathbb{Z}$, so should project to a subspace over $\mathbb{F}_2$.

Second, since the cosine of the angle between two roots of $E_8$ is always a multiple of $\frac{1}{2}$, and the points in the second shell have Euclidean length $2$, the dot product of two large $E_8$ roots must always be an even integer. Also, the large $E_8$ roots project to norm $0$ points. So all points of the large $E_8$ should project to norm $0$ points.

It’s not instantly obvious to me that large $E_8$ should project to a maximal flat, but it clearly does.

So I’ll assume each $E_8$ corresponds to a maximal flat, and generally that everything that I’m going to talk about over $\mathbb{F}_2$ lifts faithfully to Euclidean space, which seems plausible (and works)! But I haven’t proved it. Anyway, assuming this, a bunch of stuff follows.

Total number of large $E_8$ lattices

We immediately know there are $270$ large $E_8$ lattices, because there are $270$ maximal flats in $Q_7^+(2)$, either from the formula $\prod_{i=0}^{m-1}\left(1+q^i\right)$, or immediately from triality and the fact that there are $135$ points in $Q_7^+(2)$.

Number of large $E_8$ root systems sharing a given point

We can now bring to bear some more general theory. How many large $E_8$ root-sets share a point? Let us project this down and instead ask, How many maximal flats share a given point?

Recall fact 1e:

1e. Pick a polar space $\Sigma$ of rank $m$. Pick a point $p$ in it. The space whose points are lines (i.e. $1$-flats) through $p$, whose lines are planes (i.e. $2$-flats) through $p$, etc, with incidence inherited from $\Sigma$, form a polar space of the same type, of rank $m-1$.

So pick a point $p$ in $Q_7^+(2)$. The space of all flats containing $p$ is isomorphic to $Q_5^+(2)$. The maximal flats containing $p$ in $Q_7^+(2)$ correspond to all maximal flats of $Q_5^+(2)$, of which there are $30$. So there are $30$ maximal flats of $Q_7^+(2)$ containing $p$, and hence $30$ large $E_8$ lattices containing a given point.

We see this if we fix $\left(1,1,1,1,1,1,1,1\right)$, and the maximal flats correspond to the $30$ ways of putting a Fano plane structure on $7$ points. Via the Klein correspondence, I guess this is a way to show that the $30$ Fano plane structures correspond to the points and planes of $PG(3,2)$.

Number of large $E_8$ root system disjoint from a given large $E_8$ root system

Now assume that large $E_8$ lattices with non-intersecting sets of roots correspond to non-intersecting maximal flats. The intersections of maximal flats obey rule 1c:

1c. Two maximal flats of different types must intersect in a flat of odd codimension; two maximal flats of the same type must intersect in a flat of even codimension.

So two $3$-flats of opposite type must intersect in a plane or a point; if they are of the same type, they must intersect in a line or not at all (the empty set having dimension $-1$).

We want to count the dimension $-1$ intersections, but it’s easier to count the dimension $1$ intersections and subtract from the total.

So, given a $3$-flat, how many other $3$-flats intersect it in a line?

Pick a point $p$ in $Q_7^+(2)$. The $3$-flats sharing that point correspond to the planes of $Q_5^+(2)$. Then the set of $3$-flats sharing just a line through $p$ with our given $3$-flat correspond to the set of planes of $Q_5^+(2)$ sharing a single point with a given plane. By what was said above, this is all the other planes of the same type (there’s no other dimension these intersections can have). There are $14$ of these ($15$ planes minus the given one).

So, given a point in the $3$-flat, there are $14$ other $3$-flats sharing a line (and no more) which passes through the point. There are $15$ points in the $3$-flat, but on the other hand there are $3$ points in a line, giving $\frac{14\cdot15}{3}=70$ $3$-spaces sharing a line (and no more) with a given $3$-flat.

But there are a total of $135$ $3$-flats of a given type. If $1$ of them is a given $3$-flat, and $70$ of them intersect that $3$-flat in a line, then $135-1-70=64$ don’t intersect the $3$-flat at all. So there should be $64$ large $E_8$ lattices whose roots don’t meet the roots of a given large $E_8$ lattice.

Other numbers of intersecting roots systems

We can also look at the intersections of large $E_8$ root systems with large $E_8$ root systems of opposite type. What about the intersections of two $3$-flats in a plane? If we focus just on planes passing through a particular point, this corresponds, in $Q_5^+(2)$, to planes intersecting in a line. There are $7$ planes intersecting a given plane in a line (from the Klein correspondence—they correspond to the seven points in a plane or the seven planes containing a point of $PG(3,2)$). So there are $7$ $3$-flats of $Q_7^+(2)$ which intersect a given $3$-flat in a plane containing a given point. There $15$ points to choose from, but $7$ points in a plane, meaning that there are $\frac{7\cdot15}{7}=15$ $3$-flats intersecting a given $3$-flat in a plane. A plane has $7$ points, so translating that to $E_8$ lattices should give $7\cdot16=112$ shared roots.

That leaves $135-15=120$ $3$-flats intersecting a given $3$-flat in a single point, corresponding to $16$ shared roots.

$\array{\arrayopts{\collayout{left}\collines{solid}\rowlines{solid}\frame{solid}} \mathbf{\text{intersection dim.}}&\mathbf{\text{number}}&\mathbf{\text{same type}}\\ 2&15&No\\ 1&70&Yes\\ 0&120&No\\ -1&64&Yes }$

A couple of points here related to triality. Under triality, one type of maximal flat gets sent to the other type, and the other type gets sent to singular points ($0$-flats). The incidence relation of “intersecting in a plane” gets sent to ordinary incidence of a point with a flat. So the fact that there are $15$ maximal flats that intersect a given maximal flat in a plane is a reflection of the fact that there are $15$ points in a maximal flat (or, dually, $15$ maximal flats of a given type containing a given point).

The intersection of two maximal flats of the same type translates into a relation between two singular points. Just from the numbers, we’d expect “intersection in a line” to translate into “orthogonal to”, and “disjoint” to translate into “not orthogonal to”.

In that case, a pair of maximal flats intersecting in a (flat) line translates to $2$ mutually orthogonal flat points—whose span is a flat line. Which makes sense, because under triality, $1$-flats transform to $1$-flats, reflecting the fact that the central point of the $D_4$ diagram (representing lines) is sent to itself under triality.

In that case, two disjoint maximal flats translates to a pair of non-orthogonal singular points, defining a hyperbolic line.

Fixing a hyperbolic line (pointwise) obviously reduces the rank of the polar space by $1$, picking out a $GO_6^+(2)$ subgroup of $GO_8^+(2)$. By the Klein correspondence, $GO_6^+(2)$ is isomorphic to $GL_4(2)$, which is just the automorphism group of $PG(3, 2)$—i.e., here, the automorphism group of a maximal flat. So the joint stabiliser of two disjoint maximal flats is just automorphisms of one of them, which forces corresponding automorphisms of the other. This group is also isomorphic to the symmetric group $S_8$, giving all permutations of the coordinates (of the $E_8$ lattice).

(My guess would be that the actions of $GL_4(2)$ on the two maximal flats would be related by an outer automorphsm of $GL_4(2)$, in which the action on the points of one flat would match an action on the planes of the other, and vice versa, preserving the orthogonality relations coming from the symplectic structure implied by the orthogonal structure—i.e. the alternating form implied by the quadratic form.)

Nearest neighbours

We see this “non-orthogonal singular points” $\leftrightarrow$ “disjoint maximal flats” echoed when we look at nearest neighbours.

Nearest neighbours in the second shell of the $E_8$ lattice are separated from each other by an angle of $cos^{-1}\frac{3}{4}$, so have a mutual dot product of $3$, hence are non-orthogonal over $\mathbb{F}_2$.

Let us choose a fixed point $\left(2,0,0,0,0,0,0,0\right)$ in the second shell of $E_8$. This has as our chosen representative $\left(1,1,1,1,1,1,1,1\right)$ in our version of $PG(7,2)$, which has the convenient property that it is orthogonal to the all-integer points, and non-orthogonal to the half-integer points. The half-integer points in the second shell are just those that we write as $\left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)^\star$ in our notation, where the $*$ means that we should replace any $\frac{1}{2}$ by $-\frac{3}{2}$ or replace any $-\frac{1}{2}$ by $\frac{3}{2}$ to get a corresponding element in the second shell of the $E_8$ latttice, and where we require $1$ or $3$ minus signs in the notation, to correspond two points in the lattice with opposite signs in all coordinates.

Now, since each reduced isotropic point represents $16$ points of the second shell, merely saying that two reduced points have dot product of $1$ is not enough to pin down actual nearest neighbours.

But very conveniently, the sets of $16$ are formed in parallel ways for the particular setup we have chosen. Namely, lifting $\left(1,1,1,1,1,1,1,1\right)$ to a second-shell element, we can choose to put the $\pm2$ in each of the $8$ coordinates, with positive or negative sign, and lifting an element of the form $\left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)^*$ to a second-shell element, we can choose to put the $\pm\frac{3}{2}$ in each of the $8$ coordinates, with positive or negative sign.

So we can line up our conventions, and choose, e.g., specifically $\left(+2,0,0,0,0,0,0,0\right)$, and choose neighbours of the form $\left(+\frac{3}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)$, with an even number of minus signs.

This tells us we have $64$ nearest neighbours, corresponding to the $64$ isotropic points of half-integer form. Let us call this set of points $T$.

Now pick one of those $64$ isotropic points, call it $p$. It lies, as we showed earlier, in $30$ maximal flats, corresponding to the $30$ plane flats of $Q_5^+(2)$, and we would like to understand the intersections of these flats with $T$: that is, those nearest neighbours which belong to each large $E_8$ lattice.

In any maximal flat, i.e. any $3$-flat, containing $p$, there will be $7$ lines passing through $p$, each with $2$ other points on it, totalling $14$ which, together with $p$ itself form the $15$ points of a copy of $PG(3,2)$.

Now, the sum of two all-integer points is an all-integer point, but the sum of two half-integer points is also an all-integer point. So of the two other points on each of those lines, one will be half-integer and one all-integer. So there will be $7$ half-integer points in addition to $p$ itself; i.e. the maximal flat will meet $T$ in $8$ points; hence the corresponding large $E_8$ lattice will contain $8$ of the nearest neighbours of $\left(2,0,0,0,0,0,0,0\right)$.

Also, because the sum of two half-integer points is not a half-integer point, no $3$ of those $8$ points will lie on a line.

But the only way that you can get $8$ points in a $3$-space such that no $3$ of them lie on a line of the space is if they are the $8$ points that do not lie on a plane of the space. Hence the other $7$ points—the ones lying in the all-integer subspace—must form a Fano plane.

So we have the following: inside the projective $7$-space of lattice elements mod $2$, we have the projective $6$-space of all-integer elements, and inside there we have the $5$-space of all-integer elements orthogonal to $p$, and inside there we have a polar space isomorphic to $Q_5^+(2)$, and in there we have $30$ planes. And adding $p$ to each element of one of those planes gives the $7$ elements which accompany $p$ in the intersection of the isotropic half-integer points with the corresponding $3$-flat, which lift to the nearest neighbours of $\left(2,0,0,0,0,0,0,0\right)$ lying in the corresponding large $E_8$ lattice.

Posted by: Tim Silverman on January 27, 2016 10:56 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Oops! This:

$GO_6^+(2)$ is isomorphic to $GL_4(2)$

is wrong. $GL_4(2)$ is the same as $PSL_4(2)$. $GO_6^+(2)$ includes also the outer automorphism that swaps points and planes of $PG(3,2)$.

So the subgroup that preserves each of two mutually disjoint maximal flats should be $\Omega_6^+(2)$—that is, $A_8$. Maybe $S_8$ includes swapping the two flats.

Posted by: Tim Silverman on January 28, 2016 7:06 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

We should apply this approach to the Leech lattice construction.

First, an observation which will make calculations easier. If we have two disjoint maximal flats, with bilinear form $B$, then (if I am not mistaken) we can always find a symplectic basis

$\mathbf{e}_i$, $\mathbf{f}_i$, $1\le i\le4$

with

$B\left(\mathbf{e}_i,\mathbf{e}_j\right) = B\left(\mathbf{f}_i,\mathbf{f}_j\right) = 0$
$B\left(\mathbf{e}_i,\mathbf{f}_j\right) = B\left(\mathbf{f}_i,\mathbf{e}_j\right) = \delta_{i j}$

such that $\left\{\mathbf{e}_i\right\}$ is a basis of one flat and $\left\{\mathbf{f}_i\right\}$ is a basis of the other.

Basics

Now, the Leech construction that Egan derived from Wilson says that if we have two large $E_8$ lattices, $L_1$ and $L_2$, with disjoint root systems, in a (small) $E_8$ lattice $L_0$, then we get a Leech lattice from triples $\left(a,b,c\right)$ such that

$\array{ a,b,c\in L_0\\ a+b, a+c, b+c\in L_1\\ a+b+c\in L_2 }$

Reducing mod $2$, we will start working with vector spaces rather than projective spaces, because we will need to consider the $0$ vector. So the above construction, mod $2$, says that if we have two maximal flat vector subspaces, $M_1$ and $M_2$, inside ${\mathbb{F}_2}^8$ equipped with plus-type quadratic form $Q$ and corresponding bilinear form $B$, such that $M_1\cap M_2=\left\{0\right\}$, then some kind of reduced version of the Leech lattice consists of triples $\left(a,b,c\right)$ such that

$\array{ a,b,c\in {\mathbb{F}_2}^8\\ a+b, a+c, b+c\in M_1\\ a+b+c\in M_2 }$

Now, let’s draw out what this means, bearing in mind that we are working over $\mathbb{F}_2$.

1) $\left(a+b\right)+\left(b+c\right)=\left(a+c\right)$

Hence either the three points in $M_1$ lie on a projective line, or one of them is $0$ and the other two are equal to each other.

Let’s define

$\array{\arrayopts{\collayout{left}} x=a+b+c\\ p=b+c\\ q=a+c\\ r=a+b }$

Then

$\array{ a=x+p\\ b=x+q\\ c=x+r }$

and

$p+q+r=0$

Now, we have

$Q(x)=Q(p)=Q(q)=Q(r)=0$

$B(p,q)=B(p,r)=B(q,r)=0$

(because they lie in flats of the quadratic form).

Hence $B(a,b)=B(x+p,x+q)=B(x,x)+B(x,p+q)+B(p,q)=B(x,r)$

and similarly for the other pairs.

Also, $Q(a)=Q(x+p)=Q(x)+Q(p)+B(x,p)=B(x,p)$

and similarly for the other pairs.

Also, since $r=p+q$, also $B(x,r)=B(x,p)+B(x,q)$

and similarly for other pairs, implying that if two of these inner products have the same value, then the third one is $0$.

So of the three points $a$, $b$ and $c$, either all three have norm $0$ or else one of them has norm $0$ and the other two have norm $1$.

Euclidean norms

Now let’s have a go at constructing the first shell of the Leech lattice, that is the points with minimal positive norm.

First, we’ll think about the possible values of the Euclidean norm-squared for the lifts of points in ${\mathbb{F}_2}^8$. Since we’re only trying to construct the first shell of the Leech lattice, we’ll only consider the first few shells in $E_8$.

1) The $0$ vector lifts to a point in $2E_8$. The minimal-norm points in here are the Euclidean $0$ vector, of norm-squared $0$, and double a point in the first shell, e.g. $\left(2,2,0,0,0,0,0,0\right)$, of norm-squared $8$.

2) Vectors of norm $1$ lift to points in the first shell of $E_8$, of norm-squared $2$, or the third shell, of norm-squared $6$.

3) Non-zero vectors of norm $0$ lift to points in the second shell of $E_8$, of norm-squared $4$, or points in the fourth shell, of norm-squared $8$.

To summarise:

$\array{\arrayopts{\collayout{left}\collines{solid}\rowlines{solid}\frame{solid}} \mathbf{\text{vect. type}}&\mathbb{F}_2\text{ norm}&\mathbf{\text{Eucl. norm-sq 1}}&\mathbf{\text{Eucl. norm-sq 2}}\\ \text{zero}&0&0&8\\ \text{norm 1}&1&2&6\\ \text{non-zero norm 2}&0&4&8 }$

It will turn out that we won’t need all of these.

Points in the maximal flats

Let’s pick a symplectic basis as described above, with $\left\{\mathbf{e}_i\right\}$ spanning $M_1$, and $\left\{\mathbf{f}_i\right\}$ spanning $M_2$, and write the coordinates like $c_1c_2c_3c_4\vert d_1d_2d_3d_4$ where the $c_i$ correspond to $\mathbf{e}_i$ and the $d_i$ correspond to $\mathbf{f}_i$.

That is, $M_1$ consists of vectors like $c_1c_2c_3c_4\vert0000$ and $M_2$ consists of vectors like $0000\vert d_1d_2d_3d_4$.

Now, pick a non-zero vector $x\in M_2$. For the sake of specificity and convenience, suppose this point is $0000\vert1000$.

Now in $M_1$, there are, apart from the $0$ vector, $7$ vectors orthogonal to this one, viz. those of the form $0c_2c_3c_4\vert0000$, while the remaining $8$ vectors have product $1$ with $x$.

If we have $p,q,r\neq0$, then they lie on a projective line in the projective $3$-space corresponding to $M_1$. There are $35$ projective lines in a projective $3$-space over $\mathbb{F}_2$.

If all three of $p$, $q$ and $r$ are orthogonal to $x$, they must form a line in $0c_2c_3c_4\vert0000$, and since this is just a Fano plane, there are $7$ such lines.

If only one of them is orthogonal to $x$, the line must be one of the $28$ other lines.

This is enough information to construct the following table (key below):

$\array{\arrayopts{\rowlines{solid}\collines{solid}\frame{solid}} \mathbf{x}&\mathbf{#x}&\mathbf{p q r}&\mathbf{B}&\mathbf{#p q r}&\mathbf{a b c}&\mathbf{Q(a b c)}&\mathbf{\text{perms}}&\mathbf{\text{min Euclid.}}&\mathbf{\text{#lift}}\\ x&15&p q r&000&7&x+p,x+q,x+r&000&6&4+4+4=12&16\cdot16\cdot16=4096\\ x&15&p q r&011&28&x+p,x+q,x+r&011&6&4+2+2=8&16\cdot2\cdot2=64\\ 0&1&p q r&000&35&p,q,r&000&6&4+4+4=12&16\cdot16\cdot16=4096\\ x&15&0 q q&000&7&x,x+q,x+q&000&3&4+4+4=12&16\cdot16\cdot16=4096\\ x&15&0 q q&011&8&x,x+q,x+q&011&3&4+2+2=8&2\cdot2\cdot16=64\\ 0&1&0 q q&000&15&0,q,q&000&3&0+4+4=8&1\cdot16\cdot16=256\\ 0&1&000&000&1&0,0,0&000&3&0+0+8=8&1\cdot1\cdot240=240 }$

Key:

In the $x$ column, we indicate whether we have a zero ($0$) or non-zero ($x$) element of $M_2$.

The $#x$ column gives the number of possible points of this type in $M_2$.

In the $p q r$ column, we indicate whether we have all non-zero ($p q r$) or one zero and two equal non-zero ($0 q q$) or all zero ($000$) points in $M_1$.

The $B$ column gives the values of $B(x,p)$, $B(x,q)$ and $B(x,r)$.

The $#p q r$ column gives the number of choices of $p$, $q$ and $r$ subject to the conditions in the previous columns.

The $a b c$ column tells us the result of summing the above.

The $Q(abc)$ column gives the values of the quadratic form on $a$, $b$ and $c$.

The perms column tells us the number of distinct ways to permute $a$, $b$ and $c$ to get different points of the Leech lattice.

The min Euclid. column tells us the minimum positive Euclidean norm-squared of a point in the Leech lattice coming from the given choice of $a$, $b$ and $c$. (This is twice the norm-squared of the standard normalisation of the Leech lattice.)

The $\text{#lift}$ column gives the number of lifts to $E_8$, bearing in mind that:

The zero vector lifts either to one vector (zeroth shell) or $240$ vectors ($4$th shell).

Vectors of norm $1$ lift to $2$ vectors (first shell).

Non-zero vectors of norm $0$ lift to $16$ vectors (second shell).

If we pick only the rows with Euclidean norm-squared of $8$, corresponding to the first shell of the Leech lattice, and count $\text{#x}\cdot\text{#pqr}\cdot\text{perms}\cdot\text{#lift}$ we get

$\array{\arrayopts{\rowlines{solid}\collines{solid}\frame{solid}\collayout{left}} \mathbf{x}&\mathbf{p q r}&\mathbf{B(x,p q r)}&\mathbf{\text{total}}\\ x&p q r&011&15\cdot28\cdot6\cdot64=161280\\ x&0 q q&011&15\cdot8\cdot3\cdot64=23040\\ 0&0 q q&000&1\cdot15\cdot3\cdot256=11520\\ 0&000&000&1\cdot1\cdot3\cdot240=720 }$

The total $161280+23040+11520+720=196560$.

Since $196560$ is the number of points in the first shell of the Leech lattice, this naive counting method would appear to be correct.

Posted by: Tim Silverman on January 28, 2016 1:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Oh, and just one more little note on trialty. The fact that a pair of disjoint maximal flats translates under triality to a pair of non-orthogonal points is actually quite straightforward. Since the incidence of a point and a maximal flat translates under triality to the incidence of a maximal flat (of the other type) and a point, it follows that a pair of maximal flats of the same type which share no points translates into a pair of points that are not both contained in any maximal flat (of one type, but that doesn’t affect the result). This implies they share no flat lines, which is precisely to say that they are not orthogonal to each other.

Posted by: Tim Silverman on January 29, 2016 9:12 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

There is another beautiful relation between the Leech lattice and $E_8$: The 196560=819*240 minimal norm vectors of the Leech lattice contain 819 disjoint copies of the 240 minimal norm vectors of $E_8$ (scaled to norm 4), forming a generalized hexagon with 819 points: See

Elkies, N.D. and Gross, B.H., 2001. Cubic rings and the exceptional Jordan algebra. Duke Mathematical Journal, 109(2), pp.383-410.

Hoggar, S.G., 1982. t-Designs in projective spaces. European Journal of Combinatorics, 3(3), pp.233-254.

Posted by: Arnold Neumaier on November 24, 2017 9:21 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

Very interesting—thanks! 819 is not a number I instantly recognized, but Geoffrey Dixon and Robert Wilson showed that the relation between the $E_8$ lattice and Leech lattice gives the formula

$196560 = 3 \times 240 \times (1 + 16 + 16^2)$

so the point of 819 is that it’s thrice

$1 + 16 + 16^2 = \frac{16^3 - 1}{16 - 1} = \frac{4095}{15} = 273$

Posted by: John Baez on November 25, 2017 7:15 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

The occurrence of $E_8$ and of Hoggar in the same comment reminded me of some old notes I had written, which I have foolishly gone ahead and posted in the Geometric McKay Correspondence thread. The paper by Hoggar cited above mentions some of the structures I discuss there: His examples 5 and 8 are SICs in dimensions 3 and 8. Also, his example 16 is related to the Hesse SIC, as it turns out to be what the quantum information people call a complete set of Mutually Unbiased Bases which is easily derived from the Hesse SIC.

Posted by: Blake Stacey on November 25, 2017 8:10 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 11)

This seems like as good a place as any to point to this MathOverflow question that an acquaintance apprised me of. It’s about the icosians, the binary tetrahedral and octahedral groups, and the $E_8$ lattice.

Posted by: Blake Stacey on March 8, 2020 5:13 AM | Permalink | Reply to this

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