Weight filtration of MHSs

This is probably a very stupid question, but could someone explain to me where the weight filtration of mixed Hodge structures come from and why we actually need it? If the Hodge-to-de Rham spectral sequence degenerates at $E_1$, then why does that not define a pure Hodge structure? I sort of understand that we need this for open varieties, but it seems to me that people talk about MHSs of smooth projective varieties as well. Is there any use of them that does not follow from pure Hodge structures? Thanks!

flag	asked Jun 25 2011 at 20:07 Alfonz 161●5

"people talk about MHSs of smooth projective varieties as well" - no.:) – Mikhail Bondarko Jun 26 2011 at 8:43

You may want to read some of Deligne's Hodge I, II, III, and his ICM talk in 1974. To see where the weight filtration comes from, you may try to compute the MHS on $H^1(C)$ for an open or singular curve $C.$ For why we need it, one reason seems to be that any map between the underlying vector spaces which arises from (alg.) geometry must preserve this filtration (strictly), and this has many applications. Note that for open varieties, the Hodge-de Rham spec. seq. does not necessarily degenerate: the classic. argument fails b/c $\int_X$ makes no sense, and $X$ may not be Kaehler. – shenghao Jun 26 2011 at 10:13

In fact the de Rham side, being isomorphic to the Betti cohom (let's work with smooth varieties), are finite-dimensional vector spaces, whereas the Hodge-side could be infinite-dimensional for open varieties. – shenghao Jun 26 2011 at 10:23

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