# Matrices belonging to restricted Lorentz group $SO^+(1,3)$

My Professor says that all members of the restricted Lorentz group $$SO^+(1,3)$$ may be written as $$e^\Gamma$$, where $$\Gamma^{\mu}{}_{\nu}=\Lambda^{\mu \rho} \eta_{\rho \nu}.$$ Here $$\Lambda$$ is an antisymmetric Matrix, and $$\eta$$ is the standard Minkowski metric [$$diag(1,-1,-1,-1)$$]. I want to prove a weaker statement. That all matrices of the said form belong to $$SO^+(1,3)$$. I have been able to show that they belong to $$SO(1,3)$$. However, the orthochronous bit is troubling me. Any help is appreciated.

Progress so far: $$(\Lambda \eta)^T=-\eta \Lambda$$ $$\implies (e^{\Lambda \eta})^T=e^{-\eta \Lambda}$$ $$\eta e^{-\eta \Lambda} \eta=e^{-\Lambda \eta}$$
$$\eta e^{-\eta \Lambda} \eta e^{\Lambda \eta}=I$$ $$e^{-\eta \Lambda} \eta e^{\Lambda \eta}=\eta$$ $$\implies (e^{\Lambda \eta})^T\eta e^{\Lambda \eta}=\eta.$$

Hence $$e^{\Gamma} \in O(1,3)$$. Further, Suppose $$\Lambda$$ is written as $$\begin{bmatrix} 0&\vec{\lambda}\\ -\vec{\lambda}&R\\ \end{bmatrix}$$ Where $$R$$ itself is antisymmetric. Block Multiplication on the right by $$\eta$$, gives $$\Lambda \eta$$ to be, $$\begin{bmatrix} 0&-\vec{\lambda}\\ -\vec{\lambda}&-R\\ \end{bmatrix}.$$ Clearly Trace of $$\Lambda \eta$$ is 0. Hence $$Det(\Lambda \eta)=1$$. Therefore $$e^{\Gamma} \in SO(1,3)$$.

• Hint: all such matrices can be continuously connected to the identity by multiplying Gamma by a factor lambda between zero and one. The determinant of these matrices is a continuous function of lambda that equals 1 at zero, and can only equal 1 or -1, and that proves membership in SO(1,3). A similar argument holds for orthochronality, changing the determinant for a suitable function. Dec 6, 2015 at 11:33

1. That the exponential map $$\exp: o(1,d) ~\to~ O(1,d)$$ has image $$\exp(o(1,d))~\subseteq~ SO^+(1,d)$$ inside the restriced Lorentz group $$SO^+(1,d)~:=~\{ \Lambda \in SO(1,d) | \Lambda^0{}_0>0 \}$$ follows from the facts that the image of a connected set under a continuous map must again be connected, cf. above comment by Emilio Pisanty. (One more hint: No Lorentz matrix $\Lambda$ can have zero determinant $\det(\Lambda)=0$ or zero 00-entry $\Lambda^0{}_0=0$, cf. e.g. this Phys.SE post.)
2. The non-trivial fact that the exponential map $$\exp: o(1,d) ~\to~ SO^+(1,d)$$ is surjective is discussed in this Phys.SE post.