What do non-principal divisors in a Picard group look like

The way Divisors on Elliptic Curves are motivated in cryptography is to say it's a convenient way to represent rational functions by keeping track of multiplicities of the zeros and poles of a rational function. This of course forms an abelian group since polynomial multiplication => addition of exponent of poles and zeros. Then one defines the principal divisors as those degree-zero divisors which actually correspond to a rational function and Pic$$^0$$(C)=Div$$^0$$(C)/Princ(C), where Div$$^0$$ is the subgroup of degree 0 divisors of C.

I can easily visualize what Princ(C) looks like (they are rational functions), but I'm having hard time visualizing what non-principal divisors in Div$$^0$$(C) look like? Clearly, they are not rational functions by definition, so what are they?

Also, is the Abel-Jacobi divisor ([P] - [0]) a principal divisor on an Elliptic Curve? (Me thinks not, but how do I prove it?)

Note: I'm not a mathematician (more like an applied cryptographer) and I'm assuming this site is not just for experts to impress other experts, but also to answer dumb questions that lesser mortals might have :-)

• You'd be better off asking here math.stackexchange.com under the algebraic geometry tag. Short answer is that every function on an elliptic curve has to have at least two poles and two zeros. If ([P] - [0]) was principle, that would mean the elliptic curve had a 1-1 map to P^1, and hence be P^1, which it isn't. The link above can fill in details if this is too terse.. – meh Feb 8 '19 at 19:43
• I think I 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. – MachPortMassenger Feb 8 '19 at 22:09
• I personally think of such a thing algebraically, not geometrically. The closest I can come to helping you is to note that if one identifies an elliptic curve as a plane cubic, then (algebraic) functions can be realized as the quotient of two homogeneous polynomials. So the simplest function is the quotient of two linear functions. Any line will intersect the cubic in three points and that will be the 'zero set'. – meh Feb 9 '19 at 14:14
• Your question would be even better without claims on what "experts" want to show off on this site. – YCor Feb 10 '19 at 4:48
• On an elliptic curve a degree $0$ divisor $\sum_j P_j-Q_j$ is principal iff $\sum_j P_j = \sum_j Q_j$ in the group law. On an elliptic curve $E:y^2 = f(x)$ with $u(x,y) = x-a$ then $div(u) = P+(-P)-2O, P = (a,\sqrt{f(a)})$ so $div(\frac{1}{u(.+Q)}-\frac{1}{u(R)}) = ...$ – reuns Feb 10 '19 at 5:02

I do not pretend that to be the "right" answer and at least it is not conventional one, but nevertheless it helped me suffering from similar problems many years ago. Indeed, when you begin to study elliptic curves (or more general curves algebraic geomery) you may feel that there are many things that beginner cannot "play/touch by hands" - that exactly are the things about divisors, Riemann-Roch theorem etc...

The way round which I found quite instructive at least for me - is to consider "degenerate" singular curves. It might be strange - usually singularities creates problems, but in that case it is not like that - singular rational curves - gives you a play ground where everything can be "touched by hand".

Consider general elliptic curve y^2 = x^3+ax+b, it migt be degenereted to cusp curve y^2=x^3. And that curve might described in a little different way - consider the polynomials C[z] and impose additional condition f'(0) = 0 . So you get polynomials where $$z$$ is missed - that ring is generated by z^3, z^2, which obviously satisfy the relation y^2=x^3, so it is the same cusp curve.

Now let us turn to your questions for cusp curve:

The "principal" divisors on cusp curve - are those multisets of points which correspond to zeros of such polynomials $$f'(0)=0$$ (and corresponding rational functions).

Exercise try to describe non-principal divisors.

You may see that linear functions do not satisfy the condition f'(0) = 0, so any point P is not principal divisor (or "P- infinity" if one wants to work with projective version). So as a set (and actually group) non-principal divisors forms complex plane $$C$$. That is similar to elliptic curve - non-principal divisors corresponds to points on the curve (or "P- infinity" if one wants to work with projective version).

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.