Talk:Casimir goes to Casimir

This article was nominated for deletion on 5 Nov 2007. The result of the discussion was redirect to Casimir invariant.

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Latest comment: 16 years ago1 comment1 person in discussion

Any element in $U({\mathfrak {g}})$ commutes with $f$ by definition. What makes the Casimir operator special? SL(2,R) 01:36, 22 July 2007 (UTC)Reply

Don't ask me about notation!