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A vector bundle $E\to X$ is globally generated if there exists global holomorphic sections $s_1,\dots,s_n$ such that $E_x$ is spanned by $s_1(x),\dots,s_n(x)$ for all $x\in X$.

Consider the projection map $p:\mathbb{C}^{n+1}\to\mathbb{P}^n$. It can be shown that $p$ is a smooth submersion. Furthermore, given $n+1$ linear functions $l_i:\mathbb{C}^{n+1}\to\mathbb{C}$, with $i=0,\dots,n$, and taking the coordinates in $\mathbb{C}^{n+1}$ to be $(z_0,\dots,z_n)$, it can be shown that the global vector field $$ \sum_{i=0}^nl(z)\frac{\partial}{\partial z_i} $$ in $\mathbb{C}^{n+1}$ can be pushed forward to a global vector field of $\mathbb{P}^n$. Motivated by Ted Shifrin's answer in this question I believe that all vector fields in $\mathbb{P}^n$ must be of this form, implying that $T\mathbb{P}^n$ is globally generated.

Explicitely, (I believe, a confirmation of the following would be nice) the pushforward of this vector field is: $$ p_*\left(\sum_{i=0}^nl(z)\frac{\partial}{\partial z_i}\right)= \sum_{i=0}^nl(z)p_*\left(\frac{\partial}{\partial z_i}\right), $$ where, for instance in the coordinate chart $U_0$ of $\mathbb{P}^n$ with coordinates $w_k=z_k/z_0$, $$ p_*\left(\frac{\partial}{\partial z_0}\right)=-\sum_{i=1}^n \frac{z_i}{z_0^2}\frac{\partial}{\partial w_i}, $$ and $$ p_*\left(\frac{\partial}{\partial z_k}\right)=\frac{1}{z_0}\frac{\partial}{\partial w_k}. $$ In particular this calculation in coordinates shows the need of multiplying by a linear function to get a well defined vector field.

My problem is I don't see why every vector field in projective space must be of this form, locally and/or globally.

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user347489
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  • Why does the calculation "show the need of multiplying by a linear function to get a well defined vector field"? Also, (probably not, but) should $l(z)$ be $l_i(z)$? (i.e. have a subindex $i$. Very small typeface here on the comments). – JKEG Aug 12 '20 at 23:02

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$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$Yes, $T\Cpx\Proj^{n}$ is globally generated. Note that it doesn't matter whether or not all global holomorphic vector fields on $\Cpx\Proj^{n}$ are of the stated form; what matters is:

  1. At some point $p$ (and hence every point, by homogeneity) there exist vector fields of the stated form spanning the tangent space at $p$.

  2. The zero set of a generic vector field is a finite set (of $(n+1)$ points, if it matters).

(Incidentally, it appears you're using $n$ to denote two distinct integers: The complex dimension of the projective space, and the number of holomorphic vector fields needed to span the tangent bundle globally.)

Andrew D. Hwang
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  • Hi Andrew, thanks for your reply. A couple of questions: 1) We also need to show that the set of these vector fields is finite, right? To this end I think the vector fields I produce could do the trick. Perhaps using the presented construction I could find explicitly how many global sections are needed or is there a more direct argument? About 2), can we see that every generic vector field vanishes at $n+1$ points using the construction I'm using? (1/2) – user347489 Jul 01 '16 at 01:41
  • I came across this question trying to show there is a vector bundle epimorphism $\mathcal{O}(1)^{\oplus(n+1)}\to T\mathbb{P}^n$ given by taking $n+1$ linear functionals and sending them to the stated vector fields. As of now my argument is to show both bundles are globally generated, so showing the epimorphism at the level of global functions yields the result. (2/2) – user347489 Jul 01 '16 at 01:54
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    on the parenthetical and if the OP isn't then its already false for $\Bbb CP^1$. – PVAL-inactive Jul 01 '16 at 02:07
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    To me, the natural picture of a vector field on $\Cpx\Proj^{n}$ is to let $A$ be $(n+1) \times (n+1)$, and to differentiate the one-parameter group $t \mapsto \exp(tA)$. The eigenspaces of $A$ are the fixed points of the resulting vector field; seeing global generation (and getting a small number of generating fields) reduces to straightforward linear algebra. (That doesn't directly address your comments, but I hope helps.) – Andrew D. Hwang Jul 01 '16 at 02:08
  • @PVAL Sorry, but I can understand half of your comment. Would you mind rephrasing? I think you're saying my comment (2/2) is false even for $\mathbb{P}^1$, right? – user347489 Jul 01 '16 at 02:18
  • $T\Bbb CP^1$ does not have a non-vanishing section, so the sheaf of sections cannot be generated by a single section. – PVAL-inactive Jul 01 '16 at 02:20
  • @AndrewD.Hwang Cool, thanks! It got lost in the large comments I wrote, but any idea on how to argue that the set of sections you mention in 1) is finite? – user347489 Jul 01 '16 at 02:21
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    @PVAL agreed, but I don't see which of arguments contradicts this fact. – user347489 Jul 01 '16 at 02:28
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    To show global generation of the tangent bundle $T\Cpx\Proj^{n}$ by sections, pick $N$ "generic" vector fields; each vanishes at finitely many points (by 2.), so these fields generate except at finitely many points. Appending finitely many vector fields gives global generation. (Separately, PVAL's point is, the tangent bundle of $\Cpx\Proj^{n}$ is not globally generated by $n$ vector fields when $n = 1$; turns out the same is true for all $n > 0$.) [...] – Andrew D. Hwang Jul 01 '16 at 09:47
  • Incidentally, to show there's a bundle epimorphism $\mathscr{O}(1)^{\oplus(n+1)} \to T\Cpx\Proj^{n}$, it may be more natural to think of the tautological subbundle of the trivial $(n+1)$-plane bundle; it turns out there's a short exact sequence$$0 \to \mathscr{O}(-1) \to \Cpx\Proj^{n} \times \Cpx^{n+1} \to T\Cpx\Proj^{n}(-1) \to 0$$for geometric reasons. Tensoring with $\mathscr{O}(1)$ gives the desired epimorphism. – Andrew D. Hwang Jul 01 '16 at 09:59
  • @AndrewD.Hwang Awesome! Together with the answer these last two comments are really helpful! Thanks. – user347489 Jul 01 '16 at 11:25
  • @AndrewD.Hwang Can't help but realize the exact sequence you wrote is very similar to the Euler exact sequence twisted by $\mathcal{O}(-1)$, does this imply that $$ \mathcal{O}(1)^{\oplus(n+1)}\cong(\mathbb{P}^n\times\mathbb{C}^{n+1}) \otimes\mathcal{O}(1)?$$ – user347489 Jul 01 '16 at 22:19
  • Yes: Over an arbitrary base, the tensor product of a line bundle $L$ with the trivial $n$-plane bundle is the direct sum of $n$ copies of $L$. – Andrew D. Hwang Jul 02 '16 at 02:16
  • @AndrewD.Hwang Oh, I should've known that! Many thanks! – user347489 Jul 02 '16 at 07:10