Is the weight filtration a topological invariant? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:08:56Z http://mathoverflow.net/feeds/question/62406 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62406/is-the-weight-filtration-a-topological-invariant Is the weight filtration a topological invariant? shenghao 2011-04-20T13:06:28Z 2011-04-21T12:29:28Z <p>This question is somehow related to (but different from) the following MO question and the one linked from there</p> <p><a href="http://mathoverflow.net/questions/42744/diffeomorphic-kahler-manifolds-with-different-hodge-numbers" rel="nofollow">http://mathoverflow.net/questions/42744/diffeomorphic-kahler-manifolds-with-different-hodge-numbers</a></p> <p>Let $X$ and $X'$ be two complex algebraic varieties that are diffeomorphic to each other. One may or may not assume they are irreducible. Then:</p> <ol> <li><p>(Weak form) For each pair of integers $(i,n),$ do we always have $$\dim Gr^W_iH^n(X,\mathbb Q)=\dim Gr^W_iH^n(X',\mathbb Q),$$ where $W$ denotes the weight filtration on the mixed Hodge structures?</p></li> <li><p>(Strong form) Let $f:X\to X'$ be a diffeomorphism. Does $f^*:H^n(X')\to H^n(X)$ respect the $W$-filtrations?</p></li> </ol> <p>Variants: One can ask similar questions with "diffeomorphic" replaced by "homeomorphic" or "complex analytically isomorphic". For instance, does the weight filtration see the difference between tangential intersections and transverse intersections? One can also add coefficients.</p> <p>At the first sight, it's not clear to me how to approach the question, since there seems to be no relation between the resolutions of $X$ and $X'.$ </p> <p><strong>Edit:</strong> As algori pointed out to me, this is false without the compactness assumption, even for "complex analytic isomorphism". So I will assume $X$ and $X'$ are compact.</p> http://mathoverflow.net/questions/62406/is-the-weight-filtration-a-topological-invariant/62408#62408 Answer by algori for Is the weight filtration a topological invariant? algori 2011-04-20T13:31:16Z 2011-04-20T13:59:49Z <p>If one does not assume $X$ and $X'$ compact, then there are relatively easy counter-examples: there is a vector bundle over an elliptic curve with total space analytically equivalent to $\mathbb{C}^*\times\mathbb{C}^*$, see e.g., Peters-Steenbrink, Mixed Hodge structures, p. 102.</p> <p>There are positive results as well, but the only one I can think of is this: if $X$ and $X'$ are (possibly singular) compact algebraic surfaces, the answer to question 2 is positive, see Steenbrink-Stevens, Indag. Math., 46, 1984, no. 1, p. 63-76.</p> http://mathoverflow.net/questions/62406/is-the-weight-filtration-a-topological-invariant/62466#62466 Answer by SGP for Is the weight filtration a topological invariant? SGP 2011-04-20T22:05:27Z 2011-04-20T22:05:27Z <p>Not an answer, but related somewhat is an important topological property of the weight filtration. Namely if $f: X \to S$ is morphism of complex algebraic varieties such that $R^if_*Q$ are local systems (e.g. all fibers are homeomorphic) on $S$, then the weight filtration assembles to give sub-local systems. So the weight filtration is locally constant, in families. </p> <p>An arithmetic version (over Spec $Z$) is also available. </p> <p>See <a href="http://mathunion.org/ICM/ICM1974.1/Main/icm1974.1.0079.0086.ocr.pdf" rel="nofollow">Deligne's ICM 1974 address</a>, especially Theorems 2 and 14.</p>