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The Bianchi type IV Lie algebra (do they call it L(3,3)? ), $$ [y,z] = 0, \qquad [x,y] = y, \qquad [x, z] = y + z , $$ has the evident adjoint representation by 3×3 matrices.

Would it further have a 2-dimensional rep, 2×2 matrices—analogous to Pauli's for su(2)?

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  • $\begingroup$ See this paper with references to my papers :) The authors also have done this for dimension $3$. I think the nimmal dimesnion of a faithful representation is $3$. $\endgroup$ Commented Jan 14, 2020 at 16:17
  • $\begingroup$ The mistitled Mubarakzyanov ref [7] appears to cover this, but I haven't accessed a translation of the Russian, yet. $\endgroup$ Commented Jan 14, 2020 at 17:57
  • $\begingroup$ You can do this yourself easily. Write operators $f_x,f_y,f_z$ as $2\times 2$ matrices with these relations in unknowns and just compute. Then $f$ will not be faithful. $\endgroup$ Commented Jan 14, 2020 at 19:03

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No. Indeed, after conjugation, the image would be exactly the Lie algebra of upper triangular matrices, which is not isomorphic to yours since it has a nontrivial center.

Actually, over the complex numbers every Lie subalgebra of $\mathfrak{gl}_2$ is isomorphic to $\{0\}$, $\mathfrak{a}$ the 1-dimensional abelian Lie algebra, $\mathfrak{b}$ (the 2-dim non-abelian Lie algebra), $\mathfrak{sl}_2$, or the direct product of one of these three by $\mathfrak{a}$.

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