# About Proof of Engel's Theorem

• A
• HDB1
In summary, the Proof of Engel's Theorem (3.3), p. 13 states that for any Lie algebra L, where L consists of ad-nilpotent elements, we can use induction on the dimension of L to show that L/Z(L) is nilpotent. This is done by taking an element x+Z(L) in L/Z(L) and applying the ad operation, which results in (ad(x+Z(L)))^n(y)=0+Z(L) if n is large enough. The opposite direction of Engel's theorem is also true, as shown by the example of the Heisenberg algebra.
HDB1
in the Proof of Engel's Theorem. (3.3), p. 13:

please, how we get this step:

##L / Z(L)## evidently consists of ad-nilpotent elements and has smaller dimension than ##L##.
Using induction on ##\operatorname{dim} L##, we find that ##L / Z(L)## is nilpotent.

We know that ##L## consists of ad-nilpotent matrices. (Condition of Engel's theorem.)
This means ##(\operatorname{ad}x)^n=0## for every ##x\in L## and some ##n.##

For the induction step, we need ad-nilpotent matrices of a smaller size.
(i) ##Z(L)\neq 0.## (theorem 3.3)
(ii) Since ##(i)## holds we have ##\dim \left(L/Z(L)\right)<\dim (L)##

Proof: Take an element ##x+Z(L)\in L/Z(L).## Then ##\operatorname{ad}(x+Z(L))(y+Z(L))=[x,y]+Z(L).## Thus
\begin{align*}
\end{align*}
Since ##x\in L## is an ad-nilpotent element, we end up with ##(\operatorname{ad}(x+Z(L)))^n(y)=0+Z(L)## if ##n## is only large enough. But that means that ##x+Z(L)\in L/Z(L)## is ad-nilpotent so we can apply the induction hypothesis.

(iv) Induction hypothesis: ##L/Z(L)## is nilpotent.
(v) ##L## is nilpotent by proposition 3.2 (b)

HDB1
Thank you so much, @fresh_42 ,

please, is the opposite direction of Engel's theorem, true? do you have any example of this theorem, please?

HDB1 said:
please, is the opposite direction of Engel's theorem, true?
I think so, let's see. Engel says: all ##\operatorname{ad}X## with ##X\in L## nilpotent, then ##L## is nilpotent.

This is definitely the stronger part because it says that from ##[X,[X,[X,\ldots[X,A]\ldots]]]=0## we can conclude that ##[X,[Y,[Z,\ldots[W,A]\ldots]]]=0.## So turning the direction seems to be trivial.

If ##L## is nilpotent and ##X\in L## then ##\{0\}=L^n=[L,[L,[L,\ldots[L,L]\ldots]]]## and in particular ##[X,[X,[X,\ldots[X,A]\ldots]]]=(\operatorname{ad}^n(X))(A)=0## for all ##A\in L.##

You must learn to use the definitions of those terms. Then many answers will come in naturally.

HDB1 said:
do you have any example of this theorem, please?
Consider the Heisenberg algebra ##\left\{\begin{pmatrix}0&a&b\\0&0&c\\0&0&0\end{pmatrix}\right\}.## Set
$$A=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\, , \,B=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\, , \,C=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}$$
as basis vectors.

HDB1

• Linear and Abstract Algebra
Replies
9
Views
1K
• Linear and Abstract Algebra
Replies
7
Views
1K
• Linear and Abstract Algebra
Replies
15
Views
2K
• Linear and Abstract Algebra
Replies
5
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
1K
• Linear and Abstract Algebra
Replies
19
Views
2K
• Linear and Abstract Algebra
Replies
19
Views
2K
• Linear and Abstract Algebra
Replies
3
Views
1K
• Linear and Abstract Algebra
Replies
4
Views
975
• Linear and Abstract Algebra
Replies
24
Views
2K