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Let $\iota:X \hookrightarrow \mathbb{P}^n_k$ be a closed embedding of an irreducible non-singular projective $k$-scheme. For any hyperplane $H$ of $\mathbb{P}^n_k$ which doesn't contain $X$, since $X$ is reduced and $H$ doesn't contain $X$, one can pull-back the Cartier divisor defined by $H$ on $X$, see Lemma 1.29, p.260, Algebraic geometry and Arithmetic Curves by Qing Liu . This Cartier divisor on $X$ corresponds to a Weil divisor (via the isomorpihsm between Cartier divisors and Weil divisors on $X$), which we denote it by $H \cap X$. We have an isomorphism of invertible sheaf $\mathcal{O}_{\mathbb{P^n_k}} (H) \cong \mathcal{O}_{\mathbb{P}^n_k}(1)$ and $\mathcal{O}_X(H \cap X) \cong \mathcal{O}_X(1):= \iota^*(\mathcal{O}_{\mathbb{P}^n_k}(1))$, hence we have $\Gamma(\mathbb{P}^n_k, \mathcal{O}_{\mathbb{P}^n_k} (H) ) \rightarrow \Gamma(X, \mathcal{O}_X(H \cap X)) $. But we know that $\Gamma(\mathbb{P}^n_k, \mathcal{O}_{\mathbb{P}^n_k} (H) ) = L(H)$ and $\Gamma(X, \mathcal{O}_X(H \cap X)) = L(H \cap X)$, so we have $L(H) \rightarrow L(H \cap X)$, here $L(D)$ is the usual Riemann-Roch space for a Weil divisor $D$.

Now comes my question: If $f \in L(H) \subseteq K(\mathbb{P}^n_k)$ such that $ \mathrm{Supp }(\mathrm{div} (f) ) $ doesn't contain $X$, then one can pull-back $f$ on $X$, which is an element in the function field $K(X)$ of $X$ and turns out to be in $L(H \cap X)$. But if $ \mathrm{Supp }(\mathrm{div} (f) ) $ contains $X$, we can't pull-back $f$ on $X$ (as least, I don't know a way to do this). So it looks weired that we have a map $L(H) \rightarrow L(H \cap X)$ which is obtained from the corresponding invertible sheaf and on the other hand, we can't describe it naively by pull-back (only works for most $f$, but not all $f$ ). So could we describe $L(H) \rightarrow L(H \cap X)$ naturally?

I am thinking this question when I consider the following question :

Let $D$ be an effective Weil divisor on $X$ such that $\mathcal{O}_X(D) \cong O_X(1)$, and assume that $\Gamma(\mathbb{P}^n_k, \mathcal{O}_{\mathbb{P}^n_k} (1) ) \rightarrow \Gamma(X, \mathcal{O}_X(1)) $ is surjective, then could we find a hyperplane $H$ on $\mathbb{P}^n_k$ such that $H \cap X = \mathrm{Supp} (D) $. If the above map $L(H) \rightarrow L(H \cap X)$ can be described by pull-back for all $f \in L(H)$, then this could be done.

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Did you think about using the short exact sequence $0\to I_X(1)\to \mathcal{O}_{\mathbb{P}^n}(1)\to \mathcal{O}_X(1)\to 0$? Besides, the subjectivity $\Gamma(\mathcal{O}_{\mathbb{P}^n}(1))\to\Gamma( \mathcal{O}_X(1))$ tells us that $D$ can be extended to a hyperplane in $\mathbb{P}^n$. –  Fei YE Sep 12 '11 at 8:46
    
@ Fei YE, could you explain why surjectivity implies that $D$ can be extended to a hyperplane in $\mathbb{P}^n$, this is just what I would like to know. –  user565739 Sep 12 '11 at 10:23
1  
By extension, I mean that each section $\mathcal{O}_X$ is the image of a section of $\mathcal{O}_{\mathbb{P}^n}$. In other words, a section of $\mathcal{O}_X$ is the restriction of a section of $\mathcal{O}_{\mathbb{P}^n}$ on $X$. –  Fei YE Sep 12 '11 at 11:23

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