TITLE: Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?
QUESTION [9 upvotes]: In a Lie algebra $\mathfrak{g}$ the Jacobi identity $\newcommand{\bracket}[2]{\left[#1\,#2\right]} \bracket{x}{\bracket{y}{z}} + \bracket{z}{\bracket{x}{y}} + \bracket{y}{\bracket{z}{x}} = 0$ holds. In the quantized enveloping algebra $\mathrm{U}_q(\mathfrak{g})$ where we define $\bracket{x}{y}_q := xy-qyx$ is there a "nice" most general $q$-analogue to the Jacobi? I've found that
$$
    \bracket{x}{\bracket{y}{z}_q} 
    \;\;=\;\;
    \bracket{\bracket{x}{y}}{z} _q
    + \bracket{y}{\bracket{x}{z}}_q
    \quad\text{ and }\quad
    \bracket{x}{\bracket{y}{z}_q} 
    + \bracket{z}{\bracket{x}{y}_q}
    + \bracket{y}{\bracket{z}{x}_q}
    =0
$$
(which work of any power of $q$ really) but in general, allowing for twistings by different powers of $q$, the cleanest equation I've been able to write is that for any $m,n,k \in \mathbf{Z}$,
$$
    \bracket{x}{\bracket{y}{z}_{q^n}}_{q^m}
    + q^{m+k}\bracket{z}{\bracket{x}{y}_{q^{n-k}}}_{q^{-m-k}}
    + q^{m}\bracket{y}{\bracket{z}{x}_{q^{n-m-k}}}_{q^k}
    =0\,.
$$
But that's not very pretty or symmetric! Is there a better way to write this thing? Or a better way to think about what this $q$-analogue should be? (I'll keep playing with this off-and-on, and edit if I figure anything out)

REPLY [4 votes]: This question might be better understood in the context of the well-known problem of whether there exists a ``quantum Lie algebra'' inside $U_q(\frak{g})$. See this old M.O. question for a discussion.
While there is no consensus about what a ``quantum Lie algebra'' should be, one interesting proposed solution is Majid's notion of a braided Lie algebra, as discussed for example in his Quantum Groups Primer book. Braided Lie algebras are also discussed in the old answer M.O. linked above.