By an automorphy factor (or a factor of automorphy) for $\mathrm{SL}_2(\mathbb{R})$ on the upper half-plane $\mathbb{H}$, I mean a continuous map $$j \colon \mathrm{SL}_2(\mathbb{R}) \times \mathbb{H} \to \mathbb{C}^\times$$ such that the following conditions are satisfied:
(i) For each $g \in \mathrm{SL}_2(\mathbb{R})$, the resulting map $\mathbb{H} \to \mathbb{C}$, $z \mapsto j(g,z)$ is smooth.
(ii) $j$ is a 1-cocycle in the sense that $$j(g_1 g_2, z) = j(g_1, g_2 z) \cdot j(g_2, z)$$ for all $g_1, g_2 \in \mathrm{SL}_2(\mathbb{R})$ and $z \in \mathbb{H}$, where $\mathrm{SL}_2(\mathbb{R})$ acts on $\mathbb{H}$ by Möbius transforms.
As the title of my question indicates, I would like to classify all such automorphy factors. Of course, there is a trivial way to obtain automorphy factors (so-called 1-coboundaries), which needs to be excluded: For any smooth function $f \colon \mathbb{H} \to \mathbb{C}^\times$, we obtain the automorphy factor $j_f(g,z) := f(gz)/f(z)$. A non-trivial example of an automorphy factor is given by $$j(\Big( \begin{array}{cc} a & b \\ c & d \end{array} \Big), z) := \frac{c\overline{z}+d}{|cz+d|}$$ as well as by its integral powers, which appear in the definition of Maass forms (of a given weight $k$).
Now I tend to conjecture that the powers of $j$ are already all automorphy factors (up to 1-coboundaries), but I am not yet able to prove this. However, what I can show is that for any automorphy factor $j \colon \mathrm{SL}_2(\mathbb{R}) \times \mathbb{H} \to \mathbb{C}^\times$ there exists an integer $k \in \mathbb{Z}$ such that $j(r(\theta),i) = \exp(ik \theta)$ for every $r(\theta) = \Big( \begin{array}{cc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array} \Big) \in \mathrm{SO}_2(\mathbb{R}) \subseteq \mathrm{SL}_2(\mathbb{R})$. Indeed, since $\mathrm{SO}_2(\mathbb{R})$ is the stabilizer of $i \in \mathbb{H}$, property (ii) says that $$j(g_1g_2,i) = j(g_1,i) \cdot j(g_2,i)$$ for all $g_1,g_2 \in \mathrm{SO}_2(\mathbb{R})$, hence that $\Phi \colon \mathbb{R} \to \mathbb{C}^\times$ defined by $\theta \in \mathbb{R} \mapsto j(r(\theta),i) \in \mathbb{C}^\times$ is a continuous group homomorphism, thus a morphism of Lie groups. By Lie theory, there exists a complex number $\alpha \in \mathbb{C}$ such that $\Phi(\theta) = \exp(\alpha \theta)$ for every $\theta \in \mathbb{R}$. But $\Phi$ being $2\pi$-periodic implies that $2\pi \alpha \in 2 \pi i \mathbb{Z}$, hence that $\alpha \in i \mathbb{Z}$. Write $\alpha = ik$ for some $k \in \mathbb{Z}$.
Given this, it would suffice to prove that any automorphy factor $j$ with the property that $j(g,i) = 1$ for every $g \in \mathrm{SO}_2(\mathbb{R})$ is actually a 1-coboundary, but I do not see how to prove this (and this could also possibly be false). Are there extra conditions which I have to add in order to make this classification work? Any help is appreciated, also references to the literature are welcome!
Related questions:
I have just seen this question a couple of days ago. I believe the answer to your question is yes. First note that if we take $$ h(z) = \sqrt{\mathrm{Im}(z)}, $$ then $$ j_h(z)j(g,z)=\frac{1}{cz+d}, $$ so we can assume that $j(g,z)=(cz+d)^{-1}$ (this is a little easier to manipulate).
Now, to each automorphy factor $f:\mathrm{SL}_2(\mathbb{R})\times \mathbb{H}\to \mathbb{C}^\times$ we associate a $\mathrm{SL}_2(\mathbb{R})$-equivariant complex line bundle on $\mathbb{H}$ as follows: as a manifold, $$ L_f = \mathbb{H}\times \mathbb{C} $$ with the obvious projection onto $\mathbb{H}$. The action of $\mathrm{SL}_2(\mathbb{R})$ on $L_f$ is given by the formula $$ g\cdot (z,w) = (gz,f(g,z)w). $$ It is easy to see that if $f_1$ and $f_2$ are automorphy factors such that $L_{f_1}$ is isomorphic (as equivariant line bundles) to $L_{f_2}$, then $f_1$ and $f_2$ are equal up to coboundary. Also, it is easy to see that for every pair $f_1,f_2$ of automorphy factors, $$ L_{f_1}\otimes L_{f_2} \cong L_{f_1 f_2}. $$ Thus all we need to prove is that every equivariant line bundle is isomorphic (equivariantly) to $L_j^{\otimes k}$ for some integer $k$.
Let $L$ be any equivariant line bundle on $\mathbb{H}$. For any $z\in\mathbb{H}$ its stabilizer is a conjugate of $\mathrm{SO}_2$ and thus this group acts on the fiber $L_z$, which is one-dimensional. Thus $L_z$ is a continuous (even smooth) representation of $\mathrm{SO}_2$ and so its character is of the form $u\mapsto u^k$ for some integer $k$. Then you can prove that $L\cong L_j^{\otimes k}$.