If you take a subtraction-free rational identity like $(xxx+yyy)/(x+y)+xy=xx+yy$ and replace $\times$,$/$,$+$,$1$ by $+$,$-$,min,$0$, do you always get a valid min,plus,minus identity like min(min($x+x+x,y+y+y$)$-$min($x,y$),$\:x+y$)$\ =\ $min($x+x,y+y$)?
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asked Apr 10, 2013 at 16:23
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$\begingroup$ Colin and Will's answers are both convincing; thanks! One thing more I'd appreciate is a literature reference that I can cite if I publish something that makes use of this transfer principle. Given how straightforward the proof is (at least in hindsight), it's likely that some version of this result appears in some existing book or article. $\endgroup$– James ProppCommented Apr 10, 2013 at 19:06
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$\begingroup$ There would be an existing tag 'tropical-arithmetic'. I just would like to check if your created the new tag specifically and would like to have it in addition or if you were fine with changing to the other tag. Thanks in advance. $\endgroup$– user9072Commented Apr 11, 2013 at 17:42
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$\begingroup$ @quid: I've changed tropical-mathematics to tropical-arithmetic. Thanks for pointing out the existence of the tropical-arithmetic tag. $\endgroup$– James ProppCommented Apr 12, 2013 at 2:07
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2 Answers
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It suffices to show that whenever $F$ is a function $\mathbb R_{\geq 0}^k\to\mathbb R_{\geq 0}$ defined using $\times,/,+,1$, and $f,g$ is the corresponding tropicalization $\mathbb R^k\to\mathbb R$, for all real $x_1,\dots,x_k$ we have $$F(\exp(-\beta x_1),\dots,\exp(-\beta x_k))^{1/\beta}\to \exp(-f(x_1,\dots,x_k)).$$ as $\beta\to+\infty$.
This follows by structural induction on the formula defining $f$, using the following lemma.
Lemma: Let $F$ be one of the operations $\times,/,+$, let $f$ be the corresponding tropical operation, and let $G_1$ and $G_2$ be functions $\mathbb R_{> 0}\to\mathbb R_{> 0}$ such that $G_1(\beta)^{1/\beta}$ tends to some limit $e^{-x_1}$ as $\beta\to\infty$, and likewise $G_2(\beta)^{1/\beta}\to e^{-x_2}$. Then $$F(G_1(\beta),G_2(\beta))^{1/\beta}\to \exp(- f(x_1,x_2)).$$ as $\beta\to+\infty$.
Proof: The operations $\times$ and $/$ are easy - the only non-trivial step is $$(G_1(\beta)+G_2(\beta))^{1/\beta}\to e^{-\beta \min(x_1,x_2)}.$$
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1$\begingroup$ Relevant lemma: Say $u_1,u_2,\dots$ and $v_1,v_2,\dots$ are sequences of positive reals with $u_n^{1/n} \rightarrow u$ and $v_n^{1/n} \rightarrow v$ for some $u,v>0$. Assume wlog that $\max(u,v) = v$. Since $u_n + v_n \geq v_n$, $\liminf (u_n+v_n)^{1/n} \geq \liminf v_n^{1/n} = v$. Also, for all $\epsilon > 0$, there exists $N$ s.t. for all $n \geq N$, $u_n^{1/n} < v(1+\epsilon)$ and $v_n^{1/n} < v(1+\epsilon)$. Then $(u_n+v_n)^{1/n} < v(1+\epsilon)2^{1/n}$, and since $\limsup 2^{1/n} = 1$, $\limsup (u_n+v_n)^{1/n} \leq v(1+\epsilon)$. Hence also $\limsup (u_n+v_n)^{1/n} \leq v$. $\endgroup$ Commented Aug 9, 2018 at 20:35
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1$\begingroup$ Thus, if $u_n^{1/n} \rightarrow u$ and $v_n^{1/n} \rightarrow v$, then $(u_n+v_n)^{1/n} \rightarrow \max(u,v)$. $\endgroup$ Commented Aug 9, 2018 at 20:38
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Yes. Replace $x$ with $e^{-Na}$, $y$ with $e^{-Nb}$, etc. Then take the log, then divide by $N$. One gets a new identity where $\times$ is replaced by $+$, $/$ by $-$, $1$ by $0$, and $u+v$ by $-\ln (e^{-N u} + e^{-N v}) / N= \min(u,v) - \ln\left( 1+ e^{-N |u-v|}\right)/N = \min(u,v) + O(1/N)$. Then take the limit as $N$ goes to $\infty$. You now have a tropcal identity.
This fits with the idea of tropical geometry as the limit of classical algebraic geometry as variables get very large.
answered Apr 10, 2013 at 17:17
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1$\begingroup$ "divide by $N$" -> "divide by $-N$". (Nice argument!) $\endgroup$ Commented Aug 9, 2018 at 12:14