I'm not sure which is the exact definition of a Casimir operator.
In some texts it is defined as the product of generators of the form: $$X^2=\sum X_iX^i$$
But in other parts it is defined as an operator that conmutes with every generator of the Lie group.
Are these definitions equivalent? If the answer is yes, how could I prove it (I'm thinking in using Jacobi's identity)?
Answer
I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC.
Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then $$ [t_a,t_b] = C_{ab}{}^c t_c$$ defines the structure constants. If you define $$ g_{ab} = C_{ad}{}^e C_{be}{}^d,$$ then this gives you an inner product $$(X,Y) := g_{ab} x^a y^b, \quad X = x^a t_a \text{ and } Y = y^b t_b.$$ Indeed this "Killing form" is related to the adjoint representation, as $$(X,Y) = \text{tr}(\text{ad } X \text{ ad} Y)$$ (exercise!). Similarly, $$g_{ab} =\text{tr}(\text{ad } t_a \text{ ad } t_b).$$ In this language, the Casimir $c_2$ is given by $$ c_2 = g^{ab} t_a t_b, \qquad \text{ so}$$ $$[c_2,t_e] = g^{ab} [t_a t_b,t_e].$$ Now you need to do some basic work (expand the first factor of the commutator, work out the resulting brackets) and you'll see that this gives you $$ \ldots = g^{ab} g^{dk} C_{bek} \{ t_a,t_d \}.$$ This vanishes (why?), so you're done!
Edit (regarding Peter Kravchuk's remark): when you write $c_2 \sim t_a t_b$, it's not really part of the Lie algebra. The only multiplication that "works" in Lie algebras is the commutator $[t_a,t_b]$. So these guys live in some richer structure, which is called the "universal enveloping algebra." Indeed you often hear that "the Casimir is a multiple of the identity matrix," but the identity matrix is seldom part of the Lie algebra (the identity in a Lie algebra is 0). In practice everything is self-evident, because you do calculations in some vector space.
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