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I'm trying to prove that the identity component of the Lie group $$\mathrm{SO}(2,1) = \left\{ A \in \mathrm{SL}(3,\mathbb R) : A^T \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right)A = \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right) \right\}$$ is isomorphic to the Lie group $\mathrm{PSL}(2, \mathbb R) = \mathrm{SL}(2,\mathbb R)/\left\{I_2, -I_2\right\}$. My strategy is to construct a surjective Lie group homomorphism $\phi : \mathrm{SL}(2,\mathbb R) \to \mathrm{SO}(2,1)$ whose kernel is $\{I_2, -I_2\}$. I've constructed a Lie algebra isomorphism $\Phi : \mathfrak{sl}(2,\mathbb R) \to \mathfrak{so}(2,1)$ and would like to somehow show that this can give us the surjection I'm looking for. Any suggestions?

(I know there is some object known as the Killing form that can be used to help us out in this problem, but we haven't gotten to it yet in my Lie theory class so I want to avoid using it. Is there an equivalent but perhaps simpler formulation we can use?)

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You can use the fact that a map of Lie algebras $\mathfrak{g} \to \mathfrak{h}$ exponentiates to a map $G \to H$ of the corresponding simply connected Lie groups, but this fact is considerably more difficult to prove than proving the basic properties of the Killing form. In addition, $SL_2(\mathbb{R})$ is not simply connected.

So I would really advocate for the Killing form as the most natural way to understand this result. The story goes like this: you want to find a $3$-dimensional real vector space on which $SL_2(\mathbb{R})$ not only acts, but acts preserving a symmetric bilinear form of signature $(2, 1)$. There is a very natural candidate for such a $3$-dimensional real vector space, namely the Lie algebra $\mathfrak{sl}_2(\mathbb{R})$: one thing that's very nice about this representation is that it's easy to see that its kernel is the center of $SL_2(\mathbb{R})$, so we get with very little work that this is even a faithful representation of $PSL_2(\mathbb{R})$.

Now the question is why the adjoint representation preserves a symmetric bilinear form of signature $(2, 1)$. The answer is that the adjoint representation of a semisimple Lie algebra always preserves a symmetric bilinear form called the Killing form, and it's straightforward in this case to verify that the signature of the Killing form is $(2, 1)$ as desired.

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  • $\begingroup$ I'm still not sure I understand the argument. If $SL(2,\mathbb R)$ acts on the Lie algebra by conjugation (the adjoint representation, I suppose), why is the kernel the center (since the center is the subgroup that commutes with everything in $SL(2,\mathbb R)$, not $\mathfrak{sl}(2,\mathbb R)$)? And why is it a faithful representation of $PSL(2,\mathbb R)$? And even once we have all of this, what does this have to do with finding the surjection $SL(2,\mathbb R) \to SO(2,1)$ I'm looking for? $\endgroup$ – D Ford Sep 30 '17 at 15:26
  • $\begingroup$ @D Ford: 1) a connected Lie group $G$ is generated by the image of the exponential map, so an element of $G$ lies in the center iff it commutes with every exponential iff it acts trivially in the adjoint representation. 2) Because the kernel is exactly the center. 3) $SO(2, 1)$ is the connected component of the identity of the automorphism group of a real $3$-dimensional vector space equipped with a symmetric bilinear form of signature $(2, 1)$, and the Killing form on $\mathfrak{sl}_2(\mathbb{R})$ is such a form. $\endgroup$ – Qiaochu Yuan Oct 2 '17 at 6:48

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