# Understanding the Second Direction in Semi Simple Lie Algebra

• A
• HDB1
In summary, Cartan's Criterion states that if the trace of all matrix products in a Lie algebra is zero, then the Lie algebra is solvable. This can be seen in the example of the Heisenberg algebra, where the trace of the matrix product is always zero. This criterion is useful in determining the solvability of a Lie algebra, especially when it is defined by matrices.
HDB1
TL;DR Summary
L is semi simple if it is killing form is nondegenerate
Please, I need some clarifications about second direction, in the file attached,

$$\text { Then ad } x \text { ad } y \text { maps } L \rightarrow L \rightarrow I \text {, and }(\text { ad } x \text { ad } y)^2 \text { maps } L \text { into }[I I]=0 \text {. }$$Thank you in advance,

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

Let ##X\in I, Y,Z\in L.## Then by the Jacobi identity and the fact that ##I## is an ideal we get
$$A:=_{def}\operatorname{ad}(X)\operatorname{ad}(Y)(Z)=[X,[Y,Z]]=-\underbrace{[Y,\underbrace{[Z,X]}_{\in I}]}_{\in I}-\underbrace{[Z,\underbrace{[X,Y]}_{\in I}]}_{\in I} \in I$$
Furthermore, since ##A,X\in I## and ##[I,I]=0## per assumption
\begin{align*}
\end{align*}

To see that a nilpotent transformation ##\varphi ## has trace zero, consider that ##\varphi^n=0## so its characteristic polynomial is ##t^n+0\cdot t^{n-1}=0.## Since the trace is the second-highest coefficient in the characteristic polynomial, it has to be zero.

Was that your question?

Btw., I have the book (James E. Humphreys, Introduction to Lie Algebras and Representation Theory, GTM 9), so it is sufficient to quote the page, chapter, and theorem. No uploads necessary.

HDB1
fresh_42 said:
To see that a nilpotent transformation ##\varphi ## has trace zero, consider that ##\varphi^n=0## so its characteristic polynomial is ##t^n+0\cdot t^{n-1}=0.## Since the trace is the second-highest coefficient in the characteristic polynomial, it has to be zero.
Thank you, @fresh_42 , thank you, I swear I love you, you helped me a lot, I really appreciate everything you did,

please, could you explain more on above,

also please, I get confused about the first direction, I mean in:
$$\text { According to Cartan's Criterion (4.3), ad }{ }_L S \text { is solvable, hence } S \text { is solvable. }$$

my last question, please, when we say
##
##

is that means there is no simple ideal satisfy:
##
[I, I]^n=0##

thank you so much in advance,

HDB1 said:
please, could you explain more on above,
That would be

fresh_42 said:
To see that a nilpotent transformation ##\varphi ## has trace zero, consider that ##\varphi^n=0## so its characteristic polynomial is ##t^n+0\cdot t^{n-1}=0.## Since the trace is the second-highest coefficient in the characteristic polynomial, it has to be zero.

A nilpotent linear transformation ##\varphi ## is one for which there is a natural number ##k## such that ##\varphi^k=0.## This means ##\varphi (\varphi (\varphi (\ldots (\varphi (v)))\ldots) =0## for every ##v\in L.## Starting with a vector that needs the most ##\varphi ## to get killed, then one with the second most, etc., we can build a basis of ##L## such that ##\varphi ## is represented by a strict upper triangular matrix. Therefore, all entries on the diagonal of the matrix of ##\varphi ## are zero. That means that the trace of ##\varphi ## is zero and that ##\det(\varphi -t\cdot I)=(-t)^n## is the characteristic polynomial of ##\varphi.## (The trace is also always the coefficient of the second highest term of the characteristic polynomial, which in case of nilpotent transformations like ##\varphi ## doesn't exist, i.e. equals zero.) Long story short:
$$\varphi \text{ nilpotent } \Longrightarrow \varphi^n=0 \Longrightarrow \operatorname{trace}(\varphi )=0$$
HDB1 said:
also please, I get confused about the first direction, I mean in:
$$\text { According to Cartan's Criterion (4.3), ad }{ }_L S \text { is solvable, hence } S \text { is solvable. }$$
Are you confused about how to apply Cartan's criterion, or about Cartan's criterion itself?

Cartan's criterion says: Consider a Lie algebra ##L## which consists of matrices of a fixed finite size. Assume that the trace of all matrix products ##x\cdot y## is zero, where ##y\in L## is any matrix of ##L## and ##x\in [L,L],## i.e. ##x## can be written as ##x=\sum a_{ij}[a_i,a_j]## for some ##a_k\in L.## Then ##L## is solvable, means ##[\ldots,[[[L,L],[L,L]],[[L,L],[L,L]]]\ldots]=0.## This is the sequence we get if we multiply ##L## with itself, then the product of it with itself, then the result of that product with itself, and so on. Solvable means that this process ends in ##\{0\}.##

Examples are:
the Borel subalgebra of ##\mathfrak{sl}(2)## (solvable)
$$\mathfrak{B(sl}(2))=\left\{\begin{pmatrix}a&b\\0&0\end{pmatrix}\, : \,a,b\in \mathbb{R}\right\}$$
or the Heisenberg algebra (nilpotent and thus in particular solvable)
$$\mathfrak{H}=\left\{\begin{pmatrix}0&a&b\\0&0&c\\0&0&0\end{pmatrix}\, : \,a,b,c\in \mathbb{R}\right\}\\$$

So Cartan's criterion for the Heisenberg algebra says: If
\begin{align*}
\operatorname{trace}&\left(\left[\begin{pmatrix}0&a&b\\0&0&c\\0&0&0\end{pmatrix},\begin{pmatrix}0&u&v\\0&0&w\\0&0&0\end{pmatrix}\right]\cdot \begin{pmatrix}0&x&y\\0&0&z\\0&0&0\end{pmatrix}\right)\\
&=\operatorname{trace}\left(\begin{pmatrix}0&a&b\\0&0&c\\0&0&0\end{pmatrix}\cdot \begin{pmatrix}0&u&v\\0&0&w\\0&0&0\end{pmatrix}\cdot \begin{pmatrix}0&x&y\\0&0&z\\0&0&0\end{pmatrix}-
\begin{pmatrix}0&u&v\\0&0&w\\0&0&0\end{pmatrix}\cdot\begin{pmatrix}0&a&b\\0&0&c\\0&0&0\end{pmatrix}\cdot \begin{pmatrix}0&x&y\\0&0&z\\0&0&0\end{pmatrix}
\right)\\
&=0
\end{align*}
for all ##a,b,c,u,v,w,x,y,z## then ##L=\mathfrak{H}## is solvable.

In short: If ##\operatorname{trace}([A,U]\cdot X)=0## then ##L## is solvable.

Cartan's criterion gives us a sufficient condition for Lie algebras built by matrices - and all finite-dimensional Lie algebras over fields of characteristic zero (##\mathbb{Q},\mathbb{R},\mathbb{C},##etc.) are matrix algebras (Ado's theorem) - to determine solvability by looking if the traces of
$$\operatorname{trace}\left([A,U]\cdot X\right)=\operatorname{A\cdot U\cdot X- U\cdot A\cdot X}=0$$
vanish for any choice of ##A,U,X.##

Hence, given a Lie algebra defined as matrices, e.g. simple and not solvable ##\mathfrak{sl}(2)## or solvable ##\mathfrak{H},## compute that monster ##A\cdot U\cdot X- U\cdot A\cdot X## and add the diagonal entries. I would use https://www.symbolab.com/solver/matrix-calculator for that, although the pre-phone layout of this website was better.

That is the Criterion. I leave it as that in order to answer your question about theorem 5.1. In case you have a question about Cartan's criterion, please open a new thread.

In the proof of theorem 5.1 (1st direction), we assume that ##L## is semisimple, i.e. its (solvable) radical ##\operatorname{Rad}(L)=\{0\}## is zero. But forget this for a moment. We start new by considering
$$S:=_{def}\operatorname{Rad}(K)=\left\{x\in L\, : \,K(x,y)=0\text{ for all }y\in L\right\}$$
If we choose an ##x\in S## then ##K(x,y)=0## for all ##y\in L## per definition. Hence we have
$$0=K(x,y)=\operatorname{trace}(\operatorname{ad}x \cdot \operatorname{ad}y)$$
for all ##y\in [L,L]\subseteq L.## But ##\operatorname{ad}z## are matrices, building the Lie algebra ##\operatorname{ad}L## of all such matrices with elements ##z\in L.## But for matric algebras, Cartan's criterion applies. We have ##K(x,y)=0## for all ##y\in L## so in particular for all ##y\in [L,L].## Thus, by Cartan, we have that the Lie algebra of matrices ##\operatorname{ad}_L S## is solvable. The index ##L## only notes that we still multiply in ##L## which determines e.g. the size of the matrices ##\operatorname{ad}_L x,## but ##x\in S## and ##\operatorname{ad}_L S## is solvable by Cartan's criterion. (We silently assumed that ##S## is a Lie algebra. We even need in a moment that it is an ideal of ##L##. You might want to check this!)

Now go back to what solvable means and what ##\operatorname{ad}## means. We have with the Jacobi identity
\begin{align*}
\end{align*}
So ##\operatorname{ad}## is a Lie algebra homomorphism, ##\operatorname{ad}[x,y]=[\operatorname{ad}x,\operatorname{ad}y],## i.e. it walks into and out of the brackets.

Since ##\operatorname{ad}_L S## is solvable, we know that
$$[\ldots,[[[\operatorname{ad}_L S,\operatorname{ad}_L S],[\operatorname{ad}_L S,\operatorname{ad}_L S]],[[\operatorname{ad}_L S,\operatorname{ad}_L S],[\operatorname{ad}_L S,\operatorname{ad}_L S]]]\ldots]=\{0\}$$
This means by pulling out ##\operatorname{ad}## that
$$\operatorname{ad}_L[\ldots,[[[S,S],[S,S]],[[S,S],[S,S]]]\ldots]=\{0\}$$
\begin{align*}
(\operatorname{ad}_L[\ldots,[[[S,S],[S,S]],[[S,S],[S,S]]]\ldots])(y)&=0 \text{ for all }y\in S\\
&\Longrightarrow \\
[[\ldots,[[[S,S],[S,S]],[[S,S],[S,S]]]\ldots],y]&=\{0\}\text{ for all }y\in S\\
&\Longrightarrow \\
[[\ldots,[[[S,S],[S,S]],[[S,S],[S,S]]]\ldots],y]&=\{0\}\text{ for all }y\in [[\ldots,[[[S,S],[S,S]],[[S,S],[S,S]]]\ldots]\\
[[\ldots,[[[S,S],[S,S]],[[S,S],[S,S]]]\ldots] &=\{0\}
\end{align*}
and ##S## is solvable per definition of solvability. Next we need that ##S\subseteq L## is an ideal. I hope you have checked it

As a solvable ideal of ##L## it is contained in the maximal solvable ideal of ##L,## which is the radical of ##L.## Now we go back to our initial assumption that ##L## is semisimple. This means the maximal solvable ideal of ##L## is zero. So ##S\subseteq \operatorname{Rad}(L)=\{0\}.##

Finally, by definition of ##S,## the Killing-form of ##L## is non-degenerate.

(I wonder how many tiny steps are included in these proofs. I will answer the rest in a separate post.)

HDB1 said:
my last question, please, when we say
##
##

is that means there is no simple ideal satisfy:
##
[I, I]^n=0##

HDB1
HDB1 said:
my last question, please, when we say
##
##

is that means there is no simple ideal satisfy:
##
[I, I]^n=0##
Yes, but drop simple. It means there is no ideal satisfying ##I^{(n)}=\{0\}.##

The crucial point however is how ##I^n ## and ##I^{(n)}## are defined. These are different! Humphreys defines nilpotency by ##I^n=\{0\}## and solvability by ##I^{(n)}=\{0\}.##

In detail:
\begin{align*}
L^0=L\, , \,L^1=[L,L]\, , \,L^2=[L,[L,L]]\, , \,L^n=[L,L^{n-1}]&\quad \text{ nilpotent }\\
[L,[L,[L,[\ldots,[L,L]\ldots]]]]=\{0\}&\\
&\\
L^{(0)}=L\, , \,L^{(1)}=[L,L]\, , \,L^{(2)}=[[L,L],[L,L]]\, , \,L^n=[L^{(n-1)},L^{(n-1)}]&\quad \text{ solvable }\\
[\ldots,[[[L,L],[L,L]],[[L,L],[L,L]]]\ldots]=\{0\}&
\end{align*}

HDB1
Please, @fresh_42 , in theorem 5.2 page: 23, about semi simple:
how we apply Cartan criterian on ##I \cap I^{\perp}##,

Thank you very much,

HDB1 said:
Please, @fresh_42 , in theorem 5.2 page: 23, about semi simple:
how we apply Cartan criterian on ##I \cap I^{\perp}##,

Thank you very much,
It's a typo. It should be ##I\cap I^\perp.##

I will use the letters ##I,J,L,I^\perp## and write e.g. ##[L,I]\subseteq I## instead of "Let ##x\in L\, , \,y\in I ## be arbitrary elements, then ##[x,y]\in I## since ##I## is an ideal in ##L.##" So ##[J,J]## means ##[a,b]## for all ##a,b\in J.##

Let ##I \trianglelefteq L## be an ideal and ##I^\perp=\{x\in L\,|\,K(x,I)=0\}=\{x\in L\,|\,K(x,y)=0\text{ for all }y\in I\} .##

We have for the Killing-form that ##K([x,y],z)=K(x,[y,z])## holds for any ##x,y,z \in L.## If ##x\in I^\perp## then ##K([x,L],I)=K(x,[L,I])=K(x,I)=0## and ##[x,L]\subseteq I^\perp## proving that ##I^\perp \trianglelefteq L## is an ideal and therewith ##J:=_{def}\;I \cap I^\perp \trianglelefteq L## an ideal, too.

We have in particular that ##0=K([J,J],J)=\operatorname{trace}(\operatorname{ad}[J,J]\circ\operatorname{ad}J)## since ##J## is part of ##I## as well as of ##I^\perp.## Now, we can apply the Corollary 4.3. of Cartan's criterion and conclude that ##J## is solvable. Hence ##J\subseteq Rad(L)## is included in the maximal solvable ideal of ##L## which, by definition of semisimplicity is ##\{0\}.##

We already know (theorem 5.1.) that the Killing-form of ##L## is nondegenerate. This means that ##\dim I+\dim I^\perp =\dim L## and ##L=I\oplus I^\perp ## is a direct sum of ideals (their intersection ##J## is zero).

HDB1

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