Characterizing the adjoint representation

[Luck0]

Let U ∈ SU(N) and {t_a} be the set of generators of su(N), a = 1, ..., N² - 1. The action of the adjoint representation of U on some generator t_a can be written as

Ad(U)t_a = Λ(U)_abt_b

I want to characterize the matrix Λ(U), i. e., I want to see which of its elements are independent. It's known that it belongs to a subspace of SO(N² - 1), and we can derive some constraints by saying that the adjoint action respects the Lie bracket. Here https://mathoverflow.net/questions/179032/characterising-the-adjoint-representation-of-sun it says that the most general characteristic of the adjoint representation is that it preserves some 3-form, but I can't find the details (I don't really know what am I looking for). Does you guys know some place where I can find details about this kind of characterization of the adjoint representation for compact groups? Thanks!

 

[fresh_42]

Let U ∈ SU(N) and {t_a} be the set of generators of su(N), a = 1, ..., N² - 1. The action of the adjoint representation of U on some generator t_a can be written as

Ad(U)t_a = Λ(U)_abt_b

I want to characterize the matrix Λ(U), i. e., I want to see which of its elements are independent. It's known that it belongs to a subspace of SO(N² - 1), and we can derive some constraints by saying that the adjoint action respects the Lie bracket. Here https://mathoverflow.net/questions/179032/characterising-the-adjoint-representation-of-sun it says that the most general characteristic of the adjoint representation is that it preserves some 3-form, but I can't find the details (I don't really know what am I looking for). Does you guys know some place where I can find details about this kind of characterization of the adjoint representation for compact groups? Thanks!

Why do you want to know a matrix with $O(n^4)$ entries and $O(n^2)$ free parameters? The more as you get basically $SU(n)$ again?

 

[Luck0]

Why do you want to know a matrix with $O(n^4)$ entries and $O(n^2)$ free parameters? The more as you get basically $SU(n)$ again?

It turns out that doing my calculations with Λ(U) is way more easy than doing it with U. The only problem is that the difficulty now is identifying the independent components

 

[fresh_42]

It turns out that doing my calculations with Λ(U) is way more easy than doing it with U. The only problem is that the difficulty now is identifying the independent components

$\operatorname{Ad}\; : \; SU(n) \longrightarrow GL(\mathfrak{su}(n))$ is a group homomorphism and $\operatorname{Ad}(u)$ is a regular matrix for any $u \in SU(n)$, which means all $n^2-1$ columns (or rows) of $\operatorname{Ad}(u)$ are linear independent. $\operatorname{Ad}$ itself isn't linear, so no matrix here.