Schubert decomposition of a Grassmannian

Question

I'm going through Sheldon Katz's Enumerative Geometry and String Theory, and a few things regarding the Grassmannian $G(2,4)$ (lines in projective space) are bothering me:

1. How can I compute the subspaces of $\mathbb{C}^4$ that correspond to lines disjoint from a given line in $\mathbb{P}^3$? For example, using coordinates $[x_0:x_1:x_2:x_3]$ for $\mathbb{P}^3$, let $\ell$ be the line given by $x_0 = x_1 = 0$. The book says that the lines disjoint from $\ell$ are given by $\langle (1,0,a,b), (0,1,c,d) \rangle$.

I can see that $\ell$ corresponds to $\langle (0,0,1,0),(0,0,0,1) \rangle$ and I've tried showing that, given a subspace $\langle (a,b,c,d),(e,f,g,h) \rangle$, there is no non-trivial linear combination of $(a,b,c,d)$ and $(e,f,g,h)$ that is of the form $(0,0,\alpha,\beta)$, but I end up having to check a few cases and it's definitely not straightforward.

2. The book goes on to compute the general form of elements of the Schubert cycle $\sigma_1(L) = \{\ell \in G(2,4) \mid \ell \cap L \neq \emptyset\}$, which apparently is given by $\langle (1,a,0,b),(0,0,1,c) \rangle$. How so?

Again this is me encountering difficulties working with the projectivization of subspaces, so a few tips would be much appreciated.

3. The book then says that the elements $\langle (1,a,0,b), (0,0,1,c) \rangle$ form a cell whose closure is $\sigma_1(L)$. How is this a cell (a cell is the difference of two consecutive strata in a cellular decomposition), and how is it not the whole $\sigma_1(L)$, from the above consideration?

Answer 1

A line in $\mathbb{P}^3$ corresponds to a two-dimensional subspace of $\mathbb{C}^4$.

Given two two-dimensional subspaces $V, W \subset \mathbb{C}^4$, there are three possibilities for $\dim V \cap W$: it is either 0 (the two lines are disjoint), 1 (the two lines meet in a point), or 2 (the two lines are in fact the same line).

These correspond to three possibilities for $\dim(V + W)$: it is either 4, 3, or 2, respectively.

How do we calculate the dimension of $V+W$? Well, if $V$ is spanned by $e_3$ and $e_4$ and $W$ is spanned by $w_1$ and $w_2$ then

$$\dim(V + W) = \operatorname{rank}\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ w_{11} & w_{12} & w_{13} & w_{14} \\ w_{21} & w_{22} & w_{23} & w_{24} \end{pmatrix}.$$

If $V$ and $W$ represent lines that don't meet, then this matrix has rank 4.

Now if the matrix has rank 4 then $w_{11}$ and $w_{21}$ can't both vanish. Changing the basis for $W$ by row operations if necessary, we can make $w_{11} = 1$ and $w_{21} = 0$. Proceeding in a similar fashion, we can arrange for $w_{12} = 0$ and $w_{22} = 1$. At this point all our choices are made, and so the matrix looks like

$$\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & a & b \\ 0 & 1 & c & d \end{pmatrix}$$

for some $a,b,c,d$. And we see indeed that this matrix has rank four for any choice of $a,b,c,d$ whatsoever.

You can do all similar problems the same way by Gaussian elimination. Just take into account that a matrix has rank less than $r$ iff all its $r \times r$ minors vanish.

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In the last problem, be careful about what the word "general" means in algebraic geometry. It does not mean "in full generality," but rather "for most points" (to be precise, it means "on a dense open set.")

So a general point of $\sigma_1(L)$ is a line of the form

$$\begin{bmatrix} 1 & \ast & 0 &\ast \\ 0& 0& 1 &\ast \end{bmatrix}$$

But of course $L$ meets itself, i.e. $L \in \sigma_1(L)$. And $L$ is not a line of this form.

Answer 2

Note two matrices have the same row space iff they are row equivalent iff they have the same reduced echelon form. So there is a one one correspondence between 2 dimensional subspaces of C^4 and (2 by 4) rank 2 matrices in reduced echelon form. The six schubert cells correspond precisely to the 6 choices of the 2 pivot columns.

The pivot columns are by definition the lexicographically earliest pair of axes that span a coordinate 2 plane to which the given row space projects isomorphically. Since projection is an isomorphism iff the row space misses the center of the projection, this has a characterization in terms of the row space not intersecting the given center.

E.g the 1st two columns are pivots iff the row space projects isomorphically to the (X0,X1) coordinate space iff the row space has trivial intersection with X0=X1=0. This completely answers question 1, i.e. the two vectors you gave are exactly the rows of a matrix whose first two columns are pivots. Note there are 4 free parameters here, i.e. this is an open 4 cell.

To answer questions 2,3 I need to know what L is. Ok, I assume L is X0=X1=0.

Now you want the form of a "general" line that DOES meet L. This will have 3 free parameters, and will be represented by a reduced echelon form matrix that has pivots other than the first two columns. The only choice that has 3 parameters is to have pivots in the 1st and 3rd columns, i.e. [1 a 0 b], [0 0 1 c]. These form an open 3 cell whose closure is apparently the cycle of all matrices whose row space does meet L. (It is the only open 3 cell.)