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

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!

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fresh_42
Mentor
Let U ∈ SU(N) and {ta} be the set of generators of su(N), a = 1, ..., N2 - 1. The action of the adjoint representation of U on some generator ta can be written as

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?

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
Mentor
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.