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It's fairly well-known that an intersection with the integer cube grid (i.e., the vertices of the grid are in $\mathbb{Z}^3$) with the plane $u-v+w=0$ creates a hexagonal grid (usually people use $u+v+w=0$, but I need it this way for other reasons).

it is also well known that the grid vertices of a hexagonal grid lie on the Eisenstein integers $a+b\omega$ where $a,b \in \mathbb{Z}^2$ and $\omega = \frac{-1 + i\sqrt{3}}{2}$ (notice it's a complex number). Is there a canonical way to parameterize the plane $u-v+w=0$ s.t. every $u,v,w \in \mathbb{Z}^2$ would produce $a,b \in \mathbb{Z}$? I didn't find anything concrete---people just either use either set of coordinates.

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Given $\,u,v,w \in \mathbb{Z}\,$ with $\, u-v+w=0, \,$ then $\,v + w\,\omega\,$ is one candidate for the corresponding Eisenstein integer. Here $\,(1,1,0) \,$ maps to $\,1\,$ and $\,(-1,0,1)\,$ maps to $\,\omega.\,$ These two vectors form a linear basis for the plane and moreover they have the same magnitude and have angular difference of $\,2\pi/3.\,$ There are other choices which have the same properties. For example, swap $\,u\,$ and $\,w.\,$ Another way is to negate $\,u,v,w.\,$

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  • $\begingroup$ How do you obtain this though? what are the basis vectors for the plane that make this happen? $\endgroup$ – Amir Vaxman Sep 28 '18 at 13:27
  • $\begingroup$ @AmirVaxman I just drew a diagram. $\endgroup$ – Somos Sep 28 '18 at 13:30
  • $\begingroup$ Interesting. So map to $(a,b)$ with non-orthogonal basis, rather than find one explicitly. $\endgroup$ – Amir Vaxman Sep 28 '18 at 13:40

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