How do we prove that $H$ is a Lie group?

Question

We define $H$ as below: $$H=\Bigg\{\begin{pmatrix} a & b\\ 0 & 1 \end{pmatrix} a>0, b\in\mathbb{R}\Bigg\}$$

I know $H$ has group structure, but how do we show it is a smooth manifold? Not quite familiar with this part so I need some help.

Answer 1

For a smooth manifold, we need a differentiable structure (a system of neighborhoods and the concept of a linear approximation), e.g. given by the norm topology. A smooth multiplication and a smooth inversion are necessary to be a Lie group. Multiplication is $$ \left(\begin{pmatrix}a&b\\0&1\end{pmatrix}\, , \,\begin{pmatrix}a'&b'\\0&1\end{pmatrix}\right)\longmapsto \begin{pmatrix}aa'&ab'+b\\0&1\end{pmatrix} $$
so all coordinate functions are smooth: $$ m(a,b,0,1;a',b',0,1)=(aa',ab'+b,0,1) $$ The same is true for the inversion: $$ \begin{pmatrix}a&b\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}a^{-1}&-a^{-1}b\\0&1\end{pmatrix} $$ with coordinate functions $$ \iota(a,b,0,1)=(a^{-1},-a^{-1}b,0,1) $$ because $a>0$, we do not have to bother about the possible singularity of $a^{-1}.$

For the manifold aspect of the question, we need an atlas, i.e. a collection of smooth homeomorphisms (charts) from open neighborhoods of $h\in H$ to $\mathbb{R}^n$ $$ \varphi \, : \,U_h(\varepsilon )\longrightarrow \mathbb{R}^n $$ such that all maps coincide on their overlappings. If $\psi \, : \,U_k(\varepsilon' )\longrightarrow \mathbb{R}^n$ is another chart and $U_h(\varepsilon )\cap U_k(\varepsilon')\neq \emptyset$ then it is required that $$ \left.\varphi\right|_{U_h(\varepsilon )\cap U_k(\varepsilon')}\equiv \left.\psi\right|_{U_h(\varepsilon )\cap U_k(\varepsilon')} $$ The idea behind it is that we start in a neighborhood $h\in U_h(\varepsilon ),$ move over to the chart (in $ \mathbb{R}^n $), then perform the differentiation since we know how it is done in $\mathbb{R}^n$ and pull back the result to $H.$ This way, we have defined a tangent at $h\in H$ by what we computed on the chart.

It is just like a street map of a city. The city ($H$) isn't flat, the charts are (pages in the atlas in $\mathbb{R}^n$). And two overlapping maps should be the same on their overlapping part. I have chosen $n=4$ in the above description of multiplication and inversion, but it is sufficient to choose $n=2$ since we do not really have to carry the constant coordinates $0,1$ with us. This may change if we consider $H$ in the context of other four-dimensional matrix groups.

Long talk short: we need charts! Say we have $h=\begin{pmatrix}a&b\\0&1\end{pmatrix}$ and $\varepsilon =\dfrac{1}{2}a$ then $$ U_h(\varepsilon )=\left\{\begin{pmatrix}x&y\\0&1\end{pmatrix}\, : \,|x-a|<\varepsilon \wedge |y-b|<\varepsilon \right\}\subseteq H $$ are open neighborhoods for elements of $H$ and the chart maps are simply $\begin{pmatrix}x&y\\0&1\end{pmatrix}\longmapsto (x,y).$

In this specific case $H$ we can define only one chart $$ H\longrightarrow \mathbb{R}^2\, , \,\begin{pmatrix}a&b\\0&1\end{pmatrix}\longmapsto (a,b) $$ or $$ H\longrightarrow \mathbb{R}^4\, , \,\begin{pmatrix}a&b\\0&1\end{pmatrix}\longmapsto (a,b,0,1) $$ which already fulfills all requirements.

Answer 2

If you think of the $(a, b)$ plane, $H$ is just the (open) right half-plane, which is a smooth manifold basically by definition. The group operation is given by

$$(a, b) \cdot (c, d) = (a c, a d + b)$$

and hopefully you believe this is smooth, e.g. because the individual coordinates are polynomials. (And similarly for inversion.)