3
$\begingroup$

Suppose $X$ is a smooth projective variety. Whether $X$ is a complete intersection or not when viewed as a subvariety of some projective space $\mathbb P^n$ is dependent on the specific choice of the projective embedding $X\hookrightarrow\mathbb P^n$. For example, taking $X=\mathbb P^1$, we can embed it into $\mathbb P^2$ and $\mathbb P^3$ as a conic and a twisted cubic respectively (the first is a complete intersection while the second one is not).

Is there a simple example of a smooth projective $X$ which is not a complete intersection with respect to any projective embedding $X\hookrightarrow\mathbb P^n$?

$\endgroup$
6
  • 7
    $\begingroup$ Curves of many genera are not realisable as complete intersections, see mathoverflow.net/questions/179688/…. $\endgroup$
    – pbelmans
    Commented Nov 14, 2018 at 16:09
  • 9
    $\begingroup$ A smooth variety of dimension $\ge 2$ which is a complete intersection must be simply connected by the Lefschetz hyperplane theorem, so you get lots of counterexamples (e.g., for abelian varieties). $\endgroup$ Commented Nov 14, 2018 at 16:11
  • 5
    $\begingroup$ Also a hyperelliptic curve of any genus $\ge 2$ works, since the canonical bundle of a complete intersection curve would be (very) ample. $\endgroup$ Commented Nov 14, 2018 at 16:15
  • 5
    $\begingroup$ Complete intersections are a very special class of varieties. The Lefschetz hyperplane theorem gives strong restrictions on the cohomology of a complete intersection. $\endgroup$ Commented Nov 14, 2018 at 16:18
  • 1
    $\begingroup$ Thanks for all the comments, this answers my question. $\endgroup$ Commented Nov 14, 2018 at 16:50

0

You must log in to answer this question.