Timeline for On the positivity of the second Segre class of ample vector bundles
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 12, 2023 at 17:54 | comment | added | Armando j18eos | In this way the Kleiman's reasoning is clear! I thank you. | |
| Feb 12, 2023 at 7:39 | comment | added | abx | $(c_1^2-c_2)\cdot \xi ^{r-1}= c_1^2\cdot \xi ^{r-1}-c_1\cdot \xi ^r+\xi ^{r+1}=$ $c_1\cdot \xi ^{r-1}\cdot (c_1-\xi )+\xi ^{r+1}$. The first term is zero by (3.2), and $\xi ^{r+1}>0$. | |
| Feb 11, 2023 at 11:14 | comment | added | Armando j18eos | @abx My attempt: $\displaystyle0<\int_Hi^{*}\xi^2=\int_Hi^{*}(\xi-c_1(E)+c_1(E))^2=\int_Hi^{*}(\xi-c_1(E))^2+i^{*}c_1(E)^2=\int_Hi^{*}(\xi-c_1(E))i^{*}\xi+i^{*}c_1(E)^2$ if $r=2$ I find $\displaystyle0<\int_Hi^{*}c_1(E)^2-i^{*}c_2(E)=\int_Hi^{*}s_2(E)$. As I have wrote: I don't understand the Kleiman's reasoning. | |
| Feb 11, 2023 at 8:21 | comment | added | abx | Kleiman: "by (3.1) and (3.2) with $a=c_1$". | |
| Feb 11, 2023 at 7:19 | comment | added | Armando j18eos | @abx Yes, it's true. My trouble is the last step: I don't understand how to prove $\displaystyle\int_Hi^{*}\xi^2>0\Rightarrow\int_Hs_2(E)>0$! | |
| Feb 11, 2023 at 7:13 | comment | added | abx | The steps are explained quite clearly in Kleiman's paper. What is it you don't understand? | |
| Feb 11, 2023 at 5:26 | history | asked | Armando j18eos | CC BY-SA 4.0 |