Let $(E,g)$ be the space decribed in exam. 1. For all $X\in\Sl(2,\C)$ the linear map $A\mapsto XAX^*$ is an isometry of the Lorentz space $(E,g)$. 2. The map $\Phi:X\mapsto(A\mapsto XAX^*)$ from $\Sl(2,\C)$ into the group $G$ of isometries of $(E,g)$ is a homomorphism, i.e. $\Phi(XY)=\Phi(X)\Phi(Y)$. 3. For $X\in\SU(2)$ $\Phi(X)$ has a time-like eigenvector with eigenvalue $1$. 4. For $X$ positive $\Phi(X)$ is a boost.
1. If $X\in\Sl(2,\C)$, then $\det X=1$ and thus for all $A\in E$: $$ g(XAX^*,XAX^*) =-\det(XAX^*) =-\det(X)\det(A)\cl{\det(X)} =-\det(A)=g(A,A)~. $$ 2. For all $A\in E$: $$ \Phi(XY)(A)=XYA(XY)^* =XYAY^*X^* =\Phi(X)(YAY^*) =\Phi(X)(\Phi(Y)(A)) $$ 3. For $X\in\SU(2)$ we have $XX^*=\s_0$ and thus the identity matrix $\s_0$ is an eigen vector with eigen value $1$. Since $g(\s_0,\s_0)=-\det(\s_0)=-1$, $\s_0$ is a time-like eigen vector.
4. Suppose $X$ is positive and $\det X=1$, then $X=UDU^*$ for some $U\in\SU(2)$ and $D=diag\{\l,1/\l\}$, $\l > 0$. $$ \Phi(X)=\Phi(UDU^*)=\Phi(U)\Phi(D)\Phi(U)^{-1} $$ By 3. $\Phi(U)$ is a rotation in $\s_0^\perp$ and thus we simply need to compute $\Phi(D)$: $$ \Phi(D)A =DAD^* =\left(\begin{array}{cc} \l^2 x_0&x_2+ix_3\\ x_2-ix_3&x_1/\l^2 \end{array}\right) =\frac{\l^2x_0+x_1/\l^2}2\s_0 +\frac{\l^2x_0-x_1/\l^2}2\s_3 +x_2\s_1 +x_3\s_2 $$ and in particular $$ \Phi(D)\s_0=\frac{\l^2+1/\l^2}2\s_0+\frac{\l^2-1/\l^2}2\s_3,\quad \Phi(D)\s_3=\frac{\l^2-1/\l^2}2\s_0+\frac{\l^2+1/\l^2}2\s_3,\quad \Phi(D)\s_1=\s_1,\quad \Phi(D)\s_2=\s_2~. $$ This shows that $\Phi(D)$ is a boost in the direction $\s_3$ with $2\cosh(\vp)=\l^2+1/\l^2$. Hence $\Phi(X)$ is a boost in the direction $\Phi(U)\s_3$.