At the moment, I'm struggling with the following definitions i) and ii). I'd like to know why they are equivalent:

Let $\mathcal{L}$ be an invertible sheaf on a variety X.

i) $\mathcal{L}$ is called very ample, if there is an embedding $i\colon X \rightarrow \mathbb{P}^n$ such that $\mathcal{L}$ is isomorphic to $i^{∗}\mathcal{O}_{\mathbb{P}^n}(1)$.

ii) For a basis $f_0, \dots ,f_r$ of $\Gamma(X, \mathcal{L})$ we define $$\varphi \colon X \rightarrow \mathbb{P}^r, \quad x \mapsto [f_0(x), \dots , f_r(x)].$$ The invertible sheaf $\mathcal{L}$ is called very ample, if it is base-point-free and $\varphi$ is a immersion.

Thanks in advance!


It's not hard to guess that the embedding in i) is the same as the that of ii). Thus the only problem is to show, under the embeddment described in ii), the pullback of $\mathcal{O}_{\mathbb{P}^n}(1)$ is the line bundle $\mathcal{L}$. If you got the Griffiths&Harris at hand, just read the corresponding section (1.4). it's way more better than I can do.

A simple explanation is, in the aspect of divisors, pullback of an immersion is taking intersection. We can take the hyperplane as $x_0=0$. Thus the divisor of pullback is the zero locus of $f_0(x)$. This gives the line bundle $\mathcal{L}$

  • $\begingroup$ Thanks for your help! The problem is that I still don't understand why $\varphi^*(\mathcal{O}_{\mathbb{P}^n}(1))$ is the line bundle $\mathcal{L}$. The transition functions of $\mathcal{O}_{\mathbb{P}^n}(1)$ are given by $\frac{T_j}{T_i}$ so if $\varphi^*(\mathcal{O}_{\mathbb{P}^n}(1)) \cong \mathcal{L}$ holds, the transition functions of $\mathcal{L}$ should be $\frac{f_j}{f_i}$. But unfortunately I don't see why the latter holds... $\endgroup$ – claudi Oct 30 '14 at 20:33
  • $\begingroup$ I'm also having no idea on how the line bundle looks like with those patches. And another way to understand the fact is: the section $x_i\in\Gamma(\mathcal{O}_{\mathbb{P}^n}(1),{P}^n)$ would corresponds to $f_i$. $\endgroup$ – Peter Wu Nov 1 '14 at 5:13
  • $\begingroup$ I understand that $\varphi^*(x_i)=f_i$ holds. Since $ \Gamma(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(1))$ is generated by $x_0, \dots , x_n$, we have $\Gamma(X, \mathcal{L}) \cong \Gamma(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(1))$ (via $f_i \mapsto x_i$). But to prove $\mathcal{L} \cong \varphi^*(\mathcal{O}_{\mathbb{P}^n}(1))$, we need $\Gamma(U, \mathcal{L}) \cong \Gamma(U, \varphi^*(\mathcal{O}_{\mathbb{P}^n}(1)))$ for alle open sets $U\subseteq X$, do we? $\endgroup$ – claudi Nov 2 '14 at 15:25

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