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Jan 8, 2019 at 12:36 comment added Paul Broussous This is true for any smooth representation $\pi$: you do not need to suppose it is irreducible. Cf. my post below.
Jan 8, 2019 at 12:35 history edited Paul Broussous
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Jan 8, 2019 at 1:41 vote accept Cooler Panda
Jan 7, 2019 at 20:11 answer added Paul Broussous timeline score: 5
Jan 7, 2019 at 16:03 history edited Cooler Panda CC BY-SA 4.0
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Jan 7, 2019 at 15:47 history asked Cooler Panda CC BY-SA 4.0