Dual Jacobi identity for Lie bialgebra I'm studying Lie bialgebras: https://en.wikipedia.org/wiki/Lie_bialgebra.
I'm a bit confused about the way of writing the so called "dual Jacobi identity". 
On Majid's book "a Quantum group Primer" it is written as
$$\tag{1}
(\delta \otimes 1 )\circ \delta (X_i) + cyclic =0,\,\,\,\,\,X_i \in g 
$$where $\delta:g \to g\otimes g $ is the skew symmetric co-commutator map, $g$ a Lie algebra. 
My question is very simple: 
what is "$+cyclic$" in (1)? I understand the basis dependent identity but I'm unable to write down the other 2 bits in the basis independent version.. I guess one is simply:  $(1 \otimes \delta )\circ \delta (X_i)$ but what can one rotate next? (1) needs to be applied to a single Lie algebra element $X$..
Thank you!
 A: another way to write co-acobi equaton is the following: Consider $g$ as vector space ad $\Lambda g$ (the exterior algebra on $g$) as algebra. From a linear map $\delta:g\to g\wedge g$, you can extend it as a degree 1 derivation (notice $\Lambda g$ is super commutative and free as such, so any super-derivation is determined by its values on $g$. Also, since the degree is given by the tensor degree, $\delta:g\to\Lambda^2g$ is a degree 1 linear map).
So, what is the meaning of extending by superderivation? write using Sweedler-type notation
$$
\delta x = x_1\wedge x_2 
$$
then, the extended derivation (let us call $\delta$ again) is given by
(for $a,b\in g$)
$$
\delta(a\wedge b)=\delta( a)\wedge b - a\wedge \delta b\in\Lambda^3 g
$$
Notice also (always $a,b\in g$)
$$
\delta( a)\wedge b -  a\wedge \delta( b)
=\delta( a)\wedge b -  \delta( b)\wedge a
$$
so, you can write the above formula in terms of $\delta\otimes 1$ and anti-symmetrization maps. 
The full anti-symmetrization map (from $g^{\otimes 3}$ to $\Lambda^3 g$) has 6 terms, but if you already know some partial antisymmetry (e.g. $\sum "a\otimes b"=\delta x\in\Lambda^2 g$ is antisymmetric) then the six terms are actually three terms twice. 
The same happens if you write Jacoby identity as
$$
\sum_{\sigma\in S_3}(-1)^\sigma[[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(3)}] 
=2([[x_1,x_2],x_3]+[[x_2,x_3],x_1]+[[x_3,x_1]x_2])
$$
just because $[x_i,x_j]=-[x_j,x_i]$.
As a final remark, instead of using the description of Majid's book, I prefer the definition
"co-Jacoby is the condition $\delta^2=0$, where $\delta$ is the unique super-derivation on $\Lambda g$ with restriction to $g$ equal to the cobracket."
