An explicit isomorphism between $\mathbb{R}^+ \times {\rm Spin}^c(3,1)$ and ${\rm GL}^+(2,\mathbb{R})\times {\rm GL}^+(2,\mathbb{R})$? I am interested in the following isomorphism
$$
\begin{align}
\mathbb{R}^+\times {\rm Spin}^c(3,1)& \cong \mathbb{R}^+\times {\rm Spin}(3,1) \times {\rm U}(1) \tag{1}\\
&\cong \mathbb{C}^+\times {\rm Spin}(3,1)  \tag{2}
\\&\cong \mathbb{C}^+\times {\rm SL}(2,\mathbb{C}) \tag{3}\\
&\cong {\rm GL}^+(2,\mathbb{C}) \tag{4}\\
&\cong {\rm GL}^+(2,\mathbb{R}) \times {\rm GL}^+(2,\mathbb{R}) \tag{5}
\end{align}
$$
I get 8 parameters on each side. There should therefore be an isomorphism between the two groups.
 A: In the real case we have:
$~$ (a) $\,\mathrm{SL}_2\mathbb{R}\cong\mathrm{Spin}(2,1)$
$~$ (b) $\,\mathrm{GL}_2^+\mathbb{R}\,=\,\mathbb{R}^+\mathrm{SL}_2\mathbb{R}\,\cong\,\mathbb{R}^+\!\times\mathrm{SL}_2\mathbb{R}$
Here, by $HK$ I mean the elementwise product $\{hk\mid h\in H,k\in K\}$ when $H$ and $K$ are subsets of a group. In these examples, each of $H$, $K$, $HK$ will be groups, and elements from $H$ and $K$ will commute (one of the conditions for $HK$ to be a direct product), but $H\cap K$ is $\cong\mathbb{Z}_2$ in many of the following examples (in which case $HK$ is not direct).
In the complex case we have:
$~$ (w) $\,\mathrm{SL}_2\mathbb{C}\,\cong\,\mathrm{Spin}(3,1)$
$~$ (x) $\,\mathrm{GL}_2^+\mathbb{C}\,=\,\mathbb{R}^+\mathrm{SL}_2\mathbb{C}\,\cong\,\mathbb{R}^+\!\times\mathrm{SL}_2\mathbb{C}$
$~$ (y) $\,\mathrm{Spin}^{\mathbb{C}}(3,1) \,\cong\, S^1\mathrm{SL}_2\mathbb{C} \,\cong\, S^1\times_{\mathbb{Z}_2}\mathrm{SL}_2\mathbb{C} \,\cong\, \mathrm{U}(1)\times_{\mathbb{Z}_2}\mathrm{Spin}(3,1)$
$~$ (z) $\,\mathrm{GL}_2\mathbb{C}=\mathbb{C}^\times\mathrm{SL}_2\mathbb{C}\cong\mathbb{C}^\times\times_{\mathbb{Z}_2}\mathrm{SL}_2\mathbb{C}$.
You should think of the middle two, $\mathrm{GL}^+\mathbb{C}$ and $S^1\mathrm{SL}_2\mathbb{C}$, as being enhancements of $\mathrm{SL}_2\mathbb{C}$ by $\mathbb{R}^+$ and $S^1$ respectively, and the last being the "full" enhancement by $\mathbb{C}^\times=\mathbb{R}^+S^1\cong\mathbb{R}^+\times S^1$.
Note $H\times_{\mathbb{Z}_2}K$ means the quotient $(H\times K)/\mathbb{Z}_2$. The context is that $H$ and $K$ have distinguished (central) copies of $\mathbb{Z}_2$ (i.e. $\{\pm1\}$ or $\{\pm I_2\}$), which gives us a "diagonal" copy of $\mathbb{Z}_2$ in $H\times K$ we can quotient by.
In theory it's possible for $H\times_{\mathbb{Z}_2}K$ and $H\times K$ to be isomorphic via a really weird and non-obvious isomorphism. I don't think that's the case in (y) or (z) above, but I don't see a clean argument as to why at the moment.
Those in (w) have real dimension 6, those in (x) and (y) have real dimension 7 (but are those in (x) are not isomorphic to those in (y), or vice-versa), and those in (z) have real dimension 8.
There is no $\mathbb{C}^+$, and $\mathrm{GL}_2\mathbb{R}\times\mathrm{GL}_2\mathbb{R}$ is not isomorphic to anything previously mentioned. I suspect your (4) was supposed to be $\mathrm{GL}_2\mathbb{C}$, which on the lie algebra level has $\mathfrak{gl}_2\mathbb{C}=\mathfrak{gl}_2\mathbb{R}\oplus i\mathfrak{gl}_2\mathbb{R}$ as vector spaces but not as lie algebras, so $\mathrm{GL}_2\mathbb{C}\cong\mathrm{GL}_2\mathbb{R}\times\mathrm{GL}_2\mathbb{R}$ is very badly false.
