Let's consider the space V of pairs of N*N traceless hermitian matrices. For any pair {A,B} of matrices, we can associate the commutator [A,B]=AB-BA .
Let's define the group G as the group of all transformations $V \to V$ that preserves the commutator. That is, for any $g \in G$, and any pair $\{A,B\} \in V$, we have
$$ [A_g , B_g ] = [A,B]$$
where $\{A_g, B_g\} = g(\{A,B\})$
What is this group G? (What is it's dimension? Does it have a well-known name?)
Here is my attempt at answering the question:
We can decompose A and B into a complete basis of hermitian traceless matrices $A= a_n T^n$, $B = b_n T^n$, where $Tr(T_n)=0$ and $Tr(T_n T_m)= \delta_{n m}$, such that $a_n=Tr(T_n A)$.
At the infinitesimal level, I believe then that the transformations are given by
$$ a_n \to a_n + \partial_{b_n} Tr(f(A,B)), \quad b_n \to b_n - \partial_{a_n} Tr(f(A,B)) $$
Where f(A,B) is an arbitrary polynomial in the matrices A and B.
If this is true, I believe that $N^2-1$ independent traces can be formed using these two matrices, such that this group would be of dimension $N^2-1$.
Is this correct? And if yes, what group does it correspond to?
I hope the problem is well-defined enough and I thank you for your help!