How to Tropicalize a Polynomial in Two Variables? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:03:46Z http://mathoverflow.net/feeds/question/45880 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45880/how-to-tropicalize-a-polynomial-in-two-variables How to Tropicalize a Polynomial in Two Variables? John Mangual 2010-11-13T00:14:35Z 2010-11-13T18:20:59Z <h3>Trying to draw the Amoeba</h3> <p>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 <strong>1 + x + y</strong> when <strong>k = 1</strong>. Then by plotting when $k \to \infty$, these graphs converge to the tropical polynomial </p> <p><img src="http://www.freeimagehosting.net/uploads/7a0cc0b6f3.jpg" width="150"></p> <p>However, I meant to draw <strong>Min(1,x,y)</strong> which I fixed by shifting my coordinates <strong>x,y</strong> by the vector <strong>(1,1)</strong>. This is a bit ad-hoc and I am probably not understanding how constant functions "tropicalize".</p> <h3>Main question</h3> <p>I would like to "fatten" <strong>Min[x, y, 1, x + y + 1]</strong> into its amoeba, so I thought the right curve should be <strong>1 + x + y + xy</strong>. However, my "neck" is disappearing in the scaling limit. <em>How should I scale the coefficients correctly to get my amoeba?</em></p> <p>Following Zeb's suggestion (but a few second before he posted it) I came up with this imge <img src="http://www.freeimagehosting.net/uploads/b612de3bf9.gif" width="150"></p> <p>However, this "dequantization" procedure doesn't always produce the whole tropical curve. Here's the curve I drew to "requantize" <strong>Min[1, x , y , x+ y + 1, -2 + 2x , 2y]</strong>. A line has to be missing b/c of the zero tension condition, as in <a href="http://arxiv.org/abs/math/0408099" rel="nofollow">Tropical Mathematics</a> by David Speyer and Bernd Sturmfels.</p> <p><img src="http://www.freeimagehosting.net/uploads/50219e2142.gif" width="150"></p> <p>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.</p> <pre><code>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} ] </code></pre> <p>Ideally, I want to take any tropical curve and fatten it into its amoeba. Tropical conics and cubics seem the best starting point.</p> <p><hr> In anticipation of comments, by "amoeba" here I think I mean the boundary of the 2D region which is usually called "amoeba".</p> http://mathoverflow.net/questions/45880/how-to-tropicalize-a-polynomial-in-two-variables/45899#45899 Answer by zeb for How to Tropicalize a Polynomial in Two Variables? zeb 2010-11-13T05:54:57Z 2010-11-13T06:45:25Z <p>For the first amoeba you mentioned, I think your equations should be $e^{-kx}\pm e^{−ky}=\pm e^{-k}$, not $e^{-kx}\pm e^{−ky}=\pm e^{0}$.</p> <p>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.</p> <p>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.</p> <p>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 $e^{-k}\pm e^{-kx}\pm e^{-ky} \pm e^{-k(x+y+1)} \pm e^{-k(2x-2)} \pm e^{-k(2y)} = 0$.</p> <p>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:</p> <p>$e^{-k}+ e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} - e^{-k(2y)} = 0$ $e^{-k}- e^{-kx}+ e^{-ky} - e^{-k(x+y+1)} + e^{-k(2x-2)} - e^{-k(2y)} = 0$ $e^{-k}+ e^{-kx}- e^{-ky} - e^{-k(x+y+1)} - e^{-k(2x-2)} + e^{-k(2y)} = 0$</p> <p>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.</p>