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.

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    $\begingroup$ 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


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 $$\pi_n(Y)=\pi_n(S^n)=\mathbf{Z}.$$


[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

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    $\begingroup$ 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
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    $\begingroup$ 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

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