Take any smooth projective irreducible algebraic curve $X$, any vector bundle $E$ on it and any two sub-bundles $E\_1$ and $E\_2$. Over each point $x$ of $X$ we can consider the sum $F\_x$ of the vector spaces on the fibers of $E\_1$ and $E\_2$. (in general, this will not be a direct sum)


Then:


1) is $F$ a subbundle of $E$?


2) how is related $H^0(F)$ with $H^0(E\_1)$ and $H^0(E\_2)$?



Given 2 sub-bundles of a given bundle, is their sum again a subvector bundle?
Another example when a sum of two subbundles is not a subbundle is the following. Let $V$ be a vector space (say 2-dimensional) and $X = P(V)\times P(V)$. Then we have two subbundles $O(-1,0) \subset V\otimes O$ and $O(0,-1) \subset V\otimes O$ on $X$ (the pullbacks of the tautological subbundles from the factors). However, it is easy to see that summing up the embeddings one gets an exact sequence
$$
0 \to O(-1,0) \oplus O(0,-1) \to V\otimes \Delta\_\*O(1) \to 0,
$$
where $\Delta:P(V) \to P(V)\times P(V)$ is the diagonal embedding. In particular, the sum $F$ (which is the image of the first map) although isomorphic to the direct sum is not a subbundle (since the rank of the map drops at the diagonal).


As for the cohomology, there is an exact sequence
$$
0 \to E\_1 \cap E\_2 \to E\_1 \oplus E\_2 \to F \to 0,
$$
which gives a long exact sequence of cohomology. So, to know $H^0(F)$ you need to know first $H^0(E\_1\cap E\_2)$ and $H^1(E\_1\cap E\_2)$. Note however, that if $E\_1$ and $E\_2$ do not intersect generically, then $E\_1 \cap E\_2$ is a zero rank subsheaf of a torsion free sheaf $E\_1 \oplus E\_2$, hence $E\_1 \cap E\_2 = 0$. So, in this case $F = E\_1 \oplus E\_2$, hence $H^0(F) = H^0(E\_1) \oplus H^0(E\_2)$.


