HDB1 said:
Dear,
@fresh_42 , I am so sorry, but I have a question here:
about: (b), I need example of it, and I found: upper triangular matrix, let it ##A##, so if we bracket ##A## with itself, we will get strictly upper matrix, which is nilpotent, and then: solvable ideal, but what about the quotient of upper with strictly? what will be the outcome, please,
also please, do you have example of (a) or (c)..
Thanks in advance,
Let's take ##n=4## in your example, i.e. the Lie algebra of upper ##4\times 4## triangular matrices, and its ideal ##I## of strictly upper ##4\times 4## triangular matrices.
$$
\left\{\begin{pmatrix}0&x_{12}&x_{13}&x_{14}\\0&0&x_{23}&x_{24}\\0&0&0&x_{34}\\0&0&0&0\end{pmatrix}\right\} = I \trianglelefteq L = \left\{\begin{pmatrix}x_{11}&x_{12}&x_{13}&x_{14}\\0&x_{22}&x_{23}&x_{24}\\0&0&x_{33}&x_{34}\\0&0&0&x_{44}\end{pmatrix}\right\}
$$
Every element in ##L## can be written as ##X=D+S## where ##D## is a diagonal matrix, and ##S\in I.## You correctly noted that a) ##I## is a nilpotent, and therewith solvable ideal in the solvable algebra ##L.##
a) ##I\trianglelefteq L\;:##
$$
[X,T]=[D+S,T]=\underbrace{[D,T]}_{\in I}+\underbrace{[S,T]}_{\in [I,I]\subseteq I} \text{ for any } X=D+S\in L\, , \,S,T\in I
$$
b) ##I^3=[I,[I,[I,I]]]=\{0\}\,:##
Let ##e_{pq}## be the matrix with a ##1## in position ##(i,j)## i.e. ## i ##-th row and ##j## th column and zeros elsewhere. Then ##[e_{12},[e_{12},[e_{12}+e_{23},e_{23}+e_{34}]]] = [e_{12},[e_{12},e_{13}+e_{24}]]=[e_{12},e_{14}]=0## is the longest expression we can get.
c) ##I^{(2)}=\{0\}\,:##
$$
\left[\begin{pmatrix}0&x_{12}&x_{13}&x_{14}\\0&0&x_{23}&x_{24}\\0&0&0&x_{34}\\0&0&0&0\end{pmatrix},\begin{pmatrix}0&y_{12}&y_{13}&y_{14}\\0&0&y_{23}&y_{24}\\0&0&0&y_{34}\\0&0&0&0\end{pmatrix}\right]=\begin{pmatrix}0&0&z_{13}&z_{14}\\0&0&0&z_{24}\\0&0&0&0\\0&0&0&0\end{pmatrix}
$$
which is abelian, so ##[[I,I],[I,I]]=\{0\}.## I haven't calculated the values for ##z_{ij}## as we are only interested in the shape of the matric, not its values.
d) ##L/I \cong \left\{\begin{pmatrix}x_{11}&0&0&0\\0&x_{22}&0&0\\0&0&x_{33}&0\\0&0&0&x_{44}\end{pmatrix}\right\}## is abelian because ##[L,L]\ni [D+S,D'+S']=\underbrace{[D,D']}_{=0}+\underbrace{[D,S'] +[S,D']+[S,S']}_{\in I}.##
Note that ##I## is the zero in ##L/I## so ##L/I## is abelian, and therefore nilpotent, and therefore solvable.
e) Putting all these together we have:
$$
[L,L] \subseteq I \,\Longrightarrow\, \underbrace{[\underbrace{[\underbrace{[L,L]}_{\subseteq I},\underbrace{[L,L]}_{\subseteq I}]}_{\subseteq [I,I]},\underbrace{[\underbrace{[L,L]}_{\subseteq I},\underbrace{[L,L]}_{\subseteq I}]}_{\subseteq [I,I]}]}_{\subseteq [[I,I],[I,I]]=\{0\}}
$$
The idea behind (b) of the theorem is the following: We can write an element ##X\in L## as a sum of an element ##\bar X\in L/I## and ##S\in I.## ##\bar X## is the diagonal matrix ##D## I began with. Thus
$$
[L,L]=[L/I + I\, , \,L/I +I]=\underbrace{[L/I,L/I]}_{\subseteq L/I} + \underbrace{[L/I,I]+[I,I]}_{\subseteq I}
$$
Since ##L/I ## is solvable, continued multiplication by itself will end up in zero, which is ##I##. But then we are left with an expression that is completely in ##I.## However, ##I## is solvable, too, so continued multiplication by itself will end up in ##\{0\}.##
I'm not quite sure if this answers your question. If we take a higher value of ##n## then only the chains get longer, but the result will be the same. If we take ##n=2## or ##n=3## then the chains are shorter.