MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.

share|cite|improve this answer
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. – Campello Mar 4 '13 at 16:23
Sorry, I made it up. – Gerry Myerson Mar 5 '13 at 11:37

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 $\mathbb{Z}^n$.

share|cite|improve this answer
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 $v^\perp_m \cap \mathbb{Z}^n$. – J. Martel Mar 2 '13 at 5:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.