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Apr 25, 2017 at 17:37 vote accept Ninnat Dangniam
Apr 25, 2017 at 3:13 comment added Victor Protsak Yes, although to be completely precise, you should rephrase to "subspace consisting of $K$-fixed vectors means that it is actually a trivial representation". I was glad to have been able to help.
Apr 25, 2017 at 2:42 comment added Ninnat Dangniam Thank you. I see. The trivial $U(n)$-representation (the vacuum) and the $U(n)$-representation on the highest exterior power are not the same, and a subspace being $K$-fixed means that it is actually a trivial representation and not merely any 1-dimensional representation. Is that correct?
Apr 25, 2017 at 2:17 history edited Victor Protsak CC BY-SA 3.0
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Apr 25, 2017 at 1:40 history answered Victor Protsak CC BY-SA 3.0