I am reading "Group theory and physics" by Sternberg. Ch. 1.2 deals with homomorphism between ${\rm SL}(2,\mathbb{C})$ and the Lorentz group ${\rm L}$, respectively ${\rm L}_0$, the restricted Lorentz group. One defines identification of every point of Minkowski space ${\rm M^4}$ with a group of 2 by 2 Hermitian matrices by $$ \begin{equation*} X = \begin{pmatrix} x_0 + x_3 & x_1-i \,x_2 \\ x_1 + i \,x_2 & x_0 - x_3 \end{pmatrix} \leftrightarrow x= \begin{pmatrix} x_0 \\ x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} \end{equation*} $$
and it is quite simple to show the existence of homomorphism $\phi:{\rm SL}(2,\mathbb{C})\rightarrow L_0$ which is into but I am stuck at showing why it is onto.
The book provides a Lemma below but proving that, I am stuck at their claim that what they denote by $R_2$ is a rotation. (it is obviously transformation of $\mathbb{R}^3$, but how to see that it is orthogonal?)
One idea would be, if there existed some $D\in{\rm SL}(2,\mathbb{C})$, such that its action on the group of hermitian matrices would induce the transformation $\phi(M_r) S B$ on ${\rm M^4}$, then it would necessarily be unitary ($D\in{\rm SU}(2)$ since $\phi(M_r) S B e_0=e_0$) then all is well because unitary matrices induce orthogonal transformation (that i can show). But this would require the knowledge that the homomorphism is onto which is, however, the consequence of the Lemma.
Addenda:
Most of the relevant pages from the book can be found at google books preview.
The lemma
Every proper Lorentz transformation, $B$, can be written as:
$$B = R_1 L_u^z R_2\,,$$
where $R_1$ and $R_2$ are rotations, and $L_u^z$ is a suitable Lorentz boost in the z direction.
Proof:
We can write: $$ \let\oldhat\hat \renewcommand{\vec}[1]{\mathbf{#1}} \renewcommand{\hat}[1]{\oldhat{\mathbf{#1}}} $$ $$ B e_0 = \begin{pmatrix} x_0 \\ \vec{x} \\ \end{pmatrix} \,,$$ where $\vec{x}=x_1 e_1 +x_2 e_2 + x_3 e_3$ and $x_0^2 -||\vec{x}||=1$. We can find a rotation S which rotates the vector $\vec{x}$ to the positive $z$ axis, so $$ SB e_0 = \begin{pmatrix} x_0 \\ 0 \\ 0 \\ ||\vec{x}|| \end{pmatrix} \,.$$
The self-adjoint matrix that corresponds to $S B e_0$ is thus $$ \begin{pmatrix} x_0 + ||\vec{x}|| & 0\\ 0 & x_0 - ||\vec{x}|| \end{pmatrix}\,. $$
Now choose $r$ so that $r^2 = (x_0 + ||\vec{x}||)^{-1} = x_0 - ||\vec{x}||$. (Remember that
$$ (x_0 + ||\vec{x}||)(x_0 - ||\vec{x}||) = x_0^2 - ||x||^2 = 1 $$ .)
Then applying $M_r$ gives $\phi(M_r)SBe_0 = e_0$. Thus $\phi(M_r)SB$ is a rotation; call it $R_2$. We thus have
$$\phi(M_r)SB=R_2$$ or $$B = S^{-1} [\phi(M_r)]^{-1} R_1\,.$$
In Section 1.6 we will prove a theorem due to Euler which asserts that every rotation $R$ in three-dimensional space can be written as a product $$R = R_\theta^z R_\phi^y R_\psi^z$$ that is, as a rotation about the $z$ axis, followed by a rotation about the $y$ axis, followed by a rotation about the $z$ axis again. (The angles $\theta,\,\phi,\,\psi$ are called the Euler angles of the rotation $R$.) Combined with the above lemma, we conclude that every element of the proper Lorentz group can be written as a product of elements of the form $L_u^z$, $R_\theta^z$ and $R_\phi^y$. But each of these is in the image of $\phi$. So, granted Euler's theorem, we conclude that $\phi(SL(2,\mathbb{C}))$ is all of the proper Lorentz group.
The notation:
The action of the group ${\rm SL}(2,\mathbb{C})$ on the set of Hermitian matrices $\chi$ can be defined as: $$ \tilde{X} = A X A^{-1}\,, $$ for $A\in {\rm SL}(2,\mathbb{C})$ and $X\in \chi$.
$M_{e^\tau}$ is the 2 by 2 matrix which induces a boost in $z$ axis with $v=\tanh \tau$: $$ M_{e^\tau}=\begin{pmatrix} e^\tau & 0\\ 0 & e^{-\tau} \end{pmatrix}\,. $$ In the above proof $r=e^\tau$.
