Action of Adjoint Map on Lie Algebra

Question

My question is specifically concerning the Lie algebra $su(N)$. Since this is a compact, real and semi-simple Lie algebra, for each element $X\in su(N)$ there exist some $A,B\in su(N)$ such that $$X=[A,B],$$ (see e.g. this post). Now, what I'd like to know is the following: given two elements $X,Y\in su(N)$ with $[X,Y]\neq 0$, can we always find some $A\in su(N)$ such that $X=[A,Y]$? I believe this is equivalent to asking whether $$\sigma:su(N)\times su(N)\rightarrow su(N), (A,X)\mapsto ad_A(X)=[A,X]$$ acts transitively on $su(N)$. I doubt that the answer is positive, since this should require inverting $ad_A$ at some point which shouldn't be possible as it's kernel contains at least $0$ and $A$. But I'm not sure on how to prove/disprove this and would be grateful for enlightment.

Answer 1

Given two elements $X,Y$, there is not always some $A$ with $X=[A,Y]$. For example, consider the case $X=Y$, and for convenience, $L=\mathfrak{su}(2)$, with a basis $(e_1,e_2,e_3)$ and Lie brackets $[e_1,e_2]=e_3, [e_3,e_1]=e_2,[e_2,e_3]=e_1$. Then $[A,e_3]=e_3$ for $X=Y=e_3$ is impossible, since $[L,e_3]=\langle e_1,e_2\rangle$.