Lie algebra generator relation $T^a T^b \propto T^c $ valid for any $a,b$?

Question

Given a Lie Algebra (such as $su(n), so(n))$ can I always find a set of generators + identity $\{T^a\}\cup \{id\}$ such that there exists a $c$ for any given $a,b$ such that $T^a T^b = C(a,b) T^c $ for a $a,b$-dependent function $C$ into the real/complex numbers?

Answer 1

The answer is a quick no. See this question. Your left-hand-side is in the universal enveloping algebra, not the Lie algebra itself.

You could have spared yourself the question, if you had considered the (presumed) preamble of your text, that surely reminded you of the spin-1, so, adjoint, representation of su(2), namely 3×3 traceless Hermitian matrices, hence a trivial counterexample: $$ S_z=\operatorname {diag} ~ (1,0,-1), \qquad \Longrightarrow \qquad S_z S_z = \operatorname {diag} (1,0,1). $$ Now, this square cannot be written as a linear combination of $S_z$, the identity, which you allowed, and of course $S_x,S_y$ with zero in their diagonals.