# Questions about solvable Lie algebras

• A
• HDB1

#### HDB1

TL;DR Summary
Please, in the book of Introduction to Lie Algebras and Representation Theory J. E. Humphreys p.11, I have a question:

Proposition.
Let ##L## be a Lie algebra.
(a) If ##L## is solvable, then so are all subalgebras and homomorphic images of ##L##.
(b) If ##I## is a solvable ideal of ##L## such that ##L / I## is solvable, then ##L## itself is solvable.
(c) If ##I, J## are solvable ideals of ##L##, then so is ##I+J##.

Please, in proof, (b); how we get:
##\left(L^{(i)}\right)^{(j)}=L^{(i+j)} \text { implies that } L^{(n+m)}=0##

Dear, @fresh_42 , if you could help, I would appreciate it.

Please, in the book of Introduction to Lie Algebras and Representation Theory J. E. Humphreys p.11, I have a question:

Proposition.
Let ##L## be a Lie algebra.
(a) If ##L## is solvable, then so are all subalgebras and homomorphic images of ##L##.
(b) If ##I## is a solvable ideal of ##L## such that ##L / I## is solvable, then ##L## itself is solvable.
(c) If ##I, J## are solvable ideals of ##L##, then so is ##I+J##.

Please, in proof, (b); how we get:
##\left(L^{(i)}\right)^{(j)}=L^{(i+j)} \text { implies that } L^{(n+m)}=0##

You can formally prove ##\left(L^{(i)}\right)^{(j)}=L^{(i+j)}## by induction. However, the heuristic should be sufficient to convince you. What is ##L^{(i)}##? It is ##L## multiplied by itself, then the result multiplied with itself, then this result multiplied with itself and so on, ## i ## times. The same is then done with that result another ## j ## times. But this is the same as starting with ##L## and proceeding ## i+j ## times.

In part (b) of the proof we have that ##I## is solvable and ##L/I## is solvable. The first means ##I^{(n)}=0## for some ##n## and the second means ##\left(L/I\right)^{(m)}=0.##

Now we must look at the definition of the quotient (see the other thread) and remember what the zero in ##L/I## is. It is the equivalence class ##0 + I## which is ##I##. Hence solvability of ##L/I## means that ##\left(L/I\right)^{(m)} \subseteq 0+I=I.## The multiplication in ##L/I## goes
$$[a+I\, , \,b+I]=[a\, , \,b]\, +\,I$$
Combining all that we have ##\underbrace{\left(\underbrace{\left(L/I\right)^{(m)}}_{\subseteq I}\right)^{(n)} }_{\subseteq I^{(n)}=\{0\}}.##

HDB1
Thank you so much, @fresh_42 ,

But, please, how then prove that ##L## is solvable?

Thank you so much, @fresh_42 ,

But, please, how then prove that ##L## is solvable?
##L/I## is solvable. That means multiplying it with itself and so on will end up as ##\bar 0.##
But the ##\bar 0## class in the quotient ##L/I## is ##0+I=I.## So we end up in ##I## after sufficiently many steps. Now, ##I## is also solvable, too. So we proceed with multiplying everything with itself and after another ##n## steps we will be in the real ##\{0\}## subset of ##L, ## i.e. at ##0\in L.## That makes ##L## solvable.

HDB1

Please, in the book of Introduction to Lie Algebras and Representation Theory J. E. Humphreys p.11, I have a question:

Proposition.
Let ##L## be a Lie algebra.
(a) If ##L## is solvable, then so are all subalgebras and homomorphic images of ##L##.
(b) If ##I## is a solvable ideal of ##L## such that ##L / I## is solvable, then ##L## itself is solvable.
(c) If ##I, J## are solvable ideals of ##L##, then so is ##I+J##.

Please, in proof, (b); how we get:
##\left(L^{(i)}\right)^{(j)}=L^{(i+j)} \text { implies that } L^{(n+m)}=0##

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)..

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)..

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.

(a) is given by the definition of a Lie algebra homomorphism. Say we have ##\varphi \, : \,L\longrightarrow \varphi (L).## Then
$$[\varphi (L),\varphi (L)]=\varphi ([L,L]) \; , \;[[\varphi (L),\varphi (L)],[\varphi (L),\varphi (L)]]=\varphi ([L,L],[L,L])$$
and so on. If the chain of ##L's## becomes zero, so will the chain of ##\varphi (L)'s.##

Note that the opposite is not true. If a homomorphic image is solvable, i.e. ##(\varphi(L))^{(n)}=0## then we can only conclude that ##L^{(n)} \subseteq \ker \varphi ## which is in general not zero. In such a case we need the additional condition that ##\ker \varphi =\{0\},## i.e. that ##\varphi ## is injective, or at least that ##\ker \varphi ## is solvable, too.

You can take ##\varphi =\operatorname{ad}## as an important example and check it on the two-dimensional Lie algebra with ##[H,E]=2E## as multiplication.

Subalgebras are even easier. If ##U\subseteq L## then ##[U,U]\subseteq [L,L]## and so on. If the ##L's## becomes zero, so will any subalgebra ##U \trianglelefteq L.##

An example for (c) is the same as we used for (b). Take the diagonal matrices as ##J## and the strictly upper triangular matrices as ##I.## Both are solvable and so is their sum ##L.##

Edit: The diagonal matrices are only a subalgebra. If you insist on ideals, then take only those diagonal matrices that have the same value ##x_{11}=x_{22}=x_{33}=x_{44}.##

Last edited:
malawi_glenn