Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra? 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)
 A: 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.
A: There are various deformations of the Jacobi identity that can be found scattered in the literature. As far as i know, using the definition: $[A,B]_q=AB-qBA$, one of the most general ones (though i do not now if this is "symmetric" enough for your purposes) is the following one:
$$
\big[A,[B,C]_{q_1}\big]_{q_2}+q_2\big[B,[C,A]_{q_1}\big]_{q_2^{-1}}+\big[C,[A,B]_{q_1q_2}\big]=0
$$
which is valid for arbitrary values of the parameters $q_1$, $q_2$. See [1].
Notice that both identities stated by the OP are compatible with it:

*

*The first one can be recovered for: $q_1=q$, $q_2=1$,

*while a particular case (for $k=-m$) of the second one for: $q_1=q^n$, $q_2=q^m$.

Edit: I was thinking that the second identity of the OP might be combined with the 2-parameter deformation given here to construct a further 3-parameter deformation of the Jacobi identity. But i am not sure on its exact form, i 'll try to think a little on it and come back if something interesting comes up.
Edit 2: I was thinking about the OP's last paragraph (asking for something more pretty or symmetric). Well, i did some further search and it turns out that we can get some symmetric generalization if we consider a two parameter deformation of the commutator. Using $[A,B]_{(p,q)}=pAB-qBA$ we have the following 3-parameter deformation of the Jacobi identity:
$$
{\small
\big[A,[B,C]_{(q_1,q_1^{-1})}\big]_{(q_3/q_2,q_2/q_3)}+\big[B,[C,A]_{(q_2,q_2^{-1})}\big]_{(q_1/q_3,q_3/q_1)}+\big[C,[A,B]_{(q_3,q_3^{-1})}\big]_{(q_2/q_1,q_1/q_2)}=0
}
$$
See: [2].
References:
[1]. $q$-deformed Jacobi identity, $q$-oscillators and $q$-deformed infinite-dimensional algebras, Chaichian, Kulish, Lukierski, Phys. Let. B, v.237, 3-4, p. 401-406, 1990
[2]. On $q$-deformed infinite-dimensional $n$-algebra
Lu Ding, Xiao-Yu Jia, Ke Wu, Zhao-Wen Yan, Wei-Zhong Zhao,  arXiv:1404.0464v3 [hep-th]
