Is there a Pick's Theorem for a general lattice in $\mathbb{R}^{2}$?
Question about Pick's Theorem
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$\begingroup$
geometry
integer-lattices
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10Can't you just apply a linear transformation to convert the lattice to the standard lattice, and then apply the standard Pick's theorem to the image of the polygon? – 2011-11-14
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0So if I do apply a linear transformation to convert the lattice to the standard lattice, why is it guaranteed that we have the same number of lattice points inside and on the boundary of the polygon before and after the transformation? – 2011-11-14
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0Because the transformation maps lines to lines. Also, the intersection of two lines is mapped to the intersection of the mapped images of the lines. So parallel lines are mapped to parallel lines. Each lattice point is the intersection of two lines, parallel to the lines/vectors defining the lattice. This shows the number of lattice points in the boundary is invariant. – 2011-11-14
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0A similar geometric argument (perhaps a little bit harder to make rigorous; but this is just a general fact about invertible linear transformations) shows the interior of a polygon is mapped to the interior of the image of the polygon. So also the number of lattice points in the interior is invariant. – 2011-11-14
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0is pick`s theorem a part of geometry? – 2013-01-29
1 Answers
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Certainly there is. One such version is stated as Theorem 4.1 in this paper of mine:
Theorem (Pick's Theorem): Let $\Lambda$ be a two-dimensional lattice in $\mathbb{R}^k$ with 2-volume $\delta$. Let $P$ be a $\Lambda$-lattice polygon containing $h$ interior lattice points and $b$ boundary lattice points. Then the area $A(P)$ of $P$ is equal to $\delta \cdot (h + \frac{b}{2} - 1)$.
As a reference to the proof I give a 2003 book of Erdős and Surányi. But -- especially for the $k = 2$ case that you asked about -- Jim Conant's comment is right on: the proof consists simply of making a linear change of variables to get from $\Lambda$ to $\mathbb{Z}^2$.