In what sense can $GL(2,C)$ be seen as a double copy of $GL(2,R)$? It is well known that $GL(2,R)\cong \mathbb{R}\times SL(2,R)$.
I am trying to understand the relationship between $GL(2,R)$ and $GL(2,C)$ via a group product.
For instance, can one create a lie algebra
$$
\mathfrak{gl}(2,R) + i \mathfrak{gl}(2,R)
$$
such that
$$
\exp (\mathfrak{gl}(2,R) + i \mathfrak{gl}(2,R)) = GL(2,C)
$$
?
 A: Given a real lie algebra $\mathfrak{g}$, we can speak of its complexification $\mathfrak{g}_{\mathbb{C}}$. Often $\mathfrak{g}$ is a matrix lie algebra, and if we're lucky $\mathfrak{g}\cap i\mathfrak{g}=0$ and we can set $\mathfrak{g}_{\mathbb{C}}=\mathfrak{g}+i\mathfrak{g}$. The same complex lie algebra can be the complexification of different, inequivalent real lie algebras; we call these real lie algebras the "real forms" of the complex lie algebra. For example, $\mathfrak{sl}(2,\mathbb{C})=\mathfrak{sl}(2,\mathbb{R})\oplus i\mathfrak{sl}(2,\mathbb{R})=\mathfrak{su}(2)\oplus i\mathfrak{su}(2)$ tells us that $\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{su}(2)$ are both real forms of $\mathfrak{sl}(2,\mathbb{C})$. (Some physicists sometimes will write e.g. $\mathfrak{su}(2)$ when they instead mean the complexification $\mathfrak{su}(2)_{\mathbb{C}}=\mathfrak{sl}(2,\mathbb{C})$; beware.) We call $\mathfrak{sl}(2,\mathbb{R})$ the split real form and $\mathfrak{su}(2)$ the compact real form.
Here is some topological insight. Every Lie group $G$ is topologically equivalent to $M\times\mathbb{R}^n$, where $M$ is a maximal compact subgroup (unique up to conjugation), and $\mathbb{R}^n$ is as far from being compact as you can hope for. The dimension is $\dim G=\dim M+n:=m+n$, and we can talk about "how much" of $\dim G$ comes from $m$ versus from $n$. If all of it comes from $m$, then the group is compact. If all of it comes from $n$, then it is quite "expansive." Any complex lie algebra has a compact real form, corresponding to a compact Lie group, and a split real form, corresponding to a Lie group as far from being compact as possible (among real forms for that particular complex lie algebra).
Some extra geometric insight. Consider, for instance, the lie algebra $\mathfrak{gl}(1,\mathbb{C})=\mathbb{C}$ of $\mathrm{GL}(1,\mathbb{C})=\mathbb{C}^\times$. We can say that $\mathbb{C}=\mathbb{R}\oplus i\mathbb{R}$. The image of $\mathbb{R}$ and $i\mathbb{R}$ under the exponential map is $\mathbb{R}^+$ and $S^1$ respectively, and we correspondingly have $\mathbb{C}^\times=\mathbb{R}^+S^1$, an internal direct product. The most "split" one-parameter subgroup is $\mathbb{R}^+$, and the most "compact" one-parameter subgroup is $S^1$, which winds around in a loop. For any other complex number $z$ which is neither real nor purely imaginary, the corresponding one-parameter subgroup $\exp(z t)$ traces out a spiral in the complex plane. (Things can get weird with some other 1p subgroups. Consider irrational flow on a torus $S^1\times S^1$ for example, which is when you use incommensurable angles between the two $S^1$ factors for your 1p subgroup.) But ideally, you can pick out directions in the lie algebra which are most split or most compact, and the corresponding one-parameter subgroups are "expansive" vs. compact, or stretchy vs. windy.
For another example, consider the infinitesimal generators $[\begin{smallmatrix} 0 & -1 \\ 1 & \phantom{-}0\end{smallmatrix}]\in\mathfrak{so}(2)$ and $[\begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix}]\in\mathfrak{so}(1,1)$. The first yields a circular one-parameter subgroup $\cos\theta[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}]+\sin\theta[\begin{smallmatrix}0&-1\\1&\phantom{-}0\end{smallmatrix}]$ and the second yields a hyperbolic one-parameter subgroup $\cosh\sigma[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}]+\sinh\sigma[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}]$. Their ranges, within $M_2(\mathbb{R})$ with the Frobenius / Hilbert-Schmidt norm, are literally a circle and a hyperbola. But if we multiply the lie algebra generators by $i$, the 1p subgroups become $\cosh\theta[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}]+i\sinh\theta[\begin{smallmatrix}0&-1\\1&\phantom{-}0\end{smallmatrix}]$ and $\cos\sigma[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}]+i\sin\sigma[\begin{smallmatrix}0&1\\1&0\end{smallmatrix}]$ respectively; the roles are reversed! Multiplying by $i$ on the matrix lie algebra level tends to "swap" compact/real, so to speak.
By the way, we may as well restrict to $\mathrm{SL}$ rather than $\mathrm{GL}$ for this discussion, because
$$ \begin{cases} \mathfrak{gl}(2,\mathbb{R})=\mathbb{R}\oplus\mathfrak{sl}(2,\mathbb{R}) \\ \mathrm{GL}(2,\mathbb{R})=\mathbb{R}^+\mathrm{SL}(2,\mathbb{R}) \end{cases} $$
$$ \begin{cases} \mathfrak{gl}(2,\mathbb{C})=\mathbb{C}\oplus\mathfrak{sl}(2,\mathbb{C}) \\ \mathrm{SL}(2,\mathbb{C})=\mathbb{C}^\times\mathrm{SL}(2,\mathbb{C}) \end{cases} $$
Both direct sums are as lie algebras, not just as vector spaces, by the way. The first product $\mathbb{R}^+\mathrm{SL}(2,\mathbb{R})$ is an internal direct product, but the second is only approximately so: $\mathbb{C}^\times\mathrm{SL}(2,\mathbb{C})\cong\mathbb{C}^\times\!\times_{\mathbb{Z}_2}\mathrm{SL}(2,\mathbb{C})$, which is called a central product. Now, the vector space direct sum decompositions we can talk about are:
$$ \begin{cases} \mathfrak{sl}(2,\mathbb{C})=\mathfrak{sl}(2,\mathbb{R})\oplus i\mathfrak{sl}(2,\mathbb{R}) \\ \mathfrak{sl}(2,\mathbb{C})=\mathfrak{su}(2)\oplus i\mathfrak{su}(2) \end{cases} $$
Even if these were lie algebra direct sums, we wouldn't expect $i\mathfrak{sl}(2,\mathbb{R})$ to exponentiate to something that looks like $\mathrm{SL}(2,\mathbb{R})$, or $i\mathfrak{su}(2)$ to exponentiate to anything that looks like $\mathrm{SU}(2)$, any more than we should expect xponentiating $i\mathbb{R}\subset\mathbb{C}$ would yield $\mathbb{R}^+$ (remember $\exp(\mathbb{R})=\mathbb{R}^+$ while $\exp(i\mathbb{R})=S^1$ within $\mathbb{C}$). But these vector space decompositions are not lie algebra decompositions - $i\mathfrak{sl}(2,\mathbb{R})$ and $i\mathfrak{su}(2)$ are not subalgebras themselves (not closed under bracket), so $\exp(i\mathfrak{sl}(2,\mathbb{R}))$ and $\exp(i\mathfrak{su}(2))$ will not be groups. They may be interesting topologically, though.
But before we exponentiate these vector space complements, let's talk about what we should expect. The Iwasawa decomposition gives us the decomposition $\mathrm{SL}(2,\mathbb{C})=KAN$, where $K=\mathrm{SU}(2)\cong S^3$ ('K' for 'kompact'), $A\cong\mathbb{R}^+$ has diagonal matrices with reciprocal positive reals ('A' for abelian), and $N\cong(\mathbb{C},+)$ has complex unitriangular matrices ('N' for nilpotent). This $KAN$ is a knit product, meaning the multiplication map $K\times A\times N\to \mathrm{SL}(2,\mathbb{C})$ is a diffeomorphism (but not a group homomorphism, not even close), so it tells us $\mathrm{SL}(2,\mathbb{C})\simeq S^3\times\mathbb{R}^3$ topologically. If we wanted to, we could generalize the concept of "knit products" (or "approximate" knit products, if need be) to not just Lie subgroups but also closed submanifolds of Lie groups.
As $\mathrm{SU}(2)\cong S^3$, we expect something like $\exp(i\mathfrak{su}(2))\simeq \mathbb{R}^3$ topologically. We can also do $\mathrm{SL}(2,\mathbb{R})=KAN$ with $K=\mathrm{SO}(2)\simeq S^1$, $A$ the same, and $N$ only real unitriangulars, so $\mathrm{SL}(2,\mathbb{R})\simeq S^1\times\mathbb{R}^2$ topologically. Since $S^3$ is a $S^1$-bundle over $S^2$ (the Hopf fibration; like a twisted version of $S^2\times S^1$, just as the Mobius strip is a twisted version of $S^1\times[0,1]$), we can expect $\exp( i\mathfrak{sl}(2,\mathbb{R}))$ might topologically look something like $S^2\times\mathbb{R}$. Notice how this has two "compact" dimensions and one "split" dimension, the reverse of $\mathrm{SL}(2,\mathbb{R})\simeq S^1\times\mathbb{R}^2$.
For $X\in\mathfrak{su}(2)$, we may calculate $ \exp(iX)=\cosh(\|X\|)+i\sinh(\|X\|)X/\|X\|$, where $\sigma=\|X\| $ is the Frobenius norm scaled by $1/\sqrt{2}$. However we show this we need to pay attention to how the power series for $\exp,\cos,\sin,\cosh,\sinh$ are all related, but one way is to eigendecompose $X$. Another way is to view $\mathfrak{su}(2)\cong\mathfrak{sp}(1)$ as pure imaginary quaternions and think about the complexified algebra $\mathbb{H}\otimes\mathbb{C}$ (which is how I did it mentally). Since we can solve for $X$ uniquely from $\exp(iX)$, we have $\exp(i\mathfrak{su}(2))\simeq\mathbb{R}^3$ topologically, as anticipated. So it would appear $\mathrm{SL}(2,\mathbb{C})$ is a "knit product" of $\mathrm{SU}(2)$ and $\exp(i\mathfrak{su}(2))$.
For $X=[\begin{smallmatrix}x&y-z\\y+z&-x\end{smallmatrix}]$ I get $\exp(iX)=\cosh(\sqrt{x^2+y^2-z^2})+i\mathrm{sinhc}(\sqrt{x^2+y^2-z^2})X$, which is well-defined since $\cosh$ and $\mathrm{sinhc}$ are even functions. Figuring out the what the range of this is a bit more subtle, maybe you can do it. I'm not sure if what you get looks like $S^2\times\mathbb{R}$ (or, e.g., a twisted version of it) or not.
The takeaway is that $\exp(i\mathfrak{g})$ will not be a subgroup, so for Lie groups $G_{\mathbb{C}}\cong G\times G$ is very false, and topologically the most we might be able to say about $\exp(i\mathfrak{g})$ is that it "swaps" compact/split dimensions from what $G$ has. Disclaimer: I don't know if something like this is true in greater generality, I don't recall seeing a theorem about it. One way $G_\mathbb{C}$ carries on the torch for $G$, so to speak, is that their lie algebras can have the same structure constants (extending any basis for $\mathfrak{g}$), but with more scalars so eigenstuff kicks in from linear algebra (which allows us, for instance, to classify complex semisimple lie algebras).
