8
$\begingroup$

Let $(X,x)$ be an affine variety with a normal isolated singularity and $Y$ be a smoothing of $X$ (for this we mean $Y$ can be realized as a smooth fibre of a deformation of $X$).

Question. Can we say that the topology of $Y$ is nontrivial, i.e. $\pi_i(Y)\neq \{1\}$ for some $i\geq 1$? Here the topology is induced from the analytic structure instead of the Zariski topology.

$\endgroup$
1
  • 1
    $\begingroup$ In dimension 2, I think this isn't possible: Bhupal, Stipsicz, Szabó, and Wahl classified the singularities that have a rational homology disc smoothing (which is weaker than a contractible one), and none of the singularities they found bound integer homology disc smoothings (which is still weaker than contractible). In higher dimension, the h-cobordism theorem at least tells you that the link of the singularity is a sphere and that the Milnor fibre is a ball. $\endgroup$ Dec 6, 2022 at 15:30

1 Answer 1

3
$\begingroup$

Edit. This does not answer the question, but it just provides examples of homotopically non-trivial smoothings $Y$ in every dimension $n \geq 1$. Perhaps someone might find it useful, hence I will not delete it.

Consider the nodal quadric hypersurface $$X=\{x_1^2+x_2^2+ \cdots + x_{n+1}^2=0\} \subset \mathbb{C}^{n+1},$$ admitting the smoothing family $$X_t=\{x_1^2+x_2^2+ \cdots + x_{n+1}^2-t=0\}.$$ Set $Y=X_1$, which is a smooth quadric in $\mathbb{C}^{n+1}$.

By [1], $Y$ is diffeomorphic to the tangent bundle $TS^{n}$ which is, in turn, homotopically equivalent to $S^n$, see [2]. So we get $$\pi_n(Y)=\pi_n(S^n)=\mathbf{Z}.$$

References.

[1] https://math.stackexchange.com/questions/1784898/tangent-bundle-of-sphere-as-a-complex-manifold

[2] https://math.stackexchange.com/questions/2034443/tangent-bundle-and-manifold-are-homotopy-equivalent

$\endgroup$
2
  • 3
    $\begingroup$ Wasn't the question originally asking if there is always a nonvanishing homotopy group (after your edit the question has changed)? $\endgroup$ Dec 6, 2022 at 11:06
  • 1
    $\begingroup$ I checked and you are right, thanks. I misread the question. Now it is back in its original form, and I will edit the answer accordingly. $\endgroup$ Dec 6, 2022 at 11:09

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.