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](https://mathoverflow.net/questions/136471/choosing-between-the-two-ways-to-tropicalize) and [Name and notation for a binary operation](https://mathoverflow.net/questions/135204/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](https://mathoverflow.net/questions/139215/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!)



Generalizing detropicalization
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: $r(x,y)= xy$.


