## January 31, 2016

### Integral Octonions (Part 12)

#### Posted by John Baez

guest post by Tim Silverman

“Everything is simpler mod $p$.”

That is is the philosophy of the Mod People; and of all $p$, the simplest is 2. Washed in a bath of mod 2, that exotic object, the $\mathrm{E}_8$ lattice, dissolves into a modest orthogonal space, its Weyl group into an orthogonal group, its “large” $\mathrm{E}_8$ sublattices into some particularly nice subspaces, and the very Leech lattice itself shrinks into a few arrangements of points and lines that would not disgrace the pages of Euclid’s Elements. And when we have sufficiently examined these few bones that have fallen out of their matrix, we can lift them back up to Euclidean space in the most naive manner imaginable, and the full Leech springs out in all its glory like instant mashed potato.

What is this about? In earlier posts in this series, JB and Greg Egan have been calculating and exploring a lot of beautiful Euclidean geometry involving $\mathrm{E}_8$ and the Leech lattice. Lately, a lot of Fano planes have been popping up in the constructions. Examining these, I thought I caught some glimpses of a more extensive $\mathbb{F}_2$ geometry; I made a little progress in the comments, but then got completely lost. But there is indeed an extensive $\mathbb{F}_2$ world in here, parallel to the Euclidean one. I have finally found the key to it in the following fact:

Large $\mathrm{E}_8$ lattices mod $2$ are just maximal flats in a $7$-dimensional quadric over $\mathbb{F}_2$.

I’ll spend the first half of the post explaining what that means, and the second half showing how everything else flows from it. We unfortunately bypass (or simply assume in passing) most of the pretty Euclidean geometry; but in exchange we get a smaller, simpler picture which makes a lot of calculations easier, and the $\mathbb{F}_2$ world seems to lift very cleanly to the Euclidean world, though I haven’t actually proved this or explained why — maybe I shall leave that as an exercise for you, dear readers.

N.B. Just a quick note on scaling conventions before we start. There are two scaling conventions we could use. In one, a ‘shrunken’ $\mathrm{E}_8$ made of integral octonions, with shortest vectors of length $1$, contains ‘standard’ sized $\mathrm{E}_8$ lattices with vectors of minimal length $\sqrt{2}$, and Wilson’s Leech lattice construction comes out the right size. The other is $\sqrt{2}$ times larger: a ‘standard’ $\mathrm{E}_8$ lattice contains “large” $\mathrm{E}_8$ lattices of minimal length $2$, but Wilson’s Leech lattice construction gives something $\sqrt{2}$ times too big. I’ve chosen the latter convention because I find it less confusing: reducing the standard $\mathrm{E}_8$ mod $2$ is a well-known thing that people do, and all the Euclidean dot products come out as integers. But it’s as well to bear this in mind when relating this post to the earlier ones.

#### 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 $\mathrm{PG}(n,q)$.

The full symmetry group of $\mathrm{PG}(n,q)$ is $\mathrm{GL}_{n+1}(q)$, and from that we get subgroups and quotients $SL_{n+1}(q)$ (with unit determinant), $\mathrm{PGL}_{n+1}(q)$ (quotient by the centre) and $\mathrm{PSL}_{n+1}(q)$ (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 $B(x,x)=0$ for all $x$ 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 $\mathrm{GO}_{n+1}^+(q)$, by analogy with $\mathrm{GL}_{n+1}(q)$. Similarly we have $\mathrm{SO}_{n+1}^+(q)$, $\mathrm{PGO}_{n+1}^+(q)$ and $\mathrm{PSO}_{n+1}^+(q)$. However, whereas $\mathrm{PSL}_n(q)$ is simple apart from $2$ exceptions, we usually have an index $2$ subgroup of $\mathrm{SO}_{n+1}^+(q)$, called $\Omega_{n+1}^+(q)$, and a corresponding index $2$ subgroup of $\mathrm{PSO}_{n+1}^+(q)$, called $\mathrm{P}\Omega_{n+1}^+(q)$, and it is the latter that is simple. (There is an infinite family of exceptions, where $\mathrm{PSO}_{n+1}^+(q)$ is simple.)

Symplectic structures are easier — the determinant is automatically $1$, so we just have $\mathrm{Sp}_{n+1}(q)$ and $\mathrm{PSp}_{n+1}(q)$, 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 $\mathrm{Q}_n^+(q)$.

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 forms 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 $\mathrm{SO}_{n+1}^+(q)$, polar spaces, etc.

Over $\mathbb{F}_2$, we have that $\mathrm{GO}_{n+1}^+(q)$, $\mathrm{SO}_{n+1}^+(q)$, $\mathrm{PGO}_{n+1}^+(q)$ and $\mathrm{PSO}_{n+1}^+(q)$ are all the same group, as are $\Omega_{n+1}^+(q)$ and $\mathrm{P}\Omega_{n+1}^+(q)$.

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.

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

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 $\mathrm{Q}_{2m-1}^+(q)$:

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 $\prod_{i=0}^{m-1}\left(1+q^i\right)$.

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 $\Pi$, is 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 $\Sigma$, is 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 $\mathrm{PG}(3,q)$ and $\mathrm{Q}_5^+(q)$, 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:

$\array{\arrayopts{\collayout{left}\collines{dashed}\rowlines{solid dashed}\frame{solid}} \mathbf{\mathrm{Q}_5^+(q)}&\mathbf{\mathrm{PG}(3,q)}\\ \text{point}&\text{line}\\ \text{orthogonal points}&\text{intersecting lines}\\ \text{line}&\text{plane pencil}\\ \text{plane}_1&\text{point}\\ \text{plane}_2&\text{plane} }$

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 types of plane in $\mathrm{Q}_5^+(q)$ are two families of maximal flats, and they correspond, in $\mathrm{PG}(3,q)$, to “all the lines through a particular point” and “all the lines in a particular plane”.

From fact 1c above, in $\mathrm{Q}_5^+(q)$ 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 $\mathrm{PG}(3,q)$ 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 $\mathrm{PG}(3,q)$ that any two points lie in a line, and any two planes intersect in a line.

#### Triality zips past your window

In $\mathrm{Q}_7^+(q)$, 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. $\mathrm{PG}(3,2)$ 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. $\mathrm{Q}_5^+(2)$ has $35$ points, corresponding to the $35$ lines of $\mathrm{PG}(3,2)$, and $30$ planes, corresponding to the $15$ points and $15$ planes of $\mathrm{PG}(3, 2)$. There’s lots and lots of other interesting stuff, but we will ignore it.

2c. $\mathrm{Q}_7^+(2)$ 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 $1$.

#### $\mathrm{E}_8$ mod $2$

Now we move onto the second part.

We’ll coordinatise the $\mathrm{E}_8$ 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 $\mathrm{E}_8$ by $2\mathrm{E}_8$. 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 $\mathrm{E}_8$ lattice element representing this element of $\mathrm{E}_8/2\mathrm{E}_8$.

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 $\mathrm{E}_8$ , 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:

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

Anisotropic:

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

Since the quadratic form in $\mathbb{F}_2$ comes from the quadratic form in Euclidean space, it is preserved by the Weyl group $W(\mathrm{E}_8)$. In fact the homomorphism $W(\mathrm{E}_8)\rightarrow \mathrm{GO}_8^+(2)$ is onto, although (contrary to what I said in an earlier comment) it is a double cover — the element of $W(\mathrm{E}_8)$ that reverses the sign of all coordinates is a (in fact, the) non-trivial element element of the kernel.

#### Large $\mathrm{E}_8$ lattices

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

Here is a large $\mathrm{E}_8$ 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 $\mathrm{E}_8$ 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 $\mathrm{E}_8$ 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 $\mathrm{Q}_7^+(2)$. (That is, $15$ points projectively, forming a projective $3$-space in a projective $7$-space.)

That a large $\mathrm{E}_8$ 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 $\mathrm{E}_8$ 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 $\mathrm{E}_8$ roots must always be an even integer. Also, the large $\mathrm{E}_8$ roots project to norm $0$ points. So all points of the large $\mathrm{E}_8$ should project to norm $0$ points.

It’s not instantly obvious to me that large $\mathrm{E}_8$ should project to a maximal flat, but it clearly does.

So I’ll assume each $\mathrm{E}_8$ 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 $\mathrm{E}_8$ lattices

We immediately know there are $270$ large $\mathrm{E}_8$ lattices, because there are $270$ maximal flats in $\mathrm{Q}_7^+(2)$, 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 $\mathrm{Q}_7^+(2)$.

#### Number of large $\mathrm{E}_8$ root systems sharing a given point

We can now bring to bear some more general theory. How many large $\mathrm{E}_8$ 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 $\mathrm{Q}_7^+(2)$. The space of all flats containing $p$ is isomorphic to $\mathrm{Q}_5^+(2)$. The maximal flats containing $p$ in $\mathrm{Q}_7^+(2)$ correspond to all maximal flats of $\mathrm{Q}_5^+(2)$, of which there are $30$. So there are $30$ maximal flats of $\mathrm{Q}_7^+(2)$ containing $p$, and hence $30$ large $\mathrm{E}_8$ 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 $\mathrm{PG}(3,2)$.

#### Number of large $\mathrm{E}_8$ root system disjoint from a given large $\mathrm{E}_8$ root system

Now assume that large $\mathrm{E}_8$ 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 $\mathrm{Q}_7^+(2)$. The $3$-flats sharing that point correspond to the planes of $\mathrm{Q}_5^+(2)$. Then the set of $3$-flats sharing just a line through $p$ with our given $3$-flat correspond to the set of planes of $\mathrm{Q}_5^+(2)$ 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 $\mathrm{E}_8$ lattices whose roots don’t meet the roots of a given large $\mathrm{E}_8$ lattice.

#### Other numbers of intersecting root systems

We can also look at the intersections of large $\mathrm{E}_8$ root systems with large $\mathrm{E}_8$ 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 $\mathrm{Q}_5^+(2)$, 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 $\mathrm{PG}(3,2)$). So there are $7$ $3$-flats of $\mathrm{Q}_7^+(2)$ 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 $\mathrm{E}_8$ 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 $\mathrm{GO}_6^+(2)$ subgroup of $\mathrm{GO}_8^+(2)$. By the Klein correspondence, $\mathrm{GO}_6^+(2)$ is isomorphic to $\mathrm{PSL}_4(2)$, which is just the automorphism group of $\mathrm{PG}(3, 2)$ — 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 $\mathrm{E}_8$ lattice).

(My guess would be that the actions of $\mathrm{GL}_4(2)$ on the two maximal flats would be related by an outer automorphsm of $\mathrm{GL}_4(2)$, 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 $\mathrm{E}_8$ 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 $\mathrm{E}_8$ . This has as our chosen representative $\left(1,1,1,1,1,1,1,1\right)$ in our version of $\mathrm{PG}(7,2)$, 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 $\mathrm{E}_8$ 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 $\mathrm{Q}_5^+(2)$, and we would like to understand the intersections of these flats with $T$: that is, those nearest neighbours which belong to each large $\mathrm{E}_8$ 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 $\mathrm{PG}(3,2)$.

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 $\mathrm{E}_8$ 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 $\mathrm{Q}_5^+(2)$, 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 $\mathrm{E}_8$ lattice.

Posted at January 31, 2016 1:42 AM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2865

### Re: Integral Octonions (Part 12)

For me this post has a vertical line after each bit of LaTeX. Gets a bit hard to read.

Posted by: Qiaochu Yuan on January 31, 2016 6:07 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

For me it doesn’t, and I don’t see why it would; it’s just like any other post as far as I can tell. I think it’s your browser acting up. I’m using an up-to-date Firefox on Windows.

Anyone else see those lines?

Posted by: John Baez on January 31, 2016 6:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I see those vertical lines too, using up-to-date Chrome in Ubuntu 12.04.

Posted by: Allen K. on January 31, 2016 6:50 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I also see the lines using Chrome on a Mac. I noticed the same thing the other day on the arXiv.

Posted by: Mark Meckes on January 31, 2016 7:47 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I’m also using Chrome on a Mac and don’t see the lines. (Chrome 47.0.2526.111 and OS X 10.10.5.)

Posted by: Mike Stay on February 1, 2016 6:13 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I’ve got Chrome 48 on a Mac. It could be a 48 vs 47 thing, Mike.

Posted by: David Roberts on February 3, 2016 5:48 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I see them in Chrome on Windows.

Chrome is not my primary browser but when I go there and look in the debugger I see that the nCategory is loading MathJax from a copy in a local folder:

https://golem.ph.utexas.edu/wiki/MathJax/

while the nForum (where there is no line problem) is loading it from:

cdn.mathjax.org

I’m guessing the the golem copy is out of date.

Posted by: RodMcGuire on January 31, 2016 8:14 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I’ll try to get Jacques Distler to update things and see if that helps. In the meantime, maybe it will help to read the copy of this article on my website:

Posted by: John Baez on February 1, 2016 2:01 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Number 12 does not yet appear in the main list.

http://math.ucr.edu/home/baez/octonions/integers/

Posted by: Jeffery Winkler on February 10, 2016 7:25 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks. Fixed!

Posted by: John Baez on February 10, 2016 7:34 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Jacques wrote:

Looks perfectly fine to me, in Chrome 47.0.2526.111 (64-bit, MacOSX), which is – I believe – the latest version available.

I do intend to update MathJax to 2.6 (released Dec 30, 2015), from the current 2.5.1. But I’m not sure that’s the source of peoples’ problem(s).

Posted by: John Baez on February 1, 2016 5:18 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

John, this is the same problem I wrote to you about on 22 Dec 2015, in reference to this post of yours. Here’s a screenshot of what I saw then, which is similar to what I see now. As Qiaochu says, there appears to be a blank line inserted before every line that contains Latex.

This happened before the new MathJax release that Jacques refers to. I think that post of yours was the first time I saw it happening.

My browser was up to date then, and it’s up to date now (Firefox 44.0 on Ubuntu 12.04.5 LTS).

Posted by: Tom Leinster on February 4, 2016 4:56 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Actually this seems to be a different problem than what Qiaochu reported and what I saw, which is a thin vertical black line to the right of each bit of LaTeX. I can’t provide a screen shot at the moment since I’m not at the computer where I had that issue.

Posted by: Mark Meckes on February 4, 2016 8:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Here’s a screenshot of an arXiv abstract where the problem appears. It seems to have been fixed here on the n-café

Posted by: David Roberts on February 4, 2016 10:36 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I have Tom’s problem, too, even on the nLab and the nForum. I use Firefox 44.0 on Xubuntu 14.04. I also see large vertical spaces in the MathML ‘torture test’ under items 18, 23 and 24. Tom, do you have those as well? If so, it might be a problem on our end?

Posted by: Tobias Fritz on February 5, 2016 7:24 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks for putting me straight about the two different problems. I’ve never had the problem with the vertical black lines.

Tobias, like you, I fail the torture test on items 18, 23 and 24. I guess I once did something bad to my system, and now I’m paying for it. The only thing I can remember doing on the browser typesetting front is to install the Styx fonts, years ago (is that even what they were called?).

Anyway, if I really want to get rid of the blank lines then I can use Tor. Everything looks great there. To be honest, this is one of those problems where unless someone tells me an easy fix, I’ll probably just put up with it until it’s time to get a new computer.

Posted by: Tom Leinster on February 5, 2016 5:19 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

From Google Group “MathJax Users”: Extra line in MathJax 2.2

We are using MathJax 2.2 in one of our installation.

Recently we started noting a vertical line at the end of each equation …

This is https://github.com/mathjax/MathJax/issues/1300 and was fixed in MathJax v2.6.

([HTML-CSS] [Chrome 48] visual artifacts due to Chrome now rounding up [was: rendering issue in chrome canary builds] #1300)

Posted by: RodMcGuire on February 8, 2016 5:28 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

This post is actually part 13, not 12.

Posted by: Joe Yoon on January 31, 2016 7:39 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks! Actually the old Part 12 shot off in a wildly different direction, so I renamed it:

Braided 2-groups from lattices, January 1, 2015.

Posted by: John Baez on January 31, 2016 8:09 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Tim — I’m slowly trying to absorb this, and I feel this:

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

must be a typo for something like this:

1b. The number of maximal flats is $\prod_{i=0}^{m-1}\left(1+q^i\right)$.

Posted by: John Baez on January 31, 2016 10:31 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Darn it, yes. You’re right.

Posted by: Tim Silverman on February 1, 2016 7:16 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Tim wrote:

The two types of plane in $\mathrm{Q}_5^+(q)$ are two families of maximal flats, and they correspond, in $\mathrm{PG}(3,q)$, to “all the lines through a particular point” and “all the lines in a particular plane”.

By the way, Penrose, in his work on twistors, calls these two types of planes $\alpha$-planes and $\beta$-planes. At least I think those are the same thing! Are they? Of course he’s working over the complex numbers. But apart from that…

Here’s some stuff I just found online, which may help:

I’m guessing that $\mathbf{M}$ is complexified, compactified Minkowksi spacetime, aka the Klein quadric.

Posted by: John Baez on January 31, 2016 11:17 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Yep, that looks right to me.

Posted by: Tim Silverman on February 1, 2016 7:17 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I think I understand the main claims of this post by now, though I haven’t gotten to the point of seeing why they’re all true. I can see that this is great stuff!

Here’s a somewhat digressive question. There are algebraic geometers who like to work in a ‘field-independent’ way as long as they can, for example using schemes. They can do things like define the group $SL(n)$ as an ‘affine group scheme’, simply by writing down the polynomial equation obeyed by the entries of an $n \times n$ matrix with determinant one. Then, if you hand them any field $\mathbb{F}$, they can define the ‘$\mathbb{F}$-points’ of this affine group scheme: namely, the set of solutions of this equation. These will be the points of the usual group $SL(n,\mathbb{F})$. This is an ‘affine algebraic group’.

All this is very nice, but I’m wondering how it meshes with the fact that certain things get weird in fields of nonzero characteristic.

For example, I can write down an affine group scheme which I’ll call $SO(n)$, simply by writing down the equations obeyed by a matrix with determinant one whose transpose is its inverse. If you hand me any field $\mathbb{F}$, the $\mathbb{F}$-points of $SO(n)$ will be the solutions of these equations. They’ll form a group, and I’m really tempted to call this group $SO(n,\mathbb{F})$.

However, you’re telling me that for a finite field $\mathbb{F}_q$ there are two versions of $SO(n,\mathbb{F}_q)$, called $SO(n,\mathbb{F}_q)^+$ and $SO(n,\mathbb{F}_q)^-$. So which one do we get from my construction if I plug in $\mathbb{F} = \mathbb{F}_q$? Or do I get neither?

I’m guessing I’ll get one or the other, depending on whether the ‘standard’ quadratic form

$x_1^2 + \cdots + x_n^2$

is of plus type or minus type over $\mathbb{F}_q$.

Which is it? And how do the extra nuances of characteristic 2 fit into this game?

I’m guessing now that there are a bunch of group schemes that all give isomorphic algebraic groups when we take $\mathbb{F} = \mathbb{C}$, but that look different when we take $\mathbb{F} = \mathbb{F}_q$.

This effect already happens for $\mathbb{F} = \mathbb{R}$, where we get different ‘real forms’, but thanks to my physics training that seems second nature to me, while the story for finite fields remains rather mysterious.

Among other questions: is there a nice place to read about this issue for $SO(n)$ and its close relatives?

Posted by: John Baez on February 1, 2016 8:12 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I think the default would probably be the plus type, which has an analogue over every field.

I don’t know of any good references for algebraic groups, I’m afraid.

Posted by: Tim Silverman on February 2, 2016 5:40 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

It seems to me the plus type has an analogue over every field in in even dimensions, while the minus type has an analogue over every field in odd dimensions. Let me say what I mean, just to see if I understand what’s going on.

You started out talking about a vector space with a bilinear form $B$ on it, I think:

Points $x$ for which $B(x,x)=0$ are isotropic.

Then I think you switched to talking in somewhat projective terms, as is your wont. However, I’ll remain resolutely vectorial and translate you thus:

A plane spanned by two isotropic points, $x$ and $y$ such that $B(x,y)=1$ is a hyperbolic plane. Any space with a non-degenerate bilinear (or Hermitian) form can be decomposed as the orthogonal sum of hyperbolic planes, possibly together with an anisotropic space containing no isotropic points at all.

If I take any field $\mathbb{F}$, I can put a symmetric bilinear form $B$ on $\mathbb{F}^2$ by setting

$B((x_1,x_2), (y_1,y_2)) = x_1 y_2 + x_2 y_1$

The points $x = (1,0)$ and $y = (0,1)$ are isotropic since

$B(x,x) = 1 \cdot 0 + 0 \cdot 1 = 0$ $B(y,y) = 0 \cdot 1 + 1 \cdot 0 = 0$

and they obey

$B(x,y) = 1 \cdot 1 + 0 \cdot 0 = 1$

so this $\mathbb{F}^2$ is a hyperbolic plane.

Then:

… all the orthogonal spaces we consider here will be a sum of hyperbolic lines — we say they are of plus type.

So, I think over any field I can make $\mathbb{F}^{2n} = \mathbb{F}^2 \oplus \cdots \oplus \mathbb{F}^2$ into an orthogonal space of plus type, using the bilinear form $B \oplus \cdots \oplus B$.

Since I can do this ‘systematically’ for any field, I should be able to make an affine group scheme $GO_{2n}^+$ by writing down the equations that $(2n)^2$ variables obey when they’re the entries of square matrix that preserves the form $B \oplus \cdots \oplus B$.

Over some fields $\mathbb{F}$ (like $\mathbb{C}$) this affine group scheme will give an algebraic group $GO_{2n}^+(\mathbb{F})$ that’s isomorphic to $O(2n,\mathbb{F})$. Over others fields (like $\mathbb{R}$) it gives an algebraic group that’s different from $O(2n,\mathbb{F})$. But that’s part of the fun of life.

I’m too tired to give as much detail for my other hunch, but it seems that in odd dimensions we can cook up an affine group scheme $GO_{2n+1}^-$ that for any field $\mathbb{F}$ gives the algebraic group $GO_{2n+1}^-(\mathbb{F})$. The idea is: systematically build a form of minus type by taking the previous construction and sticking on an extra ‘anisotropic line’.

So is the special feature of finite fields that we have both plus and minus types in every dimension? You almost said that, but didn’t exactly emphasize it.

Posted by: John Baez on February 2, 2016 9:05 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Oh, sorry, in odd dimension there’s only one type! I was trying to make the introduction as short as possible by ruthlessly cutting out everything I didn’t absolutely need, and seem to have introduced some confusion. Over finite fields an orthogonal anisotropic space must be $0$, $1$ or $2$ dimensional (as vector spaces). The $0$ and $2$ dimensional ones give the plus and minus type in even dimension, and the $1$ dimensional ones give odd dimensional orthogonal structures.

Posted by: Tim Silverman on February 2, 2016 9:13 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks! Wow, working over a finite field is sure different from working over $\mathbb{R}$. I have the classification of orthogonal structures over $\mathbb{R}$ worked deep into my neural net by now. What you just said is so different that at first I thought you must be wrong! But now I’m beginning to get it.

Posted by: John Baez on February 3, 2016 6:40 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Over $\mathbb{R}$, there is one anisotropic orthogonal space (up to isomorphism) in each dimension. Over $\mathbb{C}$, there’s one $0$-dimensional one, and one $1$-dimensional one. So over $\mathbb{C}$ there’s one orthogonal space in each dimension, but over $\mathbb{R}$, there’s the whole range of signatures.

It is a bit confusing what carries over from $\mathbb{R}$ and what doesn’t. Some intuitions work fine, some really don’t.

Posted by: Tim Silverman on February 3, 2016 7:30 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

So, it seems Tim has shown that:

• the 270 ways of sublattices of the $\mathrm{E}_8$ lattice formed by expanding it by a factor of $\sqrt{2}$ and rotating it a bit

correspond in a natural way to

• the 270 maximal isotropic subspaces of $\mathbb{F}_2^8$ equipped with its obvious orthogonal structure $Q(x) = \sum_{i=1}^8 x_i^2$.

I’m making a guess here: Tim said we should use an 8-dimensional orthogonal space ‘of plus type’ over $\mathbb{F}_2$, and he said something to make me guess this ‘obvious’ orthogonal structure $Q$ was of plus type, but I haven’t checked that it is! In his blog post, he used a formula for $Q$ that’s better adapted to seeing the relation to $\mathrm{E}_8$. So, if my formula for $Q$ doesn’t work, we can use his. This doesn’t affect my remaining point.

Which is:

The 270 maximal isotropic subspaces come in two kinds, ‘left-handed’ and ‘right-handed’. There are 135 of each kind.

So, the 270 large $\mathrm{E}_8$ lattices should also come in two kinds. How do we distinguish between these two kinds?

Posted by: John Baez on February 3, 2016 6:55 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Above, egged on a bit by Tim, I was guessing that the quadratic form

$Q(x) = \sum_{i = 1}^n x_i^n$

on $\mathbb{F}_2^n$ is of plus type whenever $n$ is even, meaning (I believe) that the space can be decomposed into a direct sum of hyperbolic planes: 2d spaces spanned by two vectors with $Q(x) = 0$.

But this is not true when $n = 2$. In this case we have

$Q(1,0) = 1, \quad Q(0,1) = 1, \quad Q(1,1) = 0$

so $\mathbb{F}_2^n$ is not spanned by two vectors with $Q(x) = 0$, for the simple reason that there don’t exist two nonzero vectors with $Q(x) = 0$.

So, this quadratic form on $\mathbb{F}_2^2$ is of minus type.

To give $\mathbb{F}_2^2$ a quadratic form of plus type we can use

$Q'(x) = x_1 x_2$

so that

$Q'(1,0) = 0, \quad Q'(0,1) = 0, \quad Q(1,1) = 1$

For insanely devoted readers of this blog, here’s where Tim perhaps unknowingly egged me on. I wrote:

[…] depending on whether the ‘standard’ quadratic form

$x_1^2 + \cdots + x_n^2$

is of plus type or minus type over $\mathbb{F}_q.$

Which is it? And how do the extra nuances of characteristic 2 fit into this game?

He replied:

I think the default would probably be the plus type, which has an analogue over every field.

At first I inferred from this that my so-called ‘standard’ quadratic form was of plus type over $\mathbb{F}_2$. Later I realized it obviously can’t be when $n$ is odd. Now I’m seeing it’s not when $n = 2$. I haven’t studied other even $n$ yet.

Obviously somewhere there is a textbook that explains all this in the first 10 pages, something like Math Mod 2 for Dummies. But I don’t know that book.

Posted by: John Baez on February 3, 2016 7:48 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

For any $n$, the isotropic vectors in $\mathbb{F}_2^n$ for the “standard” quadratic form all have coordinates that sum to 0, since $x_i^2 = x_i$ over $\mathbb{F}_2$. So we always have some anisotropic stuff left over.

Posted by: Layra on February 3, 2016 9:00 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks! Forgetting that $x^2 = x$ is the kind of thing that makes me wish I’d read Math Mod 2 for Dummies.

Posted by: John Baez on February 3, 2016 9:14 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Poking at it a bit more, this happens for all $q = 2^k$, since $x \rightarrow x^2$ is an automorphism of $\mathbb{F}_{2^k}$. So we get that $\sum_i x_i^2 = 0$ iff $\sum_i x_i = 0$.

Posted by: Layra on February 3, 2016 9:50 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

$\sum_{i=1}^n x_i^2$ isn’t an orthogonal structure in characteristic $2$! I can’t remember exactly where it unravels, but I don’t think orthogonal bases of non-null vectors exist in characteristic $2$.

Posted by: Tim Silverman on February 3, 2016 10:12 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Oh! Maybe Layra’s observation that $\sum x_i^2 = \sum x_i$ can help us show that this quadratic form is degenerate.

What is the condition is for a quadratic form $Q$ to be nondegenerate in characteristic 2? You mention that there’s a bilinear form

$B(x,y) = Q(x+y) - Q(x) - Q(y)$

Normally I’d say that the condition for $Q$ to be nondegenerate is that this bilinear form be nondegenerate. But if $Q(x) = \sum x_i^2 = \sum x_i$, this bilinear form is zero.

This feeds back into my earlier question: what are the affine group schemes related to the orthogonal groups? We could try to define an affine group scheme $\mathrm{O}(n)$ by taking the equations on $n \times n$ variables that say an $n\times n$ matrix preserves the quadratic form

$Q(x) = \sum_{i=1}^n x_i^2$

Over $\mathbb{R}$ this gives the usual orthogonal group, but over $\mathbb{F}_2$ this quadratic form is degenerate, so we get some funny algebraic group that apparently doesn’t deserve the name $\mathrm{O}(n,\mathbb{F}_2)$. It seems to be $\mathrm{GL}(n-1,\mathbb{F}_2)$, since it’s preserving a nonzero linear functional.

Posted by: John Baez on February 3, 2016 10:36 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Oh, I was being obtuse. From

$Q(u+v)=Q(u)+Q(v)+B(u,v)$

and

$Q(\lambda u)=\lambda^2Q(u)$

we get

$B(u,u)=2Q(u)$.

So in characteristic $2$, $B(u,u)=0$. So if we have a normalised basis, with $B(e_i,e_j)=0\forall i\neq j$, and we’re also in characteristic $2$, then $B$ is identically zero. So the quadratic form is automatically degenerate.

Posted by: Tim Silverman on February 7, 2016 10:42 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Nice argument.

Posted by: John Baez on February 8, 2016 12:33 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Actually I’m a bit confused. If we take

$Q(x) = \sum_{i = 1}^n x_i^2$

then I agree that the bilinear form $B$ with $B(x,x) = 2Q(x)$ is degenerate in characteristic two, since $B = 0$. But it seems a bit harsh to call $Q$ degenerate because of this! For example, there’s a bilinear form

$C(x,y) = \sum_{i = 1}^n x_i y_i$

with $C(x,x) = Q(x)$, and I don’t think $C$ is degenerate.

So the rules of the game seem rigged against $Q$: it gets called degenerate thanks to the sins of $B(x,x) = 2Q(x)$.

To take this out of the realm of politics and name-calling into actual mathematics, the group of transformations of $\mathbb{F}_2^n$ preserving $Q$ seems like it might be interesting. What is this group? Is it simple, or does it at least have a large simple subquotient?

(Does it have a name? In my naive youth I would have called it $\mathrm{O}(n,\mathbb{F}_2)$.)

Posted by: John Baez on February 8, 2016 7:32 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

OK, I was overstating things. What we want to avoid are non-zero points that are both orthogonal to everything (under $B$) and also have norm $0$. The basis vectors would pass, under that criterion, but everything that’s a sum of an even number of basis vectors would fail. We end up with the orthogonal sum of a $2m-1$-dimensional space on which everything in sight is zero, so the “orthogonal” transformations are all invertible linear transformations, and a $1$-dimensional orthogonal space which is preserved under all orthogonal transformations. Over a non-trivial extension of $\mathbb{F}_2$ it would at least have transformations coming from field automorphisms, but over $\mathbb{F}_2$ itself, it makes a trivial contribution to the automorphism group.

Posted by: Tim Silverman on February 9, 2016 4:39 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

As you said earlier, this will give $\mathrm{GL}_{2m-1}(\mathbb{F}_2)$.

Posted by: Tim Silverman on February 9, 2016 9:06 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Okay, here’s another question.

Tim takes the $\mathrm{E}_8$ lattice, equips it with the unique quadratic form

$\widetilde{Q} : \mathrm{E}_8 \to \mathbb{Z}$

that takes the value 1 on all the roots, and then works mod 2, getting

$Q: \mathrm{E}_8/2\mathrm{E}_8 \to \mathbb{F}_2$

This is, he argues, a quadratic form of plus type. So, there must be an isomorphism of $\mathbb{F}_2$-vector spaces-with-quadratic-form

$\mathrm{E}_8/2\mathrm{E}_8 \cong \mathbb{F}_2^2 \oplus \mathbb{F}_2^2 \oplus \mathbb{F}_2^2 \oplus \mathbb{F}_2^2$

where each copy of $\mathbb{F}_2^2$ is equipped with the quadratic form

$Q(x_1,x_2) = x_1 x_2$

So, my first question is: can we see this isomorphism explicitly in some nice way, and understand it?

Secondly: to what extent can we lift this splitting into four pieces to $\mathrm{E}_8$ itself? We can’t write $\mathrm{E}_8$ as a direct sum of four 2-dimensional lattices

$\mathrm{E}_8 \cong L_1 \oplus L_2 \oplus L_3 \oplus L_4$

where all these lattices are orthogonal to each other. But maybe we can do it in a way where $v \cdot w = 0$ mod 2 when $v \in L_i, w \in L_j$ and $i \ne j$?

Posted by: John Baez on February 3, 2016 9:33 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Incidentally, we can tell it’s of plus type by counting null and non-null vectors. An orthogonal structure of plus type over $\mathbb{F}_2$ in dimension $2m$ (as a vector space) will have $2^{m-1}\left(2^m-1\right)$ non-null vectors, and $2^{m-1}\left(2^m+1\right)$ vectors of $0$ norm including the $0$ vector. For a structure of minus type, these will be the other way round.

Also, I constructed a singular subspace too large to appear in a space of minus type.

Posted by: Tim Silverman on February 3, 2016 10:30 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Well, here is a basis I constructed earlier, if we want something concrete:

$\array{\arrayopts{\collayout{left}} \left(1,1,1,1,0,0,0,0\right)&\left(\frac{3}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\\ \left(1,1,0,0,1,1,0,0\right)&\left(\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)\\ \left(1,1,0,0,0,0,1,1\right)&\left(\frac{3}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\\ \left(0,1,1,0,0,1,1,0\right)&\left(0,1,1,0,0,1,0,1\right) }$

Posted by: Tim Silverman on February 3, 2016 10:40 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks! That explicit basis isn’t helping me much yet, but it might someday.

Here’s a question. We seem to agree that the group of transformations of $\mathbb{F}_2^8$ preserving the quadratic form

$R(x) = x_1 x_2 + x_3 x_4 + x_5 x_6 + x_7 x_8$

is the Weyl group $W(\mathrm{E}_8)$ modulo its center $\{\pm 1\}$. And that, in turn, is the same as the ‘even part’ of $W(\mathrm{E}_8)$: the part generated by products of pairs of reflections.

But this quadratic form $R$ has some obvious symmetries. First, we can permute the four pairs of variables $x_i, x_{i+1}$ (where $i = 1, 3, 5, 7$), so the symmetric group $\mathrm{S}_4$ acts as symmetries. Second, we get four copies of $\mathbb{Z}/2$ acting as symmetries, because the linear transformations of $\mathbb{F}_2^2$ preserving

$Q(x_1, x_2) = x_1 x_2$

form a copy of $\mathbb{Z}/2$.

So, we get the wreath product of $S_4$ and $\mathbb{Z}_2$ sitting inside the even part of $W(\mathrm{E}_8)$.

That’s not so impressive: I guess that wreath product has just $4! \times 2^4 = 384$ elements, while the even part of $W(\mathrm{E}_8)$ has a whopping 348,364,800 elements.

Still, there’s something very nice about ‘chopping $\mathrm{E}_8$ into four parts’ going on here, and I’d like to know what it is.

Is there some geometrically ‘obvious’ reason while the even part of $W(\mathrm{E}_8)$ should contain 4 commuting elements that can be permuted by conjugating them by elements in some $\mathrm{S}_4$ subgroup of $W(\mathrm{E}_8)$?

Naively, one might hope has something to do with a way of taking the 8 dots in the $\mathrm{E}_8$ Dynkin diagram and partitioning them into four pairs. But of course the $\mathrm{E}_8$ Dynkin diagram doesn’t have $\mathrm{S}_4$ symmetry.

A perhaps more promising avenue is Manivel’s tetrality description of $\mathfrak{so}(8,\mathbb{C})$, which starts by finding a copy of

$\mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C})\oplus \mathfrak{sl}(2,\mathbb{C})$

inside $\mathfrak{so}(8,\mathbb{C})$. There could be a version of this at the level of groups rather than Lie algebras in characteristic 2.

Posted by: John Baez on February 8, 2016 9:41 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

There is a group tetrality over $\mathbb{F}_2$ in $SO(8,R)$. Specifically, there are 4 commuting copies of $SL(2,\mathbb{F}_2) \cong S_3$ where everything preserves $R$. The four copies are arranged with two in each of a pair of $SO(4)$ blocks. Here’s an order 2 and an order 3 element from one of the copies: $\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right],\left[\begin{array}{cccc} 1 & 1 & 1 & 1\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 1\\ 1 & 0 & 1 & 0 \end{array}\right]$ The other copy is just this flipped over the antidiagonal. Swapping two $SL(2,\mathbb{F}_2)$s within a single $SO(4)$ block isn’t too bad. Swapping between the two $SO(4)$ blocks is probably a bit messier; I haven’t worked it out yet.

Posted by: Layra on February 9, 2016 9:49 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I think these $\mathrm{SL}_2(2)$ groups can also be thought of as $\mathrm{SO}_2^-(2)$, that is they involve a decomposition into $2$-spaces of minus type rather than hyperbolic planes (or, in $\mathrm{E}_8$-speak, non-orthogonal roots summing to another root). There certainly are such decompositions; I haven’t checked if your example corresponds to one.

Posted by: Tim Silverman on February 9, 2016 10:35 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I’ll say a bunch of stuff you guys probably already know, just to put it on the record.

$\mathbb{F}_2^2$ has a quadratic form of minus type given by

$Q(x,y) = x^2 + x y + y^2$

This equals 1 on all the nonzero elements of $\mathbb{F}^2$:

$Q(1,0) = 1$ $Q(0,1) = 1$ $Q(1,0) = 1$

So, the group of linear transformations preserving this is all of $GL(2,\mathbb{F}_2)$, or in other words $\mathrm{S}_3$: the group that permutes the 3 nonzero vectors. In short,

$SO^-_2(2) \cong \mathrm{S}_3$

I had been inspired by the fact that the even part of Weyl group of $\mathrm{E}_8$ is the group of linear transformations preserving quadratic form of plus type on $\mathbb{F}_2^8$. This made me want to split $\mathbb{F}_2^8$ as an orthogonal direct sum of four copies of $\mathbb{F}_2$, each equipped with a quadratic form of plus type:

$P(x,y) = x y$

This has

$P(1,0) = 0$ $P(0,1) = 0$ $P(1,1) = 1$

so the group preserving it just permutes the two vectors on which $P$ vanishes:

$SO_2^+(2) = \mathrm{S}_2$

That gave me a copy of the wreath product of $\mathrm{S}_4$ and $\mathrm{S}_2$ sitting inside the even part of $W(\mathrm{E}_8)$. This subgroup has only

$4! \times 2^4 = 384$

elements.

If we can also split $\mathbb{F}_2^8$ with its quadratic form of plus type into an orthogonal direct sum of four copies of $\mathbb{F}_2$ equipped with a quadratic form of minus type, then I can understand what Layra is doing!

This will give us a copy of the wreath product of $\mathrm{S}_4$ and $\mathrm{S}_3$ sitting inside the even part of $W(\mathrm{E}_8)$. This subgroup has

$4! \times 6^4 = 31104$

elements. That’s much nicer!

Posted by: John Baez on February 10, 2016 9:58 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Let $y_i = x_i + \sum_{j = 1}^4 x_j$ for $i \leq 4$ and $y_i = x_i + \sum_{j = 5}^8 x_j$ otherwise. We then have $P(y_i) = 1$, and $P(y_{2k-1} + y_{2k}) = 1$. So yes, we can split $(\mathbb{F}_2^8,P)$ as four copies of $(\mathbb{F}_2^2,Q)$, namely $\{0,y_{2k-1},y_{2k},y_{2k-1}+y_{2k}\}$, and the matrices I found act as $S_3$ on $(y_{2k-1},y_{2k},y_{2k-1}+y_{2k})$.

Posted by: Layra on February 11, 2016 2:32 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Wow! I’ll have to think about that calculation. I guess now you’re using $P$ to mean the quadratic form of plus type on $\mathbb{F}_2^8$ that earlier I was calling $R$—the orthogonal direct sum of four copies of the quadratic form of plus type on $\mathbb{F}_2^2$ that I recently called $P$?

Posted by: John Baez on February 11, 2016 6:34 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Err, wait. I think I’m confusing vectors and their coordinate entries. I’m so used to “change the inner product” meaning “change basis” rather than using the same basis and actually changing the inner product itself. Um. Let me remake the notation completely to clear up at least my own confusion.

We have a bilinear form $B$ that takes two vectors $\vec{x} = (x_1, x_2)$ and $\vec{y} = (y_1,y_2)$ and spits out $B(\vec{x},\vec{y}) = x_1y_2 + x_2y_1$ We also have two quadratic forms on $\mathbb{F}_2^2$. We have $x_1x_2$ which is of plus type, and $y_1^2 + y_1y_2 + y_2^2$ which is of minus type. Both of them yield $B$ as the corresponding bilinear form.

The $SO(8)$ I’m using is written with respect to the quadratic form $S$ where for $\vec{a} = (a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8)$, we have $S(\vec{a}) = a_1a_2 + a_3a_4 + a_5a_6 + a_7a_8$ which is of plus type. Restricting $S$ to vectors where only $a_{2i-1}$ and $a_{2i}$ are allowed to be nonzero yields a quadratic form of plus type. We have that, writing $T$ for the matrix form of $S$ in the standard basis, the elements $A$ of $SO(8,S)$ obey $ATA^t = T$.

We can look at a set of vectors $\begin{array}{ccc}\vec{f_1} &=& (0,1,1,1,0,0,0,0)\\ \vec{f_2} &=& (1,0,1,1,0,0,0,0)\\ \vec{f_3} &=& (1,1,0,1,0,0,0,0)\\ \vec{f_4} &=& (1,1,1,0,0,0,0,0)\\ \vec{f_5} &=& (0,0,0,0,0,1,1,1)\\ \vec{f_6} &=& (0,0,0,0,1,0,1,1)\\ \vec{f_7} &=& (0,0,0,0,1,1,0,1)\\ \vec{f_8} &=& (0,0,0,0,1,1,1,0) \end{array}$ and we get that $S(\vec{f_i}) = 1$ and that $S(\vec{f_{2i-1}} + \vec{f_{2i}}) = 1$, i.e. the span of $\vec{f_{2i-1}}$ and $\vec{f_{2i}}$ gives a subspace where $S$ restricts to minus type, and furthermore the matrices I wrote earlier act as $S_3$ on the nonzero vectors in that span.

Posted by: Layra on February 11, 2016 8:04 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks! This will make it a lot easier to understand what’s going on. I’m a bit distracted now by the most exciting physics experiment in decades, but I definitely want to keep digging into this fascinating relation between $\mathrm{O}^+(8,\mathbb{F}_2)$, tetrality and $\mathrm{E}_8$.

Posted by: John Baez on February 11, 2016 4:18 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

More exciting than Higgs boson detection? (I don’t have a good feeling for relative orders of excitement in these matters, but I thought the Higgs was pretty darned exciting; weren’t there a fair number of people betting against such a thing?)

Posted by: Todd Trimble on February 11, 2016 5:56 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

The gravitational wave search had been tantalising us for decades, with not a single detection, so there was more of a “not in my lifetime” feel to it, even though people who knew what they were talking about were telling us they were approaching the needed sensitivity.

With the Higgs, particle accelerators have been producing a massive amount of data successfully for decades, so the search just seemed less far-fetched and exotic.

In both cases, it would have been pretty weird if the phenomenon had actually not been there at all, despite a few people betting that way. They do both potentially open up a new window on stuff that we couldn’t get at any other way, so they’re equally exciting in that sense.

Posted by: Tim Silverman on February 11, 2016 9:08 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Todd wrote:

More exciting than Higgs boson detection?

This is subjective, but to me this is much, much more exciting. This is the start of gravitational wave astronomy, a new window to the Universe that with luck will teach us many new things, while the Large Hadron Collider feels like the end of the road for particle accelerators: it’s going to be very hard for people to build a better one, and the only surprise it’s delivered so far is that there have been no surprises.

I thought the Higgs was pretty darned exciting; weren’t there a fair number of people betting against such a thing?

I don’t know about people in general, but among particle physicists nobody had been able to figure out a way to get the Standard Model to work without a Higgs boson or something very Higgs-like: for example, more than one Higgs. So far it’s looking like the most boring option, the Standard Model everyone believed in already, is fitting the data fine.

In particular, superstring theory seems to have been dealt a mortal wound—at least as far as the politics of physics goes—by the failure of the Large Hadron Collider to find any superpartners. It might still find some, and of course that would change everything. Many are still clinging to this hope, but this hope keeps fading as successive experiments keep ruling out more possibilities. As a consequence, it’s getting harder for superstring theorists to get jobs. Physics departments are instead hiring cosmologists and people searching for dark matter, e.g. in the ‘beam dumps’ at the Large Hadron Collider.

I suppose this is exciting in a certain sense: if the previously popular versions of superstring theory that predict new particles at the TeV scale are wrong, we are now finally in the process of learning they are wrong! But most people involved would characterize this as depressing, in part because we have no idea of what’s right—except, of course, the Standard Model, which is old news.

Posted by: John Baez on February 11, 2016 9:35 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks for the explanation. I think I was thrown off how exciting this was by reading this bit of what you wrote at Azimuth:

The discovery of gravitational waves is completely unsurprising, since they’re predicted by general relativity, a theory that’s passed many tests already.

I unfortunately wasn’t really picking up on what was exciting you and others, when I skimmed further into your post.

My memory of Higgs boson skepticism is hazy, and maybe not worth trying to dredge up.

Posted by: Todd Trimble on February 12, 2016 4:31 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Todd wrote:

I think I was thrown off how exciting this was by reading this bit of what you wrote at Azimuth:

The discovery of gravitational waves is completely unsurprising, since they’re predicted by general relativity, a theory that’s passed many tests already.

Okay, I see. Just for the sake of people too lazy to click the link, let me quote the crucial next sentence:

But it would open up a new window to the universe—and we’re likely to see interesting new things, once gravitational wave astronomy becomes a thing.

This is why the discovery is exciting.

In fact it’s already interesting to see a 36-solar-mass black hole collide with a 29-solar-mass black hole, because these are getting close to the elusive ‘intermediate mass black holes’, bigger than those made from a single star, yet smaller than the ‘supermassive’ black holes we see at the centers of galaxies.

We’ve seen lots of supermassive black holes, which are at least 100,000 times the mass of our Sun. There are various theories of how they formed, but nobody really knows.

An obvious possibility is that black holes made from collapsing stars collide and form bigger black holes, which in turn form bigger ones, etc. But nobody has definitively seen an intermediate mass black hole, meaning one between 100 and 100,000 solar masses! So astronomers are puzzled and intrigued by this issue.

I guess I should drop some of my natural restraint and say what I really think. According to general relativity, the Universe as we know it seems to have started with one big singularity (the Big Bang) and ends with many little ones (black holes). It’s very curious how gravity becomes dominant at both the beginning and the end. Is it a coincidence, or is the beginning somehow connected to the end? If we understood quantum gravity and the true nature of spacetime, we’d know more about what these so-called singularities really are. But it would be nice to understand the whole ‘life cycle’ of our Universe. Today, we made a small but notable step towards this goal. We opened a new pair of eyes.

Posted by: John Baez on February 12, 2016 5:00 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

The wreathe product of $S_4$ and $\mathbb{Z}_2$ contains the Weyl group of $D_4$, and I think that might be why it’s there (I mean, inside a group of form $\mathrm{SO}_8(\text{something})$, where you’d expect to find it).

Posted by: Tim Silverman on February 9, 2016 10:38 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

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 $\mathrm{E}_8$ lattices, $L_1$ and $L_2$, with disjoint root systems, in a (small) $\mathrm{E}_8$ lattice $\mathrm{L}_0$, 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 $\mathrm{E}_8$.

1) The $0$ vector lifts to a point in $2\mathrm{E}_8$. 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 $\mathrm{E}_8$, 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 $\mathrm{E}_8$, 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):

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

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 contains the product of two numbers. The first tells us the number of distinct ways to permute $a$, $b$ and $c$ to get different points of the Leech lattice. The second will be used when we are lifting to higher shells, and indicates the number of ways we can assign shells to $\mathrm{E}_8$ points according to the specified shape, as we will see below.

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

The $\text{#lift}$ column gives the number of lifts to $\mathrm{E}_8$, 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).

(We will be seeing higher lifts shortly.)

All vectors in the table above give a Leech vector of norm-squared at least $8$, except for the last row. There, we can lift one of the $\mathrm{E}_8$ points to the second available shell (with norm-squared $8$) to get some more points of the first Leech shell:

$\array{\arrayopts{\rowlines{solid}\collines{solid}\frame{solid}} \mathbf{x}&\mathbf{#x}&\mathbf{p q r}&\mathbf{B}&\mathbf{#p q r}&\mathbf{a b c}&\mathbf{Q(a b c)}&\mathbf{\text{perms}}&\mathbf{\text{Euclid.}}&\mathbf{\text{#lift}}\\ 0&1&000&000&1&0,0,0&000&1\cdot3&0+0+8=8&1\cdot1\cdot240=240 }$

If we now pick all 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

$\array{\arrayopts{\rowlines{solid}\collines{solid}\frame{solid}\collayout{left}} \mathbf{x}&\mathbf{p q r}&\mathbf{B(x,p q r)}&\mathbf{\text{Euclid.}}&\mathbf{\text{#points}}\\ x&p q r&011&4+2+2=8&15\cdot28\cdot6\cdot64=161280\\ x&0 q q&011&4+2+2=8&15\cdot8\cdot3\cdot64=23040\\ 0&0 q q&000&0+4+4=8&1\cdot15\cdot3\cdot256=11520\\ 0&000&000&0+0+8=8&1\cdot1\cdot3\cdot240=720 }$

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.

Leech shell $2$

Buoyed by our success, let us tackle the second shell of the Leech lattice, which would normally have norm-squared $6$ but in our scale has norm-squared $12$.

First, our table already contains a lot of points of norm-squared $12$.

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

But there are also several rows which give a minimum Euclidean norm-squared of $8$, but which will yield a norm-squared of $12$ if we raise one of the $\mathrm{E}_8$ points to the next available shell. For this, we need to know the lift factors for shell $3$ (with norm-squared $6$) and shell $4$ (with norm-squared $8$) in $\mathrm{E}_8$.

The integer-valued points in shell $3$ are of shape $\left(\pm1,\pm1,\pm1,\pm1,\pm1,\pm1,0,0\right)$ or shape $\left(\pm2,\pm1,\pm1,0,0,0,0,0\right)$. If we have a point in the reduced lattice of normal form $\left(1,1,0,0,0,0,0,0\right)$, it will lift to a) points $\left(0,0,\pm1,\pm1,\pm1,\pm1,\pm1,\pm1\right)$ with an odd number of minus signs (the lift has non-zero components precisely where the normal form has zero components) and b) points of the form $\left(1,-1,\pm2,0,0,0,0,0\right)$, where the $\pm1$ components appear in the same positions as the $1$ components of the normal form, and must have signs opposite to each other, while the $\pm2$ component can appear anywhere else and have either sign. There are $32$ points of the first kind, and $24$ of the second, totalling $56$.

(Similarly, we can lift normal form $\left(1,1,1,1,1,1,0,0\right)$ to points like $\left(\pm1,\pm1,\pm1,\pm1,\pm1,\pm1,0,0\right)$ with an even number of minus signs, and points like $\left(\pm2,0,0,0,0,0,\pm1,\pm1\right)$ where the components $\pm1$ must have the same sign, also giving $56$ lifts. And we can lift a normal form like $\left(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2}\right)$ by picking any two components, replacing each according to the rule $\pm\frac{1}{2}\rightarrow\mp\frac{3}{2}$ (note the change of sign), and possibly switching the sign of the whole vector, again giving $56$ lifts.)

In shell $4$, apart from points of shape $\left(\pm2,\pm2,0,0,0,0,0,0\right)$ belonging to $2\mathrm{E}_8$, we have points of shape $\left(\pm1,\pm1,\pm1,\pm1,\pm1,\pm1,\pm1,\pm1\right)$ and points of shape $\left(\pm1,\pm1,\pm1,\pm1,\pm2,0,0,0\right)$, and half-integer points with either $3$ components like $\pm\frac{3}{2}$ or one component like $\pm\frac{5}{2}$ with the rest being $\pm\frac{1}{2}$.

Sitting over normal form $\left(1,1,1,1,1,1,1,1\right)$ are those points $\left(\pm1,\pm1,\pm1,\pm1,\pm1,\pm1,\pm1,\pm1\right)$ with an even number of minus signs. (Those with an odd number belong to $2\mathrm{E}_8$.) There are $128$ of these. Similarly, sitting over normal form $\left(1,1,1,1,0,0,0,0\right)$ are a) those obtained from this by switching an odd number of signs and then replacing any of the zeroes by $\pm2$ and b) those obtained by taking the complement, switching an even number of signs, and replacing any of the (new) zeroes by $\pm2$. Again, this gives $128$ lifts.

To summarise:

$\array{\arrayopts{\rowlines{solid}\collines{solid}\frame{solid}} \mathbf{\text{shell}}&\mathbf{\text{norm-sq}}&\mathbf{\text{#lift}}\\ 0&0&1\\ 1&2&2\\ 2&4&16\\ 3&6&56\\ 4&8&128 }$

Now we will do the lifts into the second shell of the Leech lattice.

Here are the rows we will be working on:

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

Here are the available raises:

$\array{\arrayopts{\rowlines{solid}\collines{solid}\frame{solid}} \mathbf{x}&\mathbf{#x}&\mathbf{p q r}&\mathbf{B}&\mathbf{#p q r}&\mathbf{a b c}&\mathbf{Q(a b c)}&\mathbf{\text{perms}}&\mathbf{\text{Euclid.}}&\mathbf{\text{#lift}}\\ x&15&p q r&011&28&x+p,x+q,x+r&011&6\cdot2&4+2+6=12&16\cdot2\cdot56=1792\\ x&15&p q r&011&28&x+p,x+q,x+r&011&6\cdot1&8+2+2=12&128\cdot2\cdot2=512\\ x&15&0 q q&011&8&x,x+q,x+q&011&3\cdot2&4+2+6=12&16\cdot2\cdot56=1792\\ x&15&0 q q&011&8&x,x+q,x+q&011&3\cdot1&8+2+2=12&128\cdot2\cdot2=512\\ 0&1&0 q q&000&15&0,q,q&000&3\cdot2&0+4+8=12&1\cdot16\cdot128=2048 }$

Putting it all together, we get

$\array{\arrayopts{\rowlines{solid}\collines{solid}\frame{solid}\collayout{left}} \mathbf{x}&\mathbf{p q r}&\mathbf{B(x,p q r)}&\mathbf{\text{Euclid.}}&\mathbf{\text{#points}}\\ x&p q r&000&4+4+4=12&15\cdot7\cdot6\cdot4096=2580480\\ 0&p q r&000&4+4+4=12&1\cdot35\cdot6\cdot4096=860160\\ x&0 q q&000&4+4+4=12&15\cdot7\cdot3\cdot4096=1290240\\ x&000&000&4+4+4=12&15\cdot1\cdot1\cdot4096=61440\\ x&p q r&011&4+2+6=12&15\cdot28\cdot12\cdot1792=9031680\\ x&p q r&011&8+2+2=12&15\cdot28\cdot6\cdot512=1290240\\ x&0 q q&011&4+2+6=12&15\cdot8\cdot6\cdot1792=1290240\\ x&0 q q&011&8+2+2=12&15\cdot8\cdot3\cdot512=184320\\ 0&0 q q&000&0+4+8=12&1\cdot15\cdot6\cdot2048=184320 }$

The total $2580480+860160+1290240+61440+9031680+1290240+1290240+184320+184320=16773120$ which is indeed the correct number.

Posted by: Tim Silverman on February 16, 2016 7:49 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Here’s something I should have done before: count the unlifted triples in the above construction. That is, total up $\text{#x}\cdot\text{#p}\cdot\text{perms}$.

Repeating our table, we have

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

So we have

$15\cdot7\cdot6+15\cdot28\cdot6+1\cdot35\cdot6=16\cdot35\cdot6$

and

$15\cdot7\cdot3+15\cdot8\cdot3+1\cdot15\cdot3=16\cdot15\cdot3$

and

$15\cdot1\cdot1+1\cdot1\cdot1=16$

and

$16\cdot\left(35\cdot6+15\cdot3+1\right)=16\cdot256=4096$

This immediately tells us, or at least strongly suggests, that we have here a disguised version of the Turyn construction of the Golay code.

I don’t have Turyn’s paper, but the construction is sketched on pp. 316–317 of SPLAG, so let me try to painfully work backwards toward it from the Wikipedia article on the Leech lattice. I’m sure the exercise will do me good.

Here is a relevant construction of the Leech lattice.

The Leech lattice can be explicitly constructed as the set of vectors of the form $2^{-\frac{3}{2}}\left(a_1,a_2,\dots,a_{24}\right)$ where the $a_i$ are integers such that

$a_1+a_2+\dots+a_{24}\equiv4a_1\equiv4a_2\equiv\dots\equiv4a_{24}\text{ }\left(\text{mod }8\right)$

and for each fixed residue class modulo $4$, the $24$ bit word, whose $1$’s correspond to the coordinates $i$ such that $a_i$ belongs to this residue class, is a word in the binary Golay code.

Let’s unpack this.

First,

$4a_1\equiv4a_2\equiv\dots\equiv4a_{24}{ }\left(\text{mod }8\right)$

just says that the $a_i$ are all equivalent to each other mod $2$, that is, they are either all even or all odd. However, that’s for the Leech lattice times $2\sqrt{2}$. For the Leech lattice times $\sqrt{2}$, it says that the coordinates are all integers or all half-integers.

Second, $a_1+a_2+\dots+a_{24}\equiv4a_1\text{ }\left(\text{mod }8\right)$ etc says that the sum of even integers is a multiple of $8$, i.e. an even multiple of $4$, while the sum of odd integers is an odd multiple of $4$. Halving again, we have that:

The coordinates are either all integers summing to twice an even number, or all half-integers summing to twice an odd number.

Then

for each fixed residue class modulo $4$, the $24$ bit word, whose $1$’s correspond to the coordinates $i$ such that $a_i$ belongs to this residue class, is a word in the binary Golay code

This sounds a bit scary because there are $4$ residue classes mod $4$, but it turns out not to be.

If Leech times $2\sqrt{2}$ is all even integers, then there are only two actual residue classes mod $4$, viz. even and odd multiples of $2$. Hence in the Leech lattice times $\sqrt{2}$, these are just even and odd integers. So the even integers correspond to the $1$s of a Golay word. And so do the odd integers—but that is not a separate condition, because the complement of a Golay word is another Golay word. So we just have that an all-integer vector, reduced mod $2$, is a Golay word.

Then the same thing happens when the Leech times $2\sqrt{2}$ is all odd integers—there are just two residue classes present, viz. $1$ and $3$. So, to summarise:

The Leech lattice times $\sqrt{2}$ can be constructed as the vectors which are either:

a) All integers, summing to twice an even integer, such that reducing them mod $2$ gives a Golay word, or

b) All half-integers, summing to twice an odd integer, such that adding $\frac{1}{2}$ to all of them and reducing mod $2$ gives a Golay word.

What kind of pair of isotropic subspaces gives something like this?

Suppose the subspace from which we pick triples is all integers, while the other subspace is spanned by three all-integer vectors and one half-integer vector. Then half (i.e. $8$) of the vectors in the latter subspace would give half-integer Leech coordinates, and the other half would give integer Leech coordinates. That sounds promising.

For our first subspace, $M_1$, pick a Fano plane structure on $7$ points. Then the subspace is something we’ve seen before:

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. $\left(0,0,0,1,0,1,1,1\right)$.

iv) $\left(0,0,0,0,0,0,0,0\right)$

This subspace is (isomorphic to) the extended Hamming code, Hamming(8,4).

For the second subspace, $M_2$, pick another Fano plane structure sharing no lines with the first, and take just the subspace

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

(This is a subspace because

a) Coordinate $0$ is always $0$.

b) Two lines of the Fano plane intersect in a single point, so their corresponding complements in $M_2$ will both have $0$ in the coordinate corresponding to the point of intersection, and both will have $1$ in coordinates corresponding to the two points which lie in neither line (and therefore lie on the third line through the point of coincidence), but will differ at the coordinates corresponding to the other $4$ points, which lie in one line but not both. Hence the sum of the two points of $M_2$ will be $0$ at the coordinates corresponding to the third line through the point of intersection, and $1$ elsewhere, i.e. will be $1$ on just the coordinates corresponding to points that lie off that third line in the Fano plane.)

(While I’m here, I’ll just briefly add something about Fano plane structures on a set of $7$ elements. There are $30$ distinct structures, corresponding to the $15$ points and $15$ lines of $PG(3,2)$. Picking one structure and identifying it with one plane of the $3$-space, there are then $7$ Fano plane structures sharing $3$ concurrent lines with the plane, corresponding to those points of $PG(3,2)$ which lie in the given plane (viz. the points of concurrency); there are $14$ Fano plane structures sharing just one line with the given plane, and corresponding to the other planes that intersect the given one in the corresponding lines; and there are $8$ structures which share no lines—every line of each of these different structures is a triangle of the given plane. These correspond to the $8$ points of $PG(3,2)$ which lie off the plane.

Each of the $8$ sets of $7$ triangles is preserved (and cyclically permuted) by one of the $7$-cycles (of which there are $8$) in $PSL(3,2)$, the automorphism group of the Fano plane. Via the isomorphism $PSL(3,2)\simeq PSL(2,7)$, each of the $7$-cycles lies in the stabliliser of one of the $8$ points of the projective line over $\mathbb{F}_7$. This is because every occurence of the number $8$ that arises in this context is somehow cleverly related to every other occurrence of the number $8$, in a fantastical orgy of octitude.)

I was going to show how we only get words with $0$, $8$, $12$, $16$ or $24$ ones in them, using the possible intersections of a triangle with a triple of mutually concurrent lines, but it was a bit boring. Instead, I’ll just point out that what we’ve got so far, with an $8$-bit word corresponding to the complement of a line in one Fano plane structure being added to words corresponding to lines or the complements of lines summing to $0$ in another Fano plane structure is precisely the Turyn construction in SPLAG.

(For those who want to look it up: suppose SPLAG’s “line code” corresponds to a Fano plane structure with the standard labelling whose lines are $124$ together the results of successively incrementing each label in $124$ by $1$ (mod $7$), giving $235$ etc, we have a correspondence with their table 11.26(b)

$\array{1&a\\2&b\\3&c\\4&a+b\\5&b+c\\6&a+b+c\\7&a+c}$

Their table 11.27(a) giving the “line code” then has the mapping

$\array{124&D\\235&B\\346&C\\457&G\\561&E\\672&F\\713&A}$

And their table 11.27(b) giving the “point code” has the mapping

$\array{146&e\\473&b\\721&f\\254&g\\567&d\\632&a\\315&c}$

Here the “point code” triangles are related by the $7$-cycle $(1472563)$, which you can check also preserves the structure of the “line code”.)

So we’ve kind of retraced the historical development in reverse order, since I think the actual development was: Turyn shows how to build the Golay code from two versions of the Hamming code; then somebody (dunno who) uses this to build the Leech lattice from two versions of the $\mathrm{E}_8$ lattice; then Wilson shows how to get hold of the two $\mathrm{E}_8$ lattices using octonion multiplication.

Which makes me wonder if the octonion multiplication projects down to something interesting over $\mathbb{F}_2$.

Posted by: Tim Silverman on February 18, 2016 8:34 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Tim wrote:

This immediately tells us, or at least strongly suggests, that we have here a disguised version of the Turyn construction of the Golay code.

Good! I was hoping you’d get into this! I believe _Spheres, Lattice Packings and Groups’ gives a construction of the 24-dimensional Leech lattice from the 8-dimensional $\mathrm{E}_8$ lattice, which they call the ‘Turyn construction’ because it’s analogous to Turyn’s construction of the 24-bit Golay code from the 8-bit Hamming code. The stuff Greg Egan did is a spinoff of this line of thought.

I still haven’t read everything you’ve written, much less understood it. But seems a delightful project to first understand geometry over $\mathbb{F}_2$ very well and then try to ‘lift’ constructions there to constructions over the integers, i.e. constructions with lattice.

Which makes me wonder if the octonion multiplication projects down to something interesting over $\mathbb{F}_2$.

I’ve thought about this a bit. If you take the Cayley integral octonions mod 2 you get an 8-dimensional vector space $V$ over $\mathbb{F}_2$ equipped with a multiplication and a quadratic form $Q$ obeying

$Q(x y) = Q(x) Q(y)$

As you’ve shown, this quadratic form has plus type, and there are 120 elements $v \in V$ with $Q(v) = 1$. So, we get a 120-element ‘loop’: the nonassociative analogue of a group. This is just the usual 240-element loop of unit-norm Cayley integral octonions modulo its center, $\pm 1$.

If this 120-element loop were associative, it would be a group, and there are a number of tempting 120-element groups that it could be. I’m pretty sure it’s not associative. However, I’ve been so busy lately that I haven’t had time to check this!

So I guess that’s my main question right now. You can probably take 3 unit-norm Cayley integral octonions $a,b,c$ that aren’t at right angles to each other and show

$(a b)c \ne \pm a(b c)$

Posted by: John Baez on February 18, 2016 9:09 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Is this the split octonion algebra over $\mathbb{F}_2$? It seems like it should be.

Posted by: Tim Silverman on February 19, 2016 9:53 AM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Tim wrote:

Is this the split octonion algebra over $\mathbb{F}_2$? It seems like it should be.

I thought so for a while, but no.

First of all, the split octonion algebra over $\mathbb{F}_2$ is the same as the octonion algebra over $\mathbb{F}_2$, since the only difference is some signs, which don’t matter in $\mathbb{F}_2$.

Recall that in the Cayley–Dickson doubling construction, we start with a vector space $A$ equipped with a multiplication and ‘conjugation’ $\ast : A \to A$. We get a new one $CD(A)$, which is $A \oplus A$ with the multiplication

$(p, q) (r, s) = (p r - s^* q, s p + q r^*)$

and conjugation

$(p, q)^* = (p^*, - q)$

If we start with the real numbers with its usual multiplication and $a^* = a$, we get

$CD(\mathbb{R}) = \mathbb{C}$ $CD(\mathbb{C}) = \mathbb{H}$ $CD(\mathbb{H}) = \mathbb{O}$

If we start with the integers with its usual multiplication and $a^* = a$, we get lattices in $\mathbb{C}, \mathbb{H}$ and $\mathbb{O}$, closed under multiplication.

For starters,

$CD(\mathbb{Z}) = \mathbb{Z} \oplus \mathbb{Z} \subset \mathbb{R} \oplus \mathbb{R} = CD(\mathbb{R}) \cong \mathbb{C}$

is the Gaussian integers:

Repeating the process we get further lattices, all boringly cubical. For example:

$CD^2(\mathbb{Z}) = \mathbb{Z}^2 \oplus \mathbb{Z}^2 \subset \mathbb{C} \oplus \mathbb{C} = CD(\mathbb{C}) \cong \mathbb{H}$

So, $CD^2(\mathbb{Z})$ consists of the quaternions of form

$a + b i + c j + d k , \qquad a,b,c,d \in \mathbb{Z}$

Conway and Smith call these the Lipschitz integers. They are much less exciting than the Hurwitz integers, a subring of the quaternions that’s twice as dense.

By now you see where things are heading:

$CD^3(\mathbb{Z}) = \mathbb{Z}^4 \oplus \mathbb{Z}^4 \subset \mathbb{H} \oplus \mathbb{H} = CD(\mathbb{H}) \cong \mathbb{O}$

consists of the octonions of the form

$a_0 + \sum_{i = 1}^7 a_i e_i, \qquad a_i \in \mathbb{Z}$

Conway and Smith call these the Gravesian integers. They are much less exciting than the Cayley integers, which form a copy of the $\mathbb{E}_8$ lattice.

Sorry!

We can also do these constructions mod 2, getting algebras $CD^n(\mathbb{F}_2)$. It turns out that $CD^n(\mathbb{F}_2)$ is, as an algebra, isomorphic to a direct sum of $2^n$ copies of $\mathbb{F}_2$.

To see this, note that when you start with a field and repeatedly do the Cayley-Dickson construction, each time you throw in an extra square root of -1, which anticommutes and either associates or anti-associates with all the square roots of -1 you already threw in.

But when your field is $\mathbb{F}_2$, -1 is 1. So each time, you throw in an extra square root of 1, which commutes and associates with everything.

However, working over any field, if you start with an algebra $A$ and throw in an extra square root of 1 that commutes and associates with everything, you get the algebra $A \oplus A$: the direct sum of two copies of $A$. The reason is that we can think of the original $A$ as elements of the form $(a,a)$, and the new square root of 1 as $(1,-1)$.

So, if we apply the Cayley-Dickson construction $n$ times to $\mathbb{F}_2$, we get a direct sum of $2^n$ copies of $\mathbb{F}_2$.

Posted by: John Baez on February 19, 2016 4:28 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

the split octonion algebra over $\mathbb{F}_2$ is the same as the octonion algebra over $\mathbb{F}_2$

You’d think so, but no. Wilson discusses this in §4.3 of The Finite Simple Groups. The split octonions can be given a basis whose multiplication table works over any field, including those of even characteristic, whereas the non-split octonions are a bit trickier. (They are isomorphic to the split octonions over $\mathbb{C}$ and over finite fields of odd characteristic, but over finite fields of even characteristic they … don’t work.)

Actually, reading on to §4.4, I see he shows that you can define the split octonions over $\mathbb{F}_2$ as the Cayley integral octonions reduced mod $2$, and everything works. I’d read that section but apparently not taken it in. I’ll have to read it again carefully.

Posted by: Tim Silverman on February 19, 2016 9:13 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Tim wrote:

You’d think so, but no.

Then Wilson is using a different definition of ‘split octonions’ over an arbitrary field than I was.

I’m sure his is better, but just to be clear, mine was this: starting from a field we do the Cayley–Dickson construction as described above twice, but on the third time we switch to using the product

$(p,q)(r,s)=(p r + s^* q, s p+q r^*)$

I’m not surprised that something else is better in characteristic two!

Posted by: John Baez on February 19, 2016 9:25 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Yeah, sorry, that was needlessly dogmatic of me. Wilson was using the octonions to define $G_2(q)$ (and other exceptional groups), so my somewhat wild guess is that the process starts with the $\mathfrak{g}_2$ Lie algebra in Chevalley basis, uses this to get hold of $G_2(2)$ (and over other fields of characteristic $2$) and somehow works back from there to a definition of the octonion algebra which is “right” for this approach. But I don’t think he really explains why the approach he describes is “right” for what he wants to do, as the book is quite terse.

Posted by: Tim Silverman on February 19, 2016 9:49 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I should be a bit more informative. Wilson starts with the usual non-split multiplication table. For finite fields of odd characteristic, he says to pick elements $a,b\in\mathbb{F}_q$ such that $b\ne0$ and $a^2+b^2=-1$, then change basis via

$\array{ 2x_1=e_5+a e_7+b e_1&2x_8=e_5-a e_7-b e_1\\ 2x_2=e_3+b e_4+a e_6&2x_7=e_3-b e_4-a e_6\\ 2x_3=e_2-b e_7+a e_1&2x_6=e_2+b e_7-a e_1\\ 2x_4=1+a e_4-b e_6&2x_5=1-a e_4+b e_6 }$

We then get a multiplication table

$\array{\arrayopts{\equalcols{true}\collines{solid none}\rowlines{solid none}} &x_1&x_2&x_3&x_4&x_5&x_6&x_7&x_8\\ x_1&0&0&0&0&x_1&x_2&-x_3&-x_4\\ x_2&0&0&-x_1&x_2&0&0&-x_5&x_6\\ x_3&0&x_1&0&x_3&0&-x_5&0&-x_7\\ x_4&x_1&0&0&x_4&0&x_6&x_7&0\\ x_5&0&x_2&x_3&0&x_5&0&0&x_8\\ x_6&-x_2&0&-x_4&0&x_6&0&x_8&0\\ x_7&x_3&-x_4&0&0&x_7&-x_8&0&0\\ x_8&-x_5&-x_6&x_7&x_8&0&0&0&0 }$

Ignoring signs, this works over $\mathbb{F}_2$.

Posted by: Tim Silverman on February 19, 2016 9:38 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Yup, they’re nonassociative even over $\mathbb{F}_2$. Modding by 2 gets rid of some signs, but, say, picking $(1+e_1+e_2+e_4)/2,(1+e_1+e_3+e_7)/2$ and $(1+e_1+e_5+e_6)/2$ for $a,b$ and $c$ gives $(e_1-e_4-e_5+e_7)/2$ and $(e_2+e_3-e_4+e_7)/2$ for $(ab)c$ and $a(bc)$ respectively. So modding definitely doesn’t make things associative.

Note: I used the version of the Cayley integers where you swap $a_0$ and $a_1$.

Posted by: Layra on February 19, 2016 9:01 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thanks for checking that, Layra! It’s been on my to-do list for at least a month.

It follows that if we take the $\mathrm{E}_8$ lattice modulo two, we can the octonion multiplication and norm to equip it with a quadratic form $Q$ and product obeying $Q(a b) = Q(a) Q(b)$, and the 120 elements with $Q(a) = 1$ form an nonassociative loop.

I conjecture that this is a ‘simple’ loop, by which I mean one without any nontrivial quotients.

I don’t know how much progress people have made on classifying finite simple loops… apart from the case of groups, that is.

Posted by: John Baez on February 19, 2016 9:29 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

The 120-element loop we’re talking about obeys the Moufang identities, so this thesis is helpful:

It has a lot of information about split octonion algebras over finite fields, too!

Posted by: John Baez on February 19, 2016 9:36 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

I want to see how Wilson’s octonionic construction of the “large” $\mathrm{E}_8$ lattices translates down over $\mathbb{F}_2$.

Let’s take a basis of roots

$\array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \mathbf{0}&\mathbf{1}&\mathbf{2}&\mathbf{3}&\mathbf{4}&\mathbf{5}&\mathbf{6}&\mathbf{7}\\ 1&1&0&0&0&0&0&0\\ 0&1&1&0&0&0&0&0\\ 0&0&1&1&0&0&0&0\\ 0&0&0&1&1&0&0&0\\ 0&0&0&0&1&1&0&0\\ 0&0&0&0&0&1&1&0\\ 0&0&0&0&0&0&1&1\\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2} }$

We can multiply on the right by $s=\frac{1}{2}\left(-1+e_1+e_2+e_3+e_4+e_5+e_6+e_7\right)$ and by $\overline{s}=\frac{1}{2}\left(-1-e_1-e_2-e_3-e_4-e_5-e_6-e_7\right)$.

Lets make a table of what these do to each of the basis elements (doubling the answers to avoid annoying factors of $\frac{1}{2}$ everywhere).

$\array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot s&0&1&2&3&4&5&6&7\\ 1&-1&1&1&1&1&1&1&1\\ e_1&-1&-1&-1&-1&1&-1&1&1\\ e_2&-1&1&-1&-1&-1&1&-1&1\\ e_3&-1&1&1&-1&-1&-1&1&-1\\ e_4&-1&-1&1&1&-1&-1&-1&1\\ e_5&-1&1&-1&1&1&-1&-1&-1\\ e_6&-1&-1&1&-1&1&1&-1&-1\\ e_7&-1&-1&-1&1&-1&1&1&-1 }\;\;\; \array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot \overline{s}&0&1&2&3&4&5&6&7\\ 1&-1&-1&-1&-1&-1&-1&-1&-1\\ e_1&1&-1&1&1&-1&1&-1&-1\\ e_2&1&-1&-1&1&1&-1&1&-1\\ e_3&1&-1&-1&-1&1&1&-1&1\\ e_4&1&1&-1&-1&-1&1&1&-1\\ e_5&1&-1&1&-1&-1&-1&1&1\\ e_6&1&1&-1&1&-1&-1&-1&1\\ e_7&1&1&1&-1&1&-1&-1&-1 }$

Using the results in the top two tables, and remembering to divide by $2$, we get

$\array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot s&\mathbf{0}&\mathbf{1}&\mathbf{2}&\mathbf{3}&\mathbf{4}&\mathbf{5}&\mathbf{6}&\mathbf{7}\\ 1+e_1&-1&0&0&0&1&0&1&1\\ e_1+e_2&-1&0&-1&-1&0&0&0&1\\ e_2+e_3&-1&1&0&-1&-1&0&0&0\\ e_3+e_4&-1&0&1&0&-1&-1&0&0\\ e_4+e_5&-1&0&0&1&0&-1&-1&0\\ e_5+e_6&-1&0&0&0&1&0&-1&-1\\ e_6+e_7&-1&-1&0&0&0&1&0&-1\\ \frac{1}{2}(\dots)&-\frac{3}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2} }\;\array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot \overline{s}&\mathbf{0}&\mathbf{1}&\mathbf{2}&\mathbf{3}&\mathbf{4}&\mathbf{5}&\mathbf{6}&\mathbf{7}\\ 1+e_1&0&-1&0&0&-1&0&-1&-1\\ e_1+e_2&1&-1&0&1&0&0&0&-1\\ e_2+e_3&1&-1&-1&0&1&0&0&0\\ e_3+e_4&1&0&-1&-1&0&1&0&0\\ e_4+e_5&1&0&0&-1&-1&0&1&0\\ e_5+e_6&1&0&0&0&-1&-1&0&1\\ e_6+e_7&1&1&0&0&0&-1&-1&0\\ \frac{1}{2}(\dots)&2&0&0&0&0&0&0&0 }$

Now we convert these to normal form (removing an even number of minus signs, taking the complement if there are an odd number of minus signs, and making $\frac{3}{2}$ positive):

$\array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot s&\mathbf{0}&\mathbf{1}&\mathbf{2}&\mathbf{3}&\mathbf{4}&\mathbf{5}&\mathbf{6}&\mathbf{7}\\ 1+e_1&0&1&1&1&0&1&0&0\\ e_1+e_2&0&1&0&0&1&1&1&0\\ e_2+e_3&0&0&1&0&0&1&1&1\\ e_3+e_4&0&1&0&1&0&0&1&1\\ e_4+e_5&0&1&1&0&1&0&0&1\\ e_5+e_6&0&1&1&1&0&1&0&0\\ e_6+e_7&0&0&1&1&1&0&1&0\\ \frac{1}{2}(\dots)&\frac{3}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{1}{2} }\;\;\;\array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot \overline{s}&\mathbf{0}&\mathbf{1}&\mathbf{2}&\mathbf{3}&\mathbf{4}&\mathbf{5}&\mathbf{6}&\mathbf{7}\\ 1+e_1&0&1&0&0&1&0&1&1\\ e_1+e_2&1&1&0&1&0&0&0&1\\ e_2+e_3&1&1&1&0&1&0&0&0\\ e_3+e_4&1&0&1&1&0&1&0&0\\ e_4+e_5&1&0&0&1&1&0&1&0\\ e_5+e_6&1&0&0&0&1&1&0&1\\ e_6+e_7&1&1&0&0&0&1&1&0\\ \frac{1}{2}(\dots)&1&1&1&1&1&1&1&1 }$

Now, for integer coordinates with $4$ ones and $4$ zeros in normal form, let us have the following notation: if the first coordinate is $0$ and $e_a$, $e_b$ and $e_c$ also have coefficient $0$, then we write $a b c$ in some order. If the first coordinate is $1$, and$e_a$, $e_b$ and $e_c$ also have coefficient $1$, then we write $\overline{a b c}$ in some order.

Then we have

$\array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot s&\text{code}\\ 1+e_1&467\\ e_1+e_2&723\\ e_2+e_3&134\\ e_3+e_4&245\\ e_4+e_5&356\\ e_5+e_6&467\\ e_6+e_7&571 }\; \array{\arrayopts{\collines{solid}\rowlines{solid}\frame{solid}} \cdot s&\text{code}\\ 1+e_1&235\\ e_1+e_2&\overline{713}\\ e_2+e_3&\overline{124}\\ e_3+e_4&\overline{235}\\ e_4+e_5&\overline{346}\\ e_5+e_6&\overline{457}\\ e_6+e_7&\overline{561} }$

Now, in the first table, we have $6$ members of a heptad of triangles related by the $7$-cycle $(1234567)$ (the triangle $612$ is missing while $467$ is repeated in the first line and line $e_5+e_6$). These triangles are not all independent as vectors, but between them they generate a Fano plane. Together with the last vector, they give a maximal isotropic space.

In the second table, we have the complements of the standard lines of the Fano plane, except that $235$ appears both directly in the first line and as a complement in line $e_3+e_4$, while $672$ is missing. Together with the last line, they give also give a maximal isotropic space, disjoint from the first one.

(In fact, by a weird coincidence, they are my two favorite such spaces in $\mathrm{E}_8$ mod $2$: one with a standard set of Fano plane lines, their complements, and $(1,1,1,1,1,1,1,1)$; the other with my favorite heptad of triangles (the one preserved by the $7$-cycle $(1234567)$) together with my favorite half-integer vector in the second shell. I thought I had actually used these as my examples in the original post, but it seems they only appear in my notes. Well, I guess it isn’t a coincidence, but it sure is weird.)

Posted by: Tim Silverman on February 19, 2016 11:06 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Hello everyone, before I make a complete fool of myself leet me mention that I’m a physicist and would require a few years of math courses to properly catch up with all the points in this discussion. Don’t hesitate to delete this comment if its too stupid for this place.

I ended up here because I am trying to understand the structure of the symplectic subgroup of $W(E_8)$. According to Buser+Sarnak(+Conway+Sloane), 1994, $E_8$ is in fact also a symplectic lattice. As such, there exists a symplectic basis of pairs of symplectically conjugate vectors. One such basis is provided by

$(0,0,0,1,-1,0,0,0)$ $(0,0,0,2,-1,-1,0,0)$ $(0,0,0,3,-1,-1,-1,0)$ $(1,1,1,1,-1,-1,-1,-1)$ $(1,-1,0,0,0,0,0,0)$ $(0,1,-1,0,0,0,0,0)$ $(0,0,1,-1,0,0,0,0)$ $(1/2,1/2,1/2,3/2,-1/2,-1/2,-1/2,-1/2),$

where each pair transforms automorphically under $\mathrm{SL}_2(\mathbb{Z})$ in $E_8/2E_8$ (and can be used to encode a qubit via the Gotteman-Kitaev-Preskill construction). I think this may relate to the earlier comments regarding the splitting of $E_8/2E_8$….

Does anybody know how to explicitly find the symplectic subgroup of $W(E_8)$? Buser, Sarnak (1994) already state that it comes with structure $4.2^4 S_6$ but it is totally unclear to me how this is derived and how it is generated and acting on $E_8$

Best, Jonathan

### Re: Integral Octonions (Part 12)

This is an interesting question, but it’s too hard for me. Maybe someone else knows more.

It seems like you’re saying $\mathbb{R}^8$ admits, in addition to its real inner product and $E_8$ lattice, a symplectic structure for which the $E_8$ lattice has generators coming in symplectically conjugate pairs. Does the Buser–Sarnak paper explain it in terms of some Riemann surface? It’s paywalled and I’m too lazy to get ahold of it so I’ll just guess from the title:

The de Rham cohomology of any oriented surface of genus 4 is an 8-dimensional symplectic vector space containing a lattice $L$ with generators coming in symplectically conjugate pairs. If we give the surface a Kähler structure this vector space also gets a real inner product. Perhaps we can do this in such a way that $L$ becomes a copy of the $E_8$ lattice?

Even if true, I’m not sure this is enough to shed much light on the intersection of $W(E_8)$ with the symplectic group, but I don’t have any other tricks for figuring this out (other than brute-force computation using the basis you described).

Posted by: John Baez on October 27, 2022 10:45 PM | Permalink | Reply to this

### Re: Integral Octonions (Part 12)

Thank you very much for this answer. Sorry, I did not notice the paywall on the article. In fact the statement I am after is discussed in an appendix to that paper written by Conway and Sloane (a part of it are available on his website [http://neilsloane.com/doc/pub.html] ), where they prove that a lattice is symplectic if it has an isoduality i: i^2=-1 which implies that a integral lattices is symplectic if and only if it is unimodular and has an automorphism i. They use this statement to show that D^+{4m}, A6^(2), E8 and the Leech lattice \Lambda{24} are symplectic. A representation in H^g is also given but I cannot see how this allows for a non-brute-force identification of the symplectic orthogonal automorphisms and hence was hoping to find more elementary arguments.

Starting at a smaller example like A2, the even subgroup of W(A2) coincides with the commutator subgroup and consists of Rot(2\pi /3) rotations in the plane. Since in 2 dimensions the rotation axis coincides with the rotation axis of the symplectic form, which is a Rot(-\pi /2) rotation this is symplectic, we thus have that the intersection of W(A2) and the symplectic group is simple the even/commutator subgroup of W(A2)..some numerical checks I did indicate that this does not as simple anymore for E_8.