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My question is located in trying to follow the argument bellow.

Given a normal algebraic variety $X$, and a line bundle $\mathcal{L}\rightarrow X$ which is ample, then eventually such a line bundle will have enough section to define an embedding $\phi:X\rightarrow \mathbb(H^0(X,\mathcal{L}^{\otimes d}))=\mathbb{P}^N$ (notice that $N$ depends on $d$). However, if the line bundle is NOT ample we can still say something about the existence of a certain map $\phi_d$; the so-called Iitaka fibration. I'll omit some details, but the argument of the construction goes (more or less) as follows. Suppose in the first place that $\mathcal{L}$ is base-point base. That is to say, that there are no points (or a set) of $X$ which all the hyperplanes of $\mathbb{P}^N$ pass through. Then such a line bundle will define a linear system $|\mathcal{L}^d|$ which gives rise to a morphism $\phi_d:X\rightarrow \phi_d(X)\subset\mathbb{P}^N$ (again $N$ depends on $d$). Such a map may not be an embedding, however $\phi_d:X\rightarrow \phi(X)$ is an algebraic fiber space. My question is the following. As I increase the value $d$ the image $\phi_d(X)\subset \mathbb{P}^N$ may change, however, as a matter of fact such an image "stabilizes" as $d$ gets larger. Meaning that if $d$ is large enough, the image if $\phi_d$ is the "same" regardless $d$.

-What is the reason for this to happen? -What is going on with all the sections of $\mathcal{L}^{\otimes d}$ that I am getting as I increase the value of $d$?. I'll appreciate any comment.

As a result, due to the fact that after a while we no longer care about the value of $d$, we can associate the space $X\rightarrow \phi(X)$ to the line bundle $\mathcal{L}$. Here, the variety $\phi(X)$ no longer depends on $d$.

Could someone comment further about the word "stabilizes"?.

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I think the point is that the $k$-algebra $\oplus_d H^0(X,L^d)$ is finitely-generated, for $L$ a basepoint-free (or more generally semi-ample) line bundle, by a theorem of Zariski. That means there exists some $d_0$ such that any section of any power $L^d$ can be obtained as a product of section of powers $L^k$, with all $k$ less than or equal to $d_0$. –  Artie Prendergast-Smith Nov 11 '10 at 10:51
(just noticed this is a very old question) –  Artie Prendergast-Smith Nov 11 '10 at 10:53
Since this was already bumped, I fixed a typo in the title. –  Karl Schwede Jan 18 at 1:30

2 Answers 2

Let $R(X,\mathcal L)=\oplus_d H^0(X,\mathcal L^{\otimes d})$ as graded rings, $s_0,\dots s_m\in H^0(X,\mathcal L^{\otimes d})$ and finally $R(X,\mathcal L,s_{\cdot})$ the (graded) subring generated by the $s_0,\dots s_m$ in $R(X,\mathcal L)$.


  1. ${s_0,\dots,s_m}$ define a rational map $\sigma: X\dashrightarrow \mathbb P^m$. If ${s_0,\dots,s_m}$ generate $H^0(X,\mathcal L^{\otimes d})$, then $\sigma$ agrees with $\phi_t$.

  2. If $\vert\{s_0,\dots,s_m\}\vert$ is a basepoint-free linear system, then $\sigma$ is a morphism. and $\sigma(X)\simeq {\rm Proj}\, R(X,\mathcal L, s_{\cdot})$.

  3. If $R(X,\mathcal L,s_{\cdot})$ has the property that for some $a\in\mathbb N$, $R(X,\mathcal L,s_{\cdot})_{ak}=R(X,\mathcal L)_{ak}$ for all $k\in \mathbb N, k\gg 0$, then ${\rm Proj}\, R(X,\mathcal L, s_{\cdot})\simeq {\rm Proj}\, R(X,\mathcal L)$ (think of the $a$-uple embedding).

  4. Since $R(X,\mathcal L)$ is finitely generated, if $s_0,\dots s_m\in H^0(X,\mathcal L^{\otimes d})$ are chosen so that they generate $H^0(X,\mathcal L^{\otimes d})$, then for large enough $d$ the property in 3) will be satisfied, so by 1) and 2) $\phi_d(X)\simeq {\rm Proj}\, R(X,\mathcal L)$ which is independent of $d$, hence it is stabilized.

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  • The sequence stabilizes because any increasing sequence of bounded integers (the dimensions of the images of $X$) stabilizes, but I assume you mean something different.

  • Suppose that $X\to|L|$ has already stabilized, then the map $X\to|2L|$ decomposes through the map $|L|\to\mathbb{P}\mathrm{Sym}^2 H^0(L)\to\mathbb{P} H^0(2L)$.

Where the first map is the Veronese, and the second is a projection.

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The dimensions of the vector spaces H^0(X,L^d) do not form a bounded sequence. Take, for example, h^0(P^1, O(d)) = d+1. –  Alberto García-Raboso Nov 22 '09 at 16:03
but the dimensions of the images of X do. They are bounded by the dimension of X –  David Lehavi Nov 22 '09 at 19:27
Sure. I did not get that from you answer though. Maybe you should edit it to clarify. –  Alberto García-Raboso Nov 22 '09 at 20:10
Another reason to edit it is that I cannot undo the downvote unless you do... –  Alberto García-Raboso Nov 22 '09 at 20:11

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