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How to show that: There is a Lie algebra $\mathfrak g$ of dimension $k = 3$ or $k≥ 5$ iff $\frak g=[g,g]$. Also, why it is possible to choose $\frak g$ such that its center is $0$. However, for the cases where $k = 5$ or $k= 7$ then $\frak g$ can't be semisimple?

Ronald
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  • This seems like an odd phrasing. Do you mean to show the dimensions for which there exist perfect Lie algebras are $3$ and everything $5$ or greater? – Tobias Kildetoft Mar 06 '16 at 20:47
  • Perhaps the following phrasing is better: Let $k\in \mathbb Z^+$. How to show that there is a Lie algebra $\frak g$ of dimension $k ≥ 1$ satisfying $\frak g = [g,g]$ iff $k= 3$ or $k≥ 5$! – Ronald Mar 06 '16 at 20:51
  • Right, that is equivalent to what I said. So you need to rule out dimensions $\leq 2$ and $4$, and find examples for the larger ones. Do you know the dimensions of the smallest couple of simple Lie algebras? – Tobias Kildetoft Mar 06 '16 at 20:53
  • mmm, not really actually, I am still learning basic stuff from Lie algebras, sorry! – Ronald Mar 06 '16 at 20:55
  • There is a hint (but I don't understand it well): for existence when k≥ 5 use the semidirect product and use the irreducible sl(2)-modules of dimension ≥ 2; for non-existence when k= 4 argue that $\frak g$ would have to be simple and then rule this case out using a root space decomposition – Ronald Mar 06 '16 at 20:58
  • Those hints suggest that you really ought to be aware of the dimensions (and in fact structure) of the smallest simple Lie algebras. For the part about semidirect product, given a Lie algebra $L$ and a representation $V$, one can define a Lie bracket on $L\times V$, and if $L$ is simple and $V$ is irreducible, the result is perfect (which you should show). – Tobias Kildetoft Mar 06 '16 at 21:01

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A Lie algebra $L$ satisfying $L=[L,L]$ is called perfect. Every semisimple Lie algebra is perfect, so $\mathfrak{sl}(2)$ is a perfect Lie algebra of dimension $k=3$.

For an example of a perfect Lie algebra that isn't semisimple, take a semisimple $L$ and a nontrivial irreducible representation $V$ of $L$, and define a bracket on $L \times V$ by $$ [(X,v),(Y,u)] := ([X,Y],Xu-Yv). $$ This turns $L \times V$ into a perfect Lie algebra with $\text{Rad}(L \ltimes V) = V$ and center $0$. For example, take $\mathfrak{sl}(2)\ltimes V(m)$ of dimension $m+3\ge 5$, to obtain the claim.

Dietrich Burde
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  • Thanks Dietrich, I have actually some questions: Isn't that in semisimple Lie algebra any representation is completely reducible? Does that contradict taking an irreducible representation $V$ of $L$? Actually, where do we need irreducibility? And why we can't take $m$ to be $1$? Thanks a lot! – Ronald Mar 07 '16 at 01:49
  • @Ronald: look at the definition of completely reducible representation, look at the definition of irreducible representation, and figure out whether they are compatible – YCor Mar 07 '16 at 02:00
  • @YCor Completely reducible: for every submodule $W$ of $V$, there is a complement $P$ such that $V = W ⊕ P$. While Irreducible representation: if $V$ has no $L$-submodules other than ${0}$ and $V$! So if it is irreducible and completely reducible what are $W$ and $V$? Moreover, why we are looking for irreducible representations? – Ronald Mar 07 '16 at 02:23
  • That's a good start: what is $W$? – YCor Mar 07 '16 at 02:39
  • Is not it an $L$-submodule, then it should be $V$ or $0$? – Ronald Mar 07 '16 at 02:49
  • Alright, I almost know the answers of my questions in first comment except Why we can't take $V$ with dim = 1? – Ronald Mar 07 '16 at 08:35
  • For $\dim V=1$ we obtain $\mathfrak{sl_2}(K)\oplus K$ which is not perfect. – Dietrich Burde Mar 07 '16 at 08:49