3
$\begingroup$

Given a family $\pi: \mathcal{X}\rightarrow\Delta$ smooth away from $0\in\Delta$, Where $\mathcal{X}$ is a smooth complex manifold, $\Delta$ is a small disk, the general fiber of $\pi$ is smooth projective variety, central fiber $Y=f^{-1}(0)=\bigcup D_i$, with reduced induced $Y_{red}$ is normal crossing.

Question: If the GCD of multiplicities of components $D_i$ is 1, does this condition imply that the monodromy is unipotent?

There is a detailed proof by steenbrink in his paper "limits of Hodge structures" proof of (2.20). However, in his book "Mixed Hodge structures", He also mentioned if the LCM is $n$, then T^n is unipotent. I am confused about the correct version of this theorem.

$\endgroup$

1 Answer 1

2
$\begingroup$

On what group are you studying the monodromy? Begin with $\mathcal{Y} = \Delta \times \mathbb{P}^n$, $n\geq 2$, with its projection $\text{pr}_1$ to $\Delta$. The central fiber is smooth of multiplicity $1$. Let $d\geq 2$ be an integer. Let $Z\subset \mathcal{Y}$ be a closed submanifold such that the restriction $\text{pr}_1|_Z:Z\to \Delta$ is a $d$-to-$1$ branched cover of the disk by the disk totally ramified over the origin with ramification profile $(d_1,\dots,d_m)$, i.e., $m$ preimages of the origin with the pullback of a coordinate vanishing to order $d_r$ at the $r^\text{th}$ preimage point ($d_1+\dots +d_m = d$). Let $\mathcal{X}'\to \mathcal{Y}$ be the blowing up along $Z$. There is a further blowing up $\mathcal{X}\to \mathcal{X}'$ with center in the special fiber so that $\mathcal{X}$ is smooth with normal crossings special fiber.

The strict transform of the special fiber of $\mathcal{Y}$ is an irreducible component with multiplicity $1$ of the special fiber of $\mathcal{X}$. Thus the GCD of the multiplicities equals $1$. Yet the monodromy action of $\pi_1(\Delta^*,\text{point})$ on $H^{2r}(X)$ is non-unipotent for $1\leq r \leq n-1$. On the other hand, the LCM of the multiplicities is divisible by the LCM $\ell$ of $(d_1,\dots,d_m)$, and the $\ell^\text{th}$ power of a generator of $\pi_1(\Delta^*,\text{point})$ does act unipotently (in fact it acts as the identity, in this case).

$\endgroup$
4
  • $\begingroup$ My monodromy acts on the middle cohomology of a fixed general fiber. So after the above procedure, you get a family $\mathcal{X}$ over $\Delta$, with GCD of the multiplicity of components in central (special )fiber is 1. But still how could you know that it is not unipotent, through above specific $Z$ you choose, or why if you choose the above $Z$, it becomes non-unipotent ? $\endgroup$
    – Feng Hao
    Apr 29, 2016 at 13:54
  • $\begingroup$ "My monodromy acts on the middle cohomology of a fixed general fiber." If $n=2m$, $m\geq 1$, in my example above, then the action of the monodromy on the middle cohomology decomposes into a trivial action on a one-dimensional subspace, and cyclic actions of order $d_1,\dots,d_m$ on subspaces of dimensions $d_1,\dots,d_m$. This follows from the description of the (additive) cohomology of a blowing up of a smooth subvariety of a smooth variety in terms of the cohomology of the original ambient variety and the cohomology of the subvariety. $\endgroup$ Apr 29, 2016 at 14:00
  • $\begingroup$ Sometime even the LCM is not 1, the monodromy can be unipotent. I am wondering is there a general fine criterion for unipotency of monodromy? or someone already did some work on that? $\endgroup$
    – Feng Hao
    Apr 29, 2016 at 17:39
  • $\begingroup$ That is a great example! I was trying to understand all the details and I understand everything except how to construct such a $Z \subset \mathcal{Y}$. Can you elaborate on how to construct such a $Z$ or point me to some reference? Thanks! $\endgroup$ Apr 30, 2016 at 16:43

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.