# Toric automorphism of P1 times P1 blown up at four pts

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?

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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.)
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
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.)
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