Diagonalization of adjoint representation of a Lie Group

[Luck0]

So, we know that if g is a Lie algebra, we can take the cartan subalgebra h ⊂ g and diagonalize the adjoint representation of h, ad(h). This generates the Cartan-Weyl basis for g. Now, let G be the Lie group with Lie algebra g. Is there a way to diagonalize the adjoint representation Ad(T) of some abelian subgroup T ⊂ G and take the resulting eigenvectors as a basis of g using the Cartan-Weyl basis?

 

[fresh_42]

What about T = exp(h)? Thus we get Ad(T)=Ad(exp(h))=exp(ad(h)). What are your conditions? You'll certainly need some restrictions of the field and maybe on the Lie algebra, too.

 

[Luck0]

What about T = exp(h)? Thus we get Ad(T)=Ad(exp(h))=exp(ad(h)). What are your conditions? You'll certainly need some restrictions of the field and maybe on the Lie algebra, too.

Yes, this is what I thought too but I'm having trouble with the following: in the Lie algebra case, we take the generators of Cartan subalgebra and diagonalize them in the adjoint representation. The other generators (let's call them k) of the algebra are given by eigenvectors of ad(h), something like ad(h)|k> = k|k>. But, since [h,k] = -[k,h], I also obtain a diagonalization of the adjoint representation of the other generators k (right?), so I have a diagonalization of ad(v), for every v ∈ g. I want the same to happen in the Lie group case, that is, I want to find a diagonalization of Ad(U), for every U ∈ G. And I'm not sure if with this exponential thing I can do that. I have to put more thought on it. About my conditions, I want to do this for SU(N).

 

[fresh_42]

About my conditions, I want to do this for SU(N).

This makes things a lot easier: complex number field, simple Lie algebra, so no need to think about characteristic two, centers or nilpotent Lie algebras etc., and the group is compact, simple and a subgroup of $SL(n,\mathbb{C})$ plus the Dynkin diagram is the simplest one: $A_{n-1}$.

However, you do not have a diagonalization for every $\operatorname{ad}X\; , \;X \in \mathfrak{g}$. We have $\mathfrak{su}(n)=\mathfrak{h} \oplus \sum_{\alpha \in \mathfrak{h}^*} \mathbb{C}\cdot E_\alpha$ with eigenvectors $E_\alpha$ for $\mathfrak{h}$. That is $\operatorname{ad}(H)=\alpha(H)\cdot E_\alpha\; , \;H\in \mathfrak{h},\alpha \in \mathfrak{h}^*$. The multilpication of $E_\alpha$ follows the rule $[E_\alpha,E_\beta] \in \mathbb{C}E_{\alpha+\beta}$ which is not diagonal.

The diagonalizable generators are precisely those in $\operatorname{\mathfrak{h}}$. $\mathfrak{h}$ is Abelian and corresponds to $T$, which I suspect stands for toral subgroup, which simply means semisimple which again means simultaneously diagonalizable. It could only be, that some care with the center of $SU(n)$ is necessary, which are the diagonal matrices $\xi \cdot I\; , \;\xi^n=1$ and thus again no big deal.

 

[Luck0]

However, you do not have a diagonalization for every $\operatorname{ad}X\; , \;X \in \mathfrak{g}$. We have $\mathfrak{su}(n)=\mathfrak{h} \oplus \sum_{\alpha \in \mathfrak{h}^*} \mathbb{C}\cdot E_\alpha$ with eigenvectors $E_\alpha$ for $\mathfrak{h}$. That is $\operatorname{ad}(H)=\alpha(H)\cdot E_\alpha\; , \;H\in \mathfrak{h},\alpha \in \mathfrak{h}^*$. The multilpication of $E_\alpha$ follows the rule $[E_\alpha,E_\beta] \in \mathbb{C}E_{\alpha+\beta}$ which is not diagonal.

Of course! I was mixing things

The diagonalizable generators are precisely those in $\operatorname{\mathfrak{h}}$. $\mathfrak{h}$ is Abelian and corresponds to $T$, which I suspect stands for toral subgroup, which simply means semisimple which again means simultaneously diagonalizable. It could only be, that some care with the center of $SU(n)$ is necessary, which are the diagonal matrices $\xi \cdot I\; , \;\xi^n=1$ and thus again no big deal.

Yes, you're probably right, I will give it a try. Thank you!