7
$\begingroup$

Ok so there's a lot of litterature about nearby cycles functor since it was introduced by Grothendieck and Deligne but I couldn't find any clear answer to the following natural question:

Problem: Let $X$ be a reduced complex analytic space, $f = (f_1,f_2) : X \to \mathbb{C}^2$ a couple of functions and $K \in D^b_c(A_X)$ a constructible complex. When do we have a natural isomorphism between iterated nearby cycles: $$ \psi_{f_1}\psi_{f_2}(K) \simeq \psi_{f_2}\psi_{f_1}(K) $$

This is well known as part of the hypercube description of perverse sheaves when $f = id$ and $K$ is constructible with respect to the strict normal crossings divisor defined by the coordinates hyperplanes.

In general, I don't think a natural map between the two sides even exists but one might look for something like morphisms $$ \psi_{f_1}\psi_{f_2}(K) \leftarrow \psi_f(K) \to \psi_{f_2}\psi_{f_1}(K). $$
where $\psi_f(K)$ would be some sort of global or simultaneous nearby cycles that would induce the iterated nearby cycles under suitable hypothesis.

Actually I'm more interested in the algebraic case where $K$ is regular holonomic D-module but this kind of problems seems to have been studied a lot more by topologists in the spirit of Thom's isotopy lemmas to I'm trying to understand the Milnor fibration viewpoint. Please correct me if I'm wrong.

Dimenson 1 base: In the case $(X,x)$ is a germ of analytic space inside $U \subset \mathbb{C}^N$ and $g: X \to \mathbb{C}$ a single analytic function, the Milnor-Le fibration theorem states that for $0<\eta \ll \varepsilon \ll 1$
$$ g: \bar{B}(x,\varepsilon) \cap g^{-1}(D^*(f(x),\eta)) \to D^*(g(x),\eta) $$ is a locally trivial fibration over the punctured disc $D^*(g(x),\eta)$. The fiber $F_{g,x} = \bar{B}(x,\varepsilon) \cap g^{-1}(\eta)$ is the local Milnor fiber of $g$ at $x$.

Almost by definition, we have $\psi_g(K)_x = R\Gamma(F_{g,x},K)$.

First question: is the fibration independant of the local embedding $X \subset \mathbb C^N$. This seems to be well known but I've never seen an actual proof. It seems to me it could be proved quite easily if one can replace the ball $\bar{B}(x,\varepsilon)$ by a polydisk as in Le's "La monodromie n'a pas de point fixe" but I haven't written it down yet.

Dimenson > 1 base: Consider $f = (f_1,f_2): X \to \mathbb{C}^2$.

In this case, Milnor's fibration theorem fails in general (classical exemples include simple blow-ups or Whitney's umbrella).

But, by iterating the usual one function construction for $f_i: X \to \mathbb{C}$, one can still define a Milnor fibration $X_{(f_1;f_2),x}(\varepsilon,\eta) \to S_{\eta_1}^1 \times S_{\eta_2}^1$ independant of $0 < \eta_1 \ll \eta_2 \ll \varepsilon$ with fiber $F_{(f_1,f_2),x}$. This is done for example in McCrory and Parusinky's "Complex monodromy and the topology of real algebraic sets". We have $$ \psi_{f_1}\psi_{f_2}(K)_x = R\Gamma(F_{(f_1,f_2),x};K) $$ But this fibration depends of the ordering we chose: $F_{(f_1;f_2),x} \neq F_{(f_2;f_1),x}$.

I expect the problem to disappear with Thom's $a_f$ condition. More precisely, in "Morphismes sans éclatement et cycles évanescent" Sabbah defines a morphism $f:X \to Y$ between stratified analytic spaces as being "sans éclatement" ("without blowup") if

  • the stratification on $Y$ satisfies Whitney's conditions,
  • for each strata $Y_\beta$ subset $Y$, the stratification on $X$ induces a Whitney stratification on $f^{-1}(Y_\beta)$.
  • Thom's $a_f$ condition is satisfied.

Let's stratify $\mathbb{C}^2$ by the coordinates hyperplanes and suppose there is a stratification $S$ of $X$ so that $K\in D^b(A_X)$ is $S$-constructible and $f:X\to \mathbb{C}^2$ is without blow-up. Then we have a locally trivial topological fibration $f: B(x,\varepsilon) \cap f^{-1}((\mathbb{C}^*)^2) \to (\mathbb{C}^*)^2$ with stratified fiber $F_{f,x}$.

Question: Am I right in thinking that the above fibration induces the iterated Milnor fibrations so that
$$ \psi_{f_1}\psi_{f_2}(K)_x = R\Gamma(F_{f,x}, K) = \psi_{f_2}\psi_{f_1}(K)_x $$

Thanks

$\endgroup$
2
  • $\begingroup$ maybe this arxiv.org/pdf/math/0605369.pdf will be helpful $\endgroup$ Aug 27, 2013 at 5:12
  • $\begingroup$ Thanks for the link, I had already glanced at it but I don't think it is relevant since they are only considering a dimension 1 base, i.e. a single function $f:X\to \mathbb C$. $\endgroup$
    – AFK
    Aug 29, 2013 at 9:50

1 Answer 1

6
$\begingroup$

Why is the Milnor fibration (for dimension 1 base) independent of the embedding? Because the embedding just determines what you are using for balls, and Lê's old proof shows that the fiber-homotopy-type of the fibration doesn't depend on which balls you're using. If $B_1$ and $B_2$ are balls in one embedding and $P_1$ and $P_2$ are pull-backs of balls from the other, such that $B_2\subseteq P_2\subseteq B_1\subseteq P_1$, then the equivalence of the fibrations using $B_1$ and $B_2$, and using $P_1$ and $P_2$, shows that the fibrations using $P_2$ and $B_1$ are equivalent.

With a base of $\mathbb C^2$ or larger, $\mathbb C^d$, the problem is worse than just needing $a_f$ to control blowing up. In general, what one hopes for is that the discriminant of $f$ is a hypersurface in $\mathbb C^d$ and that, off of this hypersurface, the fibers are the same. This is where a_f gets used. But the discriminant (or Cerf diagram) is already an obstruction to a fibration over $(\mathbb C^*)^2$.

Under some manageable conditions, the stalks of $\psi_{f_1}\psi_{f_2}K$ and $\psi_{f_2}\psi_{f_1}K$ are isomorphic, but the complexes are not naturally isomorphic; the Milnor monodromy of $f_1$ would typically act very differently on these complexes. For instance, if our domain is affine space, $f_1$ has a critical point at the origin, and $f_2$ is a generic linear form, and the relative polar curve is not empty, then the $f_1$-monodromy is very different on the two complexes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.