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Please instruct me on how to interpret $min(2x,2y,0)$ as a piecewise linear function.

I have been working up to it for a while because I really wanted insight towards computing blowups in algebraic geometry... according to nlab though it is also useful in mirror symmetry problems... if I don't find out here what else should I try!!!

Library notes: I couldn't figure out how to optimize the function along the planes like advised to me in the comment section. 1/4 2/4 3/4 4/4

Ohio skateboarder . 7
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    Just draw the planes $z=2x$, $z=2y$ and $z=0$, and look at them from below...? – Hans Lundmark Feb 06 '22 at 16:43
  • I tried Lagrangian optimization again but I didn't get if you meant to make those planes a constraint to optimize the function along or to do them individually. I put the notes in my revised question. – Ohio skateboarder . 7 Feb 07 '22 at 00:50
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    Do you need the specific details, or do you just want the qualitative fact? Surely the min (or max) of any finite set of linear functions is linear? – paul garrett Feb 07 '22 at 01:05
  • I would simply like to know the tropicalization of an arbitrary polynomial, e.g. $4x+4y-2=0$ and compute $x \oplus y$ for at least the coordinate of input $(1,1,1)$. Reach out to me if it's just too complicated to make up on the spot thanks! – Ohio skateboarder . 7 Feb 07 '22 at 01:52
  • I think it's you who are making it too complicated. You don't need Lagrangian optimization to pick the smallest of three numbers. In your example you get $\min(4+x,4+y,-2)$ when you tropicalize, and when you plug in $(x,y)=(1,1)$ (I don't know what you meant by “input $(1,1,1)$”), you get $\min(5,5,-2)=-2$. – Hans Lundmark Feb 07 '22 at 06:32
  • (Those notes are unreadable, by the way.) – Hans Lundmark Feb 07 '22 at 06:34

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Here is a way you can do it: just solve the following inequalities:

$$ 2x \leq 2y, 0\\ 2y\leq 2x, 0\\ 0\leq 2x, 2y $$ The solutions $(x,y) \in \mathbb{R}^2$ of each of these inequalities will tell you in which region the function on the left will be equal to $f(x,y) = \min\{2x,2y,0\}$. For example, solving the first inequality, we obtain that: $$ y \geq x, x \leq 0 \Rightarrow f(x,y) = 2x. $$ Solving the other two, we obtain: $$ y \leq x, y\leq 0 \Rightarrow f(x,y) = 2y,\\ y \geq 0, x \geq 0 \Rightarrow f(x,y) =0. $$ Now, we have found the expression of $f$ in the following regions:

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which cover the plane and such that the values of the involved functions agree on the intersection of these regions. In the end, we obtain the following expression for the function $f$: $$ f(x,y) = \begin{cases} 2x \qquad &\text{if } y \geq x \land x \leq 0\\ 2y &\text{if } y \leq x \land y \leq 0\\ 0 &\text{if } x \geq 0 \land y \geq 0 \end{cases}.$$

Dario Antolini
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