I'm reading some notes and I come across this computation in the context of computing Springer fibers:
Let $V$ be a $4$ dimensional vector space over $k$ with standard basis $e_i$. We have a nilpotent matrix $N: e_4 \mapsto e_2 \mapsto 0$ and $e_3 \mapsto e_1 \mapsto 0$. We are computing the space of flags $0 \subset V_1 \subset V_2 \subset V_3 \subset V$ such that $N(V_i) \subset V_{i-1}$. There are two cases, either $N(V_2) = 0$ or $N(V_2) = V_1$. Suppose we are in the latter case.
Then for any choice of line $\langle (a,b,0,0) \rangle = l = V_1$ we can parametrize the planes $V_2$ containing $V_1$ as the set of $\langle l, (0,0,a,b) + Ker(N) \rangle$ where $Ker(N) = \{(x,y,0,0)\}$, modulo the fact that $\langle l, (x,y,a,b) \rangle = \langle l, (x',y',a,b) \rangle$ iff $(x,y)$ and $(x',y')$ differ by an element of $l$, in other words the viable $V_2$'s are parametrized by $Ker(N)/l \cong k \cong \mathbb{A}^{1}$. Finally it is easy to see $V_3$ must be $\langle V_2, Ker(N) \rangle$.
From our concrete description, we know such flags are parametrized by lines $l \in \mathbb{P}^{1} = \mathbb{P}(Ker(N))$ and elements of $Ker(N)/l$. Clearly this is line bundle over $\mathbb{P}^{1}$. The notes conclude by saying this is the total space of the line bundle $\mathcal{O}_{\mathbb{P}}(2)$. How can I see that this is the case?
By your description, the flag is parameterized by $\text{Hom}(l,\ker (N)/l)$, with $l\in \mathbb P(\ker(N))\cong \mathbb P^1$. Since this $\text{Hom}$-space is naturally identified with the tangent space $T_l\mathbb P^1$ at $l$, the parameter space is the tangent bundle $$T\mathbb P^1,$$ which is isomorphic to $\mathcal{O}_{\mathbb P^1}(2).$