Can one axiomatize projective lines using the cross-ratio?

Question

I known axiomatizations of projective spaces of dimension > 2 and also of projective planes (either those obeying the axiom of Pappus, which come from fields, or those obeying the axiom of Desargues, which come from division rings, or the fully general ones, which come from planar ternary rings).

How about projective lines? One might try to axiomatize at least those coming from fields using the cross-ratio. If $k$ is a field the cross-ratio gives at least a partially-defined 4-ary function

$$ kP^1 \times kP^1 \times kP^1 \times kP^1 \to kP^1 $$

where $kP^1$ is the corresponding projective line. Is the cross-ratio well-defined even when several of the four arguments coincide? Does it give a regular map of projective varieties? Does it obey enough equational laws that we can use these as axioms for an interesting concept of projective line?

Some related ideas are discussed here:

• Mike Shulman, Synthetic projective lines, Math Overflow.
• Jacob Lurie, Action of PGL(2) on projective space, Math Overflow.

The latter turned up this nice result: as is well known, $\mathrm{PGL}(2,k)$ acts in a sharply 3-transitive way on $kP^1$, but more generally, any sharply 3-transitive group arises from a 'KT-field', which is a kind of generalized field. I don't know how KT-fields are related to planar ternary rings.

However, all this is---at least superficially---a bit different than trying to axiomatize the cross-ratio. Perhaps it's just another viewpoint on the same thing.

Answer 1

Is the cross-ratio well-defined even when several of the four arguments coincide?

No, it is not. Modeled as homogeneous coordinates over $k^2$ you get the null vector in cases where three or more points coincide, but that's not an element of your line any more.

If I understand you correctly, it would be sufficient to have a set of axioms for the projective line from which you can derive all the required axioms of the underlying field, thus demonstrating that there is indeed a field underlying the line. Which means you need to be able to perform arithmetic. At least if the characteristic of the underlying field is different from two, I'd probably approach this using harmonic throws: four points $a,b,c,d$ form a harmonic throw iff $(a,b;c,d)=-1$, so if you have cross ratio you also have harmonic throws.

So how can harmonic throws encode arithmetic operations?

• $(\infty,0;a,-a)=-1$ so you have the additive inverse (except if $a=0$ or $a=\infty$, in which case the cross ratio is undefined as three points coincide).
• $(\infty,a;0,2a)=-1$ so you can multiply or divide by two (again excepting $a=0$ and $a=\infty$).
• $(a,b;\infty,\frac{a+b}2)=-1$ so you can compute the average, and by doubling that the sum of two values. You now have addition. The special cases where $a$ and/or $b$ is zero or infinite can be handled explicitely and included here as well.
• $(-1,1;a,\frac1a)=-1$ so you have the multiplicative inverse (including division by zero resulting in infinity, but excluding $a=\pm 1$).
• $(-a,a;1,a^2)=-1$ so you can square values (unless $a=\pm1$ or $a=0$).
• Using $ab = \frac{(a+b)^2-a^2-b^2}2$ you can combine the above to obtain multiplication. The cases where $a$ or $b$ is zero, $\pm1$ or infinity can again be handled explicitely.

One could start an axiomatization with “there exists at least three distinct points on the line. Picking three and labeling them as $0,1,\infty$ …” and then use the above to define addition and multiplication, then include the axioms of a field expressed this way. The large number of special cases mentioned above may make this a bit awkward, as does the large number of intermediate steps required to e.g. represent multiplication. So I guess the resulting set of axioms could be trimmed down a bit, but it would be a starting point.

I'm currently unsure what to do about characteristic two. You don't have a $-1$ distinct from $1$ there, and you'd have some more undefined cross ratios. I guess there should be a set of axioms which covers characteristic two as well, but I don't know how to get there just now.