I am deeply confused about this. I have seen a proof of the fact: L semisimple over C implies ad L = Der L but i dont know if the converse is true or false.
Asked
Active
Viewed 300 times
1 Answers
0
The converse is already false for the $2$-dimensional non-abelian Lie algebra $L$. We may assume that the Lie brackets are given by $[x,y]=y$ for a basis $(x,y)$. A short computation shows that $Der(L)=ad(L)$, but of course $L$ is $2$-step solvable, because $[[L,L],[L,L]]=0$. In particular, $L$ is not semisimple.
One can view $L$ as the standard Borel subalgebra of $\mathfrak{sl}_2(\mathbb{C})$, i.e., as upper triangular matrices of size $2$ with trace zero. The $3$-dimensional Lie algebra $L$ of your title does not satisfy that $Der(L)=ad(L)$, because only $\mathfrak{sl}_2(\mathbb{C})$ satisfes this in dimension $3$ (this is well-known).
Dietrich Burde
- 130,978
-
Thank you for clearing the converse part up! Now, do you have any ideas on how to tackle the question in the title ? – user321814 Mar 11 '16 at 14:52
-
The matrices in the title are simply upper triangular, not of trace zero ? Sorry i dont follow – user321814 Mar 11 '16 at 14:53
-
Great! I have been trying to prove this. How would one go about contructing an element of Der(L) that is not in ad(L) ? – user321814 Mar 11 '16 at 14:59
-
By a direct computation (easy, because the Lie algebra has dimension $3$)! – Dietrich Burde Mar 11 '16 at 16:29