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?***



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


Does this help?


