Kodaira embedding theorem says that a positive line bundle is ample, i.e. high tensor powers are holomorphically embeddable into complex projective space of high dimension.

However, ampleness is not stable under blow-ups. Usually a replacement is to consider big line bundles, which is stable under blow-ups.

Is there an embedding theorem for big line bundles? One cannot hope that an embedding of high tensor powers of a big line bundle to be everywhere injective, but can we have injectivity almost everywhere?

Here's the precise question. Let $X$ be a complex compact manifold. Let $L$ be an ample line bundle over $X$. Let $f:Y\to X$ be a blow up, or series of blow ups. Is there some condition for the pullback $f^*L$ to be an embedding outside the exceptional locus of $f$?


If $X$ is normal, then the Iitaka fibration theorem implies that $L$ is big if and only if the rational map

$\phi_m \colon X \dashrightarrow \mathbb{P}H^0(X, L^{\otimes m})$

is birational onto its image for some $m >0$, see [Lazarsfeld, Positivity in Algebraic Geometry I, p. 139].

I guess this is the "embedding theorem for big line bundles" you are looking for.

Regarding your last question, since $L$ is ample by assumption and $f$ is an isomorphism outside its exceptional locus $E$, some power of $f^*L$ will surely give an embedding of $X \setminus E$.

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  • $\begingroup$ Thank you very much. This is exactly what I'm looking for. $\endgroup$ – user2529 Oct 24 '10 at 10:18


Francesco has already mentioned the "embedding theorem for big line bundles" that you were probably looking for, but I would like to add that this is pretty much the definition of big.

An alternative point one could make is that according to Kodaira's Lemma "big=ample+effective", so a high power of a big divisor is very ample on the complement of an effective divisor.

As for your last question, if $f:Y\to X$ is any proper morphism and $\mathscr L$ is an ample line bundle on $X$, then if $g$ denotes the morphism(!) defined by the global sections of very high powers of $f^*\mathscr L$, then it follows that $f$ factors through $g$. In particular, then $g$ is an isomorphism to its image (hence an embedding) wherever $f$ is (if $f$ is not birational, then this "wherever" is the empty set).

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  • $\begingroup$ Can really $f$ be "any morphism"? Don't you implicitly assume that it is proper and birational? $\endgroup$ – ACL May 24 '11 at 7:38
  • $\begingroup$ It was assumed that $X$ is compact, so I did implicitly assumed that so is $Y$, which makes $f$ automatically proper. On the other hand, birational is not needed. $\endgroup$ – Sándor Kovács May 24 '11 at 19:25
  • $\begingroup$ But, if $f$ is finite, $f^*\mathscr L$ will be ample, and the image of the morphism defined by the global sections of very high powers will by $Y$, and not $X$... $\endgroup$ – ACL May 27 '11 at 18:04
  • $\begingroup$ @ACL: Fair enough. I edited the answer so the message is the same, but I think it is now correct. Thanks for the corrections. $\endgroup$ – Sándor Kovács May 27 '11 at 18:35

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