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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.

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I suggest you consider the dual of the Euler sequence, considered as a morphism from $\mathcal{O}(-1)^{\oplus (n+1)}$ to $\mathcal{O}$. Now take the Koszul complex of this and consider the $p^\text{th}$ syzygy. How does it relate to $\Omega^p$? So how can you write $\Omega^p$ as a quotient of a direct sum of copies of $\mathcal{O}(-p-1)$? – Jason Starr Jul 13 '12 at 20:04
I knew that the Euler sequence was the only tool one could use but haven't thought about the Koszul complex. You are totally right. $\Omega$ is a quotient of $\mathcal{O}(-2)^{\binom{n+1}{2}}$. Thank you. – Muon Jul 13 '12 at 23:16
@Martin -- I agree with that, but this wasn't an answer (more like a "hint"). I will turn it into an answer now. – Jason Starr Jul 14 '12 at 11:42
up vote 5 down vote accepted

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.

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This fact is more than what I expected. Thank for the complete proof! – Muon Jul 15 '12 at 23:47
I now think that this can be generalized to weighted projective spaces, which also have "Euler sequences" (although twisting is no longer 2). Maybe even to toric varieties. – Muon Jul 16 '12 at 19:10

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 non-zero 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 non-trivial 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 $$ 0\to N_{X/\mathbb P^n}^*(2)\to\Omega_{\mathbb P^n}^1(2)|_X\to\Omega_X^1(2)\to 0. $$
By the above criterion, the cotangent bundle of any (embedded) smooth projective manifold twisted by $\mathcal O(2)$ is globally generated.

If you want more sophisticated criterions, then probably you should restrict a little bit more your second question.

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Thanks for your explicit computation. As you pointed out, this clearly explains why "2". – Muon Jul 16 '12 at 18:06
My argument also shows my "2", and more generally why "p+1". For $1\leq p \leq n$, there is an exact sequence $0 \to \Omega^p(p) \to \bigwedge^p_k V^\vee\otimes_k \mathcal{O} \to \bigwedge^{p-1}_k V^\vee\otimes_k \mathcal{O}(1)$. Taking global sections, $H^0(\Omega^p(p))$ is the kernel of the linear map $\bigwedge^p V^\vee \to \bigwedge^{p-1} V^\vee\otimes_k V^\vee$. This map is the transpose of the cup product map $\bigwedge^{p-1}V\otimes_k V \to \bigwedge^p V$, which is surjective. Thus the transpose is injective, i.e., $H^0(\Omega^p(p))$ equals $\{0\}$ – Jason Starr Jul 17 '12 at 15:55
Typo correction: "shows my" --> "shows why". – Jason Starr Jul 17 '12 at 15:56
Of course your argument also explains this! I think it's just a matter of tastes, isn't it? – diverietti Jul 17 '12 at 16:32
Yes. Both answers explain why "2" (and "p+1"). I just wanted to comment on diverietti's remark about meromorphic functions. – Muon Jul 18 '12 at 6:30

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