Consider the lattice $$N=\{(a,b,c)\in\mathbb{Z}^3\mid a+b+c\equiv 0\mod 2\}$$ and the Cone $$\sigma=Cone(e_1,e_e,e_3)\subset N_{\mathbb{R}}\cong \mathbb{R}^3$$ The associated affine toric variety $X_{\sigma}$ has as its canonical divisor $K_{X_\sigma}=\sum_{i=1}^3-D_i$, where $D_i$ are the torus orbits corresponding to the rays $Cone(e_i)\subset \sigma$. The minimal ray generators are $(2,0,0),(0,2,0),(0,0,2)$, denote them by $u_1,u_2,u_3$ respectively.

We know that for $m\in M=N^\vee$ the corresponding rational function $\chi^m:X_\sigma\to \mathbb{C}$ has: $$div(\chi^m)=\sum_{i=1}^3 \langle m, u_i\rangle D_i$$

So when we take $m:(a,b,c)\mapsto -\frac{1}{2}(a+b+c)$ then $m\in N^\vee$, and $div(\chi^m)=\sum_{i=1}^3-D_i$, showing that $K_{X_\sigma}$ is Cartier.

However, in "Toric varieties" by Cox,Little,Schenck, example 11.2.7 the authors claim that $K_{X_\sigma}$ is not Cartier, while $2K_{X_\sigma}$ is.

I don't understand this. They seem to imply that the $m$ I defined is not an element of $N^\vee$, but to me it seems clear that it is... It takes only integer values on the element of $N$, so should be a well defined element of $N^\vee$.

What am I not understanding here?

Here's another way of seeing that it should be Cartier. Taking the basis $(1,1,0),(1,0,1),(0,1,1)$ for $N$ we get an isomorphism $N\cong \mathbb{Z}^3$. Under this isomorphism $\sigma$ corresponds to the cone $$Cone((-1,1,1),(1,-1,1),(1,1,-1))\subset\mathbb{R}^3$$

Now the dual lattice is just $\mathbb{Z}^3$, and the pairing is the dot product. Then set $m=(-1,-1,-1)$.

  • $\begingroup$ I have just a small remark : are you sure about your minimal generators ? How do you obtain $(1,1,0) \in N$ from your generators ? $\endgroup$
    – user171326
    May 19, 2017 at 17:47
  • $\begingroup$ @N.H. $(1,1,0), (1,0,1),(0,1,1)$ are not contained in $\sigma$; they are just a basis for $N$ (so we can move from the weird "shifted lattice" $N$ to the standard lattice $\mathbb{Z}^3$). With respect to this basis the elements $(2,0,0),(0,2,0),(0,0,2)\in N$ are given by $(-1,1,1),(1,-1,1),(1,1,-1)\in\mathbb{Z}^3$. Does that make sense? $\endgroup$ May 19, 2017 at 17:51
  • $\begingroup$ I'm probably a bit rusty (and sorry to bother you !) but I still don't see why $(1,1,0) \notin \sigma$ ? For me since $\sigma = \text{cone}(e_1, e_2,e_3)$ $\sigma$ is the set of $ae_1 + be_2 + ce_3$ with $a,b,c$ positive real numbers right ? So $(1,1,0) = e_1 + e_2$ should be in $\sigma$ ? Sorry again to be not very useful. I think your argument looks correct, I can try to read this part and think about it. $\endgroup$
    – user171326
    May 19, 2017 at 18:51
  • $\begingroup$ @N.H. No problem! Thanks for thinking about it. You are Right that $(1,1,0)\in\sigma$, but it is not the generator for one of the 1-dimensional Faces of $\sigma$; the ray generated by $(1,1,0)$ Lies in the relative interior of the cone. On the other hand, the $(2,0,0)$ etc generated the 1 dimensional Faces of the cone. $\endgroup$ May 19, 2017 at 18:56
  • $\begingroup$ Ok I got it ! I'll try to think about it. $\endgroup$
    – user171326
    May 19, 2017 at 19:00

1 Answer 1


Indeed $K_{X_{\sigma}}$ is Cartier. Your argument is correct, and the claim stated in the book is wrong. They also claim that $Cl(X) \cong \Bbb Z/2 \Bbb Z$. But the rays generators $u_i$ verify $u_i = (f_1 + f_2 + f_3) - f_i$ where $f_i = (1,1,1) - e_i$ is the basis of your lattice. In particular, we have an exact sequence $$ 0 \to M \to \oplus_{\rho_i} \Bbb Z D_i \to Cl(X) \to 0$$ where the first map is the matrix $\alpha = \begin{pmatrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1\end{pmatrix}$

So we have $$ Cl(X) \cong \text{coker}(\alpha) = \{(a,b,c) \in \Bbb Z^3 : a = b + c, b = a + c, c = a + b\} \cong \{(a,b) : 2a = 2 b = 0\} \cong \Bbb Z/ 2 \Bbb Z \times \Bbb Z/2 \Bbb Z$$ which shows that they probably wrote the wrong lattice.

  • $\begingroup$ A little more remark : my teacher told me that the correct space to consider should be the quotient of $\Bbb C^3$ by the action of $\Bbb Z/2 \Bbb Z : t \cdot(x,y,z) = (-x,-y,-z)$. Cox gives a way of finding the lattice from the action of a cyclic group in the case of surface but I am not sure how to adapt the argument. My teacher also suggested that the good lattice was $N = \{a +b = a + c = 0 \mod 2 \}$. $\endgroup$
    – user171326
    May 23, 2017 at 19:30
  • $\begingroup$ Thanks a lot for this answer and contacting your professor. I'll see if this $N$ does indeed produce a $K_X$ that is not Cartier, while $2K_X$ is. $\endgroup$ May 24, 2017 at 9:25
  • $\begingroup$ Ok, if you find the correct answer don't hesitate to write it as an answer, I'll be very glad to see the computations. $\endgroup$
    – user171326
    May 24, 2017 at 9:35

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