# $PSL_2(\mathbb R)$ as a lie subgroup of $SL_2(\mathbb R)$

I was woundering, is $PSL_2(\mathbb R)$ a lie subgroup of $SL_2(\mathbb R)$?

I don't quite know how to check if the inclusion is an immersion because I don't know how to work with the tangent space at points $p\not=e$ (at $e$ this should be trivial, because it's simply the lie algebras, which are equal for both groups).

If the answer is yes, then $\exp:\:\mathfrak{sl_2}(\mathbb R) \to PSL_2(\mathbb R)$ should be just the exponentiation of $SL_2$ composed with the projection $SL\to PSL$, right? But if the answer is no, what's the exponentiation map?

• You have the projection the wrong way. $PSL_2(\Bbb R) = SL_2(\Bbb R)/\{\pm I\}$. – Ted Shifrin Aug 12 '17 at 15:23
• ya, just a typo though (and not something I didn't understand correctly). fixing it, thanks! @TedShifrin – The way of life Aug 12 '17 at 15:25
• When you say "the inclusion," what inclusion are you talking about? – anon Aug 12 '17 at 15:40
• @anon: that's a fair question. I suppose one option is $[g]\mapsto \det(g)\cdot g$, which is a well defined lifting. – The way of life Aug 12 '17 at 15:44
• Yes, it's weird. It's a quotient group, not a subgroup. If $G_1\to G_2$ is a covering homomorphism between Lie groups (equivalently, discrete kernel) then they share the same lie algebra $\frak g$ and, as you say, the exponential map $\exp:{\frak g}\to G_2$ is the composition of $\exp:{\frak g}\to G_1$ and $G_1\to G_2$. – anon Aug 12 '17 at 15:54

As I suggested above, if $H$ is a normal subgroup of $G$, then it's a standard well-definedness check that there is a group homomorphism $\beta$ injecting $G/H$ into $G$ (with $\pi\circ\beta$ the identity, $\pi\colon G\to G/H$ being the obvious map) if and only if there is a group homomorphism from $G$ to $H$ that is the identity on $H$. So, in our case, we ask if there is a group homomorphism from $G=SL_2(\Bbb R)$ to $H=\{\pm I\}$ that is the identity on $H$. To see that the answer is no, just note that $-I$ is a square in $G$.