Hi All,

I'm having some trouble understanding a result about calculating $D.C$ on a toric variety. The proposition I am trying to follow is from Cox, Little, Schenck. Either there is a mistake with the proposition, or I'm doing something wrong in my example - I'm hoping that someone can tell me which one it is. If people have questions about the notation that I will say what they mean, but I think everything is pretty standard.

Let $C=V(\tau)$ be the complete torus invariant curve in $X_\Sigma$ coming from the wall $\tau = \sigma \cap \sigma'$. Let $D$ be a Cartier divisor with Cartier data $m_\sigma$, $m_{\sigma'} \in M$ corresponding to $\sigma, \sigma' \in \Sigma(n)$. Pick $u \in \sigma' \cap N$ that maps to the minimal generator of $\overline{\sigma'}$ in $N(\tau)$. Then $D.C = (m_\sigma - m_{\sigma'})(u)$.

I'm running into problems through the following example. Consider the usual fan for $\mathbb{P}^1 \times \mathbb{P}^1$. Let $\sigma$ be the top right quadrant and let $\sigma'$ be the bottom right quadrant. Let C be the curve corresponding to $\tau = \sigma \cap \sigma'$. Let $D$ be the curve coming from the top vertical ray. We know that $C$ gives us $\mathbb{P}^1 \times p$ and that $D$ is the curve $q \times \mathbb{P}^1$ (maybe I got the order mixed up, but it shouldn't change the calculation). So $D.C=1$. Now let's calculate $D.C$ using the proposition. I found that $D$ has cartier data $m_\sigma =e_1^* -e_2^*$ and $m_\sigma'=e_1^*+e_2^*$, and $u=(1,-1)$ maps to a minimal generator. So then $D.C = -2 e_2^* ((1,-1)) = 2$

Thanks a lot, and apologies if this is just me making a silly blunder.


edit: I figured out my problem - my $m_\sigma'$ is incorrect. I made the same mistake 3 times in a row. Fixing the mistake gives $m_\sigma'=-e_1^*$, which then gives $D.C=1$. Not sure how to close my own thread so this doesn't keep popping up to the front.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Browse other questions tagged or ask your own question.