Timeline for Variant of Leopoldt's conjecture
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 18, 2013 at 15:43 | comment | added | Ramsey | I think what David meant was that CM fields are characterized by the property that, under any embedding into the complex numbers, the image is preserved by complex conjugation and the resulting involution on the field is independent of the embedding. | |
| Jan 18, 2013 at 15:06 | history | edited | David Loeffler | CC BY-SA 3.0 |
Cleaned up statement
|
| Jan 18, 2013 at 7:54 | comment | added | Filippo Alberto Edoardo | David: I do not understand your comment (note I wrote $\mathbb{C}$ and not $\mathbb{C}_p$). But I confess that my comment was imprecise: I simply wanted to say that if $K$ is non totally real and we consider complex-valued instead of $p$-adic logarithms, then there will always be a singular minor. I was asking because I was trying to figure out if the condition that $p$ split completely in $K$ in your computations, making the local Galois group trivial, was somehow "equivalent" to $K$ being totally real when computing the usual archimedean regulator. Following Calegari, it should not be so... | |
| Jan 18, 2013 at 7:30 | history | edited | David Loeffler | CC BY-SA 3.0 |
Added a remark of Frank Calegari's
|
| Jan 18, 2013 at 7:24 | comment | added | David Loeffler | Filippo: No, that does not work, because there is no uniquely defined notion of "conjugate" unless K is CM. | |
| Jan 18, 2013 at 1:57 | comment | added | Filippo Alberto Edoardo | David, when you say that the statement is definitely false if $K$ is CM and $d>4$ do you really need CM or totaly imaginary is enough? The argument I figure out is that for every unit $u$, the complex aboslute value of its norm is $1$ and if we decompose this as the product over all embeddings $\tau_j$ in $\mathbb{C}$ of the absolute value of $\tau_j(u)$ we get $|\tau_j(u)|=|\tau_i(u)|$ if the two are conjugate. And this is enough to kill many minors (if there is enough room, i.e. $d>4$), right? | |
| Jan 17, 2013 at 21:35 | comment | added | David Loeffler | Yes, I have checked a few examples (although only ones where p splits completely in K). | |
| Jan 17, 2013 at 20:53 | comment | added | Ramsey | Curious: have you checked in any examples? | |
| Jan 17, 2013 at 17:52 | history | asked | David Loeffler | CC BY-SA 3.0 |