The semidirect product of $\Bbb R^*$ and $\Bbb R$ is isomorphic to the group of upper-triangular real 2-by-2 matrices with determinant 1.

Question

I want to present the group of matrices $$G=\{ \begin{pmatrix} a & b \\ 0 & \frac1a \\ \end{pmatrix}; a,b\in\Bbb R, a\ne0 \}.$$ as a semidirect product of $\Bbb R^∗$ and $\Bbb R$ with respect to some action of $\Bbb R^∗$ on $\Bbb R$. I've looked at this post: Can this matrix group be obtained from $(\mathbb R,+)$ and $(\mathbb R^*,\cdot)$? But it was not totally clear for me of how I should define the action $φ$ of $\Bbb R^*$ on $\Bbb R$ and $φ_g$, and how the operation given in the definition below would look in my case.

Definition: given two Lie groups $G$ and $H$ and a smooth action $φ : G×H → H$ of $G$ on $H$ such that, for any $g ∈ G$ the action $φ_g : H →H$ is a group automorphism. The semidirect product of $G$ and $H$ is defined as follows: 1. as a manifold, it is the product $G×H$. 2. The multiplication is given by the formula $(g,h)(g',h') = (gg',φ_{g'^−1}(h)h')$.

Can you kindly explain the things I mentioned above regarding the definition?

Answer 1

Here is one-line summary. $(g,h)\in\Bbb R^*\ltimes\Bbb R$ is identified with $\pmatrix{g & gh \\ 0 &\frac1g}\in G$.

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$\Bbb R^*$ with multiplication is a Lie group.
$\Bbb R$ with addition is a Lie group.
Define $\phi:\Bbb R^*\times\Bbb R\to\Bbb R$, $(g,h)\mapsto g^2h$.
Verify that $\phi$ is smooth.
For any $g\in\Bbb R^*$, $\phi_g:\Bbb R\to\Bbb R$ is defined by $\phi_g(h)\mapsto g^2h$ for any $h\in\Bbb R$.
Verify that $\phi_g$ is a Lie-group automorphism of $\Bbb R$.
Furthermore, $\phi_{g_1g_2}=\phi_{g_1}\phi_{g_2}$ and $\phi_{1}$ is the identity map on $\Bbb R$. That is, $g\mapsto \phi(g)$ is a group homomorphism from $\Bbb R^*$ to $\text{Aut}(\Bbb R)$, the automorphism group of Lie group $\Bbb R$. $\phi$ defines a smooth group action of $\Bbb R^*$ on $\Bbb R$.

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Hence we can define a Lie group $\Bbb R^*\ltimes_\phi\Bbb R$, the (external) semidirect product of $\Bbb R$ and $\Bbb R^*$ with respect to $\phi$.

• As a manifold, it is $\Bbb R^*\times\Bbb R=\left\{(a, b): a\in\Bbb R^*,\ b\in \Bbb R\right\}$.
• The multiplication is given by $$(g,h)(g',h') = (gg',\phi_{{g'}^{−1}}(h)h')=(gg',{g'}^{−2}h+h')$$

By definition, $\Bbb R^*\ltimes_\phi\Bbb R$ is the internal semiproduct of its Lie subgroup $\Bbb R^*\times\{0\}$, which is isomorphic to $\Bbb R^*$, and its Lie subgroup $\{1\}\times\Bbb R$, which is isomorphic to $\Bbb R$.

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Define a map $$\begin{array}{ccl}\iota: &\Bbb R^*\ltimes_\phi\Bbb R&\to &G\\ &(g,h)&\mapsto &\pmatrix{g & 0 \\ 0 &\frac1g}\pmatrix{1 & h \\ 0 & 1}=\pmatrix{g & gh \\ 0 &\frac1g}\end{array}$$

Verify $\iota$ is a Lie group isomorphism.
Hence we say the semidirect product of $\Bbb R^*$ and $\Bbb R$ with respect to $\phi$ is isomorphic to $G$ as Lie groups via $\iota$.

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Since $\iota$ identifies

• $\Bbb R^*\times\{0\}$ with the Lie subgroup $H_1$ of $G$, where $H_1=\left\{\pmatrix{a & 0 \\ 0 &\frac1a}: a\in\Bbb R^*\right\}$,
• $\{1\}\times\Bbb R$ with the Lie subgroup $H_2$ of $G$, where $H_2=\left\{\pmatrix{1 & b \\ 0 & 1}: b\in\Bbb R\right\}$,

$G$ is the internal semidirect product of its Lie subgroups $H_1$ and $H_2$.

Two Easy Exercises

1. Write the inverse map of $\iota$ explicitly.

2. Is $G$ isomorphic to the direct product $\Bbb R^*\times\Bbb R$ as Lie groups?