# Showing surjectivity of exponential map.

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?

• The exponential function is not surjective in this case, see here. – Dietrich Burde Feb 18 at 20:33
• @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? – Sito Feb 18 at 21:02

## 1 Answer

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.

• 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! – Sito Feb 18 at 19:27
• 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... – Sito Feb 18 at 20:56
• The quadratic form is the determinant. $det(gMh^{-1}=det(M)$, as both $g,h$ are with determinant 1. – Thomas Feb 19 at 9:38
• 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. – Sito Feb 19 at 20:53
• $gM=Mh$ for evry $M$ implies $g=h$ ($M=id)$ Then $gMg^{_1}=M$ for every $M$ implies that $g$ is an homothety. – Thomas Feb 20 at 18:12