0
$\begingroup$

Let $\mu\in\Bbb Z^d$. Suppose that $H=\{x\in\Bbb Z^d:\langle\mu,x\rangle=1\}$ is nonempty and fix $v\in H$.

Now, let $\{a_1,\dotsc,a_{d-1}\}$ be a basis for the kernel of the map $\langle \mu,-\rangle:\Bbb Z^d\to\Bbb Z$. Form a $d\times d-1$ matrix $A$ with these vectors $$ A=\begin{bmatrix} a_1 & \dotsb & a_{d-1}\end{bmatrix} $$ and let $\phi:\Bbb Z^{d-1}\to H$ be the map $\phi(x)=Ax+v$. It's not hard to check that this map is well-defined and injective. My questions are:

  1. Is this map surjective in general?
  2. If not, what conditions can we impose on $\mu$ to ensure that $\phi$ is surjective?

Since $A$ is full rank we have a left inverse, but this is not an integer matrix so I'm not immediately convinced that the answer to (1) is yes.

My motivation here is that I have a large number of lattice polytopes whose vertices live in the hyperplane $H$. I'd like to write down an algorithm that produces an invertible affine $\Bbb Z$-linear transformation that translates these polytopes to the hyperplane defined by $\langle e_1,x\rangle=1$.

$\endgroup$
1
$\begingroup$

Let $w\in H$. Since $<\mu,w-v>=0$, there are $u_i\in\mathbb{Z}$ s.t. $w-v=\sum_{i=1}^{d-1}u_ia_i=A[u_1,\cdots,u_{d-1}]^T$, that is $w=\phi([u_1,\cdots,u_{d-1}]^T)$.

EDIT. I wrote too fast. OK, you know that $\ker(<\mu,.>)$ has a basis with $d-1$ elements because $\mathbb{Z}$ is a PID.

$\endgroup$
  • $\begingroup$ @ Brian Fitzpatrick , I see that you do not even have the courtesy to thank those who respond to your questions. This would never happen again. $\endgroup$ – loup blanc Jan 15 '16 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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