Coordinate free map to projective space

Question

Given a line bundle $\mathcal{L}=\mathcal{O}_X(D)$ on a smooth complete curve $X$ (over $\mathbb{C}$), it's quite well-known that if the space of global sections has $k+1$ base point free, linearly independent sections $s_0, \cdots ,s_k$ (assuming that $\text{dim}_{\mathbb{C}} \Gamma(X,L)=k+1$) then there is a morphism $f: X \to \mathbb{P}^k$ defined by $x \mapsto [s_0(x):\cdots:s_k(x)].$

What is the coordinate free map $X \to \mathbb{P}(\Gamma(X,\mathcal{L}))?$ does one have to always choose coordinates to define $f?$

I was thinking that the bijection between $\mathbb{P}(\Gamma(X,\mathcal{L}))$ and the complete linear system of divisors $|D|$ might be useful, but I'm losing my faith now!

score 8 · Accepted Answer · answered Feb 13 '13 at 08:39

8

No, there is no coordinate-free description of $f$.
However there does exist a canonical morphism $g:X\to \mathbb P^*(\Gamma(X,\mathcal L))$ from $X$ to the Grassmannian $\mathbb P^*(\Gamma(X,\mathcal L)$ of hyperplanes of the $\mathbb C$- vector space $\Gamma(X,\mathcal L)$.
This Grassmannian is also a projective space, non-canonically isomorphic to $\mathbb P(\Gamma(X,\mathcal L)$, just as a finite-dimensional vector space is non-canonically isomorphic to its dual.

The description of the morphism $g$ is very easy:
To $x\in X$ associate the hyperplane $g(x)=H_x$ of global sections $s$ of $\mathcal L $ vanishing at $x$ : $$H_x=\{s\in \Gamma(X,\mathcal L)\mid s(x)=0\}\subset \Gamma(X,\mathcal L)$$
That $H_x$ is a hyperplane of $\Gamma(X,\mathcal L)$ (and not the whole space $\Gamma(X,\mathcal L)$) is guaranteed by the absence of base points for the line bundle $\mathcal L$ .

[This and other considerations have led Grothendieck to redefine projective space as the set of hyperplanes of a vector space (rather than the set of lines), thereby introducing in algebraic geometry a confusion still present today :-)]

answered Feb 13 '13 at 08:39

Georges Elencwajg

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Dear @Georges, thank you for your illuminating answer. That's very helpful. – Ehsan M. Kermani Feb 13 '13 at 09:08
Dear Ehsan, it was a pleasure. – Georges Elencwajg Feb 13 '13 at 10:34
Nice answer Georges ! So the correct definition of $\mathbb P(V)$ for a vector space $V$ is $\mathrm{Proj}(\mathrm{Sym}(V))$ (the symetric algebra of $V$). The rational points correspond to $V^*$. – Feb 13 '13 at 19:48
Nice to hear from you again, @QiL (I have been absent from the Internet for some time). Have you changed your name ? – Georges Elencwajg Feb 13 '13 at 19:53
Your absence was very visible. Yes, 8 means 8 hours. :) – Feb 13 '13 at 19:58
Dear @QiL (may I still call you so, or is it too familiar?): do you mean you are present 8 hours a day on this site? – Georges Elencwajg Feb 13 '13 at 20:10
No, this would be too much :). ça veut dire que je laisse au mois 8 heures à une question avant de poster une "answer" (si j'en ai une). – Feb 13 '13 at 22:00
Ah, merci @QiL pour ces réponses à un ami et admirateur sans doute trop indiscret... – Georges Elencwajg Feb 13 '13 at 22:16
@GeorgesElencwajg: thanks ! Non il n'y rien d'indiscret. – Feb 14 '13 at 15:19

Coordinate free map to projective space

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