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In the paper "Asymptotic Behavior of a Variation of Hodge Structure" by Philip Griffiths, written by Loring Tu, in the book "Topics in Transcendental Algebraic Geometry", page 68 to 69, for a given variation of Hodge structure of weight $n$ over $\Delta^*$, we have \begin{equation} N=\log T,\,N^{n+1}=0 \end{equation} where $T$ is the monodromy matrix. $N$ induces a unique filtration of $H_{\mathbb{Q}}$,

\begin{equation} 0 \subset W_0 \subset W_1 \subset \cdots \subset W_{2n-1} \subset W_{2n}=H_{\mathbb{Q}} \end{equation}

In the remark of page 69, it says that Ron Donagi points out a general formula, set

\begin{equation} N^{p,q}= \text{im}\, N^p \cap\, \text{ker}\, N^{n-q} \end{equation}

then we would have \begin{equation} W_q=\text{span}(\sum_{r+s\leq n-q}N^{r,s}) \end{equation}

I do not really understand this formula, e.g, by this formula we have $W_1 \subset W_0$ since if $r+s \leq n-1$, then trivially $r+s \leq n$, so I am wondering whether this formula has a typo? Could anyone provide a correct general formula? Any references will also be appreciated!

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    In Illusie's paper "Autour du théorème de monodromie locale", the filtration is defined as $M_j=\sum_{k-i=j} \ker N^{k+1}\cap \operatorname{im} N^i$. But there is shift between the filtration $M_j$ and your filtration $W_q$. In fact $M_j$ is centered at 0, in other words $0=M_{-n-1}\subset M_{-n}\subset ...\subset M_n\subset M_{n+1}=H_\mathbb{Q}$ and $N$ induces an isomorphism $\operatorname{gr}^M_k\overset\sim\rightarrow\operatorname{gr}^M_{-k}$. – 2017-03-01
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    @Roland Thank you, I guess the $\leq$ in the sum I of my question should be an $=$, i.e a equation between $r$ and $s$. For a power $N^i$ in the sum of $M_j$, should $i \geq 1$? or $i \geq 0$ and let $N^0=1$? – 2017-03-01
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    You can use an inequality in the sum (the reverse one ?) since $N^{r+1,s}\subset N^{r,s}$ and similarly for $s$. All this indices gives me an headache :p I should write things down. About $N^i$, it works trivially for $i=0$ with $N^0=1$ as you said. – 2017-03-01
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    By the way, there is a typo in my first comment, this is should be $N^k$ which induces an isomorphism $\operatorname{gr}^M_k\overset\sim\rightarrow\operatorname{gr}^M_{-k}$ as you have guessed. – 2017-03-01

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