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.

  1. 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.

  1. 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?

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.

  • $\begingroup$ The kernel seems to be of dimension $(n+1)(n+2)/2 - 3$. Why did you wrote $\text{dim}E|_F - 2$ in the right hand side? $\endgroup$
    – mahavishnu
    Nov 29 '15 at 20:36
  • $\begingroup$ @mahavishnu You are right, it is $(n+1)(n+2)/2-3$, i have corrected. For the right hand side, note that $V$ is the space of homogeneous polynomials of degree 2, so its dimension is $(n+1)(n+2)/2$. $E$ being the kernel of $V\rightarrow\mathcal{O}(2)$ has dimension $(n+1)(n+2)-1$. $\endgroup$
    – Roland
    Nov 29 '15 at 20:48
  • $\begingroup$ I do not understand what do you mean by dimension of $E$. Is it rank maybe? Then it is correct - the rank of $E$ is equal to $(n+1)(n+2)/2-1$. $\endgroup$
    – mahavishnu
    Nov 29 '15 at 20:57
  • $\begingroup$ yes it is its rank $\endgroup$
    – Roland
    Nov 29 '15 at 21:08
  • $\begingroup$ Ok, thank you very much. $\endgroup$
    – mahavishnu
    Nov 29 '15 at 21:11

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