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Apr 16, 2018 at 4:11 vote accept Mehta
Apr 10, 2018 at 12:26 comment added Paul Broussous @L.Spice. OK, I deleted my comment.
Apr 8, 2018 at 6:16 comment added LSpice @PaulBroussous, by "rational character" do you mean "abstract homomorphism $G \to \mathbb C^\times$" (or maybe $G \to F^\times$?) or "homomorphism deduced from a rational map of algebraic groups $\boldsymbol G \to \mathrm{GL}_1$"? If the latter, then note that $\boldsymbol G = \mathrm{PGL}_2$ has no such characters (since, for example, the rational characters of $\mathrm{GL}_2$ are all integer powers of the determinant, none of which is trivial on the centre), so that $G^1 = G$; but that the image of $\mathrm{SL}_2(F)$ is an open, normal subgroup.
Apr 8, 2018 at 2:39 answer added nfdc23 timeline score: 7
Apr 5, 2018 at 9:06 comment added Paul Broussous @nfdc23 Your comment deserves to be an answer !
Apr 4, 2018 at 9:03 history edited YCor CC BY-SA 3.0
added 3 characters in body; edited title
Apr 4, 2018 at 8:31 answer added Uri Bader timeline score: 1
Apr 4, 2018 at 8:05 history edited Mehta CC BY-SA 3.0
edited title
Apr 4, 2018 at 7:27 review First posts
Apr 4, 2018 at 7:31
Apr 4, 2018 at 7:25 history asked Mehta CC BY-SA 3.0