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 $X$?

  • $\begingroup$ Complete interesections are quite special among projective varieties. I am doubtful as to whether there is much to say for an arbitrary projective variety. $\endgroup$ Sep 6, 2010 at 12:51
  • 5
    $\begingroup$ 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.) $\endgroup$ Sep 6, 2010 at 13:05

3 Answers 3


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 $\mathbb P^3$.


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.

  • $\begingroup$ 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.) $\endgroup$
    – user5117
    Sep 6, 2010 at 14:24
  • 1
    $\begingroup$ 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. $\endgroup$
    – Angelo
    Sep 6, 2010 at 14:27
  • $\begingroup$ 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. $\endgroup$
    – user5117
    Sep 6, 2010 at 14:33
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 6, 2010 at 15:06
  • $\begingroup$ 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 $X$. $\endgroup$ Oct 29, 2010 at 8:22
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.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.