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Consider the group $G$ consisting of real $2\times 2$ matrices of the form $\begin{pmatrix} x & 0\\ y & z \end{pmatrix}$ of nonzero determinant with multiplication as the operation. I believe that this is a Lie group because as a space it is the intersection of $GL(2,\mathbb{R})$ and the variety $V(w)$, where $w$ is the upper right matrix entry (I seem to recall a theorem saying that such intersections give Lie groups). Furthermore, if we define $Z(G)$ to be the center of $G$, consisting of matrices of the form $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}$, we have that $Z(G)$ is a normal subgroup of $G$ which is closed and thus a Lie subgroup. This makes the group $G' = G/Z(G)$ a (2-dimensional) Lie group.

My question is twofold. First, is $G$ in fact a Lie group? Second, can someone give an example of an atlas for $G'$ and the corresponding transition maps between charts?

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Instead of "seeming to recall a theorem" you can actually try to prove it is a Lie group by hand. :) – Mariano Suárez-Alvarez Apr 8 '11 at 6:45
@Mariano: I will try to do so, but my knowledge of Lie groups is very shaky. – Alex Becker Apr 8 '11 at 13:38
Well, at least confirm that there is a theorem that applies, then! :=) – Mariano Suárez-Alvarez Apr 8 '11 at 17:13

A quotient of a Lie group by a closed subgroup is always a Lie group.

Can you describe the quotient group as an abstract group? When you do that, you will have charts for it...

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Is this true even when the subgroup is not normal? – Sam Apr 8 '11 at 8:19
@Sam, in that case the quotient is only a smooth manifold. – Mariano Suárez-Alvarez Apr 8 '11 at 11:53
You might want to change the first sentence then. ;) – Sam Apr 8 '11 at 12:27
up vote 0 down vote accepted

I did not realised that the structure of this group would be as simple as it turned out to be, mostly because I had simply assumed that working with this group would be beyond me given that I've never worked with Lie groups before. After looking again, I believe I've proved that $G'$ is a Lie group and found an atlas for it.

The group $G'$ is isomorphic to the group of real $2\times 2$ matrices with nonzero determinant of the form $\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}$ as the equivalence class of $\begin{pmatrix} x & 0 \\ y & z \end{pmatrix}$ contains precisely $1$ matrix of such form, specifically $\begin{pmatrix} \frac{x}{z} & 0 \\ \frac{y}{z} & 1 \end{pmatrix}$ (note that $z\neq 0$ because $xz \neq 0$).

$G'$ can be made into a topological space by equipping it with the metric defined as $d\left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix},\begin{pmatrix} x' & 0 \\ y' & 1 \end{pmatrix}\right) = \sqrt{(x-x')^2 + (y-y')^2}$. An atlas for $G'$ is given by the charts $\alpha: G'_+ \rightarrow \mathbb{R}^2$ and $\beta: G'_- \rightarrow \mathbb{R}^2$, where $G'_+$ is the (open) set of matrices with positive determinant and $G'_-$ the (open) set with negative determinant, defined by $\alpha \left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}\right) = \begin{pmatrix} x \\ y \end{pmatrix}$ and $\beta \left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}\right) = \begin{pmatrix} x \\ y \end{pmatrix}$. These charts are non-intersecting, and this makes $G'$ a smooth manifold.

In order to show that $G'$ is a Lie group we need only show that the map $\phi: G'\times G' \rightarrow G'$ defined as $\phi \left(\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix},\begin{pmatrix} x' & 0 \\ y' & 1 \end{pmatrix},\right) = {\begin{pmatrix} x & 0 \\ y & 1 \end{pmatrix}}^{-1} \begin{pmatrix} x' & 0 \\ y' & 1 \end{pmatrix} = \begin{pmatrix} \frac{x'}{x} & 0 \\ \frac{xy'-yx'}{x} & 1 \end{pmatrix}$ is smooth. This follows easily from the fact that rational functions on $\mathbb{R}^2$ are smooth whereever their denominator is not $0$ and that $x\neq 0$ in $G'$.

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