I'm trying to understand variations of Hodge structure. I understand that this is a very broad field, and that many of the concepts have been extended to algebraic geometry over fields other than $\mathbb{C}$, and so forth. In particular, there is a version of the monodromy theorem by Grothendieck, which is rather incomprehensible to me. I am asking this question in the context of variations of Hodge structure for Calabi-Yau manifolds and mirror symmetry, so I hope that the answer can be given with that in mind.

Given a family of projective Kähler manifolds $\pi:\mathcal{X}\to\Delta^*$, where $\Delta^*$ denotes the punctured disk, we get a monodromy operator $T\in\text{GL}(H^k(X_t,\mathbb{C}))$, for $t\in\Delta^*$. This operator is obtained from the local system associated to the family, which gives a representation $\rho:\pi_1(\Delta^*)\to\text{GL}(H^k(X_t,\mathbb{C}))$, and $T$ is the image of a generator of the fundamental group under this representation. One can then show that $T$ is quasi-unipotent, meaning that there are integers $m,N$ such that $(T^m-I)^N=0$. After pulling back along the map $z\mapsto z^m$, we may as well assume that $m=1$, so that $T$ is in fact unipotent. One then defines the nilpotent operator $$N=\log T=(T-I)+\dots+(-1)^{N+1}(T-I)^N/N!$$ Associated to this operator is a weight filtration of $H^k(X_t,\mathbb{Q})$, called the monodromy weight filtration. It is uniquely defined by $$N(W_i)\subseteq W_{i-2}\quad\quad N^k:W_{N+k}/W_{N+k-1}\xrightarrow{\cong}W_{N-k}/W_{N-k-1}$$ I have two questions about this.

  1. Why do we impose this filtration on the cohomology with rational coefficients, rather than with complex coefficients? Does this make a difference, if we restrict our attention to complex manifolds and their variations of Hodge structure?
  2. What does this filtration represent, geometrically? Up until this point, there has been a nice geometric interpretation for the concepts which were introduced (e.g. the Hodge bundles, Gauss-Manin connection, the period map, etc.), but I have no idea how to think about this filtration in geometric terms.

1 Answer 1


Given a nilpotent endomorphism $N$ of a finite dimension vector space $V$, Jordan canonical form implies that we can decomponse $V$ into a sum of "blocks" on which we can find bases satisfying $Ne_1 = e_{2}, Ne_2=e_3,\ldots Ne_k=0$. To make it basis independent, pass to the filtration $$ \langle e_k\rangle\subset \langle e_k, e_{k-1}\rangle \subset \ldots$$ After suitable indexing, this is the monodromy weight filtraton $W_\bullet$.

The point I wanted to make is that $W$ may seem complicated, but it isn't that bad. Furthermore the construction works over $\mathbb{Q}$ (because $N$ has zero eigenvalues, and so has a Jordan form over this field). This is also useful further along in the story, when one gets to limit mixed Hodge structures: part of the definition of a mixed Hodge structure is that $W$ is rational.

Regarding your second question, I don't really have a good answer. It is clear that $W$ is quite natural from the point of linear algebra, and Schmid gave an analytic interpretation in terms of certain growth conditions. However, the geometric meaning seems much more elusive.

  • $\begingroup$ Thank you prof. Arapura! Your book has been of great help to me on this journey. Regarding your answer: I understand that we can do this over $\mathbb{Q}$ but I was just wondering if one loses anything by doing the same construction over $\mathbb{C}$ instead, if the intent is to stay within the complex geometry world. I would assume not, because of the universal coefficients theorem. Would it be possible to elaborate on that slightly? Either way, thanks for your time. $\endgroup$ Jul 25, 2022 at 19:57
  • 1
    $\begingroup$ Your welcome. Universal coefficients tells you the inclusions $H^i(X,\mathbb{Z}), H^i(X,\mathbb{Q})\subset H^i(X,\mathbb(C))$ give "lattices". But these lattices are extra data that you may not want to throw away. $\endgroup$ Jul 25, 2022 at 21:18
  • $\begingroup$ It escaped my mind that the rational cohomology also gives us this lattice structure. That makes sense, then - we want to preserve this structure, but the integers don't allow us to define these matrices, so we must pass to rational cohomology. I will leave the question open in case someone has an insight about the geometric interpretation of this filtration, but your answer helps a great deal. $\endgroup$ Jul 25, 2022 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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