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I was told that the upper half plane has a purely group theoretic description $$ \mathcal{H}\cong\text{SL}_2(\mathbb{R})/\text{SO}_2(\mathbb{R}) $$

I tried to construct a group homomorphism to prove it:

$$ \varphi:\text{SL}_2(\mathbb{R})\rightarrow\mathcal{H} $$

by $$ \gamma\mapsto\gamma(i) $$

Because $\text{SO}_2(\mathbb{R})$ fixes $i$.

But unfortunately I cannot come out a group structure on $\mathcal{H}$ making $i$ the unit.

So can anyone help?

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That "isomorphism" is as manifolds (or as manifolds-with-a-$SL_2(\mathbb R)$-action), not as groups in any sense. –  Mariano Suárez-Alvarez Jan 8 '13 at 3:11
2  
Also, notice that $SO_2(\mathbb R)$ is not a normal subgroup of $SL_2(\mathbb R)$! –  Mariano Suárez-Alvarez Jan 8 '13 at 3:15

1 Answer 1

up vote 5 down vote accepted

The open half plane is diffeomorphic to the plane, so it has a Lie group structure isomorphic to that of $\mathbb R^2$.

There is another, which turns it into a Lie group isomorphic to connected component of the identity in the affine group on the line, also known as $\mathbb R\rtimes\mathbb R_{>0}$.

In fact, these are all Lie group structures in the upper half plane, up to isomorphism. Indeed, the Lie algebra corresponding to such a Lie group is of dimension $2$, and there are two real Lie algebras of dimension $2$. Since there are as many real Lie algebras of a given dimension as there are simply connected Lie groups of that dimension, we have the result.

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Oh...It is far beyond what I know. Anyway, thank you! –  hxhxhx88 Jan 8 '13 at 5:54

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