Toric perspective 1

This series of posts will be about toric varieties. The author is not sure if there is a part 2, but he still calls this part 1. This post is about computing the dimensions and degrees of popular toric varieties. Readers are assumed to know linear algebra and algebraic geometry in the level of Cox-Little-O’Shea.

We begin with a $d\times n$ integer matrix $A$ of full rank. Denote by ${\rm ker}_\mathbb{Z}(A)$ the integer points in the kernel of $A$. If the binomial ideal

$I_A:=\langle x^{u_+} - x^{u_-} \ : \ u\in {\rm ker}_\mathbb{Z}(A)\rangle\subset \mathbb{C}[x_1,\cdots,x_n]$

is prime, then $I_A$ is called the toric ideal while the irreducible variety

$Z(I_A):=\{x\in\mathbb{C}^n : f(x)=0\text{ for any }f\in I_A\}$

is called a toric variety.

Theorem 1.        Let $V = Z(I_A)$ be a toric variety. Then the dimension of $V$ is the rank of $A$, while the degree of $V$ is the normalized volume: consider the polytope $P$ which is the convex hull of the columns of $A$ in $\mathbb{Z}^d$, then the normalized volume is $\delta! {\rm Vol}(P)$ where $\delta$ is the dimension of $P$.

Example 1.         Consider the twisted cubic $T$ in $\mathbb{C}^4$ or $\mathbb{P}^3$ with defining ideal

$I_T := \langle x_1 x_3-x_2^2, \ x_2x_4-x_3^2, \ x_1x_4-x_2x_3\rangle \subset \mathbb{C}[x_1,\cdots,x_4]$.

From the degrees of the generators above, we extract $u_1 = (1,-2,1,0)$, $u_2 = (0,1,-2,1)$ and $u_3 = (1,-1,-1,1)$. Now we try to construct the matrix $A$. First stack the three vectors as follows:

$\begin{pmatrix} 1 & -2 & 1 & 0\\ 0 & 1 & -2 & 1\\1 & -1 & -1 & 1\end{pmatrix}$.

This matrix has rank two, and its kernel is spanned by $(1,1,1,1)$ and $(2,1,0,-1)$. Hence $d=2$. Thus we can take

$A=\begin{pmatrix} 1 & 1& 1 & 1\\ 2 & 1 & 0 & -1\end{pmatrix}$.

Thus the dimension of $T$ is ${\rm rank}(A)=2$ as an affine variety and 1 as an projective variety. The polytope associated to $A$ is a line segment with 3 segments, so its normalized volume is three which is the degree of $T$.

Example 2.           Consider the projective space $\mathbb{P}^n$. Then we can only extract $u=0$. Hence we can take the $(n+1)\times (n+1)$ matrix

$A = \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & & \vdots \\ 1 & 1 & 1 & \cdots & 1\end{pmatrix}$.

Since ${\rm rank}(A)=n+1$ the projective dimension of $\mathbb{P}^n$ is $n$, as expected. The polytope associated to $A$ is just the $n$-simplex $\Delta_n$, with normalized volume $n! {\rm Vol}(\Delta_n) = n! \frac{1}{n!}=1$ which is the degree of $\mathbb{P}^n$.

Example 3.          Consider the variety of $(r+1)\times (s+1)$ matrices of rank one. This projective variety in $\mathbb{P}^{rs+r+s}$ can be viewed as $\mathbb{P}^r \times \mathbb{P}^s$ via the Segre embedding

$\mathbb{P}^r \times \mathbb{P}^s \rightarrow\mathbb{P}^{rs+r+s}: (a,b)\mapsto ab^\top$.

The polytope associated to $\mathbb{P}^r \times \mathbb{P}^s$ is the product simplex $\Delta_r\times \Delta_s$ with normalized volume

$(r+s)! {\rm Vol}(\Delta_r \times \Delta_s) = (r+s)! {\rm Vol}(\Delta_r) {\rm Vol}(\Delta_s) = (r+s)!\frac{1}{r!s!} = {r+s \choose r}$

which is the degree of $\mathbb{P}^r \times \mathbb{P}^s$. Let’s look at the special case $r=2$ and $s=1$. Then the defining prime ideal of $\mathbb{P}^2 \times \mathbb{P}^1\subset \mathbb{P}^5$ is

$I = \langle x_1x_5-x_2x_4, \ x_1x_6-x_3x_4, \ x_2x_6-x_3x_5\rangle \subset \mathbb{C}[x_1,\cdots,x_6]$.

From the degrees of the generators of the ideal we extract the vectors $u_1 = (1,-1,0,-1,1,0)$, $u_2 = (1,0,-1,-1,0,1)$ and $u_3 = (0,1,-1,0,-1,1)$. We stack them together to form a $3\times 6$ matrix of rank two and then compute the basis of its kernel. We stack these vectors in the basis to form a $4\times 6$ matrix $A$:

$A = \begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 & 0 & 0\\ 0 &1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1\end{pmatrix}$.

To make the life easier we add the second and third rows to the fourth row to get this new matrix $A$:

$A = \begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 & 0 & 0\\ 0 &1 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1\end{pmatrix}$.

We only need to read the first three rows of $A$, and consider the six points from reading the columns of $A$: (1,1,0), (1,0,1), (1,0,0), (0,1,0), (0,0,1), (0,0,0). It is easy to see the convex hull of them is a prism of height one while the base is a 2-simplex of length one and height one. This is precisely $\Delta_2 \times \Delta_1$. It can be viewed as a union of three $\Delta_3$, so its normalized volume is three.

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One Response to Toric perspective 1

1. tong cheung yu says:

hallmarkXmas

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寄件者: Mathematics@CUHK
寄件日期: 2015年5月23日 9:17
收件者: yiuyu4@hotmail.com
主旨: [New post] Toric perspective 1

Hon Leung posted: “This series of posts will be about toric varieties. The author is not sure if there is a part 2, but he still calls this part 1. This post is about computing the dimensions and degrees of popular toric varieties. Readers are assumed to know linear algebra”