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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?

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.

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.

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.

In more detail (but different coordinates: (X,Y,Z,W)): if two matrices are row equivalent then they have the same row space, the span of their rows. The reduced echelon form reduces the original sequence of rows to a sequence which is independent, hence forms a basis for the row space. So one point of view is that finding the reduced echelon form amounts to choosing an especially nice or simple basis for the row space.

From this perspective the process is very simple. If the rank is r, then there must exist some coordinate subspace of rank r onto which the row space projects isomorphically. If we find one, we can pull back the standard basis for this coordinate subspace to a basis of the row space via this isomorphism. This provides at most “m choose r” distinguished bases for the row space of an n by m matrix of rank r.

The only one in echelon form is the one whose choice of r coordinates among the m possible coordinates, is lexicographically minimal. This provides a unique choice of ordered basis for the row space, hence the reduced echelon form of a given matrix is unique. I.e. its pivots correspond to the unique lexicographically minimal choice of coordinate subspace onto which the row space projects isomorphically.

(It follows that the row space can also be viewed as the graph of a linear map from the coordinate space of pivot variables to that of non pivot variables, whose matrix is the transpose of the matrix of non pivot columns.)

Grassmannian point of view for G(2,4) = planes in 4 space.
G(2,4) = all planes in 4 space ≈ all possible 2 dimensional row spaces for 2 by 4 matrices ≈ all possible 2 by 4, rank 2, reduced echelon forms.

Since projection onto a given coordinate subspace is isomorphic if and only if the center of projection does not meet the row space, this has an interpretation in terms of incidence relations as discussed previously. E.g. for a 2 by 4 matrix of rank 2, we can choose as a “flag” of subspaces the spaces X=0, X=Y=0 and X=Y=Z=0.

Then projection onto the coordinate subspace of the first 2 coordinates, the X,Y subspace, is isomorphic if and only if the row space does not meet the subspace X=Y=0 non trivially, iff the first two columns are pivots. (4 parameters.) (your case #1.)

Thus the row space does meet X=Y=0 non trivially iff the first two columns are not both pivots. If the row space does not meet the line X=Y=Z=0 then the 4th column is not a pivot. If the row space does not lie in the space X=0 then the first column is a pivot, so if all three of these hold, the pivots are the 1st and 3rd columns. (3 parameters.) (your case #2.)

If the row space does lie in the space X=0 but does not contain the line X=Y=Z=0 then the row space does not have the 1st column as pivot, and does not have the 4th column as pivot, so must have the 2nd and 3rd columns as pivots. (2 parameters.)

If the row space does contain the line X=Y=Z=0 but does not lie in the space X=0, then the 1st column and also the 4th column is a pivot. (2 parameters.)

If the row space does contain the line X=Y=Z=0 and also lies in the space X=0, then the 1st column is not a pivot but the 4th is. If also the row space does not equal X=Y=0 then the 3rd and 4th column are not both pivots, so the pivots are the are the 2nd and 4th. (1 parameter.)

If the row space is X=Y=0 then the pivots are the 3rd and 4th columns. (0 parameters.)