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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.

JDecou
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1 Answers1

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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 $[L,e_3]=\langle e_1,e_2\rangle$.

Dietrich Burde
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  • Thanks for your quick reply. In this case, can we find A s.t. the above is true for X, Y not equal? – JDecou Jun 06 '19 at 19:11
  • No, we still can take, say, $Y=2X$ with $X\neq Y$ and have the same problem. Also, taking $Y=0$ and $X\neq 0$ we cannot find an $A$ with $X=[A,Y]=[A,0]=0$. – Dietrich Burde Jun 07 '19 at 08:00
  • Right, I've been imprecise here. What I meant is, can we find an A if $[X,Y]\neq 0$? I updated the question accordingly. – JDecou Jun 07 '19 at 08:28
  • Or, maybe we even have to demand $\langle X,Y\rangle=0$ (if we choose X,Y in fundamental and use Hilbert-Schmidt inner product for example). In this case you should easily find an $A$ for $su(2)$ at least, but does it work for $N>2$ as well with the above assumption? – JDecou Jun 07 '19 at 11:51