My Professor says that all members of the orthochronous Lorentz group 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 anti symmetric. 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 $e^{\Gamma} \in SO(1,3)$

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    $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Dec 6 '15 at 11:33
  • $\begingroup$ @emilio : Since the trace of $\Gamma$ is always zero, determinant of $e^{k\Gamma}$ is always 1. Is there something i am missing? $\endgroup$ – user100570 Dec 6 '15 at 11:37
  • $\begingroup$ See edited comment. $\endgroup$ – Emilio Pisanty Dec 6 '15 at 11:39
  • $\begingroup$ @graviton - Emilio has given you the answer. $\det e^{k\Gamma} = 1$ always. But any element that is not orthochronous has $\det = -1$. $\endgroup$ – Prahar Dec 6 '15 at 17:09
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    $\begingroup$ @prahar This is incorrect. There exist elements that belong to $SO(1,3)$, but not $SO^+(1,3)$. As a counter example to this claim, set $\gamma$ as $-\gamma$ in the standard lorentz transform. It belongs to $SO(1,3)$ but not $SO^+(1,3)$. $\endgroup$ – user100570 Dec 6 '15 at 17:32

Comments to the question (v2):

  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.


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