in which sense is a mixed Hodge structure an extension of pure ones?

Question

I have heard several times that mixed Hodge structures are iterated extensions of pure ones. What does it mean? Here is what I figured out.

A mixed Hodge structure $H$ comes with an increasing filtration

$W_{-1} H =(0) \subset W_0H \subset W_1H \subset \cdots \subset W_mH=H$

For each $k$, the inclusion $W_{k-1} \subset W_k$ gives an exact sequence $$ 0 \to W_{k-1}H \to W_kH \to Gr_k^W H:=W_k H/W_{k-1} H \to 0 $$ which is, I guess, compatible with the Hodge structures, so $W_kH$ is an extension of $Gr_k^WH$ (which is pure) by $W_{k-1}H$ (which is not).

In particular, $H$ is an extension of $Gr_m^W H$ by $W_{m-1}H$, $W_{m-1}H$ is an extension of $Gr_{m-1}^W H$ by $W_{m-2}H$ and so on until $W_0H$ which is a pure Hodge structure.

Is that right? Am I missing something?

Answer 1

This question is answered in the comments. This CW answer is here purely to document these answers as actual answers, and to remove the question from the unanswered list.

Answered in the comments by Sam Gunningham:

You are not missing anything. The term iterated extension means that (in your example): $W_{1}$ is an extension of $\mathrm{Gr}_{1}$ by $W_{0}$, $W_{2}$ is an extension of $\mathrm{Gr}_{2}$ by $W_{1}$ etc. So $H$ is "built up" from pure modules, in a similar way that a finite group is built as an iterated extension of finite simple groups.

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Also note the Donu Arapura's comment:

It isn't necessary that $W_{−1}=0$ but everything else is correct.