I'm reading Huybrechts's Complex Geometry, p.250, proof of the Kodaira Embedding theorem :
(Here Kodaira embedding theorem is,
"Let $X$ be a compact Kahler manifold. Then a line bundle $L$ on $X$ is positive if and only if $L$ is ample." And a line bundle $L$ on a compact complex manifold $X$ is called ample if and only if $L^{k}$ for some $k >0$ defines a closed embedding $\varphi_{L^{k}} : X \to \mathbb{P}^N$)
Here, the Hartog's theorem means (His book, exercise 2.2.6) :
"Let $L \in \operatorname{Pic}(X)$ and $Y \subset X$ a submanifold of codimension at least two. Then the restriction $H^{0}(X,L) \to H^{0}(X-Y,L|_{X-Y})$ is bijective."
Q. My question is, for $L \in \operatorname{Pic}(X)$ and any submanifold $Y \subset X$, $H^{0}(X,L) \to H^{0}(X-Y,L|_{X-Y})$ is always injective? If not, how about the case "Y has codimension one"?
This question originates from following :
I'm now understanding the "due to Hartogs' theorem(above underlined statement)" as next commutative(?) diagram :
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
H^{0}(\hat{X}, \sigma^{*}L^k) & \ra{res_E} & H^{0}(\hat{X}-E, \sigma^{*}L^{k}|_{\hat{X}-E}) \\
\ua{\sigma^{*}} & & \ua{\bar{\sigma}^{*}} \\
H^{0}(X,L^k) & \ra{res_x} & H^{0}(X-\{x\}, L^{k}|_{X-\{x\}}) \\
\end{array}
$$
, where $res_E$ and $res_{x}$ are the restriction maps, and $\sigma : \hat{X} \to X$ and $\bar{\sigma} : \hat{X} -E \cong X-\{x\}$ (really commute?)
$\bar{\sigma}^{*}$ is isomorphism(?) and $res_x$ is also bijective by the Hartog's theorem.
So, if we show that $res_E$ is injective, then we can show that $\sigma^{*}$ is bijective.
Q. That $res_E$ is injective is true? If $E$ has codimension at least two in $\hat{X}$, then we can also apply the Hartog's theorem to $res_E$ and can deduce that $res_E$ is bijective. But is that true? How large the dimension of blow up $\hat{X}$ at a point $x\in X$? Same as dimension of $X$?
If "$res_E$ is injective" is not true, then what does "due to Hartog's Theorem" he said means? How can we use the Hartog's theorem to deduce that $H^{0}(X,L^{k}) \to H^{0}(\hat{X},L^{k})$ is bijective (as the underlined statement)?
Best Answer
I think that I solved problem.
First, let $h:=\bar{\sigma}^{*} \circ res_x$. Then $h$ is an isomorphism and by the commutativity of diagram, $h=res_E \circ \sigma^{*}$ and $\sigma^{*}$ is injective.
Second, note that $res_E = \sigma^{*} \circ h^{-1}$. So, $res_E$ is injective and from this, we can show that $\sigma^{*}$ is bijective and we are done.