Let $G$ and $H$ be linear algebraic groups over the rationals. Suppose that we know that $G(\mathbb Q)$ and $H(\mathbb Q)$ are isomorphic as abstract groups. Does it under any circumstances follow that $G$ and $H$ are isomorphic as linear algebraic $\mathbb{Q}$-groups?
$\begingroup$
$\endgroup$
18
-
4$\begingroup$ There's a large literature on this, notably by J. Tits, in the 50s and 60s. $\endgroup$– YCorCommented Jan 20, 2022 at 11:49
-
2$\begingroup$ Is $G = SL_3(\mathbb{Q})$ and $H = PSL_3(\mathbb{Q})$ a counterexample? $\endgroup$– spinCommented Jan 20, 2022 at 13:21
-
2$\begingroup$ (Previous discussion of “homomorphismes abstraites”: theorem of Borel and Tits.) $\endgroup$– LSpiceCommented Jan 20, 2022 at 13:22
-
2$\begingroup$ @spin no. Because $\mathrm{PSL}_3(\mathbf{Q})$, in the sense, the $\mathbf{Q}$-points of the $\mathbf{Q}$-group $\mathrm{PSL}_3$ (which to avoid ambiguity should better be called $\mathrm{PGL}_3$), is not reduced to the image of $\mathrm{SL}_3(\mathbf{Q})$. The cokernel of the induced map $\mathrm{SL}_3(\mathbf{Q})\to\mathrm{PGL}_3(\mathbf{Q})$ is the infinite abelian group $\mathbf{Q}^*/{\mathbf{Q}^*}^3$. $\endgroup$– YCorCommented Jan 20, 2022 at 14:05
-
3$\begingroup$ Here are three non-$\mathbf{Q}$-isomorphic tori, for each of which the group of $\mathbf{Q}$-points is isomorphic to the direct product of a group of order 2 and a free abelian group of infinite countable rank: (a) the 1-dimensional split torus $\mathrm{GL}_1$ (b) the group $\mathrm{SO}_2$ (determinant 1 matrices of size 1 preserving the form $q$ — this is an anisotropic 1-dimensional torus (c) the group of similarities of $q$ — this is a 2-dimensional torus, whose $\mathbf{Q}$-points form $\mathbf{Q}[i]^*$. $\endgroup$– YCorCommented Jan 21, 2022 at 1:19
|
Show 13 more comments