Timeline for Hodge theory on complex spaces
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15, 2011 at 5:02 | vote | accept | Gunnar Þór Magnússon | ||
| Mar 14, 2011 at 5:34 | answer | added | Sándor Kovács | timeline score: 14 | |
| Mar 13, 2011 at 22:48 | comment | added | shenghao | I see. I was thinking that even for the intermediate step (between the well-known situation and singular analytic spaces), namely compact complex manifolds, the question is already interesting and not clear to me. Besides, $\Omega_X$ doesn't seem to me to be the right thing to consider for singular spaces (see Arapura's answer below for more): at least we don't have Poincare's lemma to relate the singular cohom with de Rham cohom, which seems to be the 1st step, and then one hopes for degeneration of the spectral sequence, which happens e.g. when X is Kaehler or proper algebraic... | |
| Mar 13, 2011 at 17:12 | comment | added | Gunnar Þór Magnússon | I was just trying to make clear that the space might be singular with nilpotent elements in its structure sheaf, as I've seen some people mean reduced spaces when they say "complex space". These spaces seem very mysterious to me (I've mostly studied Kahler geometry of smooth manifolds) and I just wondered to what extent the knowledge of smooth spaces carries over. | |
| Mar 13, 2011 at 15:53 | comment | added | Qfwfq | I think Gunnar wanted to mean "complex space (as opposed to complex manifold)", so a complex space may possibly be singular and noreduced. [the standard defn of a complex space is: a loc ringed space loc isom to the quotient of the sheaf of holomorph funct's on a domain of $\mathbb{C}^n$ by a sheaf of ideals] | |
| Mar 13, 2011 at 13:58 | answer | added | Donu Arapura | timeline score: 15 | |
| Mar 13, 2011 at 13:45 | comment | added | shenghao | By "singular and non-reduced" do you mean to say "non-singular and reduced"? The singular cohom doesn't see the nilpotence while $H^q(X,\Omega)$ does, and for singular algebraic varieties one has Hodge III. | |
| Mar 13, 2011 at 12:50 | history | asked | Gunnar Þór Magnússon | CC BY-SA 2.5 |