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Trying to draw the Amoeba

With Mathematica, it's possible to graph $e^{-k x} + e^{-k y} = 1,e^{-k x} - e^{-k y} = 1$ and $e^{-k x} + e^{-k y} = -1$ to get the amoeba of 1 + x + y when k = 1. Then by plotting when $k \to \infty$, these graphs converge to the tropical polynomial

http://www.freeimagehosting.net/uploads/7a0cc0b6f3.jpg

However, I meant to draw Min(1,x,y) which I fixed by shifting my coordinates x,y by the vector (1,1). This is a bit ad-hoc and I am probably not understanding how constant functions "tropicalize".

Main question

I would like to "fatten" Min[x, y, 1, x + y + 1] into its amoeba, so I thought the right curve should be 1 + x + y + xy. However, my "neck" is disappearing in the scaling limit. How should I scale the coefficients correctly to get my amoeba?

Following Zeb's suggestion (but a few second before he posted it) I came up with this imge http://www.freeimagehosting.net/uploads/b612de3bf9.gif

However, this "dequantization" procedure doesn't always produce the whole tropical curve. Here's the curve I drew to "requantize" Min[1, x , y , x+ y + 1, -2 + 2x , 2y]. A line has to be missing b/c of the zero tension condition, as in Tropical Mathematics by David Speyer and Bernd Sturmfels.

http://www.freeimagehosting.net/uploads/50219e2142.gif

Here is the code. You have to draw 4 different versions of the curve to get all the absolute values. Maybe this should somehow involve complex phases as well.

q[x_] := E^( x)
f[a_, b_, c_] := c + a + b + c a b + (1/c^2) a^2 + b^2;
{x0, y0} = { -1, -3};
L = 5;
k = 8;
ContourPlot[
    { f[q[k x], q[k y], q[k ]] == 0, f[-q[k x], q[k y], q[k ]] == 0, 
    f[q[k x], -q[k y], q[k ]] == 0, f[-q[k x], -q[k y], q[k ]] == 0}, 
    {x, x0, x0 + L}, {y, y0, y0 + L}
]

Ideally, I want to take any tropical curve and fatten it into its amoeba. Tropical conics and cubics seem the best starting point.

In anticipation of comments, by "amoeba" here I think I mean the boundary of the 2D region which is usually called "amoeba".

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  • $\begingroup$ You're really trying to "detropicalize" a polynomial, right? $\endgroup$
    – j.c.
    Nov 13, 2010 at 1:12
  • $\begingroup$ I'm not comfortable with it yet well enough to write an answer, (and perhaps you've seen this already) but the very pretty paper "Dequantization of real algebraic geometry on logarithmic paper" of Oleg Viro arxiv.org/abs/math.AG/0005163 explains the process of tropicalization in pretty good detail, though in different terms and notation from you. $\endgroup$
    – j.c.
    Nov 13, 2010 at 1:22
  • $\begingroup$ Yeah, maybe this is Maslov dequantization. The paper says the coefficients must also be dequantized. Over R you can only dequantize positive real numbers (where log is defined). Probably they coincide with the tropical curve over C. $\endgroup$
    – john mangual
    Nov 13, 2010 at 5:27
  • $\begingroup$ The links to the images all seem to be broken. $\endgroup$
    – jeq
    Nov 5, 2017 at 17:16

1 Answer 1

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For the first amoeba you mentioned, I think your equations should be $e^{-kx}\pm e^{−ky}=\pm e^{-k}$, not $e^{-kx}\pm e^{−ky}=\pm e^{0}$.

For the main question, I think you should be using an equation like $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} = 0$... so really you want a curve like $c+x+y+cxy$, where when you rescale $x$ and $y$ by raising them both to a power, you also rescale the coefficient $c$ by raising it to the same power.

Edit: Ok, for the second problem you are having, I think this is coming up because plugging in different signs of $x$ and $y$ doesn't give you all the different sign possibilities for your detropicalized polynomial.

So, if you want to get the tropical curve $Min(1,x,y,1+x+y,2x-2,2y)$, you want to use all of the equations $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} \pm e^{-k(2x-2)} \pm e^{-k(2y)} = 0$.

In fact, I think you don't need to use all $32$ of those equations, you just need to use enough of them that every pair of terms have opposite signs in one of your equations, such as the following three:

$e^{-k}+ e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} - e^{-k(2y)} = 0$ $e^{-k}- e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} + e^{-k(2x-2)} - e^{-k(2y)} = 0$ $e^{-k}+ e^{-kx}- e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} + e^{-k(2y)} = 0$

In the limit this will give you the right amoeba, but I'm cheating a bit, because really we should be doing fancy stuff with logarithms of complex numbers.

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  • $\begingroup$ Yeah, I'm trying to avoid logs of complex numbers. For any individual curve it may be possible to avoid that (just as we never really need all of $\mathbb{R}$ just individual numbers). I was afraid writing all 32 equations would give extraneous points. What do you think? $\endgroup$
    – john mangual
    Nov 13, 2010 at 6:48
  • $\begingroup$ When k is large enough, I don't think they can introduce extraneous points. The whole idea of tropical curves is that you draw the points where two of the terms in the expression are very close in absolute value and the rest of the terms are small, and sending k to infinity forces the roots to be of this form - you just need the terms to be opposite in sign so they can actually cancel each other out. $\endgroup$
    – zeb
    Nov 13, 2010 at 7:02
  • $\begingroup$ With the complex numbers you can do this for any particular pair of terms by just picking the phases right to make those terms opposite in sign... in your second example, this probably corresponds to letting x and y be pure imaginary as well as positive and negative. $\endgroup$
    – zeb
    Nov 13, 2010 at 7:02

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