Timeline for Dolbeault Cohomology of $\mathbb{P}^1$
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2010 at 18:10 | comment | added | Clay Cordova | Hi David, Griffiths and Harris chapter 3 has a section on how to define cohomology using distributions. I would start there. | |
| Apr 5, 2010 at 14:12 | comment | added | David E Speyer | (continued) Of course, $\theta_1$ is not smooth at $z=0$. You propose to push on regardless. Then $\overline{\partial} \theta_1$ is a $\delta$-function at $z=0$, while $\overline{\partial} \theta_2$ is identically zero. So our $(1,1)$ form is a $\delta$ function at $z=0$. This is really cool! Is there a book where I can learn when this sort of computation is justified? | |
| Apr 5, 2010 at 14:08 | comment | added | David E Speyer | OK, I've been thinking about this more, and I start to understand what you are doing. I'll continue using the notation from my answer. We are supposed to find smooth (1,0) forms $\theta_i$, on $U_i$, such that $\theta_i - \theta_2 = dz/z$. The (1,1) form will then be the form which, on $U_i$, restricts to $\overline{\partial} \theta_i$. Your suggestion is to take $\theta_1 = dz/z$ and $\theta_2=0$. | |
| Apr 3, 2010 at 13:17 | comment | added | David E Speyer | (3) Why didn't you pick up a second $\delta$ function at $\infty$, where $dz/z$ also has a pole? Presumably, the answer to this is related to the answer to (1). | |
| Apr 3, 2010 at 13:16 | comment | added | David E Speyer | (2) I guess it really is true that $d (dz/z) = \delta_0$ in the sense that $\int_{\partial A} dz/z = \int_A \delta_0$ for any disc $A$. So this is going to be a horrible nitpick but: In what sense am I allowed to do DeRham cohomology with measures? The theorem/definition I know is that $H^2$ is (closed smooth $2$-forms)/d(smooth $1$-forms). Is there an analogous result where "smooth forms" is replaced by measures? | |
| Apr 3, 2010 at 13:03 | comment | added | David E Speyer | (1) your construction should remember not just the form $dz/z$, but the choice of Cech cover. If we switched the order of the open sets $\mathbb{P}^1 \setminus \{ 0 \}$ and $\mathbb{P}^1 \setminus \{ \infty \}$, we should negate the cohomology class. | |
| Apr 3, 2010 at 13:03 | comment | added | David E Speyer | Can you explain what you are doing here? This looks like nonsense, but your end result does seem to be morally right in some sense. (I assume that $\delta$ is the $\delta$ measure at $0$.) Here are some of the things which confuse me: | |
| Mar 30, 2010 at 6:48 | history | answered | transvectant | CC BY-SA 2.5 |