# What is the meaning of the monodromy theorem in Hodge theory?

Let $$f : X^m \to Y^n$$ be an algebraic fiber space (between projective manifolds) whose discriminant locus is denoted by $$E$$. Let $$U$$ be a polydisk in $$\mathbb{C}^n$$ (with coordinates $$(y_1, ..., y_n)$$) such that $$U \backslash E \simeq (\Delta^{\ast})^{\ell} \times \Delta^{n-\ell}$$, where $$1 \leq \ell \leq n$$. Let $$x_0$$ be a point in $$U\backslash E$$. Associated to $$x_0$$ is the monodromy operator $$T_k : H^{m-n}(f^{-1}(x_0), \mathbb{C}) \longrightarrow H^{m-n}(f^{-1}(x_0), \mathbb{C})$$ for a loop based at $$x_0$$ around the $$k$$th copy of $$\Delta^{\ast}$$, where $$1 \leq k \leq \ell$$.

The monodromy theorem states that $$T_k$$ is quasi-unipotent, i.e., there are positive integers $$m_k$$ and $$d_k$$ such that $$(T_k^{m_k} - I)^{d_k} =0.$$ Here, $$m_k$$ is the least common multiple of the multiplicities of the irreducible components over the generic point of $$\{ y_k =0 \}$$.

Question: What does the quasi-unipotence of the monodromy transformation tell us qualitatively?

• While I don't know a great deal about Hodge or representation theory, a cursory examination of the relevant concepts suggests something to the effect of "the monodromy action on the fiber always fixes some $(m-n)$-cycle class." Sep 2 '20 at 4:34
• It says the eigenvalues are roots of unity, which means, since the coefficients of the characteristic polynomial are integers, there are only finitely many possibilities, depending on the dimension, for the Jordan normal form of the local monodromy. Is that the kind of thing you're looking for? What do you mean by "qualitative"? Sep 3 '20 at 1:18
• @WillSawin Thank you for your comment! Personally, "qualitatively" means "geometric picture". Sep 3 '20 at 2:19
• I believe this is saying that you can choose an ordered basis of the cohomology so that monodromy on a basis element gives you back the basis element (multiplied by a root of unity from the multiplicity of the irr. component) plus only basis elements that come later in the list. Think of it as a generalization of Picard-Lefschetz theory (en.wikipedia.org/wiki/Picard%E2%80%93Lefschetz_theory for a very concrete calculation) Sep 3 '20 at 3:40

I'm not completely sure what you're after, but perhaps it's simply some insight. So let make a few remarks. For simplicity, assume $$Y$$ is a curve. (In general, you need to assume that $$E$$ has normal crossings, which you did implicitly.) Replace it by a small disk. So $$f:X\to D$$ becomes a $$C^\infty$$ fibre bundle when restricted to $$D^*= D-\{0\}$$, which can be replaced by the circle, since we can work up to homotopy. To construct a such a fibre bundle, you just take a manifold $$F$$ and a diffeomorphism $$\phi:F\to F$$, and the glue the ends of $$F\times [0,1]$$ using $$\phi$$. Now ask, given a fibre $$F$$ diffeomorphic to a projective manifold and a self diffeomorphism, when can it arise from a projective family? The monodromy theorem tells you that the answer is almost always no, even when $$F$$ is $$2$$-torus, because a necessary condition is that the eigenvalues of $$\phi^*H^i(F)\to H^i(F)$$ must be roots of unity for all $$i$$. I still find this statement pretty striking.
• It's interesting that, in your comment, you are pretty much keeping inside the smooth category. That is, Ehressman's theorem gives you the smooth fiber bundle structure that you mentioned, but we do not seem to be concerned about whether the fibers are biholomorphic. Of course, $f$ is a holomorphic fiber bundle iff the fibers are all biholomorphic. There is a question somewhere in this rambling, but I can't formulate it at present. Sep 3 '20 at 22:23