special relativity - Adjoint representation of the Lorentz group

Is it possible to construct an adjoint representation for the Lorentz group?

Answer

I suspect the impeccable WetSavannaAnimal answer is too abstract and general to satisfy the OP. Let me vulgarize it a bit. The Lorentz group has 6 generators, so your adjoint rep will be a 6-dimensional rep, a set of 6x6 matrices, which satisfy the same Lie algebra (commutators) as the conventional 4x4 matrices transforming the fundamental (x,y,z,t) 4-vector. As an aside, note the curvature form (parity inv. 2-form) of the Lorentz group is also 6-dimensional.

Recall the Lorentz group may be written in a pretty form, that is rotations $J_i\equiv −\epsilon_{imn} M_{mn} /2$ and boosts, $K_i\equiv M_{0i}$, so that $[J_m,J_n] = i \epsilon_{mnk} J_k$ , $[J_m,K_n] = i \epsilon_{mnk} K_k$ , $[K_m,K_n] = -i \epsilon_{mnk} J_k$.

Might further note the important simplification $[J_m +i K_m, J_n −i K_n]$, which permits reduction of the Lorentz algebra to su(2)⊕su(2) and efficient treatment of its associated representations.

Now consider the six 6x6 matrices with the 3 J+iKs on the upper left 3x3 subspace of these 6x6 ones, and the 3 J-iKs on the lower right block, spanned by the 4,5,6 indices. Keep the indices of the upper left block to be the usual 1,2,3; and rename the indices of the lower right block from 1,2,3 to 4,5,6. The commutation relations are then manifest, and the structure constants $f_{mnl}$ are sparse, basically $\epsilon_{ijk}$ for indices (1,2,3) or (4,5,6), and zero otherwise.

So, then, as you probably learned from SU(3), these very structure constants f provide your 6x6 matrices in the adjoint, with one of the 3 indices (taking 6 values) specifying the particular generator of the Lorentz group represented.