I have an elementary question about globally generation of a vector bundle. I would like to see why $\Omega\_{\mathbb{P}^n}(2H)$ is globally generated (it seems this is well-known among experts). Here $H$ is the hyperplane class of $\mathbb{P}^n$. In general how do we prove globally generation of a vector bundle (line bundle is easy), any criteria? Thanks for your help. 



Globally generation of $\Omega_{\mathbb{P}^n}(2H)$
The projective space associated to a finite-dimensional vector space $V$ over a field $k$ is a universal pair $(\mathbb{P}V,\tilde{\gamma})$ of a $k$-scheme $\mathbb{P}V$ and a surjection of coherent sheaves 
$$ \tilde{\gamma}:V^\vee \otimes\_k \mathcal{O}\_{\mathbb{P}V} \to \mathcal{O}\_{\mathbb{P}V}(1), $$
such that $\mathcal{O}\_{\mathbb{P}V}(1)$ is an invertible sheaf (the Serre twisting sheaf).
$\textbf{NB}.$ Some people prefer to use $V$ rather than $V^\vee$ in this definition. 


Tensoring $\tilde{\gamma}$ by the identity on $\mathcal{O}\_{\mathbb{P}V}(-1)$ gives another morphism of coherent sheaves,


$$ \gamma:V^\vee\otimes\_k \mathcal{O}\_{\mathbb{P}V} (-1) \to \mathcal{O}\_{\mathbb{P}V}.$$


As a map to the structure sheaf, we can use this to form a Koszul complex $(K\_\bullet,d\_\bullet)$ where the term $K\_p$ is 


$$\bigwedge\_{\mathcal{O}}^p (V^\vee\otimes\_k \mathcal{O}\_{\mathbb{P}V}(-1) ) \cong (\bigwedge\_k^p V^\vee)\otimes\_k \mathcal{O}\_{\mathbb{P}V}(-p),$$


and where the differentials $d\_\bullet$ are the unique morphisms of coherent sheaves such that $d\_1$ equals $\gamma$, and such that $(K\_\bullet,d\_\bullet)$ is a differential graded algebra, i.e., the differentials satisfy the Leibniz rule for exterior product.


Because $\gamma$ is surjective, this complex is exact. In particular, if we define $S\_p$ to be the $p^\text{th}$ syzygy, i.e., the
kernel of $d\_p:K\_p \to K\_{p-1}$, then we can break up the complex into a sequence of short exact sequences,


$$ 0 \to S\_{p+1} \to K\_{p+1} \to S\_p \to 0. $$


Moreover, because $(K\_\bullet,d\_\bullet)$ is a differential graded algebra, there are cup product maps


$$
\bigwedge^p S\_1 \to S\_p,
$$


which turn out to be isomorphisms (easiest to check locally, where $\gamma$ splits). Finally, the Euler sequence identifies $S\_1$ as $\Omega\_{\mathbb{P}V/k}$. Therefore the short exact sequences above give


$$ 0 \to \Omega^{p+1}\_{\mathbb{P}V/k} \to (\bigwedge^{p+1}\_k V^\vee)\otimes\_k \mathcal{O}\_{\mathbb{P}V}(-p-1) \to \Omega^p\_{\mathbb{P}V} \to 0. $$


From this it follows immediately that $\Omega^p\_{\mathbb{P}V}(p+1)$ is globally generated for every $p\geq 1$. 


EDIT. The argument above is only valid for $1\leq p \leq n-1$. But the short exact sequence also proves that $\Omega^n\_{\mathbb{P}V} \cong (\bigwedge^{n+1}\_k V^\vee) \otimes\_k \mathcal{O}\_{\mathbb{P}V}(-n-1)$. So when $p$ equals $n$, also $\Omega^n\_{\mathbb{P}V}(n+1) \cong (\bigwedge^{n+1}\_k V^\vee)\otimes\_k \mathcal{O}\_{\mathbb{P}V}$, which is also globally generated.


