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)



Is there a nice q-analogue of the Jacobi identity in a quantized enveloping algebra?
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](https://www.sciencedirect.com/science/article/abs/pii/037026939091196I?via%3Dihub), Chaichian, Kulish, Lukierski, Phys. Let. B, v.237, 3-4, p. 401-406, 1990  

[2]. [On $q$-deformed infinite-dimensional $n$-algebra](https://arxiv.org/abs/1404.0464)
Lu Ding, Xiao-Yu Jia, Ke Wu, Zhao-Wen Yan, Wei-Zhong Zhao, arXiv:1404.0464v3 [hep-th]


