I'm thinking about the image under the adjoint representation $\mathrm{Ad}$ of the proper (identity connected component) Lorentz group $SO^+(1,3)$. Since this group has a trivial centre (it contains, for example, $SO(3)$ which has a trivial centre), then it must be mapped to itself by the adjoint representation.
However, if I build the image from "first principles" from the Lie algebra's structure co-efficients, I of course get $\mathrm{Ad}(SO^+(1,3))$ equal to the group comprising all finite products from $\exp(\mathfrak{g})$ where I realise the Lie algebra $\mathfrak{g}$ as the six dimensional vector space spanned by the following $6\times6$ monsters:
$$L_j=\left(\begin{array}{cc}1&0\\0&1\end{array}\right) \otimes \hat{S}_j\tag{1}$$ $$K_j=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right) \otimes \hat{S}_j\tag{2}$$
where $\hat{S}_x,\,\hat{S}_y,\,\hat{S}_z$ are the $3\times3$ basis members for the Lie algebra $\mathfrak{so}(3)$. These of course fulfill the same commutation relationships $[L_i,\,L_j] = \epsilon_{i\,j\,k}\, L_k$, $[L_i,\,K_j] = \epsilon_{i\,j\,k}\, K_k$, $[K_i,\,K_j] = -\epsilon_{i\,j\,k}\, L_k$ as the corresponding basis vectors for $\mathfrak{so}^+(1,3)$.
By my argument of the first paragraph, this "first principles" construction must be isomorphic to $SO^+(1,3)$ itself. Can someone construct an explicit isomorphism between the two?