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Sep 10, 2016 at 15:01 answer added William of Baskerville timeline score: 1
Aug 31, 2016 at 18:30 answer added Qiaochu Yuan timeline score: 2
Aug 31, 2016 at 18:19 comment added Qiaochu Yuan But if $G$ is simple in addition to being compact and connected then $\pi_1(G)$ is finite, so there also aren't any morphisms $\pi_1(G) \to \mathbb{Z}$ either...
Aug 31, 2016 at 14:12 comment added Jens Reinhold Since the circle is abelian, every morphism $G \to \mathbb T^1$ factors through the abelianization of $G$, so if $G$ is simple every such morphism is constant!
Aug 31, 2016 at 13:32 comment added user1688 It's wrong for non-compact groups, $SL_2({\mathbb R})$ being a counter example.
Aug 31, 2016 at 10:23 history asked William of Baskerville CC BY-SA 3.0