I know all real forms of ${\rm SL}(2,{\Bbb C}$). They are ${\rm SL}(2,{\Bbb R})$ and ${\rm SU}(2)$.
Moreover, ${\rm SL}(2,{\Bbb R})$ is isomorphic to ${\rm SU}(1,1)$. Thus I can say that all real forms of ${\rm SL}(2,{\Bbb C})$
are of the form ${\rm SU}(2,F\_\lambda)$, where $F\_\lambda$ is the diagonal Hermitian form on ${\Bbb C}^2$
with matrix ${\rm diag}(1,\lambda)$, $\lambda$ taking the values 1 and $-1$.



> 
> **Question.** Is it true that any ${\Bbb Q}$-form of ${\rm SL}(2,{\Bbb C})$ is isomorphic to ${\rm SU}(2,F\_{K,\lambda})$,
> where $F\_{K,\lambda}$ is the diagonal Hermitian form on $K^2$ for some quadratic extension $K/{\Bbb Q}$ with matrix ${\rm diag}(1,\lambda)$,
> for some $\lambda\in {\Bbb Q}^\times$ ?
> 
> 
> 



Forms of ${\rm SL}(2)$
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Br{Br}\DeclareMathOperator\U{U}\DeclareMathOperator\disc{disc}\DeclareMathOperator\Nm{Nm}\DeclareMathOperator\diag{diag}\DeclareMathOperator\SU{SU}\DeclareMathOperator\GL{GL}$The general theory tells us that forms of a reductive group $G$ are classified by $H^1$ of the automorphism group, which is a semidirect product $(\text{root datum automorphisms}) \ltimes (\text{inner automorphisms})$. For $\SL\_2 / k$ the root datum has no automorphisms, so the classifying object is $H^1(k, \PGL\_2(\overline{k})) \cong H^2(k, \smash{\overline{k}}^\times)[2] = \Br(k)[2]$. Hence, by standard results on the Brauer group, a form of $\SL\_2$ over a number field is uniquely determined by the set of places at which it's locally non-split, and the only constraint on this set is that it must have even size.


Now, let's look at unitary groups $\U(V)$ over local fields. If $k$ is local and $K$ is a degree 2 étale ring extension of $k$, then $K$ is either $k \oplus k$ or a field extension. In the former case, any Hermitian space wrt $K/k$ is obviously split. In the latter case, the isomorphism class of $V$ is uniquely determined by the discriminant $\disc(V) \in k^\times / \Nm(K^\times)$. For your example, with matrix $\diag(1, \lambda)$, the discriminant is $-\lambda$. When $V$ is split, then one computes that $\SU(V) \cong \SL(2)$, as in the familiar argument for $k = \mathbf{R}$ (of course $\U(V)$ is *not* $\GL(2)$ unless $K = k \oplus k$!)


Conversely: it's non-obvious, but true, that if $V$ is non-split then $\SU(V)$ is also non-split. If it were split, then the Borel subgroup of $G$ would have to have a fixed point in $\mathbf{P}(V)$ by the Borel–Morozov theorem; and the restriction of the Hermitian form to this line has to be zero, hence $V$ is split.


Going back to the global case, if $K = k(\sqrt{\alpha})$, and $\beta = -\lambda$, we have shown that $\SU(F\_{K,\lambda})$ is split locally at $v$ if and only if $(\alpha, \beta)\_{k\_v} = 1$, where $(-, -)\_{k\_v}$ is the Hilbert symbol. In other words, $\SU(F\_{K,\lambda})$ is split locally at $v$ if and only if the quaternion algebra $D = (\alpha, \beta)\_k$ splits there. So $\SU(F\_{K,\lambda})$ is isomorphic to $(D^\times)^{\Nm = 1}$.


Since every quaternion algebra is $(\alpha,\beta)\_{k}$ for some $\alpha$ and $\beta$, we conclude **the answer to the question is "Yes: if $k$ is a number field, every $\overline{k}/k$-form of $\SL\_2 / k$ is isomorphic to such an $\SU(2, F\_{K, \lambda})$ for some quadratic extension $K/k$ and $\lambda \in k^\times$".**


[PS: With a little more work, one should be able to write down explicitly an isomorphism from $\SU(F\_{k(\sqrt{\alpha}),\lambda})$ to the norm 1 elements of $D^\times$ where $D = (\alpha, -\lambda)\_k$, using the fact that $D$ is naturally a 2-dim'l vector space over $k(\sqrt{\alpha})$. But it's a Sunday morning so I'm going to be lazy and not do the exercise.]


