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Is the Casimir element of $U(sl_2)$ equal to $ef+fe+h^2/2$ or $(h+1)^2/4+fe$? Is $ef+fe+h^2/2$ equal to $(h+1)^2/4+fe$? How to compute the Casimir element? I think that $ef+fe+h^2/2 = 2fe+(h+1)^2/2-1/2$. Thank you.

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Your first expression is correct. And using the commutation relation you can show that this is also equal to $h^2/2+h+2fe$ and your other element is $1/2(h^2/2+h+2fe)+1/2=1/2(c_2+1)$.

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    thanks. But on page 196 of the book [Introduction to quantum groups](http://books.google.com/books?id=bn5GNkLfnsAC&printsec=frontcover&dq=a+guide+to+quantum+groups&hl=en&ei=jhyvTaSYFOP40gGCu5SmDw&sa=X&oi=book_result&ct=book-thumbnail&resnum=1&ved=0CC8Q6wEwAA#v=onepage&q=casimir&f=false), it is said that the Casimir element is $(h+1)^2/4 + fe$. But $(h+1)^2/4 + fe$ is not a multiple of $ef + fe +h^2/2$.2011-04-20
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    look carefully! That is for $U_q(sl_2)$ which is actually a deformation of $U(sl_2)$. These are different algebras.2011-04-21
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    over the next couple pages it related these two Casimir elements.2011-04-21