If you have a complex manifold $Z$, and if you have a $\mathbb{C}P^1$ embedded in $Z$, what are the available techniques/methods for computing the normal bundle of such a $\mathbb{C}P^1$? I am aware that any holomorphic vector bundle (of finite rank) over $\mathbb{C}P^1$ splits as a direct sum of holomorphic line bundles (is that result due to Grothendieck by the way?). So if $Z$ for instance has low dimension, computing Chern classes can come in handy, as well as using adjunction formulas. But what other methods are available in the general case, where the dimension of $Z$ could be high, say.
As another question, suppose you were able to describe the normal bundle $N$ using a short exact sequence
$0 \to E \to N \to F \to 0$
where both $E$ and $F$ are known, and their direct sum decomposition is known. I would not expect $E \oplus F$ to be the only possibility for $N$. However, it is clear that you get several restrictions, using Chern classes say. Can one describe all possibilities for $N$ using maybe cohomology? I remember stumbling on an old paper (perhaps by Sir Michael Atiyah) on this, but I don't remember the details. Perhaps I should split this into two posts, but they are related sets of questions.
Regarding your first question, there are fairly obvious facts such as the adjunction formula plus the fact that if $C$ is general in a family filling up a $k$-dimensional subvariety then at least $k - 1$ line bundle summands in the normal bundle are nonnegative. Beyond that, I'm not aware of any really general results, just specific ones in specific situations (e.g. a general rational curve of given degree in $\mathbb{P}^n$).
The other question is much easier, and is certainly answerable "in principle" though I don't know of a neat formulation. $F$ is endowed with an essentially canonical filtration with quots $\mathcal{O}(a_i)$, where $a_i \ge a_2$$\ldots$ so working inductively you can assume $F$ has rank $1$. Then, dualizing, the question becomes: when is a general map between direct sums of line bundles$$\mathcal{O}(a_1, \ldots, a_{r + 1}) \to \mathcal{O}(b_1, \ldots, b_r)$$surjective? This means exactly that $\mathcal{O}(a.)$ occurs as an extension of $\mathcal{O}(b.)$ by $\mathcal{O}\left(\sum a.- \sum b.\right)$. But that question has an easy answer: consider the matrix of coefficients $[b_i-a_j]$ and let $A$ be the matrix with zeros for each $b_i-a_j<0$ and general entries for each $b_i-a_j \ge0$. This is an echelon-type matrix with zeros on the left bottom. Then $A$ should be have rank $r$ everywhere, meaning there should be at most one of
Moreover for each of the $r$ many columns not followed by a column of the same shape, the degree at the corner should be zero. Call such a corner "essential". The polynomials at the essential corners should be constant, else they would vanish somewhere and lower the rank. There should be a neat way of phrasing this, maybe written down somewhere. But I don't have a reference.
Actually, come to think of it, there is a simple condition that the matrix $A$ above has rank $r$ everywhere: for all $i=1, \ldots, r$, we have $b_i \ge a_{i+1}$, and if $b_i<a_i$, then $b_i=a_{i+1}$.
The first condition ensures that all the corners lie on or below the "diagonal" line $(1,2)$, $\ldots$ , $(r, r+1)$; the second ensures that the corners on the diagonal correspond to constants.