7
$\begingroup$

The following question on simultaneous resolutions is a follow-up to earlier questions posed here (e.g. Resolution of singularities for flat families.). What I'm interested in is an "obstruction theory" for simultaneous resolutions. More precisely:

Let $f:X\to B$ be a flat projective (for simplicity) morphism of smooth varieties. What are obstructions that prevent the existence of a simultaneous resolution (*) of $f$?

If we further simplify and assume that $B\cong \mathbb{A}^n$ for some $n$, then an obvious obstruction is the monodromy (**) of $f^{sm}:X^{sm}\to B^{sm}$, the restriction of $f$ to its smooth locus. In some cases, this obstruction can be "killed" by a base change along a finite branched covering $\tilde{B}\to B$. For example, this is the case for the adjoint quotient $\chi:\mathfrak{g}\to \mathfrak{t}/W$ of a semisimple complex Lie algebra $\mathfrak{g}$ where the base change along the quotient $\mathfrak{t}\to \mathfrak{t}/W$ is sufficient (Grothendieck's simultaneous resolution).

Is this the only obstruction to the existence of simultaneous resolutions or are there more refined ones? Does it matter if we work in the algebraic or analytic category?

Edit: For simplicity, let us assume that we work over $\mathbb{C}$ but feel free to make comments/give answers for characteristic $p$. To further clarify my question:

(*) A simultaneous resolution of $f:X\to B$ is a pair $(\tilde{f},g)$ of a morphisms $\tilde{f}:\tilde{X}\to B$ and a $B$-morphism $g:\tilde{X}\to X$ such that the induced morphism $g_b:\tilde{X}_b\to X_b$ between the (scheme-theoretic) fibers $\tilde{X}_b=\tilde{f}^{-1}(b)$ and $X_b=f^{-1}(b)$ is a resolution of singularities ($b\in B$).

(**) More precisely, what I had in mind: The monodromy of the local systems $R^kf^{sm}_*\mathbb{Z}$.

$\endgroup$
  • $\begingroup$ Please define "simultaneous resolution" and "monodromy". Trivial monodromy of the etale cohomology of the geometric generic fiber is not sufficient. For instance, let $k$ be a field of characteristic $\neq 2$. Let $B$ be the affine $2$-plane $\text{Spec}\ k[s,t]$, Let $\mathbb{P}^3_k$ be $\text{Proj}\ k[u,v,w]$. Let $X\subset B\times_k \mathbb{P}^3_k$ be the zero scheme of $u^2 + v^2s +w^2t$ with its projection to $B$. The etale cohomology of the geometric generic fiber is invariant under monodromy, yet there are no proper modifications of $B$ and $X$ making $f$ a smooth morphism. $\endgroup$ – Jason Starr May 30 '18 at 12:07
  • $\begingroup$ Thank you, Jason, for your comment, in particular the example. I edited the question correspondingly and hope it's clearer now. Unfortunately, I'm not that familiar with etale cohomology: How does its monodromy relate to the monodromy of the local systems $R^kf_*^{sm}\mathbb{Z}$ (assuming $f:X\to B$ to be defined over $\mathbb{C}$)? $\endgroup$ – Florian May 30 '18 at 14:02
  • $\begingroup$ In the example that I wrote, $B^{\text{sm}}$ is the complement of the two coordinate axes, $X^{\text{sm}}$ is the inverse image of $B^{\text{sm}}$, and the sheaves $R^qf^{\text{sm}}_*\mathbb{Z}$, $q=0,1,2$ are, respectively, the constant sheaf $\mathbb{Z}$, the zero sheaf, and the constant sheaf $\mathbb{Z}$. There is zero monodromy. Thus, vanishing of monodromy is not sufficient. $\endgroup$ – Jason Starr May 30 '18 at 14:07
  • $\begingroup$ Thanks! Is it easy to see why it cannot admit a simultaneous resolution? Sorry, if this is obvious but I didn't have time yet to think about it more carefully. $\endgroup$ – Florian May 30 '18 at 14:19
  • $\begingroup$ The complement of $B^{\text{sm}}$ in $B$ is a divisor (the union of two prime divisors that equal the two coordinate axes). You cannot modify $X$ so that it is smooth over general points of those divisors. A precise formulation uses the Purity Theorem for the Brauer group (every modification of $X$ that is smooth over general points of the divisors extends to a smooth family over $B$), together with vanishing of the Brauer group of the affine plane. You can also directly use the Residue Exact Sequence for Brauer groups over the local ring of $B$ at the generic points of those two divisors. $\endgroup$ – Jason Starr May 30 '18 at 14:32
5
$\begingroup$

I am writing my comments as a solution. Assume that $k$ has characteristic $0$, for simplicity. First I will make precise one interpretation of "vanishing monodromy around the discriminant". Denote by $\Delta$ the complement of $B^{\text{sm}}$ in $B$. For simplicity, assume that $\Delta$ has pure codimension $1$ in $B$ (components of $\Delta$ that have codimension $\geq 2$ have no "local monodromy" by the Purity Theorem for fundamental groups). For every irreducible component $\Delta_i$ of $\Delta$, for the generic point $\eta_i$ of $\Delta_i$, for the strict Henselization of the local ring $$R_i:=\mathcal{O}^{\text{sh}}_{B,\eta_i},$$ the (profinite) Galois group of the fraction field of this local ring, $$\widehat{\pi}^{\text{inertia}}_1(B^\text{sm};\overline{\eta}_i):= \text{Aut}(\overline{\text{Frac}(R_i)}/\text{Frac}(R_i)),$$ equals the profinite completion of $\mathbb{Z}$ (this is essentially the Newton-Puiseux Theorem), i.e., it is procyclic. If $k$ equals $\mathbb{C}$, for a general $\mathbb{C}$-point $p_i$ of $\Delta_i$, for a transverse slice $N$ to $\Delta_i$ at $p_i$, for the punctured neighborhood $N^*=N\setminus\{p\}$, this procyclic group corresponds to the profinite completion of $\pi_1(N^*)\cong \mathbb{Z}$.

Denote by $X_{\overline{\eta}_i}$ the fiber product of $X\to B$ and $\text{Spec}(\overline{\text{Frac}(R_i)}) \to B.$ Then the profinite Galois group $\text{Aut}(\overline{\text{Frac}(R_i)}/\text{Frac}(R_i))$ acts continuously on the etale cohomology of the $\overline{\text{Frac}(R_i)}$-scheme $X_{\overline{\eta}_i}$. If there is a modification of $X$ that makes $f$ smooth over $\eta_i$, then this action is trivial. This is what is usually meant by vanishing monodromy on cohomology of the fiber. If $k$ equals $\mathbb{C}$, then (via the usual comparison theorems) this is equivalent to saying that the linear representation of $\pi_1(N^*,q)$ on the singular cohomology of the fiber $X_q$ over $q\in N^*$ is a trivial representation.

Vanishing of monodromy on cohomology of the fiber is not sufficient for the existence of a modification. Here is one example. Let $B$ be the projective plane $\text{Proj}\ k[r,s,t]$. Let $\mathbb{P}^2_k$ be the projective plane $\text{Proj}\ k[u,v,w]$. Let $X$ be the closed subscheme of $B\times \mathbb{P}^2_k$ with defining equation, $$f=u^2r+v^2s+w^2t.$$ Then $X$ is smooth, and the projection $f$ to $B$ is flat. Moreover, the fiber $X_{\overline{\eta}_i}$ is just $\mathbb{P}^1_{\overline{\text{Frac}(R_i)}}$. The étale cohomology of this is straightforward: $H^q(X_{\overline{\eta}_i},\widehat{\mathbb{Z}}_\ell)$ equals $\widehat{\mathbb{Z}}_\ell$ for $q=0$, it is zero for $q=1$, and it equals $\widehat{\mathbb{Z}}_\ell(-1)$ for $q=2$. Of course the action of monodromy on $H^0$ and $H^1$ is trivial. For $q=2$, observe that the first Chern class of the relative dualizing sheaf is a nonzero element that is preserved under monodromy. Since $H^2$ is procyclic and the action is continuous, it follows that the monodromy also acts trivially on $H^2$.

However, there is no modification of $X$ that makes $f$ smooth over the generic points $\eta_i$ of the three coordinate axes in $B$. If there were such a modification, then by Zariski glueing, there exists a modification of $X$ that is smooth over a dense Zariski open $B^o$ in $B$ containing both $B^{\text{sm}}$ and containing the generic point $\eta_i$ of each irreducible component $\Delta_i$. Thus, the complement of $B^o$ is a closed subset of $B$ that has codimension $\geq 2$. By the Purity Theorem for Brauer groups, Théorème 6.1 of the following, the restriction map from the Brauer group of $B$ to the Brauer group of $B^o$ is an isomorphism.

MR0244271 (39 #5586c)
Grothendieck, Alexander
Le groupe de Brauer. III. Exemples et compléments.
(French) Dix exposés sur la cohomologie des schémas, 88–188, Adv. Stud. Pure Math., 3,
North-Holland, Amsterdam, 1968.

The Brauer group of $B$ is zero. This implies that the Brauer class of $X^{\text{sm}}/B^{\text{sm}}$ is zero, and thus there exists an element of the Picard group of $X^{\text{sm}}$ whose base change to $X_{\overline{\eta}_i} \cong \mathbb{P}^1_{\overline{\text{Frac}(R_i)}}$ is a generator of the Picard group. Since $X$ is smooth, every divisor class on $X^{\text{sm}}$ is the restriction of a divisor class on $X$. The projection from $X$ to $\mathbb{P}^2_k$ is the projective bundle associated to a locally free sheaf of rank $2$, from which it is straightforward to compute the Picard group of $X$. There is no divisor class on $X$ whose base change to $X_{\overline{\eta}_i}$ is a generator of the Picard group. This contradiction proves that there is no modification of $X$ that is smooth over a generic point $\eta_i$ of $D_i$.

$\endgroup$
  • $\begingroup$ This is more than I had hoped for, thanks a lot, Jason! I need some time to digest your answer but I'll definitely accept it. $\endgroup$ – Florian May 30 '18 at 20:04

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.