# 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!)

-
Tropicalization is also known as Maslov dequantization so is de-tropicalization just "quantization"? Basically you are going backwards from $\lim_{t \to \infty} \frac{1}{t} \log(e^{ta} + e^{tb}) = \max(a,b)$. – john mangual Aug 13 '13 at 18:22
What is the definition of a homegeneous rational function? – Alexandre Eremenko Aug 15 '13 at 3:23
By a homogeneous rational function, I mean one that is a ratio of homogeneous polynomials. Alternatively, it is a rational function such that $r(tx,ty)=t^k r(x,y)$ for some positive or negative integer $k$. – James Propp Aug 15 '13 at 16:54
A 1-parameter family of solutions allowing one to continuously deform $x+y$ into $xy/(x+y)$ would be especially helpful. – James Propp Aug 20 '13 at 11:34
When I write "algebraic identity", I mean an identity using the field operations (not $n$th roots, which are problematical over {\mathbb{R}} and over {\mathbb{C}} for different reasons). – James Propp Aug 22 '13 at 13:25

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

-
I like this answer, but I'd like to see more details. Regarding the $d=0$ case: there are lots of degree-0 rational functions that aren't constant, and many of them yield commutative binary operations (if the numerator and denominator polynomials are both symmetric or both antisymmetric, say); could some of them yield binary operations that are associative as well? Regarding the $d=1$ case: I assume the "inverse function to $r$" is the inverse to $x \mapsto r(x,y)$ for fixed $y$, but I don't see why such maps must have rational inverses. (I'll defer my other questions till later.) – James Propp Aug 23 '13 at 15:36
the degree of the map from $\mathbb P^1$ to $\mathbb P^1$ is not the net degree of the rational function, but is the max of the degree of the numerator and the degree of the denominator. So all degree $0$ maps are constant, and all degree $1$ maps, being rational linear transformations, have inverses. – Will Sawin Aug 23 '13 at 16:44
Thanks for reminding me what "degree" means in this context. But I remember that for this notion, the degree of a composition of two maps is not always the product of their degrees. See for instance Example 2.12 of my article (with Hasselblatt) "Degree-growth of monomial maps" (arxiv.org/abs/math/0604521). I suspect your claim is right, but I don't see why yet. – James Propp Aug 23 '13 at 19:01
Yes, this only works for holomorphic maps. But all rational maps $\mathbb P^1 \to \mathbb P^1$ can be extended to regular maps. – Will Sawin Aug 23 '13 at 20:02
I think we can check that the degree is multiplicative here by doing some commutative algebra. Suppose a linear term, without loss of generality $x$, divides both $f(p(x,y),q(x,y))$ and $g(p(x,y),q(x,y))$, where $f,g$ and $p,q$ are two pairs of relatively prime homogeneous polynomials of equal degree. Then $f(x,y)$ must have a linear factor $ax+by$, and $g(x,y)$ must have a linear factor $cx+dy$, such that $x$ divides $ap(x,y)+bq(x,y)$ and $c p(x,y)+q(x,y)$. Because $f$ and $g$ are relatively prime, $ad-bc=1$, so $x$ divides $p(x,y)$ and $q(x,y)$, so $p$ and $q$ are not relatively prime. – Will Sawin Aug 23 '13 at 20:04