3
$\begingroup$

Let $k$ be an algebraically closed field of characteristic zero. Let $\sigma$ be a 3-dimensional simplicial cone in $\mathbb{R}^3$. Let $X=\operatorname{Spec}(k[\sigma^{\vee}\cap\mathbb{Z}^3])$ be the affine toric variety associated to the cone $\sigma$. I am trying to compute $G_0(X)$, where $G_0$ is the Grothendieck group of coherent sheaves. So far,I have shown that if the minimal generators of the simplicial cone $\sigma$ are $e_1,e_2,(0,b,c)$, then we have $G_0(X)\cong\mathbb{Z}\oplus\mathbb{Z}/\delta$, where $\delta$ is the determinant of the the matrix $M$ taking the minimal generators of the cone $\sigma$ as its columns. Now I am trying to compute the general case. I showed that there is a matrix $A$ in $GL(3,\mathbb{Z})$ such that left multiplication by $A$ maps the first column of the matrix $M$ to $e_1\in\mathbb{Z}^3$. This is where I am stuck. Is it always possible to reduce the general case to the case where the minimal generators of the simplicial cone $\sigma$ are $e_1,e_2,(0,b,c)$ via left multiplication by a finite sequence of matrices in $GL(3,\mathbb{Z})$? In other words, is it always possible to construct a matrix $B\in GL(3,\mathbb{Z})$ such that the first two columns are $B$ are the first two minimal generators of a 3-dimensional simplicial cone $\sigma$ in $\mathbb{R}^3$?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ To the specific question about $e_1$, $e_2$, $(0,b,c)$, no. Consider the cone with rays $(1,1,-1)$, $(1,-1,1)$, $(-1,1,1)$. For any two of these rays $\rho$ and $\sigma$, the point $(1/2) (\rho+\sigma)$ is in $\mathbb{Z}^3$ but not an integer combination of $\rho$ and $\sigma$. So no $GL_3(\mathbb{Z})$ matrix can transform $\rho$ and $\sigma$ to $e_1$ and $e_2$. $\endgroup$
    – David E Speyer
    Commented Aug 30, 2023 at 17:02
  • 2
    $\begingroup$ But the more general question of understanding what $G_0(X)$ DOES look like in this case seems worthwhile, so I'm not leaving an answer on the assumption that someone else will answer that question. $\endgroup$
    – David E Speyer
    Commented Aug 30, 2023 at 17:03
  • $\begingroup$ @ David E Speyer Thank you very much for your kind help. $\endgroup$
    – Boris
    Commented Aug 30, 2023 at 17:10

0

Reset to default

You must log in to answer this question.