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fresh_42

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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).

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fresh_42

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

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.

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Of course! I was mixing thingsHowever, 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.

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

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