10
$\begingroup$

In this question a variety means any subset of complex projective space $\mathbb CP^n$ that is the set of common zeroes of a set of homogeneous polynomials. Thus if $A,B\subset{\mathbb C}P^n$ are varieties so is $A\cup B$.

Suppose $V_k$ is a sequence of varieties that converge to a subset $V\subset{\mathbb C}P^n$ in the Hausdorff topology on closed subsets of ${\mathbb C}P^n$. (This means after choosing a Riemannian metric on ${\mathbb C}P^n$ that given $\epsilon>0$ there is $K$ so that if $k>K$ then $V\subset N_{\epsilon}(V_k)$ and $V_k\subset N_{\epsilon}(V)$ where $N_{\epsilon}(X)$ is an $\epsilon$-neighborhood of $X$).

Also suppose that there is $d$ such that each $V_k$ is the set of common zeroes of some polynomials of degree at most $d$.

Is it true that $V$ is always a variety ?

This is a theorem of Ed Bishop if the $V_k$ are pure-dimensional.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
-1
$\begingroup$

For each subspace $M$ of degree $d$ homogeneous polynomials let $V(M)$ be the common vanishing locus. Because Grassmannians are compact you can assume your $V_k= V(M_k)$s where the $M_k$s converge to a limit $M_{\infty}$. I claim that $V(M_k)$ converges in Hausdorff topology to $V(M_{\infty})$.

It is enough to see that, if $M, M'$ are close in the natural topology on subspaces, then $V(M)$ and $V(M')$ are close in the Hausdorff topology.

This amounts to the following assertion: if a basis of $M$ takes small values on $x \in \mathbb{C}P^n$, then $x$ must actually be very close $V(M)$. That's a standard thing - Lojasiewicz inequality - it's much easier in the algebraic context.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ But homogeneous polynomials do not "take values" on points of projective space. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 16, 2015 at 2:45
  • $\begingroup$ No, they don't. I was too lazy to normalize things, but I am sure you can manage it in your head. $\endgroup$
    – category_student
    Commented Jul 16, 2015 at 3:08
  • 5
    $\begingroup$ Ah. I am sorry. I thought you might be trying to be helpful. My mistake. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 16, 2015 at 3:11
  • 4
    $\begingroup$ category_student: Thanks for telling me about Lojasiewicz inequality. I am a topologist and do not know any algebraic geometry. However it is not true that M and M' close implies V(M) is close to V(M') e.g. $M=\langle x^2y,xy^2\rangle$ and $M_k=\langle x^2y,xy^2+ (x+y)^3k^{-1} \rangle$. Then $M_k\to M$ as $k\to\infty$ but $V(M)$ and $V(M_k)$ have different dimensions. The problem seems to be there is a constant in the Lojasiewicz inequality that does not vary continuously with the data. $\endgroup$
    – Daryl Cooper
    Commented Jul 16, 2015 at 14:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .