Consider the upper half plane $\Bbb H = \{z = x + iy \in \Bbb C ; y > 0\}$ with the Riemannian metric $g = \dfrac 1 {y^2} \langle , \rangle$ where $\langle , \rangle$ is the standard inner product on $\Bbb R^2$.
For each $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $\det A = 1$, i.e. $A \in SL(2, \Bbb R)$, consider the action $A \cdot z = \dfrac {az + b} {cz + d}$.
Show that the $x$-axis is mapped into the $x$-axis under the action of $A \in SL(2, \Bbb R)$ and the upper half plane $\Bbb H$ is mapped into the upper half plane.
So I'm not exactly sure how to go about doing this. My professor emailed the class saying we might also want to show the commutivity of the action ($A(Bz) = (AB)z$) and identity ($Iz = z$) but I'm not sure how to even show that under $A$ the real axis is mapped to the real axis. Could I just take an arbitrary $z = x + \Bbb iy$ and have $A$ act upon it such that we get $Az$, as above, and then expand $z$ into $x + \Bbb iy$ then take the real part? That's about all I can think of but I'm not sure how, mathematically, to argue that the real axis or the upper half plane gets mapped into themselves.
Yes, your guess is correct. Letting $z = x + y \Bbb i$ and $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL (2, \Bbb R)$, we have that
$$A \cdot z = \frac {az + b} {cz + d} = \frac {(az + b) \overline{(cz + d)}} {(cz + d) \overline{(cz + d)}} = \frac {(ac z \bar z + bd) + (adz + bc \bar z)} {|cz + d|^2} = \\ \frac {(ac |z|^2 + bd + adx + bcx) + \overbrace{(ad - bc)} ^{=1}y \Bbb i} {|cz + d|^2} = \frac {ac |z|^2 + bd + adx + bcx} {|cz + d|^2} + \frac y {|cz + d|^2} \Bbb i .$$
If $z$ is on the $x$-axis then $y=0$, so by the above formula $A \cdot z \in \Bbb R$, i.e. $A \cdot z$ is on the $x$-axis, too. If $z \in \Bbb H$ then $y >0$, so the imaginary part of $A \cdot z$ is $\dfrac y {|cz + d|^2}$, strictly positive, therefore $A \cdot z \in \Bbb H$.