Timeline for A question on Nekovar's paper Belinson's Conjectures
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2018 at 20:40 | comment | added | Wenzhe | @KeerthiMadapusiPera Thank you for your patience! | |
| Apr 12, 2018 at 20:36 | comment | added | Keerthi Madapusi | Another way to see this is to note directly that the first map in the purportedly exact sequence is $F^0M_{dR}\otimes\mathbb{R}\oplus M^+_B\otimes\mathbb{R}$ mapping to $M_{dR}\otimes \mathbb{R}$, and so is injective precisely when $F^0M_{dR}\otimes\mathbb{R}$ has trivial intersection with $M^+_B\otimes\mathbb{R}$. | |
| Apr 12, 2018 at 20:31 | comment | added | Keerthi Madapusi | Sorry, that doesn't make any sense. What's correct is this: The dual of $\alpha_{M^\vee(1)}$ is the map $F^0M_{dR}\otimes\mathbb{R} \to (M_{dR}\otimes\mathbb{R})/(M_B^+\otimes \mathbb{R})$. The kernel of this is the same as the kernel of $\alpha_M$. | |
| Apr 12, 2018 at 20:27 | comment | added | Keerthi Madapusi | The map $\alpha_{M^\vee(1)}$ is just the dual of the map $\alpha_M$, and so is surjective precisely when $\alpha_M$ is injective. | |
| Apr 12, 2018 at 20:23 | comment | added | Wenzhe | @KeerthiMadapusiPera Ifeel very sorry to ask you again, by which duality? Could you explain more carefully why it is equivalent to $\text{ker} \, \alpha_M=0$? | |
| Apr 12, 2018 at 20:17 | comment | added | Keerthi Madapusi | By duality, it's equivalent to $\ker \alpha_M$ being $0$; that is, to $M^+_B\otimes\mathbb{R}$ (which is contained in $M_{dR}\otimes\mathbb{R}$ under the comparison isomorphism) intersecting trivially with $F^0M_{dR}\otimes\mathbb{R}$. But in fact, $M_B\otimes\mathbb{R}$ intersects trivially with $F^0M_{dR}\otimes \mathbb{C}$ when $w<0$, as can be checked from the fact that the latter cannot contain both of any conjugate pair of weights. | |
| Apr 12, 2018 at 19:38 | comment | added | Wenzhe | @KeerthiMadapusiPera Thank you. What still confuse me is that the injectivity of the left map of this short sequence is equivalent to $\alpha_{M^\vee(1)}$ being surjective, but I don't know how to prove $\alpha_{M^\vee(1)}$ is surjective! | |
| Apr 12, 2018 at 18:58 | comment | added | Keerthi Madapusi | Just dualize the sequence and note that $F^0M_{dR}$ is the dual of $M^\vee(1)_{dR}/F^0M^\vee(1)_{dR}$. Your confusion may be because the standard indexing of the filtration on the dual ensures that $F^1M^\vee_{dR}$ is actually the annihilator of $F^0M_{dR}$. Therefore you need to twist $M^\vee$, which then makes $F^0M^\vee(1)_{dR}$ the annihilator instead. | |
| Apr 12, 2018 at 18:07 | history | asked | Wenzhe | CC BY-SA 3.0 |