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Jul 4, 2019 at 13:21 comment added Jim Humphreys An arbitrary maximal compact subgroup of a Lie group might well be reductive (almost-direct product of a semisimpe group with a compact torus) rather than semisimple. This is a drawback to using maximal compact subgrouips here.
Jul 3, 2019 at 23:04 comment added Ami I'm actually not sure why would K be semisimple...
Jul 3, 2019 at 17:01 comment added Ami Ok let me see if I got it, by Steinberg's lectures(page 126) there is a maximal compact subgroup K of a complex Chevalley group G, and K is Zariski-dense in G. And so the rational irreducible representations of G remain distinct and irreducible on restriction to K. K is a compact semisimple Lie group and so have no nontrivial rational characters and so does G(because different representations have different characters) am I correct?
Jul 3, 2019 at 11:40 history edited Jim Humphreys CC BY-SA 4.0
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Jul 3, 2019 at 11:34 history answered Jim Humphreys CC BY-SA 4.0