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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.

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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).

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  • $\begingroup$ 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 '16 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. $\endgroup$ – Jason Starr Apr 29 '16 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 '16 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! $\endgroup$ – Partha Solapurkar Apr 30 '16 at 16:43

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