What was the primary motivation for the study of the amoebas?
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$\begingroup$ Hah, that article is really interesting. But does it really have anything to do with tropical geometry? $\endgroup$– rschwiebMay 2, 2014 at 17:47
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3$\begingroup$ Maybe this Notices of the AMS will help: What is an Amoeba?. I'm surprised this isn't cited in the Wikipedia article. $\endgroup$– Dave L. RenfroMay 2, 2014 at 17:51
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2$\begingroup$ Here's a quite readable article. Amoebas are related to tropical geometry. arxiv.org/abs/math/0403015 $\endgroup$– Fredrik MeyerMay 2, 2014 at 18:06
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1$\begingroup$ @Fredrik Well, tropical geometry gives a lot of reasons to study tropical curves; tropical curves are, of course, degenerate amoebae — (which explains the relation of tropical geometry to usual one, but) that doesn't quite explain why «real» (i.e. not degenerate) amoebae are interesting... $\endgroup$– Grigory MMay 9, 2014 at 15:29
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1$\begingroup$ AFAIK, one area where amoebae appear is (random) dimer configurations (see papers of Kenyon &co — e.g. Dimers and Amoebae or (perhaps, better) Lectures on Dimers). $\endgroup$– Grigory MMay 9, 2014 at 15:34
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I heard they came from studying Laurent series in several variables. Consider Pascal's triangle of $\binom{n}{k}$ for $n \geq k \geq 0$. Can we find a generating series? Perhaps
$$ \sum_{n \geq k \geq 0} \binom{n}{k} x^k y^n = \sum_{n \geq 0}(x+1)^n y^n = \frac{1}{1 - (x+1)y}$$
This suggest maybe trying an alternative that makes $x$ and $y$ more symmetric:
$$ \sum_{n \geq k \geq 0} \binom{n}{k} x^k y^{n-k} = \sum_{n \geq 0}(x+y)^n = \frac{1}{1 - (x+y)}$$
This only works if $|x + y| < 1$, since we get convergence. Probably we also need $|x| < 1$ and $|y| < 1$ for the first sum to work.
Using Cauchy integral formula, we can recover Laurent series from this rational function. So, if $f(z) = a_0 + a_1 z + a_2 z^2 + \dots $ then
$$ a_n = \oint \frac{dz}{2\pi i z} \times z^{n+1}f(z)$$
Could this formula work in two dimensions. Maybe we can recover the binomial coefficients like so:
$$ \binom{m+n}{m} = \oint \frac{dz}{2\pi i z} \oint \frac{dz}{2\pi i z} \cdot z^{m+1}w^{n+1} \cdot \frac{1}{1- (z+w)}$$
Here $\times, \cdot$ are just multiplication over complex numbers.
This is a double contour integral and whether the "pole" $z+w =1$ lies inside the contour is related to amoebas, I think.
Then you have to solve the equation $|z+w| = |e^{2\pi i \theta} + e^{2\pi i \phi}|\leq 1$ and these are related to the sides of the triangle.
arXiv:math/0701039 - How to compute $\sum \tfrac{1}{n^2}$ by solving triangles (Mikael Passare)
EDIT I found the statement in Discrimants, Resultants and Multidimensional Determinants by Kapranov, Fomin and Zelevinsky.
Chapter 6, Corollary 1.6 Let $f(x)$ be a Laurent polynomial. All the components of the complement $\mathbb{R}^k - \log (Z_f)$ to the amoeba of $f$ are convex subsets in $\mathbb{R}^k$. They are in bijective correspondence with the Laurent series expansions of the rational function $\tfrac{1}{f(x)}$.
That is what I did here, is work through different expansions of $\tfrac{1}{1-(x+y)}$ and understand why the equation amoeba $\{ (\log|z|, \log|w|) : z+w = 1 \}$ makes an appearance.
