Splitting type of vector bundle on projective space Let $V=H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(2))$ and $\varphi:V\otimes\mathcal{O}_{\mathbb{P}^n}\to\mathcal{O}_{\mathbb{P}^n}(2)$ be an evaluation map. 


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*How to prove that $E:=\text{ker}(\varphi)$ is locally free?


This is obvious, because kernel of surjective morphism of vector bundles is vector bundle. Also, we can show, that $\mathcal{Ext}^i(E, \mathcal{O}_{\mathbb{P}^n})=0$ for all $i>0$, which shows that $E$ is vector bundle.


*Let $L\subset\mathbb{P}^n$ be an arbitrary line. Then we have a decomposition $E|_L=\bigoplus\limits_{i=1}^{\text{dim}V-1}\mathcal{O}_L(a_i(L))$ for some $a_i(L)\in\mathbb{Z}$. How to find all $a_i(L)$ in terms of $L$ explicitly?

 A: I think the following works, maybe there is an easier answer...
First note that $\det E_{|L}=\bigotimes_{i=1}^{\dim V-1}\mathcal{O}_L(a_i(L))=\mathcal{O}_L(\sum a_i(L))$. Note also that $\det E_{|L}\otimes \mathcal{O}_L(2)=\det (V\otimes\mathcal{O}_{\mathbb{P}^n})_{|L}=0$. So $\det E_{|L}=\mathcal{O}_L(-2)$ and $\sum a_i(L)=-2$.
Then, note that, because the map $E_{|L}\rightarrow (V\otimes\mathcal{O}_{\mathbb{P}^n})_{|L}$ is injective, the $a_i(L)$ must be non positive. So there is two cases either they are all 0 except one which is -2, or they are all 0 except two which are -1.
To compute the number of 0, you can simply compute the dimension of $H^0(L,E_{|L})=\ker\left(H^0(L,(V\otimes\mathcal{O}_{\mathbb{P}^n})_{|L})\rightarrow H^0(L,\mathcal{O}_L(2))\right)$. 
This map simply take a homogeneous polynomial of degree 2 and $n+1$ variables to its restriction to $L$. Assume that in suitable coordinates, $L$ is given by the equations $x_2=x_3=\dots=x_n=0$, the map become 
$$ \sum_{0\leq i\leq j\leq n} a_{ij}x_i x_j\mapsto \sum_{0\leq i\leq j\leq 1} a_{ij}x_i x_j$$
so its kernel is easily seen to be of dimension $\frac{(n+1)(n+2)}{2}-3=\dim V-3=\operatorname{rk} E_{|L}-2$.
So we found that all the $a_i(L)$ are 0 except two which are -1.
