Timeline for Generalizing detropicalization
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2013 at 13:28 | history | bounty ended | James Propp | ||
| Aug 26, 2013 at 2:21 | vote | accept | James Propp | ||
| Aug 26, 2013 at 0:33 | comment | added | john mangual | @WillSawin Maybe these correspond to the surfaces of constant curvature? sphere, plane, hyperbolic disc | |
| Aug 24, 2013 at 17:32 | comment | added | James Propp | I accept this solution to my "related question". Thanks, Will! I'd be even happier if Will or someone else could use this to show that no other ways of "pseudo-detropicalizing" are possible (my original question). | |
| Aug 24, 2013 at 13:33 | comment | added | Will Sawin | Yeah, that's incomplete. Homogeneity further implies that $\infty$ is the identity - the opposite of the usual coordinate for $\mathbb G_a$, where $0$ is the identity and $\infty$ is not in the group, so this group can be gotten from the usual additive group by applying an automorphism that switches $0$ and $\infty$, like $x \to 1/x$. | |
| Aug 24, 2013 at 4:44 | comment | added | James Propp | I think I understand the first and third cases of the last sentence, but how do you get from "$𝔾a$, 0 missing" to "$r(x,y)=1/((1/x)+(1/y))=xy/(x+y)$"? | |
| Aug 24, 2013 at 0:36 | comment | added | Will Sawin | The additive and multiplicative groups. | |
| Aug 23, 2013 at 21:42 | comment | added | James Propp | Okay, I'm ready with my next question: What are 𝔾a and 𝔾m? | |
| Aug 23, 2013 at 20:04 | comment | added | Will Sawin | 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. | |
| Aug 23, 2013 at 20:02 | comment | added | Will Sawin | Yes, this only works for holomorphic maps. But all rational maps $\mathbb P^1 \to \mathbb P^1$ can be extended to regular maps. | |
| Aug 23, 2013 at 19:01 | comment | added | James Propp | 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. | |
| Aug 23, 2013 at 16:44 | comment | added | Will Sawin | 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. | |
| Aug 23, 2013 at 15:36 | comment | added | James Propp | 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.) | |
| Aug 23, 2013 at 15:31 | history | edited | James Propp | CC BY-SA 3.0 |
Fixed a typo
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| Aug 23, 2013 at 15:19 | history | edited | James Propp | CC BY-SA 3.0 |
Fixed a typo
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| Aug 23, 2013 at 3:31 | history | answered | Will Sawin | CC BY-SA 3.0 |