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 wellknown 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.

The projective space associated to a finitedimensional 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 $d_p:K_p \to K_{p1}$, 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 $$ 0 \to \Omega^{p+1}_{\mathbb{P}V/k} \to (\bigwedge^{p+1}_k V^\vee)\otimes_k \mathcal{O}_{\mathbb{P}V}(p1) \to \Omega^p_{\mathbb{P}V} \to 0. $$ 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 $1\leq p \leq n1$. But the short exact sequence also proves that $\Omega^n_{\mathbb{P}V} \cong (\bigwedge^{n+1}_k V^\vee) \otimes_k \mathcal{O}_{\mathbb{P}V}(n1)$. 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. 


Here is another answer, probably from a slight different angle. In my opinion, the easiest way to see that $\Omega_{\mathbb P^n}^1(2)$ is globally generated (at least over $\mathbb C$) is the following direct computation. Let $\mathbb P^n=\mathbb P(\mathbb C^{n+1})$ be the projective space of lines in $\mathbb C^{n+1}$ with coordinates $(Z_0,\dots,Z_n)$ and $U_j$, $j=0,\dots,n$ be the affine open set in $\mathbb P^n$ corresponding to $Z_j\ne 0$. Finally, let $(z_{j,1},\dots,z_{j,n})$ be the corresponding affine coordinates on $U_j$. The (anti)tautological line bundle $\mathcal O(1)$ on $\mathbb P^n$ is then described by the following transition functions: $$ g_{\alpha\beta}([Z_0:\cdots:Z_n])=\frac{Z_\beta}{Z_\alpha},\quad U_\alpha\cap U_\beta. $$ Thus, at $[Z_0:\cdots:Z_n]\in U_\alpha\cap U_\beta$, the line bundle $\mathcal O(2)$ has transition functions given by $g_{\alpha\beta}^2([Z_0:\cdots:Z_n])=(Z_\beta/Z_\alpha)^2$. Now, take any nonzero vector $v$ in $\Omega_{\mathbb P^n,x_0}^1(2)$. Without loss of generality (by acting with $PGL(n)$, rotating and rescaling if necessary), you can suppose that $x_0=[1:0:\cdots:0]\in U_0$ and that $v=(dz_{0,1}\otimes\eta_0)(x_0)$, where $\eta_j$ is a local frame for $\mathcal O(2)$ on $U_j$, so that $\eta_\beta=g_{\alpha\beta}^2\eta_\alpha$ on $U_\alpha\cap U_\beta$. I claim that the section $dz_{0,1}\otimes\eta_0$, a priori defined only over $U_0$, is in fact a global holomorphic section. To see this, it suffices to check what happens when passing from $U_0$ to $U_j$, $1\le j\le n$. We have, for $j=1$, $z_{0,1}=1/z_{1,1}$ and, for $j\ge 2$, $z_{0,1}=z_{j,2}/z_{j,1}$. Therefore, $$ dz_{0,1}=\frac{dz_{1,1}}{z_{1,1}^2}=\frac{z_{j,1}dz_{j,2}z_{j,2}dz_{j,1}}{z_{j,1}^2},\quad j\ge 2. $$ On the other hand, $$ \eta_{0}=(Z_0/Z_j)^2\eta_j=z_{j,1}^2 \eta_j. $$ So, $$ dz_{0,1}\otimes\eta_0=dz_{1,1}\otimes\eta_1=(z_{j,1}dz_{j,2}z_{j,2}dz_{j,1})\otimes\eta_j,\quad j\ge 2, $$ which are actually holomorphic. This proves "by hands" the global generation of $\Omega_{\mathbb P^n}^1(2)$. Of course, with a little bit more involved computation, you can show as well by hands the global generation of twisted exterior powers $\Omega_{\mathbb P^n}^p(p+1)$. Please note, that this global generation is just a reformulation of the elementary fact that the differential of a meromorphic function with a simple pole is at worst a meromorphic $1$form with a pole of order $2$! Turning to your second question, I think that in this generality you propose one cannot say much more than the following. Any quotient of a trivial vector bundle is globally generated and, conversely, every globally generated vector bundle $E\to X$ of rank $r$ over a $n$dimensional complex manifold $X$ is isomorphic to the quotient of a trivial vector bundle of rank $\le n+r$ (this is a nontrivial fact). In particular, if you are able to fit your vector bundle $E$ into a sequence $V\to E\to 0$, where $V$ is globally generated, then $E$ is globally generated itself. Thus, for instance, if $X\subset\mathbb P^n$ is a smooth submanifold, you have a short exact sequence of vector bundles
$$
0\to T_X\to T_{\mathbb P^n}_X\to N_{X/\mathbb P^n}\to 0.
$$
Twisting by $\mathcal O(2)$ and taking duals, you obtain
$$
0\to N_{X/\mathbb P^n}^*(2)\to\Omega_{\mathbb P^n}^1(2)_X\to\Omega_X^1(2)\to 0.
$$ If you want more sophisticated criterions, then probably you should restrict a little bit more your second question. 

