TL;DR: How to show eq. (2) using eq. (1)?

I'm currently having a bit of a hard time with the following problem. In class we showed that $$\mathfrak{so}(1,3;\mathbb{C})\cong \mathfrak{sl}(2,\mathbb{C})\oplus \mathfrak{sl}(2,\mathbb{C}),\tag{1}$$ where $\mathfrak{so}(1,3;\mathbb{C})$ is the Lie algebra of the proper Lorentz group over $\mathbb{C}$ and $\mathfrak{sl}(2,\mathbb{C})$ the Lie algebra of $SL(2,\mathbb{C})$. If I understood my tutor correctly it is possible to show that $$SO(1,3;\mathbb{C})\cong SL(2,\mathbb{C})\times SL(2,\mathbb{C}),\tag{2}$$ using eq. (1). For this to be true we would need to show that $\exp$ is surjective on the Lie algebras separately, aka $$\exp: \mathfrak{g}_i\to G_i,$$ where $\mathfrak{g}_1=\mathfrak{so}(1,3;\mathbb{C}),G_1 = SO(1,3;\mathbb{C})$ and $\mathfrak{g}_2=\mathfrak{sl}(2,\mathbb{C})\oplus \mathfrak{sl}(2,\mathbb{C})$,$G_2 =SL(2,\mathbb{C})\times SL(2,\mathbb{C})$.

The problem with this is that I don't really know how to do that... I thought that I'd try first to show that $\exp:\mathfrak{sl}(2,\mathbb{R})\to SL(2,\mathbb{R})$ is surjective and then try to work out the rest from there. Unfortunately it turned out that $\exp$ isn't surjective on this Lie algebra.. Could somebody suggest some strategies to prove this?

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    $\begingroup$ The exponential function is not surjective in this case, see here. $\endgroup$ – Dietrich Burde Feb 18 at 20:33
  • $\begingroup$ @DietrichBurde I'm sorry, but I'm not incredibly familiar with topology (physics student here) so I'm having a hard time following the statements there... But since the question is flawed in itself, according to Thomas it should be $SL(2,\mathbb{C})\times SL(2,\mathbb{C})/\{\pm 1\}\cong SO(1,3;\mathbb{C})$, I would like to know if the exponential map is not sujective in this case either? $\endgroup$ – Sito Feb 18 at 21:02

The result $SO(3,1, C)=Sl(2,C)\times Sl(2,C)$ is certainly not true. The lfh is isomoprhic to $SO(4,C)$, whose center is $Z/2Z$. The right hand side has center $Z/2Z \times Z/2Z.$ Let $Z\subset Sl(2,C)\times Sl(2,C)$ the subgroup with two element $(Id,Id) ; -(Id,Id)$.

If you want to prove that $Sl(2,C)\times Sl(2,C) /Z$, is $SO(4,C)$ firts identify $C^4$ with the set of $(2,2)$ matrices with quadratic from the determinant $(ad-bc)$ (this is a quadratic form !).

Note that this quadratic form is $Sl(2,C)\times Sl(2,C) /Z$ invariant where the action is $(g,h).M=gMh^{-1}$, with kernel $Z$.

So we have a morphism $Sl(2,C)\times Sl(2,C)/Z\to SO(4,C)$.

It is quite easy to prove injectivity : if for every $M$,$M=gMh^{-1}$ then we have $g=h$ (let $M=id$), and then $g$ commute with every matrix so must be central of determinant 1.

For the surjectivity this follows from dimension plus injectivity of the analytic morphism we just have described.

  • $\begingroup$ First of all, thank you for the answer! Forget the comment before, I just understood what your notation is and how I'm supposed to read it... I will try to follow it and come back later if a question arises! $\endgroup$ – Sito Feb 18 at 19:27
  • $\begingroup$ After some trying out I just can't figure out some of your steps... For example, you wirte "Note that this quadratic form is $Sl(2,C)\times Sl(2,C)/Z$ invariant where the action is $(g,h).M=gMh^{−1}$, with kernel $Z$." Why should this be invariant? Could you help here a bit... $\endgroup$ – Sito Feb 18 at 20:56
  • $\begingroup$ The quadratic form is the determinant. $det(gMh^{-1}=det(M)$, as both $g,h$ are with determinant 1. $\endgroup$ – Thomas Feb 19 at 9:38
  • $\begingroup$ Thank you for the explanation, I think I‘m almost there... My last question is about your proof of injectivity. Can you expand that part a bit... I just can‘t figure out what steps you are doing and why. $\endgroup$ – Sito Feb 19 at 20:53
  • $\begingroup$ $gM=Mh$ for evry $M$ implies $g=h$ ($M=id)$ Then $gMg^{_1}=M$ for every $M$ implies that $g$ is an homothety. $\endgroup$ – Thomas Feb 20 at 18:12

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