Schubert decomposition of a Grassmannian

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.