Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A toric morphism between toric varieties is a morphism that is equivariant w.r.t. to the toric action, see e.g. section 3.2, Notes by H.Verrill and D.Joyner for definitions. In particular, any toric morphism comes from a morphism of the corresponding fans. For example, the toric automorphism group of $P^2$ is $D_3$. If we denote the fan for $P^2$ as generated by the rays (1,0), (0,1), (-1,-1), then other than the obvious reflection which has order 2, the element of order 3 in $D_3$ is given by the unimodular matrix

As another example, it is also easy to see that the toric automorphisms of $P^1\times P^1$ is $D_4$. Now if we blow up 4 points of $P^1\times P^1$ at such places so the fan after the blow-up has rays $(0,\pm 1), (\pm 1,0), (\pm 1, \pm 1)$, then the note I mentioned above says the toric automorphism group is $D_8$. My problem is that I can not find the element of order 8 in this group. The obvious rotation by $\frac{\pi}{4}$ wouldn't work, since we have to keep the lattice !

In other words, I want a unimodular matrix (i.e. integer matrix with determinant $\pm 1$)

to keep the lattice, and I need the ratio $(\frac{a+b}{c+d},\frac{a-b}{c-d})$=any of $\pm(1,-1),(0,\infty),(\infty,0), (0, \pm 1), (\pm 1, 0), (\infty, \pm 1), (\pm 1, \infty)$ to keep the fan. I couldn't find any order eight unimodular matrix satisfying these relations. Did I miss something here or I misunderstood any definitions?

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

There is no unimodular $2\times 2$ matrix of order $8$. This follows from the fact that the minimal polynomial (over $\mathbb{Q}$) of a primitive $8$'th root of unity is $x^4 + 1$ which is of degree $> 2$. Thus, the claim that the toric automorphism group of the example you describe is $D_8$ cannot be correct. (I don't think you have misunderstood any definitions.)

share|improve this answer
    
Actually, the group ${\rm SL}(2,\mathbb{Z})$ decomposes as a free product with amalgamation, of the form $\mathbb{Z}_4*_{\mathbb{Z}_2} \mathbb{Z}_6$. Thus, the complete list of finite orders of elements in this group is $\{1, 2, 3, 4, 6\}$. –  Alex Suciu Sep 17 '11 at 19:42
add comment
  1. The four new $P^1$s have self-intersection $-1$. The four old $P^1$s (or rather, their proper transforms) have self-intersection $-2$. So there's no automorphism that's going to switch the two sets of them.

  2. The notation for dihedral groups is notoriously author-dependent; many books -- and apparently, the one you're reading -- use $D_n$ for the automorphisms of an $n/2$-gon, apparently on the basis that the group then has $n$ elements. (As they don't then go on to use $S_{120}$ to denote the permutations of a $5$-element set, I have never found this a convincing foundation for this $D_n$ notation, but what can you do.)

share|improve this answer
    
Nice. Looking back at this example, the strict transform of a nice $P^1$ is "strict" on the top since it passes through one of the points we blow up. It is a nice exercise to write down the total transform and see it on the nose that pulling back keeps the self-intersection number 0. –  Ying Zhang Sep 17 '11 at 15:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.