Understanding the Second Direction in Semi Simple Lie Algebra: A Guide

[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,

 

[HDB1]

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

Thanks in advance,

 

[fresh_42]

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*}
\left(\operatorname{ad}(X)\operatorname{ad}(Y)\right)^2(Z)&=(\operatorname{ad}(X)\operatorname{ad}(Y))(A)=\underbrace{[X,\underbrace{[Y,A]}_{\in I}]}_{\in [I,I]}=0
\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]

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
$\operatorname{Rad} L=0$

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

thank you so much in advance,

 

[fresh_42]

please, could you explain more on above,

That would be

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:
$$
\varphi \text{ nilpotent } \Longrightarrow \varphi^n=0 \Longrightarrow \operatorname{trace}(\varphi )=0
$$

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*}
(\operatorname{ad}[x,y])(z)&=[[x,y],z]=-[z,[x,y]]=[x,[y,z]]-[y,[x,z-]]\\
&=(\operatorname{ad}x \operatorname{ad}y)(z)-(\operatorname{ad}y\operatorname{ad}x(z)=[\operatorname{ad}x,\operatorname{ad}y](z)
\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
$$
\operatorname{ad}_L[\ldots,[[[S,S],[S,S]],[[S,S],[S,S]]]\ldots]=\{0\}
$$
But $\operatorname{ad}x(y)=[x,y]$ hence
\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.)

my last question, please, when we say
$\operatorname{Rad} L=0$

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

 

[fresh_42]

my last question, please, when we say
$\operatorname{Rad} L=0$

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]

Thank you thank you, @fresh_42

 

[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,

 

[fresh_42]

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]

Thank you so much @fresh_42,