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Primitive sets of vectors are very important in the theory of point lattices, since they constitute the sets of vectors that are part of a basis for the lattice.

A set of integer vectors $v_1,\ldots,v_k$ is primitive in $\mathbb{Z}^n$ if there are $v_{k+1},\ldots,v_{n}$ such that the matrix whose columns are $v_i$ has determinant 1 (i.e., unimodular). Is there an easy way to characterize primitive vectors which are orthogonal? In the end, I want to find families of vectors $v_1(m),\ldots,v_k(m)$ which are primitive, orthogonal and the norm increases with m. Is there any such family, is it known?

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Here's a family of examples for $n=3$, $k=2$. $$\pmatrix{m&m+3&r\cr m+1&0&s\cr m+3&-m&t\cr}$$ The first two columns are orthogonal and of norm increasing with $m$. The determinant is $$-m(m+1)r+((m+3)^2+m^2)s-(m+1)(m+3)t$$ If $m$ is not a multiple of $3$, and $m+1$ is not a multiple of $5$, then the coefficients of $r,s,t$ are coprime, so $r,s,t$ can be chosen to make the determinant $1$.

There are many more examples where this one came from --- the conditions on $v_1,\dots,v_k$ don't seem to be terribly restrictive.

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  • $\begingroup$ Yes, this will do it. Is there any reference where you found this example? Based on this, I can find other ones by guesswork, but it would be nice to find a systematic way. $\endgroup$
    – Campello
    Commented Mar 4, 2013 at 16:23
  • $\begingroup$ Sorry, I made it up. $\endgroup$
    – Gerry Myerson
    Commented Mar 5, 2013 at 11:37
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Take any primitive nonzero vector $v$ in $\mathbb{Z}^n$. Then the orthocomplement $v^\perp$ in $\mathbb{R}^n$ (with respect to usual euclidean product) is a $\mathbb{Z}^n$-rational subspace of height $||v||^{-2}$. Moreover the codimension 1 subspace $v^\perp$ separates $\mathbb{R}^n$ into two half-spaces (where $v\cdot y$ is $>0$ or $<0$).

Now any nonzero lattice element $z$ in $v^\perp$ (primitive in the rank $(n-1)$-lattice $v^\perp \cap \mathbb{Z}^n$) will yield a primitive set $\{v,z\}$. Indeed suppose $y$ was a lattice vector lying within the interior of the parallelipiped spanned by $v,z$. Then we'd have $z=av+by$ for some positive $a,b \geq 0$. But then we'd have $v\cdot z > 0$, contradicting our choice of $z\in v^\perp$.

So we at least have a systematic way of generating primitive orthogonal pairs in $\mathbb{Z}^n$.

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  • $\begingroup$ To generate families $(v_m, z_m)$ of orthogonal primitive pairs with increasing norms, we can take $v_m$ to be a sequence of longer and longer primitive lattice elements and $z_m$ a sequence of longer and longer lattice elements in $v^\perp_m$ with $z_m$ primitive in $v^\perp_m \cap \mathbb{Z}^n$. $\endgroup$
    – JHM
    Commented Mar 2, 2013 at 5:57

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