I am trying to find necessary and sufficient conditions for a nondegenerate lattice in one of the real division algebras $\mathbb{K}$ to admit the structure of a ring with identity (alternative algebra with identity in the case of $\mathbb{K} = \mathbb{O}$), with addition, multiplication and their identities the same as in $\mathbb{K}$.

I have found these conditions for the simplest cases $\mathbb{K} = \mathbb{R}$ and $\mathbb{K} = \mathbb{C}$ (if I haven't made any mistake), but I am having trouble determining them for the other two cases; see the section below for more details.

In summary, my question is:

What are the necessary and sufficient conditions for a nondegenerate lattice in $\mathbb{H}$ or $\mathbb{O}$ to also be an alternative algebra with identity?

I would also appreciate any reference that deals with this question.

Additional details

The cases of $\mathbb{R}$ and $\mathbb{C}$ are easy: the only lattice in the reals which is also a ring is $\mathbb{Z}$, while in $\mathbb{C}$ it is easy to see that the lattice is generated by $1$ and the element with smallest norm that is not an integer, say $a$. Then, since $a^2$ is in the ring and hence in the lattice, we have

$$a^2 = m\cdot a + n\cdot 1,$$

with $m, n$ integers. This is a monic quadratic equation, which means that the lattice can only be an order in the quadratic imaginary field $\mathbb{Q}[\sqrt{-d}]$, with $d$ positive (else the lattice would be degenerate). Since all orders of this form generate lattices in $\mathbb{C}$ we are done. Two well-known examples with exceptionally large unit group are the Gaussian and Eisenstein integers $\mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$ with $\omega = (1+\sqrt{3}i)/2$; they generate the square and hexagonal lattices respectively.

Over the quaternions and octonions, we can again choose a basis for the lattice consisting of $1$, an element $a_1$ of smallest norm that is not a multiple of $1$, an element $a_2$ of smallest norm that is not in the plane spanned by $1$ and $a_1$, and so on. Each one of the basis elements (excluding $1$) satisfies a monic quadratic equation like the one above, since $a_i^2$ is inside $\mathbb{R}[a_i]$ (any 2D slice of $\mathbb{H}$ or $\mathbb{O}$ containing $\mathbb{R}$ is isomorphic to the complex plane). This again forces the $a_i$ to be quadratic integers, though not necessarily with the same discriminant.

In the case of $\mathbb{H}$, this condition is necessary but not sufficient for the lattice to be a ring, since the product of distinct basis elements $a_i a_j$ is in the lattice and thus imposes new conditions relating their respective discriminants; for example, from the expansion of $a_1a_2$ as a combination of basis elements, I found that the cosine of the angle between $\Im(a_1)/|\Im (a_1)|$ and $\Im(a_2)/|\Im (a_2)|$ (thought of as vectors in the unit sphere $S^2$ of purely imaginary elements) must be in $\mathbb{Q}[\sqrt{d_1 d_2}]$, where $d_1$ and $d_2$ are the respective discriminants of $a_1$ and $a_2$ as quadratic integers.

At this point the algebra gets messy and I haven't been able to find all the necessary conditions. My current guess is that all ring-lattices come from orders in quaternion algebras $\mathbb{Q}[\alpha, \beta]$ with $\alpha^2=-d_1, \beta^2=-d_2, \alpha\beta=-\beta\alpha$ (they are the only working examples I have been able to produce), but I know very little about these algebras and I haven't so far managed to prove whether this is the case. Perhaps this result is well known, but I haven't found it anywhere I have looked (though maybe I am missing the correct terms to look for). Examples with exceptional symmetry are the Hurwitz and duoprismatic integers, spanned (up to conjugacy) by $1,i,j,\frac{1+i+j+k}{2}$ and $1,\omega, j, j\omega$. They generate the 24-cell and 6,6-duoprismatic lattices respectively.

In the case of $\mathbb{O}$, every 4D slice containing $\mathbb{R}$ is isomorphic to the quaternions, but I again expect new conditions that paste together these copies of $\mathbb{H}$ to form an octonionic lattice structure. Here I don't even know where to start. As an example, Conway and Smith's book On Quaternions and Octonions describes the "octavian integers" which generate a copy of the $E_8$ lattice.

Possibly useful facts

Here are some facts I found that could be useful in answering the question.

-First, since in all cases basis elements are quadratic integers, this means that the real part of any element of the lattice is in $\mathbb{Z}[\frac12]$.

-This implies that the conjugate $\bar{a}=2\Re(a)-a$ is in the lattice, and thus the squared norm $|a|^2 = a\bar{a}$ is in the intersection of the lattice and $\mathbb{R}_{>0}$, i.e. a positive integer. This means that every lattice element is a quadratic integer, not just the basis elements, since every number in $\mathbb{K}$ satisfies the equation $a^2 = 2\Re(a)\cdot a + |a|^2 \cdot 1$.

-Thanks to the polarization identity, the previous fact implies that the scalar product of any two elements, if we think of the division algebra as an Euclidean vector space over the reals, must be in $\mathbb{Z}[\frac12]$.

-The commutator of two lattice elements and the associator of three lattice elements must also belong to the lattice, as they can be expressed as sums of products.

  • $\begingroup$ $\mathbb{H} =\mathbb{C} + j\mathbb{C}$. Won't the ring be contained in $\mathcal{O}_{\mathbb{Q}(\sqrt{-d})}+ j\,\mathcal{O}_{\mathbb{Q}(\sqrt{-d})}$ up to some rotation of the $i,j,k$ ? $\endgroup$ – reuns Aug 17 '17 at 12:03
  • $\begingroup$ (Deleted parts that a second reading of the post showed the OP to already be familiar with) If you really want to delve into this, you can study the tome Maximal orders by Reiner. $\endgroup$ – Jyrki Lahtonen Aug 17 '17 at 12:04
  • 1
    $\begingroup$ @reuns I think the Hurwitz' order $$\Bbb{Z}\oplus\Bbb{Z}i\oplus\Bbb{Z}j\oplus\Bbb{Z}(\frac{1+i+j+k}2)$$ won't split into an orthogonal direct sum of two 2-dimensional lattices, so it won't be contained in an order of the form you described. $\endgroup$ – Jyrki Lahtonen Aug 17 '17 at 12:11
  • $\begingroup$ @reuns Thanks, could you elaborate? I think that splitting doesn't happen in all cases, for example, consider the lattice-ring generated by $1, \sqrt{2}i, \sqrt{3}j, \sqrt{6}k$. If your "$j$" is allowed to square to a negative integer different from $-1$, then if I'm not mistaken I think your statement could be true, and in fact it would imply that any such lattice-ring is a quaternion algebra, which is my current hunch. $\endgroup$ – pregunton Aug 17 '17 at 12:39
  • $\begingroup$ @JyrkiLahtonen Thank you very much, I will check that reference! $\endgroup$ – pregunton Aug 17 '17 at 12:39

Based on what you see for $\mathbb R$ and $\mathbb C$, the following is a natural guess:

A full lattice $R$ of $\mathbb H$ (resp. $\mathbb O$) which is also a subring is the same as an order in a (resp. the) definite quaternion (resp. octonion) algebra over $\mathbb Q$.

(By a full lattice, I mean a $\mathbb Z$-lattice of rank 4 (resp. 8) which yields all of $\mathbb H$ (resp. $\mathbb O$) upon tensoring with $\mathbb R$.) I did not check all details, but here is a sketch.

One direction is obvious, as we can embed a composition algebra over $\mathbb Q$ into one of the same type over $\mathbb R$.

Consider the case of $\mathbb H$. Let $R$ be such a lattice in $\mathbb H$. Take a basis of the form $1, a_1, a_2, a_3$. Now each $a_i$ has a quadratic minimal polynomial, so $\mathbb Q(a_i)$ is a quadratic subfield of $\mathbb H$, i.e., a quadratic imaginary field. Further $\mathbb Z[a_i]$ must be an order in $\mathbb Q(a_i)$, as you observed. Let $B$ be the $\mathbb Q$-algebra generated $a_1, a_2, a_3$, i.e., $B = R \otimes_{\mathbb Z} \mathbb Q$. This is a 4-dimensional $\mathbb Q$-algebra, and one can check it must be simple, and thus is a quaternion algebra over $\mathbb Q$ by Wedderburn's classification theorem for central simple algebras. (In fact, $B$ is generated by $a_1$ and $a_2$---cf. the proof of Theorem 3.1.3 in my quaternion algebra notes). Since it is contained in $\mathbb H$, $B$ must be definite.

The case of $\mathbb O$ should be similar. Here you'll probably want to use the fact that any two elements in an octonion algebra commute or generate a quaternion subalgebra, and the classification of central simple alternative algebras.

  • $\begingroup$ Many thanks for your answer! I think that mostly solves my questions, I will try to work out the details in the octonionic case. $\endgroup$ – pregunton Aug 20 '17 at 19:40

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