Usually, one decomposes a tensor product whose elements are transformed under a Lie group into its trace part, traceless symmetric part and antisymmetric part to obtain an irreducible representation of the Lie group.
For example, the $\bf{3} \otimes 3$ of $\mathfrak{su}(2)$ is represented by a sum of irreducible reps, ${\bf 1, 3, 5}$. However, I don't understand this result in terms of tensor components.
Let me focus on a specific example to clarify my question. Let $L_i~(i = 1, 2, 3)$ be Hermitian generators of $\mathrm{SO}(3)$ which obey the $\mathfrak{su}(2)$ Lie algebra and $a^{\dagger}_{i}, a_{i}$ be tensor operators which transform as ${\bf 3}$ under the algebra.
In other words, $$ [L_i, a^{\dagger}_{j}] = i \varepsilon_{ijk} a^{\dagger}_{k}, \\ [L_i, a_{j}] = i \varepsilon_{ijk} a_{k}. \\ $$ One can think these $a^{\dagger}_{i}, a_{i}$ as creation/annihilation operators of a three-dimensional isotropic harmonic oscillator.
Then we can construct a symmetric tensor product with the form $a^{\dagger}_{i}a_{j} + a^{\dagger}_{j} a_{i}$.
Since $(\bf{3} \otimes \bf{3})_{\rm sym} = {\bf 5} + {\bf 1}$, we should show that we can derive two operators transform under ${\bf 5}$ and ${\bf 1}$ respectively from the symmetric product.
As I said, we usually find the trace part as ${\bf 1}$ (singlet) and indeed $$ [L_i, \sum_{j} a^{\dagger}_{j} a_{j}] = 0. $$
Thus what we have to do is showing the traceless symmetric tensor $$ \frac{1}{2} (a^{\dagger}_{i}a_{j} + a^{\dagger}_{j} a_{i}) - \frac{1}{3} \delta_{ij} \sum_{k} a^{\dagger}_{k} a_{k} $$ transforms as ${\bf 5}$.
However, calculating the commutator gives us $$ \left[L_i, \frac{1}{2} (a^{\dagger}_{j}a_{k} + a^{\dagger}_{k} a_{j}) - \frac{1}{3} \delta_{jk} \sum_{l} a^{\dagger}_{l} a_{l}\right] = \left[L_i, \frac{1}{2} (a^{\dagger}_{j}a_{k} + a^{\dagger}_{k} a_{j})\right] $$ so that we can check only this is transforms under ${\bf 3} \otimes {\bf 3}$.
Since $e^B A e^{-B} = A + [B, A] + \cdots$, I want to show this transformation leaves its traceless and symmetric properties invariant but how can I do that?
Any comments are appreciated.