Split-complixificatinon of Lie algebras and Lie groups The complexification of the Lie algebras and groups over $\mathbb{C}$ is a well-studied topic. However, I can not find any reference to the similar process for "split-complexification" over $\mathbb{D}$ (say over ${}^2\mathbb{R}$). Obviously, the results from both processes shall be different. (Please do not point me to Clifford algebras, as my interest is precisely in their specific representation via Lie groups).

*

*The general problem:

Let $\mathfrak{h}(\mathbb{R})$ be n-dimensional Lie algebra over the reals with the basis $E$ as $$e_1,e_2,..,e_n$$ with given commutator relations. Now consider the "split-complixification" of the algebra $\mathfrak{h}(\mathbb{D})$. Thus, the basis is doubled by adding $E'$: $$e_{n+1},e_{n+1},..,e_{2n}$$
such as $\quad e_{n+k}=j \:e_{k}\quad,k=0,1...n \quad (j^2=1, j \neq 1)$. The commutator relations for $E'$ appear to be the same as for $E$. Thus, algebra $\mathfrak{h}(\mathbb{D})$ is a new 2n-dimensional algebra. Hence, the new basis can generate a new split-complexified Lie group.
$E'$ with $E$ give $[e_{n+1},e_{1}]\neq 0$,$[e_{n+2},e_{2}]\neq 0 $ etc. Though, in case of "two copies" they would commute.


*As an example, the particular problem is Split-biquaternion $\mathbb{H}'(\mathbb{D})$ representation in terms of the Lie groups.

Split-quaternion $\mathbb{H}'(\mathbb{R})$ isomorphic to $Mat(2,\mathbb{R})$ or
$SL(2,\mathbb{R}) \cong SU(1,1)$. Therefore, split-biquaternion is split-complixification of $SL(2,\mathbb{R})$ i.e. matrices with entries of split-complex numbers $SL(2,\mathbb{D})$.
The split-biquaternions is also $\mathbb{H}\otimes\mathbb{D}=\mathbb{H}\oplus\mathbb{H}$, hence the resulting group is also two copies of $SU(2)$.
Then, looking at the properties of $SL(2,\mathbb{D})$ my guess is that there is a homomorphism $\phi$: $SL(2,\mathbb{D}) \to SO(2,2)$.
As always, thanks for any clue for further study.
 A: Let $\mathbb{D}=\mathbb{R}[j]/(j^2-1)$ be the split complex numbers. It is well-known that $\mathbb{D}\cong\mathbb{R}\oplus\mathbb{R}$ as rings, with the isomorphism given by linearly extending $(1,0)\leftrightarrow\frac{1}{2}(1+j)$ and $(0,1)\leftrightarrow\frac{1}{2}(1-j)$ (these are the orthogonal primitive idempotents for its Peirce decomposition).
Thus $\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{D}\cong\mathfrak{g}\otimes(\mathbb{R}\oplus\mathbb{R})\cong(\mathfrak{g}\otimes\mathbb{R})\oplus(\mathfrak{g}\otimes\mathbb{R})\cong\mathfrak{g}\oplus\mathfrak{g}$ as real lie algerbas. The isomorphism is actually the same as above: $$(X,Y)\mapsto \frac{1}{2}(1+j)X+\frac{1}{2}(1-j)Y=\frac{1}{2}(X+Y)+\frac{1}{2}(X-Y)j.$$
Let's also examine this claim:

Split-quaternion $\mathbb{H}'(\mathbb{R})$ isomorphic to $Mat(2,\mathbb{R})$ or
$SL(2,\mathbb{R}) \cong SU(1,1)$. Therefore, split-biquaternion is split-complixification of $SL(2,\mathbb{R})$ i.e. matrices with entries of split-complex numbers $SL(2,\mathbb{D})$.

First off, let's talk about language. There's a difference between a "normed" algebra and its group of elements of norm $1$. You can say $\mathbb{H}'$ is isomorphic to $M_2(\mathbb{R})$ but not to $SL(2,\mathbb{R})$, the latter isn't even an algebra! So here's the language-fixed version of your statement:

Split-quaternions $\mathbb{H}'(\mathbb{R})$ are isomorphic to $M_2(\mathbb{R})$. Therefore, the split-biquaternions $\mathbb{H}\otimes\mathbb{D}$ are the split-complixification of the split quaternions $\mathbb{H}'$.

But this is not true: the split biquaternions are the split-complexification of the normal quaternions, not of the split quaternions! The split biquaternions are isomorphic to $\mathbb{H}^2$ whereas the split-complexification of the split quaternions are isomorphic to $M_2(\mathbb{R})^2$, which is inequivalent.
That said, there is a $2$-to-$1$ surjective homomorphism $SL_2(\mathbb{D})\to SO(2,2)$. Note that $\mathbb{D}\cong\mathbb{R}^2$ so $SL_2(\mathbb{D})\cong SL_2(\mathbb{R})^2$ which acts on $M_2(\mathbb{R})$ via $(A,B)X:=AXB^{-1}$. This preserves the determinant, which is a quadratic form on $M_2(\mathbb{R})$ of signature $(2,2)$.
