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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!)

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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

1 Answer 1

up vote 4 down vote accepted

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

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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

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