This computation was given in lecture yesterday. Here we let $f: \mathcal{H} \rightarrow \mathbb{C}$ be a holomorphic function. We viewed $\mathcal{H}$ as the quotient space $G/K$ via $gK \mapsto g \cdot i$. We then defined $\widetilde{f}: G \rightarrow \mathbb{C}$ by : $ \widetilde{f} = \mu(g, i )^{2m} f(g \cdot i)$. Where g = $\begin{bmatrix} a & b \\ c & d \: \end{bmatrix}$ and $\mu(g, z) = (cz + d)$.
Furthermore, we are considering the differential operator on the manifold $G = SL(2, \mathbb{R})$ via $$ X.f(g) := \frac{d}{dt} \: \big{|}_{t=0} f(g \text{exp }(tX))$$ , Where $f$ is a differentiable function on $G$.
In this computation, $H$ is a basis element for the Lie($G$) such that $H = \begin{bmatrix} 1 & 0 \\ 0 & -1 \: \end{bmatrix}$.
In regards to the question I have; The first four lines of the computation are pretty straight forward, however, I am not immediately seeing the jump between the fourth and fifth lines, as well as the jump between the fifth and sixth lines. Any guidance would be appreciated.
Both steps are applications of the product rule of differentiation. On the 4th line, let $\varphi(t):= e^{-2mt}$ and $\psi(t)$ the other terms. Then the 4th line reads $$\frac{d}{dt}|_{t=0}(\varphi(t)\psi(t))= \varphi(0)\frac{d}{dt}|_{t=0}\psi(t) + \varphi'(0)\psi(0)$$ Now just plug in $\varphi(0)=1$, $\varphi'(0)=-2m$ and $\psi(0)$.
Te next line is doing the same trick again with $\mu$ and $f$.