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Is the statement "if an operator commutes with every generator of the Lie group, it is a Casimir operator" true? (I'm interested in the case of quadratic Casimir invariants, but any answers about higher-order ones would also be appreciated.) I understand its converse holds (for example, this post is helpful) at least for a quadratic Casimir invariant $C_2$ of a semisimple Lie algebra where $C_2$ is a product of generators contracted with a Killing form: $C_2 = g_{ab}X^aX^b$.

Let's say we choose the basis elements of $su(2)$ to be $T_i \equiv \sigma_i/2\ (i=1,2,3)$. The following commutation relations vanish \begin{align} [T_1^2, T_1] = 0 \end{align} \begin{align} [T_1^2, T_2] = T_1[T_1, T_2] + [T_1, T_2]T_1 = i(T_1T_3 + T_3T_1) = 0 \end{align} \begin{align} [T_1^2, T_3] = T_1[T_1, T_3] + [T_1, T_3]T_1 = -i(T_1T_2 + T_2T_1) = 0 \end{align} because two different Pauli matrices anticommute. I suppose $T_1^2$ is not a Casimir operator, but this result proves it is if the statement above is true so I suspect I'm missing something.

tak
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1 Answers1

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  1. The main point is here that a Casimir invariant of a Lie algebra $L$ belongs to the universal enveloping algebra $$U(L)~=~T(L)/I(L),$$ where $I(L)$ is the smallest Lie algebra ideal [in the tensor algebra $T(L)$] that contains all elements of the form $$t_a\otimes t_b -f_{ab}{}^ct_c~\in~T(L).$$

  2. In fact, by definition a Casimir invariant belongs to the center of $U(L)$.

  3. Concerning OP's example for $L=su(2)$. The quadratic element $t_1\otimes t_1~\in U(L)$ is not an invariant. E.g. the commutator $$ \begin{align} [t_3,t_1\otimes t_1]~=~&[t_3,t_1]\otimes t_1+t_1\otimes [t_3,t_1]\cr ~=~&it_2\otimes t_1+it_1\otimes t_2~\neq~ 0\end{align}$$ is not zero [in the quotient], because it does not belong to the ideal $I(L)$.

Qmechanic
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