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

Hi,

I'm trying to find all the smooth complete toric surfaces, following section 2.5 of Fulton's book. There is one exercise given that I'm hung up on, hopefully someone can help me out a bit. Perhaps this problem should also be tagged under some kind of C&O or lattice theory stuff?

The set-up is this: in $\mathbb{Z}^2$, we are given a sequence of vectors $v_0,v_1,...,v_{d-1},v_d=v_0$ in counterclockwise order, that wrap around the lattice exactly once, such that

(1) The angle between $v_i$ and $v_{i+1}$ is less than 180 degrees (ie they generate a convex cone).

and

(2) Each pair $v_i,v_{i+1}$ is a $\mathbb{Z}$-basis of $\mathbb{Z}^2$

We also know (***): that for 2 pairs $v_i,v_{i+1}$ and $v_j,v_{j+1}$ that we cannot simultaneously have $v_j$ in the angle strictly between $v_{i+1}$ and $-v_i$ and $v_{j+1}$ in the angle strictly between $-v_i$ and $-v_{i+1}$.

Here is the problem I have been unable to prove: let $d \geq 4$. Prove there exists a pair of opposite vectors in the sequence, ie for some $i,j$ we have $v_i=-v_j$. The 'hint' given is as follows. Take the longest possible sequence of consecutive vectors in the same half plane, then apply condition ***.

Here is what I have written so far; I am not sure if this is in the correct direction or not.

Applying an (orientation preserving) isomorphism of $\mathbb{Z}^2$, assume that $v_0=(1,0)$ and $v_1=(0,1)$, and that the longest sequence of consecutive vectors in the same half plane occurs in the half plane $y \geq 0$. Let $v_0,v_1,...,v_l$ denote this maximal sequence. Writing $v_i=(x_i,y_i)$, assume that $v_l$ has $y_l>0$: for otherwise we will have $x_l=-1$ for condition (2) to hold, and then the problem is proved since $v_0=-v_l$. Therefore $v_l$ is inside the angle strictly between $v_1$ and $-v_0$. By $\*\*\*$ we thus see that $v_{l+1}$ is strictly inside the 4th quadrant of $\mathbb{Z}^2$. I know want to somehow contradict the maximality of this sequence?

Some trivial bounds on $l$ in terms of $d$ come from condition (1): namely $l \geq 2$ and $l \leq d-2$.

Any help would be appreciated.

Robert

share|improve this question

1 Answer 1

Hi Robert,

If we know that $v_{l+1}$ is strictly in the fourth quadrant, the problem is that there is not enough room for $v_{d-1}$ (you can apply the condition $( * )$ to see what happens with $j=d-1$,$i=l$). Now, if $l+1 = d-1$ then there are other problems with the condition $( * )$.

-David

share|improve this answer

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.