# Does every positive-definite integral lattice admit an angle-preserving homomorphism into $\Bbb Z^n$ for some $n$?

Some initial clarifications

By lattice I mean an additive subgroup of $$\mathbb R^n$$ which is isomorphic to $$\mathbb Z^n$$ and has full rank (i.e. spans $$\Bbb R^n$$ when considered as set of vectors). A lattice $$\mathcal L$$ is integral if $$\langle v,w\rangle\in\mathbb Z$$ for all $$v,w\in\mathcal L$$, where $$\langle\cdot,\cdot\rangle$$ denotes the standard inner product.

Above terms are standard, but some of the terms I will use in the following, like "sublattice" and "lattice isomorphism", are probably not standard. Feel free to correct my terminology, as I simply do not know enough of that subject.

The actual question

I wonder whether every integral lattice $$\mathcal L$$ is isomorphic to a sublattice of $$\mathbb Z^n$$ in the following sense:

Consider some vectors $$v_1,..., v_k\in\mathbb Z^n$$. Then $$\mathcal L(v_1,...,v_k):=\mathrm{span}\{v_1,...,v_k\}\cap \mathbb Z^n$$ is an integral lattice in $$\mathrm{span}\{v_1,...,v_k\}\subseteq\mathbb R^n$$ (with the inner product inherited from $$\mathbb Z^n$$), and will be called a sublattice of $$\mathbb Z^n$$.

I care about whether I can find $$\mathcal L$$ in $$\Bbb Z^n$$ with the right angles, not necessarily the right scale. So I need the following kind of "isomorphism": two lattices $$\mathcal L_1$$ and $$\mathcal L_2$$ are isomorphic, if there is a group isomorphism $$\phi:\mathcal L_1\to\mathcal L_2$$ and a constant $$\alpha\in\mathbb R$$ with

$$\langle \phi(v),\phi(w)\rangle=\alpha \langle v,w\rangle,\quad\text{for all v,w\in\mathcal L_1}.$$

• "Is a lattice" makes no sense, you should specify: a lattice in some subspace? in some given subspace? Do you consider $\mathbf{Z}\times\{0\}$ as sublattice of $\mathbf{Z}^2$? – YCor Aug 5 at 13:17
• @Qfwfq Do my edits address your questions? What I mean by isomorphic is explained immediately after I use the term "isomorphic": $\mathcal L_1$ and $\mathcal L_2$ are isomorphic if they are isomorphic as groups (via some group isomorphism $\phi:\mathcal L_1\to\mathcal L_2$), and there exists a constant $\alpha\in\Bbb R$, so that $$\langle \phi(v),\phi(w)\rangle=\alpha \langle v,w\rangle,\quad\text{for all v,w\in\mathcal L_1}.$$ – M. Winter Aug 5 at 14:09
• @YCor My edit should address your question. Also, $\Bbb Z\times \{0\}$ does not fit my definition of a sublattice of $\Bbb Z^2$. – M. Winter Aug 5 at 14:15
• By the way, googling a bit I found a criterion for embeddability in the standard lattice of the same rank with your $\alpha=1$ (arxiv.org/pdf/1807.05098.pdf) and something related with embeddability in higher rank standard lattice but -if I have understood well the abstract- with your constant $\alpha$ fixed (jstor.org/stable/2398341?seq=1#page_scan_tab_contents). – Qfwfq Aug 5 at 15:17
• I suspect that what caused most of the confusion was omitting the word "Euclidean", as in "integral Euclidean lattice". This has been ameliorated somewhat by adding routine definitions, at the expense of readability. However, the notions of isomorphism and sublattice in the current version of the question are non-standard: it is more common to define isomorphism to be an isometry, so that, in particular, a bijective isomorphism satisfies the definition of an embedding and the image of an isomorphism ${\mathcal L}\to {\Bbb Z}^n$ is a sublattice according to the chosen definitions. – Victor Protsak Aug 5 at 21:41

Yes, this is true. There's some fancy number theory that one can apply (the Hasse-Minkowski invariant and embedding of quadratic forms), but one can see this directly without number-theoretic machinery.

First, notice that one can choose a basis for the lattice which is orthogonal. Just start with any basis and apply Gram-Schmidt orthogonalization. The new orthogonal basis for the lattice may no longer be integral, but we may multiply the inner product by an integer to clear denominators make it integral (and so that the original lattice is a sublattice up to scaling).

Now one is left with the problem of embedding the 1-dimensional quadratic form/lattice $$\mathcal{L}_m$$ given by $$\mathbb{Z}v_1, \langle v_1,v_1\rangle = m$$ into an integral lattice. This may be done using Lagrange's 4-square theorem. Let $$m=w^2+x^2+y^2+z^2$$. Then we may embed $$\mathcal{L}_m$$ into the lattice $$\mathbb{Z}^4$$ via $$v_1 \mapsto (w,x,y,z).$$ Do this for each basis vector individually, and take the orthogonal direct sum of these $$n$$ 4-dimensional lattices (so this embeds into the lattice $$\mathbb{Z}^{4n}$$ up to scaling, which is likely far from optimal in general).

Addendum: The answer above shows how to embed as a sublattice up to scaling, whereas the question asks for a saturated sublattice (intersection of $$\mathbb{Z}^n$$ with a subspace). I don't know how to obtain this in general, but in dimensions $$1$$ and $$2$$ one can make a construction by hand without scaling.

In dimension $$1$$, send the basis vector $$v_1$$ in the lattice $$\mathcal{L}_m$$ with $$\langle v_1, v_1 \rangle = m$$ to the vector $$(1,\ldots,1)\in\mathbb{Z}^m$$. This vector is primitive, and hence the image is a saturated sublattice.

In dimension $$2$$, one may do a variation on this embedding. Suppose that one has a 2-dimensional lattice with $$\langle v_1,v_1\rangle =a >0, \langle v_1,v_2\rangle =b\geq 0, \langle v_2,v_2\rangle = d >0$$, with $$ad-b^2>0$$. By reduction theory, we may in fact assume that $$a \leq d$$ and $$ad \geq 2b^2$$ (by taking $$v_1$$ to have minimal norm, and the acute angle between $$v_1$$ and $$v_2$$ to be $$\geq \pi/3$$).

Let $$b=(a-1)q+r$$, $$0\leq r < a-1$$. Then send $$v_1\mapsto (1,\ldots,1,0,\dots,0)$$ and $$v_2\mapsto (q,\ldots, q, r, 1, \ldots, 1) \in \mathbb{Z}^n$$, where $$n=a+d- (a-1)q^2 -r^2$$ (so the second vector has $$(a-1)$$ $$q$$s) . For this to work, one has to check that $$d > (a-1)q^2+r^2$$ which can be deduced from the inequalities $$ad \geq 2b^2$$ and $$d\geq a$$. Moreover, one may see that this is a saturated sublattice of $$\mathbb{Z}^n$$: if any of the last $$d-(a-1)q^2-r^2$$ coordinates is non-zero, then we may subtract off the corresponding multiple of $$v_2$$ to make these coordinates $$=0$$. What's left must be a multiple of $$v_1$$.

• Am I correct that this embedding gives a sublattice of $\Bbb Z^n$ which is not necessarily saturated? That would be okay, as I explained in a comment to my question, but I should (and want to) edit the question to make this more clear. – M. Winter Aug 6 at 6:46
• @M.Winter: right, I don't think these lattices will be saturated. I guess I misunderstood your definition of sublattice (as you say, your terminology is nonstandard). – Ian Agol Aug 6 at 7:12