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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, it is said that the Casimir element is $(h+1)^2/4 + fe$. But $(h+1)^2/4 + fe$ is not a multiple of $ef + fe +h^2/2$. – LJR Apr 20 '11 at 17:52
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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. – BBischof Apr 21 '11 at 5:44
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over the next couple pages it related these two Casimir elements. – BBischof Apr 21 '11 at 7:22

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