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The following proposition appears in Peters and Steenbrink's book on Mixed Hodge Structures.

Proposition([Peters--Steenbrink, Proposition C.11]) If $f\colon X\to\Delta$ is proper and smooth over $\Delta^\ast:=\Delta-\{0\}$, then the inclusion $X_0\hookrightarrow X$ is a homotopy equivalence.

Does anyone know a reference for the proof of this result?

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  • $\begingroup$ This is just a guess, but I think the conditions imply that $f$ is a fibration. It's a basic fact from topology that the pullback is then a homotopy pullback. If an arrow in a homotopy pullback is a homotopy equivalence, then the parallel arrow is also an equivalence. $\endgroup$ Mar 10, 2022 at 22:51
  • $\begingroup$ I think the importance of properness is that it deals with "missing points": consider taking some fibration over $\Delta$ and then deleting some point in the domain. If you do it wrong, you can end up with something which doesn't contract right: think about something like $I^2\to I$ by coordinate projection. Deleting a point turns the source from contractible and homotopy equivalent to any given fiber to homotopy equivalent to $S^1$. $\endgroup$
    – KReiser
    Mar 10, 2022 at 23:30
  • $\begingroup$ @VincentBoelens Why do you believe that to be the case? Thanks! $\endgroup$ Mar 11, 2022 at 1:22
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    $\begingroup$ @KReiser I agree intuitively. I mean, I intuitively think it should be like some sort of valuative criterion that takes a point $p$ in some $f^{-1}(t)$ for $t\in\Delta^\ast$ and 'flows it towards $0$' and properness is needed to say that this flow actually reaches $X_0$, but I don't know how to make this precise. My goal, by the way, is to try and finally understand the classical topology analogue of Grothendieck specialization where for $Y\to\mathrm{Spec}(\mathcal{O})$ proper over a Henselian DVR (although you don't even need DVR) you have that $Y$ and $Y_0$ (the special fiber) $\endgroup$ Mar 11, 2022 at 1:24
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    $\begingroup$ Admittedly I don't have an explicit proof in hand right now but I totally agree with you on how it should work in principle. I'll see if I can find something more substantive to say as an answer. $\endgroup$
    – KReiser
    Mar 11, 2022 at 2:13

1 Answer 1

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As mentioned in the comment, Clemens proved the case for normal crossing divisors. The general case will follow from this result, which I'll provide proof later.

Theorem. (Clemens, 1977) If $X$ is a smooth complex manifold and $f:X\to \Delta$ is a proper family. Suppose $f$ is smooth over $\Delta^*$, and $X_0=f^{-1}(0)$ is normal crossing divisor, then there is a strong deformation retract from the total space $X$ to the central fiber $X_0$.

Remark 1. Here $X_0$ being a normal crossing divisor means $X_0=\cup_im_iD_i$ and the $D_i$ are smooth proper varieties meeting transversely among each other and $m_i\ge 1$. The local equation of a point $x_0\in X_0$ in a neighborhood of $X$ is analytically equivalent to $t=x_1^{m_1}\cdots x_k^{m_k}$.

Note a normal crossing divisor is different from simple normal crossing divisor, in which the central divisor is reduced ($m_i=1$) and the family is called semi-stable. To get a semi-stable family from $X\to \Delta$, one need to take a base change with respect to a finite cover $p:\tilde{\Delta}\to \Delta$ where $\deg(p)$ is the l.c.m. of $m_i$. (People studying asymptotic Hodge theory prefer to work with a semi-stable family, that's why various literature, e.g., by David Morrison, on Clemens-Schmid sequence work with a semi-stable family, even though the existence of the limiting mixed Hodge structure only rely on normal crossing condition.)

Remark 2. A strong deformation retract $F:X\times I\to X$ means $F$ is a deformation retract ($F_0=F(\cdot,0)=Id_X$ and $F_1=F(\cdot,1)$ sends $X\to X_0$) and satisfies $F(x,t)=x$ whenever $x\in X_0$.

Now let's prove the general case based on Clemens' theorem.

Claim: Let $Y\to \Delta$ be a proper family and smooth over $\Delta^*$. Then there is a strong deformation retract of $Y$ onto $Y_0$.

Proof. By Hironaka, there is a log resolution $$\sigma: X\to Y,$$ where $\sigma$ is isomorphism over $Y\setminus Y_0^{sing}$ and $X_0=\sigma^{-1}(Y_0)$ is a normal crossing divisor.

We define $G:Y\times I\to Y$ by

$$ G(y,t)=\begin{cases}\sigma F(\sigma^{-1}(y),t),\: y\notin Y_0;\\ y,\:\:\:\:\:\:\:\:\:\:\:\: y\in Y_0. \end{cases} $$ I claim that $G$ is a strong deformation retract of $Y$ onto $Y_0$. In fact, it suffices to check the continuity. Since the points know where to flow outside $Y_0$, it suffices to check continuity on $Y_0$. Let's take a sequence of points $(y_n,t_n)$ on $Y\times I$ whose limit point is $(y_0,t_0)$ with $y_0\in Y_0$. Then $$\lim_{n\to \infty}G(y_n,t_n)=\sigma\lim_{n\to\infty}F(\sigma^{-1}(y_n),t_n).$$ Since $X$ is proper, $\sigma^{-1}(y_n)$ converges to a point $x_0\in \sigma^{-1}(y_0)\subseteq X_0$. By strongness of $F$, $F(\sigma^{-1}(y_n),t_n)\to (x_0,t_0)$, so $\lim_{n\to \infty}G(y_n,t_n)=(\sigma(x_0),t_0)=(y_0,t_0)$. Therefore $G$ is continuous on $Y_0\times I$. $\Box$

Final remark. One can think of the retraction as describing a flow that sends the vanishing cycles to the singularity on central fiber and send other points to smooth locus $Y_0^{sm}$. In the normal crossing case, the vanishing cycle is the boundary of a tubular neighborhood of $D_1\cap D_2$ (say two components) in $Y$, which is a circle bundle. The retraction is to flow each circle over $y\in D_1\cap D_2$ to the point $y$. To me, this is the topological picture behind the argument.

In addition to the references listed in this answer, one can refer to Clemens' original paper [Cle77] (especially Theorem 5.7) for the strong deformation retract theorem in normal crossing case. Besides, Nicolaescu's notes (e.g. Chapter 14) is also a good introduction to this subject.

[Cle77]: Clemens, C. H. Degeneration of Kähler manifolds. Duke Math. J. 44 (1977), no. 2, 215–290.

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  • $\begingroup$ Thanks for taking the time to type this up! I'll give it a careful read soon. Again, I'd prefer that we didn't have to use resolution, but again maybe that's unavoidable. $\endgroup$ Mar 12, 2022 at 1:04
  • $\begingroup$ Follow-up question: does this work when the total space $X$ itself is not smooth? Or, in that case, do you need to stratify your spaces delicately as in Goresky--MacPherson? $\endgroup$ Mar 12, 2022 at 5:58
  • $\begingroup$ @AlexYoucis In general case, the total space does not need to be smooth (log resolution always bring us to the smooth case). Clemens does work in the case of smooth total space. I edited and added the assumption. $\endgroup$
    – AG learner
    Mar 12, 2022 at 19:18
  • $\begingroup$ @AlexYoucis I don't know an argument without using resolution of singularities. Let's say central fiber has only an ordinary double point, do you know how to do it? $\endgroup$
    – AG learner
    Mar 12, 2022 at 19:24
  • $\begingroup$ I do not, at least rigorously (I can se intuitively why it's true). This is not very surprising as I haven't really made much attempt to understand the proof in Clemens beyond the basic intuition you also indicated above. In any case, it's no matter, as this is all I was really after. Thank you very much! $\endgroup$ Mar 13, 2022 at 5:42

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