Representations of $SL(2,R)$ and $PSL(2,R)$ I have a simple and maybe really dumb question, however:
How do the representations of $PSL(2,R)$ and $SL(2,R)$ differ? I know that given a representation of $SL(2,R)$ it factors to a representation of $PSL(2,R)$ if $\pi(-I)=id$. But how about the other direction? Is the representation theory of these two groups actually the same (as the intuition suggests)? 
 A: The two groups fit into a short exact sequence
$$0\to \{\pm I\} \to SL(2,\mathbb{R})\xrightarrow{q} PSL(2,\mathbb{R})\to 0$$
which is a central extension of the center-free group $PSL(2,\mathbb{R})$ by $\mathbb{Z}/2\mathbb{Z}$.  A representation $\rho:PSL(2,\mathbb{R})\to GL(V)$ is of course also a representation of $SL(2,\mathbb{R})$ by the composition $\rho\circ q$.  You have pointed out the converse already, that a representation $\phi:SL(2,\mathbb{R})\to GL(V)$ factors through to a representation of $PSL(2,\mathbb{R})$ if and only if $\{\pm I\}\subset\ker\phi$.
For finite-dimensional irreducible representations over $\mathbb{C}$, the center must act by scalars (Schur's lemma).  Since $(-I)^2=I$, this means that $-I$ acts by $\pm 1$.  The finite-dimensional irreducible representations of $PSL(2,\mathbb{R})$ are those of $SL(2,\mathbb{R})$ where $-I$ does not act by $-1$.
Let $V=\mathbb{C}^2$ be the two-dimensional representation of $SL(2,\mathbb{R})$ given by matrix-vector multiplication, which is irreducible.  First notice that $-I$ acts by $-1$, so this is not a representation of $PSL(2,\mathbb{R})$.
Define for $n\geq 0$ the space $V_n=\operatorname{Sym}^nV$, the $n$-fold symmetric power of $V$, which is irreducible.  This is usually thought of as the space of homogeneous degree-$n$ polynomials whose variables form a basis of $V$.  In particular, $V_0$ is the trivial representation and $V_1$ is $V$.  It's a quick check to see that $-I$ acts on $V_n$ by $(-1)^n$, so the finite-dimensional irreducible representations of $PSL(2,\mathbb{R})$ are $V_0,V_2,V_4,\dots$

Since $SL(2,\mathbb{R})$ is a double cover of $PSL(2,\mathbb{R})$, they have the same universal cover and Lie algebra $\mathfrak{g}$.  So, any representation of either gives a representation of $\mathfrak{g}$.  Neither is simply connected, so representations of $\mathfrak{g}$ do not necessarily correspond to representations of these Lie groups.  However, one can deduce that they have the same projective representations.  A homomorphism $SL(2,\mathbb{R})\to PGL(V)$ for finite-dimensional $V$ must have $-I$ in the kernel (Schur's lemma), so it factors through $PSL(2,\mathbb{R})$.
