Timeline for Does every equivalence class of Hecke characters contain a distinguished element?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2013 at 16:33 | vote | accept | Daniel Loughran | ||
| Jun 14, 2013 at 16:33 | comment | added | Daniel Loughran | Out of interest, I think there is a nice analytic way to think of this in terms of the functional equation of the associated L-function. Namely, $\chi$ is good if and only if the sum of the imaginary parameters of the associated Gamma factors is equal to zero. | |
| Jun 14, 2013 at 16:17 | comment | added | Daniel Loughran | Right I think I get it now. The map $i$ is "as good as" a splitting of the exact sequence $(*)$. It gives you a distinguished subgroup for you to play with. Thanks again. | |
| Jun 14, 2013 at 14:08 | comment | added | Joël | @Kconrad: you're right. $n$ is the degree of $k$ over $\mathbb Q$. I have edited my answer to say this. @Daniel: the map $i$ is canonical. The only freedom you have in defining the isomorphism of $\R$-\algebras $k \otimes_{\mathbb Q} \mathbb R \simeq \mathbb R^{r_1} \times \mathbb C^{r_2}$ is that you can let act the complex conjugacy on each of the $\mathbb C$ factors (hence you have $2^{r_2}$ such isomorphisms). But as far as the embedding of $\R$ is concern, this choice does not change anything. | |
| Jun 14, 2013 at 14:05 | history | edited | Joël | CC BY-SA 3.0 |
added 148 characters in body
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| Jun 14, 2013 at 8:59 | comment | added | Daniel Loughran | Thanks for the answer, but I just want to clarify something: Is the map $\mathbb{R}_{>0} \hookrightarrow I^{\infty}_k$ canonical? Or does it depend on a choice of isomorphism $k \otimes_{\mathbb{Q}} \mathbb{R} \cong \mathbb{R}^{r_1} \times \mathbb{C}^{r_2}$? | |
| Jun 14, 2013 at 1:54 | comment | added | KConrad | The integer $n$ is not specified anywhere. Was a condition left out? | |
| Jun 13, 2013 at 15:33 | history | answered | Joël | CC BY-SA 3.0 |