Vector subbundles of a given one in $\mathbb{CP}^1$ I apologize if this question is not suited for MathOverflow. This has been crossposted in MathStackExchange here and it is related to some open questions on that site that remain unsolved.
I would like to understand and if possible classify all vector subbundles of a fixed given one $E$ on the complex projective line, in the holomorphic category. The Grothendieck decomposition theorem says that over $\mathbb{CP}^1$
$$ E \simeq \bigoplus_{i=1}^{r} \mathcal{O}(k_i) $$
where $r$ is the rank of $E$.
This is partially answered in this question but unfortunately the general case is never covered there.
What are the general subtleties that arise (as the answer there suggests) in general? What kind of (probably cohomological) techniques could be used to tackle this problem?
To start somewhere, this is trivial for line bundles: $L' \subset L$ as holomorphic lines bundles implies that either $L' = 0$ or $L' = L$.
If this is trivial or there is some reference already addressing this, feel free to just point to that instead.
 A: $\sum_{i=1}^r \mathcal O(a_i)$ is a sub-bundle of $\sum_{i=1}^s \mathcal O( b_i)$ if and only if, for all $c$, (1) $\# \{ i \mid a_i \geq c \} \leq \# \{i \mid b_i\geq c\}$ and (2) if equality holds for one $c$ it holds for all greater $c$.
Proof of "only if": By twisting, we may assume $c=0$. Then $\sum_{i, a_i\geq 0} \mathcal O(a_i)$ is the maximal globally generated sub-bundle of $\sum_{i=1}^r \mathcal O(a_i)$, thus a sub-bundle of the maximal globally generated sub-bundle $\sum_{i, b_i \geq 0} \mathcal O(b_i)$ of $\sum_{i=1}^s \mathcal O( b_i)$. Hence we have the rank inequality (1) and, if the ranks are equal, the bundles must be isomorphic, giving (2).
Proof of "if": If $V, W$ are two vector bundles, if $V$ is a summand of $W$ then $V \oplus U$ is a summand of $W \oplus U$. Furthermore, our hypothesis holds for $V$ and $W$ (if and) only if holds for $V \oplus U$ and $W \oplus U$. Using this, we may reduce to the case when our two vector bundles have no common direct summand, i.e. when $a_i \neq b_j$ for all $j$.
We may assume the $a_i$ and $b_i$ are in nonincreasing order. In this case, $a_j \leq b_{j+1}$ for all $j$, since if $a_j > b_{j+1}$ then $\#\{ i \mid a_i \geq a_j \} \geq j \geq \# \{ i \mid b_{i} \geq a_j \}$ so by the (1) we have $\#\{ i \mid a_i \geq a_j \} =\# \{ i \mid b_{i} \geq a_j \}$ and then by (2) we have $a_i = b_i $ for all $i \leq j$, contradicting our assumption.
A map $\sum_{i=1}^r \mathcal O(a_i)  \to \sum_{i=1}^s \mathcal O(b_i)$ is a matrix $M$ whose entry $M_{ij}$ is a map from $\mathcal O(a_j)$ to $\mathcal O(b_i)$ and thus i section of $\mathcal O(b_i-a_j)$. The matrix has full rank at a point as long as some $r \times r $ minor is nonzero, so the map $V \to W$ is an inclusion of sub-bundles as long as the $r\times r$ minors have no common factors.
We choose a map $M$ such that $M_{ij}=0$ unless $i=j$ or $i={j+1}$. Then the $r\times r$ minor from the first $r$ rows is $\prod_{j=1}^r M_{jj}$ and the $r \times r$ minor from the second $r$ rows is $\prod_{j=1}^r M_{(j+1)j}$. Since  $a_j \leq b_{j+1} \leq b_j$ for all $j$, both of these are products of sections of a positive power of $\mathcal O(1)$. We can then take all the $M_{jj}$ to vanish only at $\infty$, and all the $M_{(j+1)j}$ to vanish only at $0$, ensuring these two minors have no common roots, giving an inclusion of sub-bundles, as desired.
