Lax Pairs In Integrability I am working through Dr. Beiserts notes (https://people.phys.ethz.ch/~nbeisert/lectures/IntHS16-Notes.pdf) and have difficulty obtaining the second step in (2.9):
$$\{{\rm tr}L^{k},{\rm tr}L^{\ell}\} = k\ell \hspace{2pt} {\rm tr}_{1,2}(L_{1}^{k-1}L_{2}^{\ell-1}[r_{12},L_{1}]-L_{1}^{k-1}L_{2}^{\ell-1}[r_{21},L_{2}]) = 0.\tag{2.9}$$
I assumed $r_{12}$, $L$ was a Lax pair, and I tried
$$[r_{12},L^{k}] = \frac{d}{dt}L_{1}^{k} = k \hspace{1pt} L^{k-1} \hspace{1pt} L  .$$
This would get me a single $L$-term and not the two in the second part of (2.9).  The $r$-matrix would need to act on separate spaces of $L$ since $L_{1} = L \hspace{1pt} \otimes \hspace{1pt} 1 $ and $L_{2} = 1 \hspace{1pt} \otimes \hspace{1pt} L $.  Does $r_{12}$, $L$ act as a Lax pair?  How does the $r$-matrix act on the different spaces of $L$ to get the second part of (2.9)?
 A: Hints: 


*

*Eq. (2.9) expanded into more steps reads: 
$$ \begin{align}
\frac{1}{k\ell}\{ {\rm tr}L^k,{\rm tr}L^{\ell} \} 
~=~& \frac{1}{k\ell}{\rm tr}_1{\rm tr}_2\{ L_1^k,L_2^{\ell} \} \cr
~=~& {\rm tr}_1{\rm tr}_2\left(L_1^{k-1} \{ L_1,L_2 \} L_2^{\ell-1}\right) \cr
\stackrel{(2.8)}{=}& {\rm tr}_1{\rm tr}_2\left(L_1^{k-1}( [r_{12}, L_1] - [r_{21}, L_2])L_2^{\ell-1}\right) \cr
~=~& {\rm tr}_2\left(L_2^{\ell-1}{\rm tr}_1(L_1^{k-1} [r_{12}, L_1] )\right)\cr
&- {\rm tr}_1\left(L_1^{k-1} {\rm tr}_2([r_{21}, L_2]L_2^{\ell-1})\right) \cr
~=~& \frac{1}{k}{\rm tr}_2\left(L_2^{\ell-1}{\rm tr}_1( [r_{12}, L_1^k] )\right)\cr
&- \frac{1}{\ell}{\rm tr}_1\left(L_1^{k-1} {\rm tr}_2([r_{21}, L_2^{\ell}])\right) \cr
~=~&0.
\end{align} \tag{2.9}
$$

*The answers to OP's other questions, such as a Lax formulation, 
$$ \frac{dL}{dt_n}~=~\{H_n, L\}~=~\{M_n,L\}, $$
$$H_n~:=~{\rm tr}(L^n), \qquad M_n~:=~-n~{\rm tr}_1(L_1^{n-1}r_{21}), $$
can be found in Ref. 2.
References:


*

*N. Beisert, Intro to Integrability, lecture notes, 2017; eq. (2.9).

*O. Babelon, D. Bernard & M. Talin, Intro to Classical Integrable Systems, 2003; sections 2.5 & 2.6.
