Let $V$ be a vector space over $\mathbb C$ of dimension $4$ and consider the Grassmanian $Gr(2,V)$. Fix an index $I=(1,3)$ and a complete flag of $V=\langle e_1, e_2, e_3, e_4 \rangle$ given by $ F =\langle e_1 \rangle \subset \langle e_1, e_2 \rangle \subset \langle e_1, e_2, e_3 \rangle \subset V$.
The associated Schubert variety $X_I(F)$ is given by the span of the matrices
$$ \begin{bmatrix} 1 & 0 \\ 0 & * \\ 0 & 1 \\ 0 & 0 \\ \end{bmatrix} \cup \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix} $$
Since it is $\mathbb C \cup \mathbb C^0$ I'm wondering if it is a copy of $\mathbb P^1(\mathbb C)$ inside the Grassmanian in some way, or if this is not exact.
Thanks!
Yes, this Schubert cycle is the set of $2$-planes containing a fixed line and contained in a fixed $3$-space. So it is, in particular, the set of $2$-planes in $\Bbb C^3$ containing a fixed line. This is, alternatively, the set of $\Bbb P^1$'s in $\Bbb P^2$ containing a fixed point, and that is easily seen to be a $\Bbb P^1$. (For example, choose a $\Bbb P^1\subset\Bbb P^2$ not containing the fixed point $p$. Each line through $p$ meets that $\Bbb P^1$ in a unique point, so the pencil of lines through $p$ is isomorphic to that $\Bbb P^1$.)