$SU(2)$ is the set of $2\times 2$ complex matrices $A$ satisfying $AA^*=I$ and $\det(A)=1$ where $A^*$ denotes the conjugate transpose of $A$ and $I$ is the identity matrix. I've seen everywhere that the elements of $SU(2)$ can be represented as $\begin{pmatrix}\alpha&\beta\\-\bar\beta&\bar\alpha\end{pmatrix}$ where $\vert\alpha\vert^2+\vert\beta\vert^2=1$, but I've been having a stupidly hard time working out the arithmetic for this claim. I was able to show that if $\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$ is in $SU(2)$ (or even just $U(2)$), then $\vert\alpha\vert=\vert\delta\vert$ and $\vert\beta\vert=\vert\gamma\vert$, but I'm stuck from there. I know I must use the equation $\alpha\delta-\beta\gamma=1$ to obtain the desired result, but I'm failing to do so.
General form of elements of $SU(2)$
lie-groupslinear algebra
Related Solutions
The condition literally says that $A$ commutes with every traceless Hermitian matrix. Since every Hermitian matrix $H$ is the sum of a traceless Hermitian part and a scalar part (i.e. $H=\left(H-\frac{\operatorname{tr}(H)}2I\right)+\frac{\operatorname{tr}(H)}2I$), $A$ must also commute with every Hermitian matrix. It follows that $A$ commutes with every complex matrix $B$, because $B$ can always be written as a complex linear combination of Hermitian matrices: $B=\frac12(B+B^\ast)-\frac i2(iB-iB^\ast)$.
Yes, the $\rho$ you constructed is an isomorphism and it is also true when $\Bbb R$ is replaced with any field of characteristic different from $2$.
This is in fact a well known result which can be proved using quaternion algebras. A reference is Arithmetique des algebres de quaternions by Marie-France Vigneras, Chapitre I, theoreme 3.4, 1). The book is in French though.
In short, for $K$ a field of characteristic different from $2$, we consider the quaternion algebra $Q = \operatorname M_2(K)$ and its trace zero space $Q^0$.
The reduced norm map $n\colon Q^0 \rightarrow K$ is a quadratic form on $Q^0$. Under the basis $$i = \begin{pmatrix} & 1\\ 1 & \end{pmatrix}, j = \begin{pmatrix}-1 & \\ & 1\end{pmatrix}, k = \begin{pmatrix}& 1\\-1&\end{pmatrix}$$ of $Q^0$, the quadratic form $n$ has matrix $\begin{pmatrix}-1 & &\\& -1 &\\& & 1\end{pmatrix}$.
Therefore $\operatorname{SO}_{1, 2}(K)$ can be identified with the group of isometries of the quadratic space $(Q^0, n)$ with determinant $1$, which is again isomorphic to $Q^\times / K^\times$, by a general result on quaternion algebras.
This last group is nothing but $\operatorname{GL}_2(K)/K^\times$, which is $\operatorname{PGL}_2(K)$.
Best Answer
If $A=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}\in SU(2)$ then its determinant is $1$ and so its inverse, which must also equal $A^H$(the transpose of its complex conjugate) is $A^{-1}=\begin{pmatrix}\delta&-\beta\\-\gamma&\alpha\end{pmatrix}$. Therefore $\delta =\overline\alpha$ and $\gamma=-\overline\beta$. But the determinant is $1=\alpha\delta-\beta\gamma=\alpha\overline\alpha+\beta\overline\beta= |\alpha|^2+|\beta|^2$.