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I am just computing some examples of Mahler volume $M(P_{n}) = \text{vol}(P_{n})\text{vol}(P_{n}^{\circ})$ of regular polygons $P_{n}$ and I'm dumbfounded by my nonsensical results, but seemingly reasonable derivation. Can anyone find my mistake(s)? I am under the impression that the polar (dual) of a regular polygon is an inscribed regular polygon with vertices placed on the midpoints of the edges of the regular polygon. Leading to the following calculations where $s^{\circ}$ denotes the edge length of the dual with $s$ the edge length of the regular polygon.

Assume $\text{area}(P_{n}^{\circ}) = \frac{1}{4}n s^{\circ^2} \cot(\pi/n)=1$, which implies by the law of cosines that $$s^{\circ^2} = \frac{4 \tan(\pi/n)}{n} = \frac{s^2}{2}(1 - \cos(\theta_{n}))$$ where $\theta_{n} = \frac{\pi(n-2)}{2}$ is the interior angle of a regular n-gon. Therefore, $$s^2 = \frac{8\tan(\pi/n)}{n(1- \cos(\theta_{n}))}$$ implies $$\text{area}(P_{n}) = \frac{1}{4}ns^2 \cot(\pi/n) = \frac{2}{1- \cos(\theta_{n})}=M(P_{n}).$$ This formula does not check out as $M(P_{4}) = \frac{4^2}{\Gamma(3)} = 8 \neq 2.$ What went wrong? Is there dual of a regular polygon not an inscribed polygon with vertices placed the midpoints of the edges?


I also tried calculating it a different way without assuming the area of the dual is 1, and just left the expression for arbitrary area and started with a regular polygon then calculated towards the dual inscribed on the inside. Namely, $$\text{area}(P_{n}) = \frac{1}{4}ns^{2}\cot(\pi/n) \Rightarrow s^{2} = \frac{4 \text{area}(P_{n})\tan(\pi/n)}{n}$$ gives us by the law of cosines that $$s^{\circ^{2}} = \frac{s^{2}}{2}(1 - \cos(\theta_{n})) = \frac{2 \text{area}(P_{n})\tan(\pi/n)(1-\cos(\theta_{n}))}{n}.$$ From this we obtain $\text{area}(P_{n}^{\circ}) = \frac{1}{4}n s^{\circ^2} \cot(\pi/n) = \frac{\text{area}(P_{n})(1-\cos(\theta_{n}))}{2}$ and hence $$M(P_{n}) = \text{area}(P_{n})\text{area}(P_{n}^{\circ}) = \frac{\text{area}(P_{n})^2(1-\cos(\theta_{n}))}{2}.$$ This formula also doesn't make sense as the Mahler volume is an affine invariant and should not depend on $\text{area}(P_{n})$. I think I'm conceptually misunderstanding the idea of duality or Mahler volume, which is very disconcerting because I've thought about these topics a lot.

For n=6

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I think you may be misunderstanding the concept of duality, seeing that your picture shows $P^\circ$ inscribed in $P$. This is not what the duality of convex bodies means.

Let's take a square of sidelength $2L$. It is the convex hull of the points $(\pm L,\pm L)$. So, the dual polygon is bounded by the lines $\pm Lx\pm Ly= 1$. This implies that the dual polygon is described by $|x|+|y|\le 1/L$. This is a square with diagonal $2/L$. Its sidelength is $\sqrt{2}/L$.

Areas: the area of original square is $4L^2$. The area of its dual is $2/L^2$. The Mahler volume is $8$, independent of $L$.

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  • $\begingroup$ Oh my. I can't believe I've been so mistaken, I realize all of the work I have done is still valid (I'm writing a few papers and working on a book project) but I just need to replace the word dual with rectification everywhere, and none of it is relevant to Mahler volume anymore. Thank you for the clarification. $\endgroup$ – Samuel Reid May 15 '14 at 3:49

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