Here's the combinatorial picture which might be useful to your understanding.

Think of a Riemann surface as a graph. The genus of a Riemann surface is the minimal number of circles you need to cut out before the surface is contractible and the genus of a graph is the minimal number of edges you need to cut open before the graph is contractible (i.e. a tree). (I.e. it is precisely the number of edges in the complement of a spanning tree.)

On the Riemann surface side, we specify a finite set of zeroes and a finite set of poles (with multiplicity) and we ask the Mittag-Leffler problem:

Given a Riemann surface $X$ and (multi-)sets $Z$ and $P$ of zeroes and poles. Is there a meromorphic function on $X$ whose zero set is $Z$ and whose poles are $P$?

This has an algebraic version:

Given an algebraic curve $X$ and (multi-)sets $Z$ and $P$ of zeroes and poles. Is there a rational function on $X$ whose zero set is $Z$ and whose poles are $P$?

And there's a discrete version as well:

Given a (metric) graph $G$ and a finite set $B$ with multiplicities (possibly negative), is there a piecewise-linear function with integer slopes on $G$ whose bend locus is $B$?

Here the bend locus of a piecewise-linear function $f$ is the set $$B = \{\text{points of $G$ at which $f$ is non-linear}\}$$

and the multiplicity at a point $x$ is the sum of the incoming slopes at $x$. So if $x$ is in the middle of an edge, this is just how much the slope of $f$ changes.

For example, model an elliptic curve as a genus 1 graph (e.g. a circle). So we take the interval $[0,1]$ with the endpoints identified (and we need $f(0) = f(1)$ for our functions). Now here is an example of a piecewise-linear function on $G$:

The bend locus is $[1/5] - [2/5] - [3/5] + [4/5]$.

Maybe you can see from this graph-model that you can't have a piecewise-linear function on a circle whose bend locus has only 2 points. In fact:

If $G$ is a metric graph, then there is a piecewise-linear function whose bend locus is $[P] - [Q]$ if and only if $G$ is a tree.

This corresponds to the classical result (e.g. Hartshorne, Example 6.10.1):

If $X$ is a smooth, projective curve then the divisor $[P] - [Q]$ is principal if and only if $X \cong \mathbf P^1$.

There is a very strong connection between how divisors work on Riemann surfaces and algebraic curves to how they work on metric graphs(†). So if you want, you can in fact think of divisors on Riemman surfaces and algebraic curves as sets of points on a metric graph with multiplicities(‡). At the very least, this can provide some intuition as to why $[P] - [0]$ isn't principal on an elliptic curve.

(†) See e.g. *Specialization of linear systems from curves to graphs* by Matthew Baker for specifics (https://arxiv.org/abs/math/0701075).

(‡) See Sections 1.3. and 1.4. of the above paper for some more definitions and references.

thinkI understand the proof (will have to write it down for my own benefit), but I'm still not clear how to actually visualize ([P] - [0]). Maybe it doesn't correspond to anything except that it's an abstract element of Pic$^0$(E). I can live with that, but just wanted to know if there was something that one could visualize. $\endgroup$ – MachPortMassenger Feb 8 at 22:09