I recently read that all infinitely sheeted Riemann surfaces can be related to exponentials, which I took to mean that they occur when you have logarithmic branch points.
Is it possible to have an infinitely sheeted Riemann surface constructed with only algebraic branch points?
For example a sheet with a square-root branch point connecting it to a sheet above, which in turn is connected via a compact branch cut to a sheet above, which in turn is connected by a square-root branch cut to a sheet above that, and so on infinitely many times.
The answer is positive but you need to know some elementary algebraic topology (fundamental groups and covering spaces) in order to understand it. Let $Z\subset {\mathbb C}$ be any infinite discrete subset (i.e. a subset with no accumulation points in the complex plane). Then for $X= {\mathbb C}- Z$, we have that $\pi_1(X)$ is a free group on countably many generators $s_k$, one for each point $z_k\in Z$. Define the homomorphism $$ \phi: G=\pi_1(X)\to H:=\oplus_{Z} {\mathbb Z}_2, $$ where ${\mathbb Z}_2={\mathbb Z}/2{\mathbb Z}$. There are many such homomorphisms, I will send each generator $s_k$ of $G$ to the generator of the $k$th factor of $H$ (the one indexed by $z_k$). Let $p:\tilde{X}\to X$ denote the (connected) cover associated with the kernel $K$ of $\phi$. I will equip $X$ with the complex structure obtained by pull-back of the complex structure on $X$, making $p$ a holomorphic covering map. Then $p$ has infinitely many sheets (since the group $H$ is infinite and, hence, $K$ has infinite index in $G$). For each $z_k\in Z$, take a small disk $D_k$ centered at $z_k$ and disjoint from the rest of $Z$. Set $D^*_k:= D_k-\{z_k\}$. Then each $p^{-1}(D_k^*)$ is conformal to a (countably infinite) disjoint union of once punctured disks $\tilde{D}_k^*$ such that the restriction of $p$ to each $\tilde{D}_k^*$ is a 2-fold cover. (This cover is conformally isomorphic to the map $z\mapsto z^2$ on the unit disk punctured at the origin). Hence, we can conformally embed $\tilde{X}$ in a Riemann surface $\tilde{C}$ by "filling in" the punctures of the disks $\tilde{D}_k^*$. (This is very standard can be made rigorous if you like.) The map $p$ will extend to a surjective holomorphic map $q: \tilde{C}\to {\mathbb C}$ which is an infinitely-sheeted branch cover of the complex plane, ramified (with local degree 2) only at the points of $Z$.
Lastly, you can make the point of $Z$ to be, say, integers to make them algebraic numbers (if this is what you have meant by "algebraic branch points").
One can modify this construction making the number of branch-points in the (extended) complex plane finite. Namely, start with the infinite group $H$ which has the presentation $$ <a,b,c| a^3, b^3, c^3, abc>. $$ Next, let $Z\subset {\mathbb C}\subset S^2={\mathbb C}\cup\{\infty\}$ be a 3-point subset, whose elements are $A, B, C$. As in the above construction, set $X:= S^2 -Z$; then $G=\pi_1(X)$ has presentation $$ <\alpha, \beta, \gamma| \alpha \beta \gamma>, $$ each generator corresponds to a small circle around the respective point $A, B, C$. You then have an epimorphism $$ \phi: G\to H, \phi(\alpha)=a, \phi(\beta)=b, \phi(\gamma)=c. $$ Let $K$ be the kernel of $\phi$; it is an infinite index subgroup of $G$ since $|H|=\infty$. Now, repeat the construction I described earlier, constructing an infinite cover $p:\tilde{X}\to X$, etc. The only difference will be that you have is that the local ramification numbers have degree 3 rather than 2.