Deriving cross product from angular momentum algebra Is it possible to derive:
\begin{equation}
\hat{L}=\hat{r}\times \hat{p}
\end{equation}
from the angular momentum algebra:
\begin{equation}
[\hat{L}_i,\hat{L}_j]=i\ \hbar\ \epsilon_{ijk}\hat{L}_k\ ?
\end{equation}
I have read somewhere that commutation relations of the form
\begin{equation}
[a_i,b_j]=\epsilon_{ijk} c_k
\end{equation}
admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. 
 A: It's not possible to derive the orbital angular momentum $L = r \times p$ from the $\mathfrak{so}(3)$ commutation relations alone, since the spin operator $S$ also fulfills the same commutation relations, but certainly is different from $r \times p$.
A: 
I have read somewhere that commutation relations of the form
  \begin{equation} [a_i,b_j]=\epsilon_{ijk} c_k \end{equation} admit a
  "natural rewriting in terms of cross products", but there weren't any
  details about this statement.

This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is:
$$
\vec J \times \vec J = i\vec J
$$
I.e., the "natural rewriting" refers to rewriting the commutation relations themselves in terms of a cross product, not rewriting a single angular momentum in terms of cross products of other quantities. 
Explicitly, for angular momentum:
$$
\vec J \times \vec J = i\vec J
$$
Thus:
$$
\epsilon_{ijk}J_jJ_k=iJ_i
$$
Thus:
$$
\epsilon_{ilm}\epsilon_{ijk}J_jJ_k=i\epsilon_{ilm}J_i
$$
Thus:
$$
J_lJ_m-J_mJ_l=i\epsilon_{ilm}J_i
$$
I.e.:
$$
[J_l,J_m]=i\epsilon_{lmi}J_i
$$
In terms of your "a", "b", and "c", the analogous "natural rewriting" is:
$$
\vec a\times \vec b + \vec b \times \vec a=2\vec c
$$
You can see this by considering:
$$
(\vec a \times \vec b)_i
=\epsilon_{ijk}a_jb_k=\epsilon_{ijk}(b_ka_j+[a_j,b_k])
$$
$$
=-\epsilon_{ikj}b_ka_j+\epsilon_{ijk}[a_j,b_k]
$$
$$
=-(\vec b \times \vec a)_i+\epsilon_{ijk}\epsilon_{jkl}c_l
$$
$$
=-(\vec b\times\vec a)_i+2\delta_{il}c_l
$$
$$
=(-\vec b\times \vec a +2c)_i
$$
A: $$\begin{align} 
\left[\hat{A}_{i}, \hat{B}_{j} \right] & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm]
\hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i} & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm]
\epsilon_{ijn}\left( \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i}\right) & = \epsilon_{ijn}\epsilon_{ijk}\hat{C}_{k}, \\[3mm]
\epsilon_{ijn}\hat{A}_{i}\hat{B}_{j} - \epsilon_{ijn}\hat{B}_{j}\hat{A}_{i} & = 2\hat{C}_{n}, \\[3mm]
\left(\hat{\vec{A}} \times \hat{\vec{B}}\right)_{n} + \left(\hat{\vec{B}} \times \hat{\vec{A}}\right)_{n} & = 2\hat{C}_{n}, \\[3mm]
\frac{1}{2}\left(\hat{\vec{A}} \times \hat{\vec{B}} + \hat{\vec{B}} \times \hat{\vec{A}} \right) & = \hat{\vec{C}}.
\end{align}$$
due to the fact that $\epsilon_{ijn}\epsilon_{ijk} = 2\delta_{nk}$.
