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Let $M = \mathrm{GL}(2, \mathbb{R})$ and define an action of $\mathbb{R}$ on $M$ by the formula $$\theta(t,A)=\begin{pmatrix}1&t\\ 0&1\end{pmatrix}A$$ for $A \in M$. Find the infinitesimal generator.

I think if we denote $A = \begin{pmatrix}a&b\\ c&d\end{pmatrix}$ we have that $$\begin{pmatrix}1&t\\ 0&1\end{pmatrix}\begin{pmatrix}a&b\\ c&d\end{pmatrix}=\begin{pmatrix}a+ct&b+dt\\ c&d\end{pmatrix}$$ and we could identify this with a point $(a+ct, b+dt, c, d)$ in $\mathbb{R^4}$ so would the infinitesimal generator for $(a,b,c,d)\mapsto(a+ct, b+dt, c, d)$ be the vector field $$\frac{d}{dt}(a+ct) \frac{\partial}{\partial a} + \frac{d}{dt}(b+dt) \frac{\partial}{\partial b} + \frac{d}{dt}(c) \frac{\partial}{\partial c} + \frac{d}{dt}(d) \frac{\partial}{\partial d} = c\frac{\partial}{\partial a} + d \frac{\partial}{\partial b} ?$$

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    $\begingroup$ The infinitesimal generator of a one-parameter subgroup {g(t)} of a Lie group is found by taking the derivative with respect to t, and then evaluating at t = 0. $\endgroup$
    – Dan Asimov
    Commented May 26, 2023 at 17:21
  • $\begingroup$ Wouldn't it give the same result as we don't have any $t$'s left? @DanAsimov $\endgroup$
    – Olander
    Commented May 26, 2023 at 17:26
  • $\begingroup$ Sorry, but I have no idea what you are referring to. $\endgroup$
    – Dan Asimov
    Commented May 26, 2023 at 17:27
  • $\begingroup$ Can you help me see where I'm going south here? @DanAsimov $\endgroup$
    – Olander
    Commented May 26, 2023 at 18:23
  • $\begingroup$ A one-parameter subgroup in GL(2,ℝ) is always of the form exp(t M) where M is an arbitrary 2×2 real matrix. Its derivative at t = 0 is M. $\endgroup$
    – Dan Asimov
    Commented May 27, 2023 at 16:36

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As Dan Asimov pointed out in the comment section, the generator $G$ of the transformation $\theta(t) = e^{tG}$ is given by its derivative at the point $t=0$, i.e. $$ G = \left.\frac{\mathrm{d}\theta}{\mathrm{d}t}\right|_{t=0} = \left.\frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\right|_{t=0} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$

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