Timeline for Classification of Tori of GL2, up to conjugation
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2013 at 19:44 | answer | added | George McNinch | timeline score: 2 | |
| Feb 17, 2013 at 9:59 | vote | accept | Jérémy Blanc | ||
| Feb 16, 2013 at 16:04 | answer | added | Jim Humphreys | timeline score: 3 | |
| Feb 16, 2013 at 8:27 | comment | added | user30379 | If $T \subset {\rm{GL}}_n$ is a maximal $k$-torus then by $k_s$-rational conjugacy of maximal $k_s$-tori we see $T(k_s)$ generates a Galois-stable etale $k_s$-subalgebra of ${\rm{Mat}}_n(k_s)$ that descends to an etale maximal commutative $k$-subalgebra $A \subset {\rm{Mat}}_n(k)$. Clearly $T$ is contained in the $k$-torus $\underline{A}^{\times}$ of units of $A$, and this is an equality for dimension reasons. Thus, $A \mapsto \underline{A}^{\times}$ is a bijection between the sets of maximal $k$-tori of ${\rm{GL}}_n$ and the etale maximal commutative $k$-subalgebras of ${\rm{Mat}}_n(k)$. | |
| Feb 16, 2013 at 7:23 | answer | added | Shripad | timeline score: 5 | |
| Feb 16, 2013 at 6:44 | answer | added | Will Sawin | timeline score: 16 | |
| Feb 16, 2013 at 1:11 | comment | added | Venkataramana | I believe Bryant's answer is correct; another way of saying this: let $l/k$ be a quadratic extension with $char k\neq 2$. Treat $l$ as a two dimensional vector space over $k$; then $l^*$ the multiplicative group of $l$ acts on $l$ by multiplication, and is $k$ linear. Hence $l^*$ lies in $GL_2(k)$. There is the norm map $l^*\rightarrow k^*$ and the kernel $N^1_{l/k}$ is a torus which over the algebraic closure, is $K^*$ but over $k$, it is not isomorphic to $k^*$. | |
| Feb 16, 2013 at 0:52 | comment | added | Robert Bryant | I think (though I'm not completely sure) that (at least over a field of characteristic not $2$) the nondiagonalizable tori of the type you mention are the subgroups of $SL2$ that stabilize a quadratic form that does not factor. These correspond to the nontrivial elements of the group $k^\ast/(k^\ast)^2$, i.e., such quadratic forms are, up to a change of basis (i.e., conjugation in $SL2$), of the form $x^2+\lambda y^2$ where $\lambda$ is not a square in $k$, and two such forms $x^2+\lambda_i y^2$ are equivalent (up to change of basis) if and only if $\lambda_1/\lambda_2$ is a square in $k$. | |
| Feb 16, 2013 at 0:29 | history | asked | Jérémy Blanc | CC BY-SA 3.0 |