I'm working with a Lie algebra that has 3 generator, $t_i$, that satisfy the following rules:
$$[t_2, t_3] = it_2, \quad [t_3, t_1] = it_1, \quad [t_1, t_2] = it_3 \tag1$$
A 3D representation is easy if we choose the adjoint representation, but I want a 2D representation. I'm trying computing the commutators (with general $2\times 2$ matrices) but I got non-sense results like parameter equal its negative or a generator with a zero in all entries. Does anyone know a more efficient method to compute this or know a group with this algebra?
It deals with SU(2). Take for the $t_k$ a representation by Pauli matrices $\sigma_k$ as well explained in the following document :{http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_3.pdf} where $\epsilon_{ijk}$ is the classical alternate tensor ($\epsilon_{ijk}=1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$, $\epsilon_{ijk}=-1$ otherwise). The presence of this tensor explains why the three identities you have given do not look connected by a circular permutation of indices.