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S Jan 9, 2020 at 20:41 history edited F. C. CC BY-SA 4.0
improved formatting in title
S Jan 9, 2020 at 20:41 history suggested Alejandro Tolcachier CC BY-SA 4.0
improved formatting in title
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S Jan 9, 2020 at 20:41
Dec 17, 2009 at 19:00 comment added Samuel Coskey Thanks much for your responses! So the conjugator is in $GL_m(Q_p)$. But does my added hypothesis about carrying $SL_n(Z)$ into $Sl_m(Z)$ help us say anything about the conjugator? Thanks again.
Nov 30, 2009 at 20:26 comment added moonface You're completely right, of course -- what I said above is FALSE and only true "virtually". Thank you for the correction!
Nov 30, 2009 at 19:47 comment added Kevin Buzzard @moonface: regarding editing comments---you can delete them, so you can edit them by cut-and-pasting, editing, and then deleting the original. Regarding mathematics: SL_2(Z_p) maps onto SL_2(Z/pZ) which will I guess have a faithful representation into some SL_N(Z). Now tensor up with the identity map SL_2(Z_p)-->SL_2(Z_p) and don't I have a counterexample to your claim that all embeddings are algebraic (or did I miss something)? My guess is that the Lie algebra argument only proves they're locally algebraic.
Nov 30, 2009 at 17:19 comment added moonface That should be $Q_p$-algebraic groups in the first line, and $GL_m(Q_p)$ in the third, sorry. Can't edit comments.
Nov 30, 2009 at 17:09 comment added moonface Any such is induced from a map of p-algebraic groups $SL_n \rightarrow SL_m$ . You can see this by passage to the Lie algebra. Given such a map, the image of $SL_n(Z_p)$ is conjugate by $GL_m(Z_p)$ to a subgroup of $SL_m(Z_p)$ . However, there can be multiple conjugacy classes of maps inducing the same algebraic morphism.
Nov 30, 2009 at 16:54 history asked Samuel Coskey CC BY-SA 2.5