# Surjective homomorphism between ${\rm SL}(2,\mathbb{C})$ and the restricted Lorentz group ${\rm L}_0$

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.

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:


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 $$r=e^\tau$$.

• Review the standard spinor map. Related. – Cosmas Zachos Jan 6 at 22:32
• What is $M_r$? It is not defined in your post and I do not own the book right now. – DanielC Jan 7 at 2:15
• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. – ACuriousMind Jan 7 at 17:22
• Isn't it enough to say that $R_2$ is a Lorentz transformation that leaves $e_0$ alone to get that it's a rotation? – d_b Jan 7 at 19:02
• @d_b Oh, but maybe it is that simple. Since $C$ represents a Lorentz transformation, it has to preserve the norm. Now just showing that if it preserves the norm, it preserves dot product would imply that it is orthogonal... – leosenko Jan 7 at 19:28