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Given a smooth complete intersection $X=D_{1} \cap D_{2} \cap \cdots \cap D_{k} \subset \mathbb{P}^{n}$ with ${\rm deg}\; D_i=d_i$, one can easily show that $\omega_{X} \simeq \mathcal{O}_{X}(\sum_{i=1}^{k} d_{i} -n-1)$, using induction on the number of hypersurfaces and the usual conormal sequence.

Here is the question.

Suppose $X \subset \mathbb{P}^{n}$ is a smooth projective variety of degree $d$, not necessarily a complete intersection. How to understand $\omega_{X}$ in terms of the embedding? Is it even necessarily true that $\omega_{X}$ is restricted from a line bundle on $\mathbb{P}^{n}$?

Similarly, how to work out the cohomology of $\mathcal{O}_X$ and $\omega_X$? Does this only depend on the degree of $X$?

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Complete interesections are quite special among projective varieties. I am doubtful as to whether there is much to say for an arbitrary projective variety. –  Daniel Loughran Sep 6 '10 at 12:51
In general $omega_X$ is not the restriction of a line bundle on $\mathbb{P}^n$. An example is the twisted cubic curve in $\mathbb{P}^3$. In this case the canonical bundle has degree -2, whereas every line bundle obtained by restriction has degree divisible by 3. You can have a look on the final section of chapter IV of Hartshorne. He gives a discussion which pairs (d(C),g(C)) are possible for smooth space curves C. In particular, it is shown that for fixed d there are many possibilities for the genus of C. (Provided that d is not 1 or 2.) –  Remke Kloosterman Sep 6 '10 at 13:05
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3 Answers

up vote 10 down vote accepted

Smooth (or Gorenstein) subvarieties in $\mathbb P^n$ whose canonical bundle is a restriction from $\mathbb P^n$ are known as subcanonical, and are very special. A rational twisted cubic in $\mathbb P^3$ is not subcanonical, for obvious reasons of degree.

It is most certainly not true that the cohomologies of $\mathcal O_X$ and $\omega_X$ only depend on the degree: for example, consider a twisted cubic as above and a plane cubic embedded in $\mathbb P^3$.

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Im not sure if this counts as a full answer, but it is a nice example which will hopefully shed light on some of your questions.

The canonical bundle $\omega_X$ of an Enriques surface $X$ satisfies $\omega_X \otimes \omega_X=\mathcal{O}_X$, but $\omega_X\neq \mathcal{O}_X$ in the Picard group. It follows that $\omega_X$ is not the restriction of any line bundle in $\mathbb{P}^n$, as these can't be non-zero torsion.

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That argument seems a bit unclear as written --- you seem to be saying that the image under a homomorphism f: A -> B of a non-torsion element in an abelian group A must be a non-torsion element in B. (Maybe I'm misreading your intention, in which cases apologies; if so, maybe the answer could be rewritten a little for clarity. The idea is obviously correct, in any case.) –  Artie Prendergast-Smith Sep 6 '10 at 14:24
The point is that the restiction of a line bundle from $\mathbb P^n$ is either ample, anti-ample, or zero, and only in this last case it can be torsion. –  Angelo Sep 6 '10 at 14:27
Right. My point was that the phrase "as these can't be non-zero torsion" is ambiguous --- on first reading "these" seems to refers to line bundles on P^n, or at least it did to me. –  Artie Prendergast-Smith Sep 6 '10 at 14:33
Just to build slightly on Angelo's comment, I was implicitly using the fact that if $X \subset \mathbb{P}^n$ is a non-singular projective variety which is not contained in a hyperplane, then the natural map $\mathbb{Z} \cong Pic(\mathbb{P}^n) \to Pic(X)$ is an injection. Hopefully this clears up my answer. –  Daniel Loughran Sep 6 '10 at 15:06
Perhaps an even simpler argument is that, as Angelo points out, every non-trivial line bundle on $\mathbb P^n$ is either ample or anti-ample. The restriction of these, by definition, remain ample or respectively anti-ample, in particular non-torsion. The only remaining case is $\mathcal O_{\mathbb P^n}$ which restricts to $\mathcal O_X$ on any $X$. –  Sándor Kovács Oct 29 '10 at 8:22
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Take any curve at all, of any genus g, and any divisor of degree d > 2g. This embeds the curve into projective space with degree d, and a generic projection embeds it in P^3 also with any degree d > 2g. So d and n determine almost nothing about the curve.

On the positive side, interestingly, the nice counterexample given for the original question, a rational cubic in P^3, although not determined by its degree, is completely determined by its degree and the fact that (unlike the plane cubic) it spans P^3. (Rational normal curves are about the only examples I can think of, spanning but not a complete intersection, where d,n do determine all the invariants.)

I guess you could give an inequality at least for the genus (i.e. h^1(O)) of curves in P^3, since a curve of degree d in P^3 projects to a plane curve of degree d-1, hence has genus bounded above by that of a general such plane curve. Indeed Castelnuovo has a famous such inequality.
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