Let $\lambda\in\text{Ext}^1(\mathcal{O}_{\mathbb{P}^1}(2),\mathcal{O}_{\mathbb{P}^1}(-2))$ and $E_\lambda$ be a vector bundle on $\mathbb{CP}^1$ which is given by the exact sequence \begin{equation}0\to\mathcal{O}_{\mathbb{P}^1}(-2)\to E_\lambda\to\mathcal{O}_{\mathbb{P}^1}(2)\to0,\,\,\,\,(1)\end{equation} and corresponds to $\lambda$.
Then, as it was discussed here, $E\cong\mathcal{O}_{\mathbb{P}^1}(a_\lambda)\oplus\mathcal{O}_{\mathbb{P}^1}(-a_\lambda)$ for some $a_\lambda\in\{0,1,2\}$.
For each $\lambda$ I want to find the explicit value of $a_\lambda$. Can anyone help me?
I tried the following explicit construction.
Let $P=\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus4}$. There is the surjective map $P\to\mathcal{O}_{\mathbb{P}^1}(2)$, which is given by the evaluation map $H^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(3))\otimes\mathcal{O}_{\mathbb{P}^1}\to\mathcal{O}_{\mathbb{P}^1}(3)$ twisted by $-1$. Let $F=\mathcal{O}_{\mathbb{P}^1}(-2)^{\oplus3}$ be the kernel of this map so we have the exact sequence \begin{equation}0\to F\to P\to\mathcal{O}_{\mathbb{P}^1}(2)\to0.\,\,\,\,(2)\end{equation} Note that $\text{Ext}^1(P,\mathcal{O}_{\mathbb{P}^1}(-2))=0$. Thus applying $\text{Hom}(-,\mathcal{O}_{\mathbb{P}^1}(-2))$ to $(2)$ we obtain the surjective connecting map $$\delta:\text{Hom}(F,\mathcal{O}_{\mathbb{P}^1}(-2))\to\text{Ext}^1(\mathcal{O}_{\mathbb{P}^1}(2),\mathcal{O}_{\mathbb{P}^1}(-2)).$$ Now from the surjectivity of $\delta$ there exists $f\in\text{Hom}(F,\mathcal{O}_{\mathbb{P}^1}(-2))$ such that $\delta(f)=\lambda$ and $E_\lambda$ is given as the push-out of $F\to P$ and $F\to\mathcal{O}_{\mathbb{P}^1}(-2)$.
I don't know how to compute $f$ explicitly for a given $\lambda$ and then how to compute the push-out.
Maybe there exists a different approach to solve this problem?
I think that you can use duality (Hartshorne, III Thm 7.1),
$$ \mathrm{Ext}^1(\mathcal{O}_{\mathbb{P}^1}(2),\mathcal{O}_{\mathbb{P}^1}(-2)) \simeq \mathrm{H}^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(2))^{\vee} $$
Identify $\mathcal{O}_{\mathbb{P}^1}(-2) = \mathcal{O}_{\mathbb{P}^1}(-p -q)$, where $p$ and $q$ are distinct points. Let $A_{\lambda}$ be a homogeneous quadratic polynomial corresponding to $\lambda$.
You can use $p$, $q$ and $A_{\lambda}$ to construct $E_{\lambda}$.
If $A_{\lambda}(p)=A_{\lambda}(q)=0$, then $a_{\lambda}=0$.
If $A_{\lambda}(p)=0 $ and $A_{\lambda}(q)\neq 0$, then $a_{\lambda}=1$.
If $A_{\lambda}(p\neq 0 $ and $A_{\lambda}(q)\neq 0$, then $a_{\lambda}=2$.
Note that choosing $p$ and $q$ is the same as choosing a basis for $\mathrm{H}^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(1))$, which determines a basis for $\mathrm{H}^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(2))$ that you have to fix to construct a isomorphism $$ \mathrm{H}^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(2)) \simeq \mathrm{H}^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(2))^{\vee} $$