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Mar 27, 2019 at 10:35 comment added YCor There's no obvious reduction from connected Lie groups to the easy case of compact connected Lie groups, because taking the abelianization does not commute with "passing to a maximal compact subgroup". To be more concrete, if $G=\mathrm{SL}_2(\mathbf{R})$ then abelianization induces $\mathbf{Z}\to 0$ on $\pi_1$, but for the maximal compact subgroups, it induces the identity of $\mathbf{Z}$.
Mar 27, 2019 at 10:24 history answered Neil Strickland CC BY-SA 4.0