Explicit isomorphism between ${\rm SL}(2,{\Bbb R})$ and ${\rm SU}(1,1)$

Question

I have heard that there exists an isomorphism of real algebraic groups as in the title. I am asking for an explicit isomorphism.

Motivation: I need such an isomorphism for a calculation of Galois cocycles (with values in some other group).

Answer 1

The elements of $SU(1,1)$ are the matrices of the form$$\begin{bmatrix}\alpha+\beta i&\gamma+\delta i\\\gamma-\delta i&\alpha-\beta i\end{bmatrix}$$with $\alpha,\beta,\gamma,\delta\in\Bbb R$ such that $\alpha^2+\beta^2-\gamma^2-\delta^2=1$. Map each such matrix into$$\begin{bmatrix}\alpha-\delta&-\beta+\gamma\\\beta+\gamma&\alpha+\delta\end{bmatrix}\in SL(2,\Bbb R).$$That will give you your isomorphism.

Answer 2

Chances are, you used to know this (long time ago) when you took a complex analysis class.

You likely knew that every conformal automorphism of the unit disk in the complex plane has the form $$ f(z)= \frac{az+b}{\bar{b}z+ \bar{a}}. $$ If you stare at this formula, you realize that this is just the group $PU(1,1)$. You can also see this by observing that the group $U(1,1)$ preserves the real quadric $$ \{(z,w): z\bar z - w \bar w=0\} $$ Passing to the projectivization, we set $w=1$ and, hence, obtain the equation $z\bar z=1$. Thus, $PU(1,1)$ preserves the unit circle in the complex plane.

You also probably knew that the upper half-plane is conformal to the unit disk and that the group of conformal automorphisms of the former is $PSL(2, {\mathbb R})$, linear-fractional transformations with real coefficients.

Now, it remains to realize that $SU(1,1)$ is a nontrivial 2-fold covering space of $PU(1,1)$, ditto $SL(2, {\mathbb R})$ for $PSL(2, {\mathbb R})$.