# A compact complex manifold admits an ample line bundle if and only if it is projective

Given a holomorphic line bundle $L$ on a complex manifold $X$, a point $x\in X$ is called a base point of $L$ if $s(x)=0$ for all $s\in H^0(X,L)$ (the space of global holomorphic sections of $L$). The base locus $\operatorname{Bs}(L)$ is the set of all base points of $L$. Given a basis $s_0,\dots,s_N$ of $H^0(X,L)$, one can define a map \begin{equation} \phi_L:X\backslash\operatorname{Bs}(L)\rightarrow \mathbb{CP}^n \end{equation} sending a point $x$ to $[s_0(x):\dots:s_N(x)]$. This is well-defined, since not all $s_i(x)$ are zero (we exclude $\operatorname{Bs}(L)$ in the domain) and since changing trivialisation scales all components $s_i(x)$ by the same amount.

I am told that the line bundle $L$ is called very ample if for any such basis, the associated map $\phi_L$ is an embedding. $L$ is called ample if there exists a positive integer $m_0$ such that $L^m$ is very ample for all $m\geq m_0$.

In Huybrechts' book Complex Geometry, he says that by definition, a compact complex manifold is projective (i.e. embeds as a complex submanifold of $\mathbb{CP}^n$) if and only if it admits an ample line bundle. But I do not understand how this follows? Do we not only get an embedding of $X\backslash \operatorname{Bs}(L)$ into $\mathbb{CP}^n$, rather than the whole of $X$? I would agree with this statement if $L$ was also assumed to be globally generated, i.e. satisfies $\operatorname{Bs}(L)=\emptyset$, but this is not assumed. Any help would be much appreciated!

• Every definition of very ample that I've seen includes the assumption that $\text{Bs}(L)=\emptyset$. – Jack Lee Aug 21 '18 at 23:30
• @JackLee Thank you, perhaps it's been left out in the notes I've been reading. In Huybrechts, however, the definition of very ample is avoided, but a line bundle $L$ is called ample if for some $m>0$ and some linear system in $H^0(X,L^m)$, the associated map $\phi_L$ is an embedding. Again, this doesn't seem to assume $L$ satisfies $\operatorname{Bs}(L)=\emptyset$, so how do I know the associated map $\phi_L$ embeds $X$ rather than $X\backslash\operatorname{Bs}(L)$? – jl2 Aug 21 '18 at 23:39
• Not to be tooooo picky, but a manifold isn't ample; a line bundle on it is (or isn't). So your statement should be that a compact complex manifold is projective if and only if there is some ample line bundle over it. A very ample line bundle will, of course, have no base locus. That's part of the proof: A sufficiently high power of an ample line bundle will have no base locus. – Ted Shifrin Aug 21 '18 at 23:39
• The version of Huybrechts that I have says (p. 248) "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 $\varphi_L:X\hookrightarrow\mathbb P^N$." This says specifically that the domain of $\varphi_L$ is all of $X$. – Jack Lee Aug 22 '18 at 0:01
• I agree that the definition on page 86 could be interpreted as meaning only that $\varphi_L$ is an embedding of $X\smallsetminus \text{Bs}(L)$. But it's obvious from what comes later that he really means to assume that it's defined on all of $X$. – Jack Lee Aug 22 '18 at 14:37