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 integer matrix of full rank. Denote by the integer points in the kernel of . If the binomial ideal
is prime, then is called the toric ideal while the irreducible variety
is called a toric variety.
Theorem 1. Let be a toric variety. Then the dimension of is the rank of , while the degree of is the normalized volume: consider the polytope which is the convex hull of the columns of in , then the normalized volume is where is the dimension of .
Example 1. Consider the twisted cubic in or with defining ideal
From the degrees of the generators above, we extract , and . Now we try to construct the matrix . First stack the three vectors as follows:
This matrix has rank two, and its kernel is spanned by and . Hence . Thus we can take
Thus the dimension of is as an affine variety and 1 as an projective variety. The polytope associated to is a line segment with 3 segments, so its normalized volume is three which is the degree of .
Example 2. Consider the projective space . Then we can only extract . Hence we can take the matrix
Since the projective dimension of is , as expected. The polytope associated to is just the -simplex , with normalized volume which is the degree of .
Example 3. Consider the variety of matrices of rank one. This projective variety in can be viewed as via the Segre embedding
The polytope associated to is the product simplex with normalized volume
which is the degree of . Let’s look at the special case and . Then the defining prime ideal of is
From the degrees of the generators of the ideal we extract the vectors , and . We stack them together to form a matrix of rank two and then compute the basis of its kernel. We stack these vectors in the basis to form a matrix :
To make the life easier we add the second and third rows to the fourth row to get this new matrix :
We only need to read the first three rows of , and consider the six points from reading the columns of : (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 . It can be viewed as a union of three , so its normalized volume is three.