Generalizing detropicalization Given an identity in max,plus arithmetic, are there ways to turn it into an ordinary algebraic identity it other than by replacing addition by multiplication and replacing max by series-plus or by parallel-plus, where the series sum of $x$ and $y$ is $x+y$ and the parallel sum is $xy/(x+y)$?  (See the related posts choosing between the two ways to tropicalize and Name and notation for a binary operation.)
As a related question, I ask: What are all the two-variable homogeneous rational functions over $\mathbb{C}$ such that $r(x,y)=r(y,x)$ and $r(r(x,y),z)=r(x,r(y,z))$?  (Here I intend equality in the formal sense; e.g., I call $x/x$ the same rational function as 1, even though as functions they are not the same, since $x/x$ is not defined at 0.)
The only examples I know of are $r(x,y) = c$ (with $c$ an arbitrary constant), $r(x,y)=x+y$, $r(x,y)=xy/(x+y)$, and $r(x,y)=xy$, but I suspect that there are others I'm overlooking.  Perhaps the theory of formal groups has an answer for me, but from what little I've seen, homogeneity does not play a role there.  See the related post
Commutative associative rational binary operations.  Now that I've played a bit with operations like $(x+y)/(1-xy)$ (thanks, Alexandre Eremenko!), I'm suspecting that I should be limiting myself to homogeneous operations.
(Note: The original version of the post used the word "detropicalize'' in a non-standard and unclear way, so I've changed the title of the post and the wording of the first paragraph accordingly.  Thanks, John Mangual!)
 A: Those are all.
Given a function $r$, by restricting to a particular value of $y$ (barring finitely many), we get a rational function, hence a map $\mathbb P^1 \to \mathbb P^1$. For all but finitely many values of $y$, this map will have the same degree, $d$. Assuming $r$ is nonconstant, let $y_1$ and $y_2$ be two such typical values such that $r(y_1,y_2)$ is also typical. Then the degree of $x => r(x,y_1) => r(r(x,y_1),y_2)$ is the degree of $x => r(x,r(y_1,y_2))$, so $d^2=d$,so $d=0$ or $1$. Clearly the $d=0$ case is the constant case.
For $d=1$, we get a rational inverse function to $r$, giving us an algebraic group structure on $\mathbb P^1$ minus finitely many points. (Or for each $y$-value but finitely many, we get an automorphism of $\mathbb P^1$, which is an element of $PGL_2$, so we have a curve in $PGL_2$ that is almost closed under composition. The closure of such a curve is always a subgroup.) There are only two of these: $\mathbb G_a$ or $\mathbb G_m$. But we still have to decide which isomorphism between the closure of these groups and $\mathbb P^1$ to take. The homogeneity means that the missing points can only be $0$ and $\infty$, which gives us three options:
$\mathbb G_a$, $\infty$ missing: $r(x,y)=x+y$
$\mathbb G_a$, $0$ missing: $r(x,y) = 1/((1/x)+(1/y))=xy/(x+y)$
$\mathbb G_m$, both $0$ and $\infty$ missing: $r(x,y)= xy$.
