MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
"people talk about MHSs of smooth projective varieties as well" - no.:) – Mikhail Bondarko Jun 26 '11 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 '11 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 '11 at 10:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.