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I'm now studying Daniel Huybrechts, Complex Geometry. But I can't understand some defitnions ;

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Q.1) What is the natural restriction map $H^{0}(X,L) \to L(x)$ ? $s \mapsto s(x)$? If so, why the surjectivity of the maps is equivalent to $BS(L) = \varnothing $ ?

Q.2) What is the $L(x)$ in the short exact sequence of sheaves (really sheaves?)?

Q.3) What is $\mathcal{I}_{\{x\}}$ ? There is no definition of the ideal sheaf of the point in the Huybrechts's book.

Anyone helps? If so, thanks!.

Plantation
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1 Answers1

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Q1) Yes, recall that $x\in BS(L)$ if $s(x)=0$ for any global section of $L$. So no point is in the base locus precisely when at every point there is some $s$ that does not vanish at that point.

Q2) It is the skyscraper sheaf at $x$ with fiber $L(x)$. It is kind of an abuse of notation because the same notation is used to mean both the fiber and the sheaf. But essentially, it is defined by $L(x)(U)= L(x)\cong\mathbb C$ whenever $x\in U$. And otherwise $0$.

Q3) It is the ideal sheaf at $x$. Defined by $\mathcal I_{\{x\}}(U)=\{f:U\to\mathbb C:\,f\,\text{ is holomorphic and vanishes at }x\}$.

Basically, the exact sequence is given by twisting the ideal sheaf sequence by $L$. It is like a formal way of saying that if you restrict sections of $L$ to a single point, then the kernel is given by those sections that vanish at the point.

lEm
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  • Yes, Thanks! :) But I can't still understand the Q1) completely. Why the surjectivity is equivalent to $BS(L)$ is empty? Can you explain more detaily? – Plantation Nov 08 '21 at 12:52
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    @Plantation Sure. Be careful the requirement is surjectivity at all $x$. Basically the claim is that if that map is surjective at a certain $x$, then $x\not\in BS(L)$. This is because the map $\varphi_L(x)=[s_0(x):\cdots:s_k(x)]$ can be defined, since not all $s_i(x)$ are 0. – lEm Nov 08 '21 at 12:58
  • Yes, the restriction map depends x. Although still can't understand the last sentence you give, but I will try to understand, and will ask again later. Anyway Thank you. – Plantation Nov 08 '21 at 13:51
  • @Plantation Maybe you can tell me which part did you not understand. So I can address your confusion. For starter, do you understand the definition of base locus? – lEm Nov 09 '21 at 02:10
  • My confusing point is, first, let's define a map $ H^{0}(X,L) \to L(x)$ by $s \mapsto s(x)$. Second, assume that $x\notin BS(L)$. Let $l \in L(x)$. Then can we find $s\in H^{0}(X,L)$ s.t. $s(x)=l$? ; i.e. the map is surjective? And the converse is also true? – Plantation Nov 09 '21 at 12:41
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    @Plantation I think you are confused about $L(x)$. It is supposed to mean the fibre of the line bundle at $x$. So really $L(x)\cong \mathbb C$. But since we are looking at linear maps, surjectivity is just saying that the image is not $0$, in other words we are requiring that not all sections vanish at $x$. – lEm Nov 09 '21 at 13:24
  • O.K. From my understanding, argument is that 1) Let $V:=H^{0}(X,L)$ and $W := L(x)$. 2) Let $T : V \to W$ by $s \mapsto s(x)$. If $T$ is linear transformation, since $W$ is one-dimensional, the surjectivity of $T$ is im$T \neq 0$. And we are done. From your point of view, this argument is correct? $T$ may be linear transformation? Anyway, Thanks!. – Plantation Nov 12 '21 at 04:09
  • @Plantation Yep, that's correct. – lEm Nov 12 '21 at 04:17
  • O.K. Thank you :) – Plantation Nov 12 '21 at 04:32