Consider a polynomial $f \in \mathbb C[x_1,\dots ,x_n]$. An atypical value of $f$ is a complex number about which $f:\mathbb C^n\to \mathbb C$ is not a topological fiber bundle. Writing $\mathrm{Atyp}(f)$ for the atypical values of $f$ we thus have a fiber bundle $$\mathbb C^n\setminus f^{-1}(\mathrm{Atyp}(f))\to \mathbb C\setminus \mathrm{Atyp}(f).$$

A book I'm reading now says the following.

The fundamental group $\pi_1(\mathbb C\setminus \mathrm{Atyp}(f))$ acts therefore on the generic fiber $f^{-1}(t_0)$ up to isotopy. The image of this action is called the geometric monodromy group. The pullback of $f$ along an admissible loop (image doesn't hit atypical values) is a fiber bundle which yields an automorphism of the fiber, called the geometric monodromy.

I am confused by this. We have for any Hurewicz fibration $E\to B$ the transport functor $\pi_1(B)\to \mathsf{hTop}$ which lands in the homotopy category. This is the only meaning I can think of for "acts up to isotopy". On the other hand, this does not give an automorphism of the fiber in the topological category.

Question. Is the geometric monodromy defined as the homotopy type of the continuous map between fibers given by the homotopy lifting property? Is the geometric monodromy group(oid) given by the image $\pi_1(B)$ in $\mathsf{hTop}$?

  • $\begingroup$ Take every closed curve $\gamma(r), r \in [a,b], \gamma(a) = t_0$ and every lift $\phi(r) \in \text{Isom}(f^{-1}(t_0),f^{-1}(\gamma(r)), \phi(a)=\text{Id} \in \text{Aut}(f^{-1}(t_0))$ and make a group from all the obtained $\phi(b) \in \text{Aut}(f^{-1}(t_0))$, it will contain the automorphisms isotopic to the identity plus some other isotopy classes. $\endgroup$
    – reuns
    Jun 23, 2019 at 1:43

1 Answer 1


As you say, a Hurewicz fibration defines a map from $\pi_1(B)$ to the automorphisms of the fiber in the homotopy category.

But a topological fiber bundle has more structure than a Hurewicz fibration. In fact, a topological fiber bundle defines a map from $\pi_1(B)$ to the mapping class group of the fiber (i.e. the component group of the group of topological automorphisms of the fiber). This is what "acts up to isotopy" means.

To see this, maybe one way is to observe that the set of pairs of a point in $B$ and an isomorphism between the fiber over that point and $F$ is an an $\operatorname{Aut}(F)$-torsor, and modding out by the identity component gives a $\pi_0(\operatorname{Aut}(F))$-torsor, which gives a homomorphism $\pi_1(B) \to \pi_0(\operatorname{Aut}(F))$.

  • $\begingroup$ Dear @Will, thanks for your answer! I am pretty lost here, so first of all I wanted to ask for a reference for this mapping class group business with fiber bundles (and the claims last pair of paragraphs). Second, I still don't understand - what is the geometric monodromy (automorphism of the fiber) and what is the geometric monodromy group? Thanks for your patience! $\endgroup$
    – Arrow
    Jun 23, 2019 at 20:15
  • $\begingroup$ @Arrow In this context, the geometric monodromy is the homomorphism from $\pi_1(B)$ to $\pi_0(\operatorname{Aut}(F))$, and the geometric monodromy group is its image. (In other contexts, these terms will mean slightly different things). $\endgroup$
    – Will Sawin
    Jun 23, 2019 at 21:02
  • $\begingroup$ @Arrow With regards to a reference, I don't know one, and wasn't able to find one with Google. But I don't really understand what needs a reference. The second paragraph makes claims, which the third paragraph proves. I can add a little more detail: To check this set is a torsor, we can work locally over an open set $U$ where the fiber bundle trivializes. Then we must check that the set of pairs of a point in $U$ and an isomorphism between $F$ and $F$ is an $\operatorname{Aut}(F)$-torsor, which is obvious. $\endgroup$
    – Will Sawin
    Jun 23, 2019 at 21:13

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