How do we prove that a nonzero nilpotent Lie algebra has a nontrivial center?

In summary: We start with the case that ##L^{(0)}=L^{0}##, which is the case that ##Z(L)In summary, the book states that if a Lie algebra is nilpotent, then it is solvable. The proof is by induction, starting with the case that the Lie algebra is solvable.
  • #1
HDB1
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7
Please, in the book of Introduction to Lie Algebras and Representation Theory J. E. Humphreys p.12, I have a question:

Proposition. (3.2)
. Let ##L## be a Lie algebra.

(c) If ##L## is nilpotent and nonzero, then ##Z(L) \neq 0##.

how we prove this,

Thanks in advance,
 
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  • #2
dear @fresh_42 , if you could help, I would appreciate that, :heart:🥹
 
  • #3
Let's start a little tutoring so that you get used to these calculations. Start with what you have.

##L## is nilpotent. Means what?
 
  • #4
if ##
\mathfrak{g}^0=\mathfrak{g}, \mathfrak{g}^1=[\mathfrak{g}, \mathfrak{g}], \mathfrak{g}^2=\left[\mathfrak{g}, \mathfrak{g}^1\right], \ldots, \mathfrak{g}^i=\left[\mathfrak{g}, \mathfrak{g}^{i-1}\right]
##
A Lie algebra ##\mathfrak{g}## is called a nilpotent if ##\mathfrak{g}^n=0##,
 
  • #5
HDB1 said:
if ##
\mathfrak{g}^0=\mathfrak{g}, \mathfrak{g}^1=[\mathfrak{g}, \mathfrak{g}], \mathfrak{g}^2=\left[\mathfrak{g}, \mathfrak{g}^1\right], \ldots, \mathfrak{g}^i=\left[\mathfrak{g}, \mathfrak{g}^{i-1}\right]
##
A Lie algebra ##\mathfrak{g}## is called a nilpotent if ##\mathfrak{g}^n=0##,
Yes. And that means if we write it out
$$
[\mathfrak{g},\underbrace{[\mathfrak{g},[\mathfrak{g},[\mathfrak{g},[\ldots[\mathfrak{g},\mathfrak{g}]\ldots ]]]]}_{=:\mathfrak{g}^{n-1}}]=0
$$
If we pick the smallest ##n## such that ##g^{n}=0## then ##g^{n-1}\neq 0## and we can pick a vector ##x\in \mathfrak{g}^{n-1}\backslash \{0\}.##

What can we say about ##x##?
 
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  • #6
fresh_42 said:
Yes. And that means if we write it out
$$
[\mathfrak{g},\underbrace{[\mathfrak{g},[\mathfrak{g},[\mathfrak{g},[\ldots[\mathfrak{g},\mathfrak{g}]\ldots ]]]]}_{=:\mathfrak{g}^{n-1}}]=0
$$
If we pick the smallest ##n## such that ##g^{n}=0## then ##g^{n-1}\neq 0## and we can pick a vector ##x\in \mathfrak{g}^{n-1}\backslash \{0\}.##

What can we say about ##x##?
##x## will be in the centre, means ##Z(g)\neq 0##?

thanks a lot, :heart:
 
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  • #7
Please, @fresh_42 , what is about this :
If ##\mathfrak{g} / Z(\mathfrak{g})## is nilpotent, then ##\mathfrak{g}## is nilpotent .
I know the idea, but how we can move from to ##\mathfrak{g} / Z(\mathfrak{g})## to ##\mathfrak{g}## to prove it is nilpotent,

Thanks in advance, :heart:
 
  • #8
Please, @fresh_42 , in general how we can prove that every nilpotent is solvable?

the opposite direction is not true, because the upper triangular matrix is nilpotent but not solvable.

Thanks in advance, :heart:
 
  • #9
HDB1 said:
Please, @fresh_42 , what is about this :
If ##\mathfrak{g} / Z(\mathfrak{g})## is nilpotent, then ##\mathfrak{g}## is nilpotent .
I know the idea, but how we can move from to ##\mathfrak{g} / Z(\mathfrak{g})## to ##\mathfrak{g}## to prove it is nilpotent,

Thanks in advance, :heart:
This is again about the quotient's multiplication rules. We have that ##\mathfrak{g} / Z(\mathfrak{g})## is nilpotent, so ##\left(\mathfrak{g} / Z(\mathfrak{g})\right)^n=\{\bar 0\}.## So what is ##\bar 0 \in \mathfrak{g}/Z(\mathfrak{g})##? It is the set ##\bar 0 =0+ Z(\mathfrak{g})=Z(\mathfrak{g}).## We also have
$$
[\mathfrak{g}/Z(\mathfrak{g})\, , \,\mathfrak{g}/Z(\mathfrak{g})]=[\mathfrak{g}\, , \,\mathfrak{g}]+Z(\mathfrak{g})
$$
This means that ##0+Z(\mathfrak{g})=\bar 0=\left(\mathfrak{g} / Z(\mathfrak{g})\right)^n= \mathfrak{g}^n+ Z(\mathfrak{g})## or in other words ##\mathfrak{g}^n\subseteq Z(\mathfrak{g}).## But if ##\mathfrak{g}^n## is in the center of ##\mathfrak{g}## then
$$
\mathfrak{g}^{n+1} =[\mathfrak{g}\, , \,\mathfrak{g}^n]\subseteq [\mathfrak{g}\, , \,Z(\mathfrak{g}^n)]=\{0\}
$$
and ##\mathfrak{g}## is nilpotent.
 
  • #10
HDB1 said:
Please, @fresh_42 , in general how we can prove that every nilpotent is solvable?

the opposite direction is not true, because the upper triangular matrix is nilpotent but not solvable.

Thanks in advance, :heart:
We prove that ##L^{(n)}\subseteq L^{n}.## If the RHS gets zero (nilpotency) for some ##n## so does the LHS (solvability).

The proof is formally by induction.
 
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