Hodge theory on complex spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:36:30Z http://mathoverflow.net/feeds/question/58330 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58330/hodge-theory-on-complex-spaces Hodge theory on complex spaces Gunnar Magnusson 2011-03-13T12:50:47Z 2011-03-15T12:33:49Z <p>If $X$ is a compact Kahler manifold, then Hodge theory says that its cohomology decomposes as a direct sum</p> <p>$$H^{p+q}(X,\mathbb C) = \bigoplus_{p,q} H^{p,q}(X,\mathbb C)$$</p> <p>where $H^{p,q}(X,\mathbb C) = H^q(X,\Omega_X^p)$ are the Dolbeault cohomology groups and $H^{p,q}$ and $H^{q,p}$ are conjugate isomorphic. One can prove that the same holds for a Fujiki manifold, i.e. a complex manifold bimeromorphic to a Kahler manifold.</p> <p><strong>Q:</strong> What happens for compact complex spaces?</p> <p>Here "complex space" should mean singular and non-reduced. Is there a notion of a "Kahler" complex space, on which some predictions of Hodge theory hold? The question should make sense, as we can define $H^{p+q}(X,\mathbb C)$ topologically and the $H^q(X,\Omega_X^p)$ exist for a complex space $X$, but do we even know that the latter are subgroups of the former for some special spaces $X$?</p> <p>One can imagine naively defining such spaces as those which admit a Fujiki desingularisation (Kahler to begin with, but desingularisations are not unique so we need to compare them somehow), and then hoping that the "upstairs" decomposition induces one "downstairs". Of course, if it were that simple someone would have done it ages ago.</p> http://mathoverflow.net/questions/58330/hodge-theory-on-complex-spaces/58336#58336 Answer by Donu Arapura for Hodge theory on complex spaces Donu Arapura 2011-03-13T13:58:45Z 2011-03-15T12:33:49Z <p>The answer is that if $X$ is compact complex space of class $C$ in Fujiki's sense (i.e. if it is dominated by compact Kaehler manifold) then $H^*(X)$ carries a natural mixed Hodge structure. This is gotten by using resolution of singularities to express the cohomology of $X$ in terms of the cohomology of simplicial Kaehler manifold, and then applying the methods of Deligne's Hodge II &amp; III (as Shengao points out) to give this a mixed Hodge structure.</p> <p>In fact, Fujiki worked this out long ago in "Duality for Mixed Hodge structures..." RIMS 1980. </p> <p><hr> I realize there was more to your question. The precise relationship between $H^*(X,\mathbb{C})$ and $H^q(X,\Omega_X^p)$ for singular spaces is, to put it mildly, complicated. At least in the algebraic category, Du Bois has shown that there exists objects $\tilde \Omega_X^p$ in the derived category, such that $$H^i(X,\mathbb{C}) = \bigoplus_{p+q=i} H^q(X,\tilde \Omega_X^p)$$ There are maps $\Omega_X^p\to \tilde \Omega_X^p$. The question of when these are isomorphisms is not well understood; except perhaps for $p=0$, where isomorphism characterizes the so called Du Bois singularities.</p> <p><hr> I thought I'd add a few rather speculative comments. Our answers (mine and Sándor's) are taking this in somewhat homological direction -- a lot of Hodge theory tends to get that way. This is perhaps a bit unfortunate, because for many the initial attraction to the area stems from its analytical aspects. I've often wondered <em>is there a purely analytic approach to mixed Hodge theory?</em> I remember having a conversation with Saper, long ago, who thought it might be feasible to do this as some sort of weighted $L^2$ cohomology for a suitably chosen metric on the smooth part. I'm being purposely vague here. Anyway, I don't think anyone has ever carried out anything like this. It would be really interesting if someone did.</p> <p>Let me try to make this a bit more precise, although it will still be pretty vague. Suppose that $X$ is a singular compact complex space which can be blown up along the singular locus <code>$X_{sing}$</code> to a Kaehler manifold. Let <code>$U = X-X{sing}$</code>. Then there are various choices of complete Kaehler metric on $U$ for which the space of harmonic forms satisfying $\int \alpha\wedge *\alpha &lt; \infty$ coincides with intersection cohomology $IH^i(X)$. This gives a pure Hodge structure on it by the Kaehler identities. Although this is not quite what I'm asking for, it points in the right direction (note that I'm pretty sure that $IH^i(X)$ is a pure subquotient of $H^i(X)$). The question is how to modify this picture so that one gets the mixed Hodge structure on $H^i(X)$? A weaker question is how to describe the pure subquotients $W_kH^i(X)/W_{k-1}H^i(X)$ by analytic means? One can ask something like this for the Du Bois complex as well. I don't want to get much more specific. But perhaps I can point out, at least one useful reference: Saper, "$L^2$ cohomology on Kaehler varieties with isolated singularities" JDG (1992).</p> http://mathoverflow.net/questions/58330/hodge-theory-on-complex-spaces/58384#58384 Answer by Sándor Kovács for Hodge theory on complex spaces Sándor Kovács 2011-03-14T05:34:01Z 2011-03-15T05:05:46Z <p>Let me add to Donu's mentioning Du Bois's Hodge decomposition. First of all, many feel that part of the credit is due to Deligne as Du Bois built heavily on his ideas. Then again that is probably true for many things in Hodge theory.</p> <p>Anyway, Du Bois's main idea was that one can do Deligne's construction "one step earlier" in the sense that Deligne used simplicial resolutions to build his Hodge structure on the cohomology groups, and Du Bois does this in the derived category of coherent sheaves (with some conditions...) so he obtains a filtered complex (now mostly called the <em>Deligne-Du Bois</em> complex) which is quasi-isomorphic to the constant sheaf $\mathbb C$ and whose associated graded quotients (these are actual quotients in the category of complexes, before one passes to the derived category) are the objects $\underline{\Omega}_X^{p}$ Donu mentions. </p> <p>It turns out that it follows directly from the construction that there is a Hodge-to-de-Rham (a.k.a. Frölicher) spectral sequence and from Deligne's work that it degenerates at the $E_1$ level. So, a lot of things actually work out the same way as in the smooth case if one uses $\underline{\Omega}_X^{p}$ in place of ${\Omega}_X^{p}$. For instance, the Kodaira-Akizuki-Nakano vanishing theorem also holds (as well as the existence of the Gysin map, etc) that can be used to prove a singular version of the Lefschetz hyperplane theorem <a href="http://mathoverflow.net/questions/57744/lefschetz-hyper-plane-theorem-for-singular-projective-varieties/57771#57771" rel="nofollow">cf. here</a>.</p> <p>As Donu mentioned, it is not known in general when the natural map ${\Omega}_X^{p}\to\underline{\Omega}_X^{p}$ is an isomorphism, except for $p=0$ which is the definition of Du Bois singularities. I think in general (that is, for arbitrary $p$) it is true that for toroidal singularities this is an isomorphism, or at least there is a good description of each object and the map between them. We have a pretty good understanding of Du Bois singularities, although not complete and there are still many interesting open questions. Rational singularities are Du Bois <a href="http://journals.cambridge.org/action/displayAbstract?aid=308740" rel="nofollow">by this</a>, log canonical singularities are Du Bois <a href="http://www.ams.org/journals/jams/2010-23-03/S0894-0347-10-00663-6/home.html" rel="nofollow">by this</a>. For an intro to Du Bois singularities and related stuff you can try <a href="http://arxiv.org/abs/0909.0993" rel="nofollow">this</a>.</p> <p>There is also an intriguing connection to singularities defined via the action of Frobenius in positive characteristic. For more on this see <a href="http://muse.jhu.edu/login?uri=/journals/american_journal_of_mathematics/v131/131.2.schwede.html" rel="nofollow">this</a> and other works of Karl Schwede.</p> <p>Regarding the question on having something that reflects the non-reduced structure, I am afraid that the topological $H^i$ do not see the non-reduced structure, so I don't think you can expect any reasonable Hodge theory that remembers that.</p>