# SL$(2, \mathbb{C})$ double cover of SO${}^{+}$(1, 3)

I'm trying to understand why SL$$(2, \mathbb{C})$$ double cover SO$${}^{+}$$(1, 3). I have read the proof in the book Zee "Group theory" (pag 453) and the wikipedia https://en.wikipedia.org/wiki/Lorentz_group, they use very similar reasoning.

In the wikipedia is stated the following:

"These maps are surjective, and kernel of either map is the two element subgroup ±I. By the first isomorphism theorem, the quotient group PSL(2,C) = SL(2,C) / {±I} is isomorphic to SO+(1,3)."

To use the "first isomorphism theorem" first of all you have to provide a surjective homomorphism

$$\phi:$$ SL$$(2, \mathbb{C})\to$$SO$$^{+}(1, 3)$$

However, both in the book and in the wikipedia they doesn't explain how is such homomorphism, and it's not proved that is surjective. They have only proved that the action of SL$$(2, \mathbb{C})$$ on the Minkowski space-time preserves the Minkowski metric.

• yes, if how know how is the map. But they don't provide the map $SL(2,C) \to SO(1,3)$, the map that they mention is the action of SL(2,C) on Minkowsky space-time, which doesn't help to get a mapping $SL(2,C) \to SO(1,3)$ Oct 4, 2021 at 14:37
• Related: physics.stackexchange.com/q/28505/2451 and links therein. Oct 4, 2021 at 15:26
• A proof should be given even in Weinberg Vol 1 chapter 2 Oct 4, 2021 at 23:29
• I have read that part in Weinberg and it seems to me that he doesn't prove the mapping is surjective, he only shows that $SL(2,\Bbb C)$ provide linear transformations that are Lorentz transformations, but he doesn't prove that it provides ALL the transformations in $SO^+(1,3)$ Oct 5, 2021 at 10:45

First of all, we have to find a homomorphism of the form

$$\varphi:\mathrm{Sl}(2,\mathbb{C})\to\mathrm{O}(1,3),$$

which is what you are asking for, if I understand correctly. For this, we first of all define the set of all hermitian $$2\times 2$$ matrices by

$$\mathbb{H}:=\{M\in\mathbb{C}^{2\times 2}\mid M=M^{\dagger}\}.$$

This space is isomorphic to the Minkowski space $$\mathbb{R}^{1,3}$$ via the map

$$\mathbb{R}^{1,3}\to\mathbb{H},(t,x,y,z)\mapsto\begin{pmatrix}t+z&x-iy\\ x+iy&t-z\end{pmatrix}.$$

Using this, we define a map

$$\varphi:\mathrm{SL}(2,\mathbb{C})\to \mathbb{H}^{\mathbb{H}},X \mapsto (M\mapsto XMX^{\dagger}),$$

where $$\mathbb{H}^{\mathbb{H}}$$ denotes the set of all maps from $$\mathbb{H}$$ to $$\mathbb{H}$$. Now, observe that this map preserves the determinant, i.e.

$$\mathrm{det}(XMX^{\dagger})=\mathrm{det}(X)\mathrm{det}(M)\mathrm{det}(X^{\dagger})=\mathrm{det}(M).$$

But now, using the map $$\mathbb{R}^{1,3}\to\mathbb{H}$$ desribed above, you see that $$\mathrm{det}$$ in $$\mathbb{H}$$ is one and the same as the norm squared in $$\mathbb{R}^{1,3}$$, i.e.

$$\Vert (t,x,y,z)\Vert_{1,3}^{2}=t^{2}-x^{2}-y^{2}-z^{2}=\mathrm{det}\begin{pmatrix}t+z&x-iy\\ x+iy&t-z\end{pmatrix}$$

and hence, the map $$\varphi$$ maps elements of $$\mathrm{SL}(2,\mathbb{C})$$ into orthogonal maps from $$\mathbb{R}^{1,3}$$ to $$\mathbb{R}^{1,3}$$, which means that we can view $$\varphi$$ as a map of the form

$$\varphi:\mathrm{SL}(2,\mathbb{C})\to \mathrm{O}(1,3).$$

Furthermore, since this map is continuous, it maps the identity component to the identity component, which means that we have that $$\varphi(\mathrm{SL}(2,\mathbb{C}))\subset\mathrm{SO}^{+}(1,3)$$. In the end, we have found a map of the form

$$\varphi:\mathrm{SL}(2,\mathbb{C})\to \mathrm{SO}^{+}(1,3).$$

What is left to show is surjectivity. For this, you just have to check that all basis elements of $$\mathrm{SO}^{+}(1,3)$$ are obtained by some element of $$\mathrm{Sl}(2,\mathbb{C})$$. As an example, going through all the steps of the definition, you see that

$$\varphi\begin{pmatrix}e^{i\varphi/2}&0\\0&e^{-i\varphi/2}\end{pmatrix}=\begin{pmatrix}1&0&0&0\\0&\cos\varphi&\sin\varphi&0\\0&-\sin\varphi&\cos\varphi&0\\0&0&0&1\end{pmatrix}.$$

It is not too hard to construct the others.

To show that the map $$\varphi$$ is a double cover, we have to use the isomorphism theorem, which states that for a homomorphism $$f:G\to H$$ of groups we have that

$$\mathrm{im}(f)\cong G/\mathrm{ker}(f).$$

In the case $$f$$ is surjective, it follows that

$$H\cong G/\mathrm{ker}(f).$$

An element $$X\in\mathrm{ker}(\varphi)\subset\mathrm{Sl}(2,\mathbb{C})$$ has to satisfy $$MX=XM$$ for all hermitian $$M\in\mathbb{C}^{2\times 2}$$ and the only possibility for this is that $$M\in \{1,-1\}$$. To sum up, we have found the following isomorphism of groups:

$$\mathrm{SO}^{+}(1,3)\cong\mathrm{Sl}(2,\mathbb{C})/\{\pm 1\},$$

which concludes the proof.

By the way, this is exactly the same proof as for the claim that $$\mathrm{SU}(2)$$ douple covers $$\mathrm{SO}(3)$$, which you can find in many textbooks (I guess), when you change some definitions accordingly.

• I only don't get how you concluded that $MX=XM$, if $X \in ker(\phi)$ then $XMX^\dagger = M$ for all hermitian M. From that how you get that $MX=XM$ ? Oct 5, 2021 at 8:23
• My way to prove that X=+-1 is that $XMX^\dagger = M$ in particular mast be true for $M = \begin{pmatrix}r&0\\0&s\end{pmatrix}$ where $r,s \in \Bbb R$. Set $X=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$ and substituting you get the matrix equation can be true only if $\lvert\alpha\rvert = \lvert\delta\rvert = 1$, $\gamma=\beta=0$. Moreover, $det(X)=\alpha\delta=1$ so $\alpha=\delta=1$ or $\alpha=\delta=-1$ Oct 5, 2021 at 8:48
• I just stated the result. Your proof is of course totally fine. Oct 5, 2021 at 15:01
• Regarding surjectivity, $SL(2,C)$ and $SO^+(1,3)$ have the same dimension and are both connected, is that sufficient to conclude the homomorphism is surjective? Feb 22 at 9:36