Timeline for Is the group $\mathrm{Gal}(\mathbb{C}/\overline{\mathbb{Q}})$ known?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2022 at 18:39 | comment | added | YCor | Also one known thing is that this group is torsion-free. Indeed, the group $\mathrm{Aut}(\mathbf{C})$ has only one non-identity conjugacy class of elements of finite order, and it consists of elements of order 2 (and this also forms the only nontrivial conjugacy class of finite subgroups). These elements map nontrivially on $\mathrm{Aut}(\bar{\mathbf{Q}})$, hence the kernel $\mathrm{Aut}_{\mathbf{Q}\text{-alg}}(\mathbf{C})$ is torsion-free. | |
| Jun 16, 2022 at 18:32 | comment | added | YCor | But "big and messy" doesn't mean there's nothing to say. For instance one can wonder whether it's simple ($\mathrm{Aut}(\mathbf{C})=\mathrm{Aut}_{\mathbf{Q}\text{-alg}}(\mathbf{C})$ is not simple because it admits $\mathrm{Aut}(\bar{\mathbf{Q}})$ as a quotient). A natural related question is whether it has a trivial abelianization. | |
| Jun 16, 2022 at 18:31 | comment | added | YCor | This group has cardinal $2^c=2^{2^{\aleph_0}}$ | |
| Jun 16, 2022 at 15:48 | comment | added | LSpice | Does $\operatorname{Gal}(\mathbb C/\overline{\mathbb Q})$ just mean $\operatorname{Aut}(\mathbb C/\overline{\mathbb Q})$? I am not accustomed to the ‘$\operatorname{Gal}$’ notation for non-algebraic extensions, though I know there are different conventions. | |
| Jun 16, 2022 at 15:25 | comment | added | user484137 | $\text{Gal}(\mathbb{C}/\overline{\mathbb{Q}})$ is as big and messy as $\text{Gal}(\mathbb{C}/\mathbb{Q})$, and for precisely the same reason: there's a load of algebraically independent transcendental elements in $\mathbb{C}$ which you can essentially permute as you which. Not sure if my comment is very satisfying as an answer, maybe there's more to be said (?) | |
| Jun 16, 2022 at 15:06 | history | asked | THC | CC BY-SA 4.0 |