Let $H$ be a braided Hopf algebra. The multiplication on $H \otimes H$ is defined by $(a \otimes b)(c \otimes d) = a \Psi(b \otimes c) d$, $a,b,c,d \in H$. 


Let $H = T(V)$. There is a algebra map $\Delta: T(V) \to T(V) \otimes T(V)$ such that $\Delta(v) = 1 \otimes v + v \otimes 1$. Therefore
\begin{align}
\Delta(xy) = 1 \otimes xy + xy \otimes 1 + x \otimes y + x\_{(-1)}.y \otimes x\_{(0)}.
\end{align}


Consider the braided commutator in $T(V)$. We have
\begin{align}
x \otimes y - \Psi(x \otimes y) = x \otimes y - x\_{(-1)}.y \otimes x\_{(0)}.
\end{align}


I want to check that $x \otimes y - \Psi(x \otimes y)$ is a primitive element or not.


We have
\begin{align}
& \Delta( x \otimes y - \Psi(x \otimes y) ) \\
& = 1 \otimes xy + xy \otimes 1 + x \otimes y + x\_{(-1)}.y \otimes x\_{(0)} - 1 \otimes (x\_{(-1)}.y) x\_{(0)} \\
& \quad - (x\_{(-1)}.y) x\_{(0)} \otimes 1 - x\_{(-1)}.y \otimes x\_{(0)} - ( x\_{(-1)}.y )\_{(-1)}.x\_{(0)} \otimes ( x\_{(-1)}.y )\_{(0)} \\
& = 1 \otimes xy + xy \otimes 1 + x \otimes y - 1 \otimes (x\_{(-1)}.y) x\_{(0)} \\
& \quad - (x\_{(-1)}.y) x\_{(0)} \otimes 1 - ( x\_{(-1)}.y )\_{(-1)}.x\_{(0)} \otimes ( x\_{(-1)}.y )\_{(0)}
\end{align}


It seems that $x \otimes y \neq ( x\_{(-1)}.y )\_{(-1)}.x\_{(0)} \otimes ( x\_{(-1)}.y )\_{(0)}$. Therefore $x \otimes y - \Psi(x \otimes y) $ is not a primitive element of $T(V)$? Thank you very much.



Are braided commutators primitive elements of a braided Hopf algebra?
Sidenote: In the situation you describe, $V$ has to be a Yetter--Drinfeld module over $H$, i.e. $$h\_{(1)}x\_{(-1)}\otimes h\_{(2)}.x\_{(0)}=(h\_{(1)}.x)\_{(-1)}h\_{(2)}\otimes (h\_{(1)}.x)\_{(0)}$$
has to be satisfied for all $x\in V$.


It is in general not true that braided commutators are primitive elements in a braided Hopf algebra. To examine the expression you derive more closely, we can see that for degree reasons ($x,y$ are degree one elements), the braided commutators are primitive elements in the braided Hopf algebra if and only if
$$x\otimes y=( x\_{(-1)}.y )\_{(-1)}.x\_{(0)} \otimes ( x\_{(-1)}.y )\_{(0)}.$$
This condition means precisely that $\Psi^2(x\otimes y)=x\otimes y$, which is not true for all elements in a general braided Hopf algebra.


For example, for a diagonal braiding $\Psi(x\_i\otimes x\_j)=q\_{ij}x\_{j}\otimes x\_i$, this means that $q\_{ij}q\_{ji}=1$ which is generally not assumed. If there exists a division $I=I\_1 \cup I\_2$ or the index set $I$ for a diagonal basis of $V$ such that $q\_{ij}q\_{ji}=1$ for all $i\in I\_1$ and $j\in I\_2$, then $T(V)$ can be written as a braided tensor product $T(V\_1)\underline{\otimes} T(V\_2)$ for $V\_1$ being the YD-module generated by $x\_i$ for $i\in I\_1$, and $V\_2$ generated by the $x\_j$ for $j\in I\_2$. (see [AS, 4.2])


So having braided commutators as primitive elements is a special situation.


[AS]: Andruskiewitsch, N. and Schneider, HJ.: Finite quantum groups and Cartan matrices, Adv. in Math. 154 (2000), 1–45.


