2
$\begingroup$

Let $X\subseteq \mathbf P^n(\mathbb C)$ be a non-singular projective variety over the complex numbers. Suppose that $X$ is given by the vanishing of homogeneous polynomials $f_1, \dots, f_r$. Is it true that for sufficiently small variations of the coefficients of the polynomials $f_1, \dots, f_r$, the resulting variety $X'$ is isomorphic as a topological manifold to $X$?

Thanks!

$\endgroup$
4
$\begingroup$

If $r=\operatorname{codim} X$, I think this is likely to be true, maybe by continuity. But if $X$ isn't a complete intersection, the question can't possibly be true, as perturbing the defining equations should (generically) give rise to something of codimension $r$.

For example, take the twisted cubic in $\mathbb{P}^3$. It requires three defining quadratic equations, but three general quadratic equations will just give 8 points.

So the answer is "no", at least to the question as stated.

$\endgroup$
  • $\begingroup$ Thank you Jake, that is very helpful. It hadn't occurred to me that non complete intersections were "singular" in this sense, but now it is very clear. Could something weaker still be true, say that they have the same homotopy type? $\endgroup$ – Bruno Joyal Oct 13 '14 at 2:38
  • $\begingroup$ Hi Bruno, the counterexample I gave doesn't have the same homotopy type (eight points, versus one P^1). I think the idea of deforming the variety by being its equations isn't quite the right approach. (Though it's fine for one equation). Vaguely, you would want something like a flat deformation of the ideal sheaf - then I think the nearby deformations of a nonsingular variety would have the properties you're going for. $\endgroup$ – Jake Levinson Oct 13 '14 at 2:51
  • $\begingroup$ (I don't think anything is likely to be true about deforming the equations for an arbitrary low-dimensional variety in a high-dimensional embedding, unfortunately. For example, the solution set could well become empty! What it comes down to is that defining equations are a little subtle for projective varieties. Unfortunately I'm not sure how to 'fix' the question in general!) $\endgroup$ – Jake Levinson Oct 13 '14 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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