Consider the lattice $N=\Bbb{Z}^d$ spanned by $e_1,\dots,e_d$ and the cone $$\sigma=\text{Cone}\{e_1,\dots,e_k\}, \quad k<d.$$ I am trying to understand why the toric variety $V_\sigma$ obtained is $\Bbb{C}^k \times (\Bbb{C}^\ast)^{d-k} $.

This would be the case if I were able to show that the coordinate ring of $V_\sigma$ has to be $$\Bbb{C}[S_\sigma]\simeq\Bbb{C}[x_1,\dots,x_k,y_1,y_1^{-1},\dots,y_{d-k},y_{d-k}^{-1}].$$ Here $S_\sigma$ is the semigroup $\check{\sigma}\cap\Bbb{Z}^d$, where $\check{\sigma}$ is the dual cone. I think one way to show this is by proving the following

Claim: $\check{\sigma}=\text{Cone}\{m_1,\dots,m_k,\pm u_1,\dots,\pm u_{d-k}\}$ where $m_j=e_j^\ast$ (dual vectors) and $u_1,\dots,u_{d-k}$ is a basis of the vector space $L_\sigma^\perp$, the orthogonal complement of $L_\sigma\simeq\Bbb{R}^k$ (the smallest vector space which contains $\sigma$).

Any ideas for proving this?


1 Answer 1


This is in Fulton's book on toric varieties. In fact, some relevant comments are made on page 11, about two-thirds down the page. This is probably sufficient, but maybe a better reference for the convex geometry is Ewald's book. The point is that the dual cone is always spanned by some generators which annihilate the generators of the original cone, along with lifts of the generators of the dual cone considering $\sigma$ as a cone in the space it spans (top of page 12).

For example, if $N=\mathbb Z^2,$ and $\sigma=\operatorname{cone}(e_1),$ then the dual vectors collinear with $e_2^*$ obviously annihilate $e_1.$ Since $e_1$ is the only generator of $\sigma,$ we pick $\pm e_2^*$ as generators. Moreover, we observe that $e_1^*(ce_1)=c>0$ for any $ce_1\in\sigma.$ Thus, $e_1^*$ also lies in the dual cone. (It generates the dual of $\sigma$ considered as a cone in $\mathbb R^1.$)

Thus, we conclude that $e_1^*,\pm e_2^*$ generate $\check\sigma.$ In particular, $\mathbb C[S_{\sigma}]=\mathbb C[X_1,X_2,X_2^{-1}]$ as desired.


Here is how this works for a general proof. Let $W$ denote the vector space spanned by $\sigma.$ Since we have generators $e_1,\ldots, e_k,$ we know $W=\mathbb R^k.$ Considering $\sigma$ as a cone in $W,$ we easily find that $\check\sigma\subseteq W^*$ is spanned by $e_1^*,\ldots, e_k^*.$ Now, to find $\check\sigma\subseteq N_{\mathbb R}^*=\mathbb R^n,$ we observe that $\pm e_{k+1},\ldots,\pm e_n\in\check\sigma,$ since they are all orthogonal to $\sigma.$ The claim now is that we have found all the generators of $\check\sigma.$ But this is easy, for, if we assume that some $v\in\check\sigma$ has a negative coefficient $c$ in front of any $e_i^*$ for $i\in\{1,\ldots,k\}$ (which is the only other possibility), then we can simply evaluate $v(e_i)=c<0,$ which is a contradiction since $v(u)$ must be nonnegative for all $u\in\sigma$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .