# What other lattices are obtainable from this noncommutative ring?

Here I will regard $$SU(2)$$ as the multiplicative group of unit quaternions.

There are just three conjugacy classes of finite subgroups $$G < SU(2)$$ where $$[G:C] > 2$$ for all cyclic subgroups $$C < G$$:

(i) If $$|G| = 24$$, $$G \cong SL_{2}(3)$$. This $$G$$ has a normal subgroup of order 8 which can be taken to be $$Q_{8} = \{ \pm{1}, \pm{i}, \pm{j}, \pm{k} \}$$. The elements of $$G$$ outside this subgroup of index 3 can be taken to be $$\{ \pm{\frac{1}{2}}\pm{\frac{i}{2}}\pm{\frac{j}{2}}\pm{\frac{k}{2}} \}$$, where the signs are taken every possible way.

(ii) If $$|G| = 120$$, $$G \cong SL_{2}(5)$$. The elements of this group can be described in terms of $$\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$$, but I won't go into that here.

(iii) If $$|G| = 48$$, $$G$$ is a double cover of $$S_{4}$$ sometimes called the binary octahedral group. This $$G$$ has a subgroup of order 24 which can be taken to consist of the elements described in (i). The extra elements of $$G$$ can be taken to be $$\frac{u+v}{\sqrt{2}}$$, where $$u$$ and $$v$$ are non-proportional elements of $$Q_{8}$$.

The Hurwitz ring of integral quaternions is the ring consisting of finite sums of elements of the group described in (i). Wilson, in The Finite Simple Groups, calls this ring the tetrian ring.

The ring consisting of finite sums of elements of (ii) is sometimes called the icosian ring. Wilson describes how to describe the Leech lattice as a 3-dimensional lattice over this ring, and SPLAG describes how to identify this ring with the $$E_{8}$$ lattice.

The ring I want to ask about here is the ring consisting of finite sums of elements of the group in (iii), which I've called the octian ring. Here, I'll use $$\mathbb{O}$$ to denote the octian ring (also used elsewhere for the octonions, but all rings discussed here are associative, so there's no danger of confusion). If I have computed correctly, it can be identified with the $$E_{8}$$ lattice as follows:

For the purpose of this construction, we define the Euclidean norm of an element of $$\mathbb{Q}(\sqrt{2})$$ by $$EN(u+v\sqrt{2}) = u+v$$, where $$u,v \in \mathbb{Q}$$. The quaternion norm is the usual $$QN(a + bi + cj + dk) = a^{2}+b^{2}+c^{2}+d^{2}$$, where $$a,b,c,d \in \mathbb{R}$$. In fact, the quaternion norm of an element of $$\mathbb{O}$$ is always in $$\mathbb{Z}[\sqrt{2}]$$, but establishing that takes a little more work.

For $$x,y \in \mathbb{O}$$, the inner product that identifies $$\mathbb{O}$$ with the $$E_{8}$$ lattice is

$$\langle x,y\rangle = EN(QN(x+y)) - EN(QN(x)) - EN(QN(y)).$$

An integral basis for $$\mathbb{O}$$ can be given by:
$$\begin{array}{ccl} v_{1} &=& \frac{1}{2}+\frac{i}{2}+\frac{j}{2}+\frac{k}{2} \\ v_{2} &=& \frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \\ v_{3} &=&i \\ v_{4} &=&\frac{i}{\sqrt{2}}+\frac{j}{\sqrt{2}} \\ v_{5} &=&j \\ v_{6} &=&\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}} \\ v_{7} &=&k \\ v_{8} &=&k\sqrt{2}. \end{array}$$

With respect to this basis, the Gram matrix of the lattice with which $$\mathbb{O}$$ becomes identified is (if I computed correctly)
$$\begin{pmatrix} 2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 2 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 2 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 2 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 & 2 & 1 & 2 \\ 1 & 0 & 0 & 0 & 0 & 1 & 2 & 2 \\ 1 & 0 & 0 & 0 & 0 & 2 & 2 & 4 \end{pmatrix}$$ To verify that this identifies $$\mathbb{O}$$ with the $$E_{8}$$ lattice, it suffices to verify that the lattice thus obtained is positive definite, even, and unimodular.

Positive definiteness: Any element of $$\mathbb{O}$$ has a quaternion norm which is a sum of 4 squares of elements of $$\mathbb{Q}(\sqrt{2})$$. The Euclidean norm of $$(s+t\sqrt{2})^{2}$$ is $$s^{2}+2st+2t^{2}$$, which is positive for $$s,t \in \mathbb{Q}$$ except when $$s=t=0$$. Therefore any nonzero element of $$\mathbb{O}$$ has positive inner product with itself, as claimed.

Evenness: All of the entries of the Gram matrix are integers, and all of the diagonal entries are even. This is sufficient to verify evenness.
A proof which does not depend on the computation of the Gram matrix notes that $$ = EN(QN(2x)) - 2EN(QN(x)) = 2EN(QN(x))$$. Since $$QN(x) \in \mathbb{Z}[\sqrt{2}]$$ for all $$x \in \mathbb{O}$$, $$EN(QN(x)) \in \mathbb{Z}$$ and evenness follows.

Unimodularity: If I entered this correctly into a determinant calculator, the determinant of this Gram matrix is 1.
A proof which doesn't depend on calculation of an 8x8 Gram matrix determinant can be given as follows:
First, consider the sublattice spanned by $$\{ 1, \sqrt{2}, i, i\sqrt{2}, j, j\sqrt{2}, k, k\sqrt{2} \}$$.
The planes spanned by $$\{ 1, \sqrt{2} \}$$, $$\{ i, i\sqrt{2} \}$$, $$\{ j, j\sqrt{2} \}$$, and $$\{ k, k\sqrt{2} \}$$ are mutually perpendicular, so the volume of a fundamental region of the sublattice we are considering is the product of the areas of the fundamental regions of the indicated 2-dimensional sublattices within those planes.
It is an easier computation that the Gram matrix of any of those 2-dimensional sublattices is $$\begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix}$$
The determinant of that Gram matrix is 4, which means the area of a fundamental region of any of those 2-dimensional sublattices is 2. Then the volume of a fundamental region of that 8-dimensional sublattice is $$2^{4}=16$$.
$$\mathbb{O}$$ is obtained by extending that sublattice by sums of evenly many of $$\{ \frac{1}{\sqrt{2}}, \frac{i}{\sqrt{2}}, \frac{j}{\sqrt{2}}, \frac{k}{\sqrt{2}} \}$$, and also by $$\frac{1}{2} + \frac{i}{2} + \frac{j}{2} + \frac{k}{2}$$. These reduce the volume of a fundamental region by a factor of $$2^{3} \cdot 2 = 16$$, so the final volume of a fundamental region is 1.

I hope the computation I detailed above isn't erroneous in a serious way, but I also want to ask:

What other discrete additive subgroups of $$\mathbb{R}^{8n}$$ can be identified (up to congruence) with free modules over $$\mathbb{O}$$?

Addendum: The minimal vectors of $$E_{8}$$

I want to describe how the minimal vectors of $$E_{8}$$ can be seen in this construction.

Recall that an algebraic number field is totally real if every embedding of it into $$\mathbb{C}$$ lands in $$\mathbb{R}$$. Within a totally real field, an element may be called totally positive if it and all its Galois conjugates are positive real numbers. Any nonzero square, or nonzero sum of squares, of elements of a totally real field will be totally positive.

$$\mathbb{Q}(\sqrt{2})$$ is a totally real field. Therefore quaternion norms of nonzero elements of $$\mathbb{O}$$ are totally positive elements of $$\mathbb{Z}[\sqrt{2}]$$. The minimal vectors of the lattice obtained this way from $$\mathbb{O}$$ are the $$x$$ such that $$2 = \langle x,x\rangle = EN(QN(2x)) - 2EN(QN(x)) = 2EN(QN(x))$$, so they come from $$\alpha \in \mathbb{Z}[\sqrt{2}]$$ such that $$\alpha$$ is totally positive, and $$EN(\alpha) = 1$$. There are just three such $$\alpha$$: $$1$$, $$2-\sqrt{2}$$, and $$3-2\sqrt{2}$$.

Quaternion norm $$1$$:

These are just the unit quaternions in $$\mathbb{O}$$, which form the binary octahedral group. There are 48 of these.

Quaternion norm $$2-\sqrt{2}$$:

Some of these have the form $$(1-\frac{1}{\sqrt{2}})u + \frac{v}{\sqrt{2}}$$, where $$u,v \in Q_{8}$$ are non-proportional elements as before. There are $$4 \cdot 3 \cdot 2^{2} = 48$$ of these. The others are the images of $$1-\frac{1}{\sqrt{2}} +(1-\frac{1}{\sqrt{2}})i + \frac{j}{2} + \frac{k}{2}$$ under the signed permutations of $$\{ 1,i,j,k \}$$. There are $$\frac{4! \cdot 2^{4}}{2^{2}} = 96$$ of these.

Quaternion norm $$3-2\sqrt{2}$$:

Some of these are of the form $$(1-\frac{1}{\sqrt{2}})u + (1-\frac{1}{\sqrt{2}})v$$, where again $$u$$ and $$v$$ are non-proportional elements of $$Q_{8}$$. There are $$\binom{4}{2}\cdot 2^{2} = 24$$ of these. Others are of the form $$\pm (1-\frac{1}{\sqrt{2}}) \pm (1-\frac{1}{\sqrt{2}})i \pm (1-\frac{1}{\sqrt{2}})j \pm (1-\frac{1}{\sqrt{2}})k$$. Since the ambiguous signs can be chosen every possible way, there are $$2^{4} = 16$$ of these. The last ones are the set $$(\sqrt{2}-1)Q_{8}$$. There are $$8$$ of these.
(More generally, all of the minimal vectors whose quaternion norm is $$3-2\sqrt{2}$$ are elements of the binary octahedral group, scaled by a factor of $$\sqrt{2}-1$$. Possibly this should replace the three cases above.)

Taken together, these account for $$48+48+96+24+16+8=240$$ vectors, which is the full count of minimal vectors of the $$E_{8}$$ lattice.

• Should the Euclidean norm of an element of $\mathbb{Q}(\sqrt{2})$ be $|u|+|v|$, or $u^2 + v^2$? I only ask, since otherwise it's not non-negative or non-degenerate. – David Roberts Mar 21 '20 at 6:50

I don't know what kind of answer you're looking for, since you have characterized the desired class of lattices $$L$$ pretty well already as those containing the designated quaternion order inside their endomorphism ring $$\mathrm{End}(L)$$.

To recover many of our favorite lattices over (possibly noncommutative) rings along the lines that you have suggested for $$E_8$$, see Martinet, Perfect lattices in Euclidean spaces; chapter 8 considers "Hermitian structures" on lattices (including the noncommutative case), and section 8.2 recovers $$E_8$$.

Perhaps another way to think about this is to think about things just in terms of automorphisms? Are you looking for a characterization in terms of the integral representation theory of the binary octahedral group $$G=2O$$? In 11.5.4 and Proposition 32.7.1(b) of my book (http://quatalg.org), what you are asking for is pretty close to asking for $$L$$ to be a $$\mathbb{Z}[\sqrt{2}][G]$$-module in a specified way?

I wrote an expository post asking people to check this construction here:

So far one person has checked it with the help of a computer.

• In your blog post, you asked if this had been known/done before. This paper from 1999 uses the terms "octian" and "octian ring" (so the naming is consistent): google.com/… But these authors' main concern is hyperbolic 4- and 5-space, not Euclidean 8-space. – DavidLHarden Aug 10 '20 at 22:28

A post in which John Baez cites Greg Egan describes how to obtain the Leech lattice from the $$E_{8}$$ lattice using an orientation-preserving rigid motion $$R: \mathbb{R}^{8} \to \mathbb{R}^{8}$$ such that:

$$< r_{i} , \sqrt{2}R r_{i} > = < r_{i} , \sqrt{2}R^{-1} r_{i} > = 1$$

for all simple roots $$r_{i}$$ of the $$E_{8}$$ lattice.
(This depends on the choice of a set of simple roots among the minimal vectors of the lattice, but the reasoning that follows, which applies to all minimal vectors, obviates the need for such a choice.)
If we wish to describe such a motion in octian terms, then it's good to note the following:
Lemma. Let $$\beta \in \mathbb{Q}\otimes\mathbb{O}$$ be such that $$QN(\beta)=1$$. Then the map $$x \to \beta \cdot x$$ is a rigid motion from $$\mathbb{Q}\otimes\mathbb{O}$$ to itself, in the sense defined by the octian inner product.
Proof. $$< \beta\cdot u, \beta\cdot v> = EN(QN(\beta\cdot u + \beta\cdot v)) - EN(QN(\beta \cdot u)) - EN(QN(\beta \cdot v))=\ EN( QN(\beta)QN(u+v) ) - EN(QN(\beta)QN(u)) - EN(QN(\beta)QN(v)) =\ EN(QN(u+v)) - EN(QN(u)) - EN(QN(v)) = $$. (The same argument applies to right-multiplications by a fixed unit quaternion in the skew-field, so it's omitted.)
One possible $$\beta$$ that allows the Baez-Egan construction to be applied here is
$$\beta = \frac{1}{\sqrt{8}} + i(\frac{1}{2} + \frac{1}{\sqrt{8}}) + j(\frac{1}{2} - \frac{1}{\sqrt{8}}) + \frac{k}{\sqrt{8}}$$, for which we have
$$\beta^{-1} = \frac{1}{\sqrt{8}} - i(\frac{1}{2} + \frac{1}{\sqrt{8}}) - j(\frac{1}{2} - \frac{1}{\sqrt{8}}) - \frac{k}{\sqrt{8}}$$.
Before proceeding further, note that $$\sqrt{2}\beta \in \mathbb{O}$$, so $$x \to \sqrt{2}\beta \cdot x$$ carries $$\mathbb{O}$$ into itself.
As for showing that the inner products work out as required by Baez and Egan, we show that $$< \sqrt{2}\beta u, u > = 1$$ for any $$u \in \mathbb{O}$$ such that $$< u,u > = 2$$ (or, equivalently, $$EN(QN(u)) = 1$$):
$$<\sqrt{2}\beta u, u> = -< \sqrt{2}\beta u, -u> = -[ EN(QN( \sqrt{2}\beta u- u )) - EN(QN( \sqrt{2}\beta u )) - EN(QN( -u )) ] =\ -EN(QN( \sqrt{2}\beta u - u )) + EN(QN( \sqrt{2}\beta u )) + EN(QN(u)) =\ -EN(QN( \sqrt{2}\beta - 1)QN(u)) + EN(QN( \sqrt{2}\beta )QN(u)) + EN(QN(u))$$.
Note that $$Re(\sqrt{2}\beta) = \frac{1}{2}$$, so it's easy to see that $$QN( \sqrt{2}\beta ) = QN( \sqrt{2}\beta - 1 )$$.
Then the preceding expression for $$< \sqrt{2}\beta u, u >$$ simplifies to
$$EN(QN(u)) = 1$$, and we are done.
It remains to show that $$x \to \beta \cdot x$$ is orientation-preserving, as a rigid motion of $$\mathbb{Q}\otimes\mathbb{O}$$.
An easy way to see it is this:
If $$\beta$$ were orientation-reversing, its $$-1$$-eigenspace would be odd-dimensional and therefore remain nonempty when the zero vector is removed. A nonzero unit vector in this eigenspace could be approximated arbitrarily well by a unit (in the $$QN$$ sense) element of $$\mathbb{Q}\otimes\mathbb{O}$$. Then we would have $$< \beta u, u >$$ being negative for some $$u \in \mathbb{Q}\otimes\mathbb{O}$$ with $$< u, u> = 1$$.
But actually $$< \beta u, u > = < \beta u u^{-1}, u u^{-1} > = < \beta, 1 > = 2 Re(\beta) = \frac{1}{\sqrt{2}} > 0$$, for a contradiction.