MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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$)?

share|cite|improve this question
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. – James Propp Apr 10 '13 at 19:06
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. – user9072 Apr 11 '13 at 17:42
@quid: I've changed tropical-mathematics to tropical-arithmetic. Thanks for pointing out the existence of the tropical-arithmetic tag. – James Propp Apr 12 '13 at 2:07
up vote 10 down vote accepted

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 $$(G_1(\beta)+G_2(\beta))^{1/\beta}\to e^{-\beta \min(x_1,x_2)}.$$

share|cite|improve this answer

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 $\ln (e^{N u} + e^{N v})= \min(u,v) + \ln\left( 1+ e^{-N |u-v|}\right)/N = \min(u,v) + O(1/N)$. 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.