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I've been learning about homeomorphisms on the circle and was watching this excellent algebraic topology video covering some basics of projective geometry, and describing a group structure on the circle which is underpinned by Pascal's theorem.

The group product $X\star Y=Z$ is arrived at by drawing a chord through $X$ and $Y$ and then drawing a parallel chord through the identity, and where that chord meets the circle is the product $Z$. It's not immediately obvious (at least to me) that drawing a unit circle in the complex plane and setting the group identity to be $1+0i$ makes $(S^1,\star)$ the standard multiplicative unit circle group in the complex plane $(S^1,\times)$. But picking sample products such as $i\star i=-1$ and in fact any of the points of the compass, does yield matching results for $\star$ and $\times$.

I was able to satisfy myself that results matched for a good selection of elements of the Prufer 2-group, which being dense in the circle and would appear to order-embed, would determine this must be the same group. Is it the exact same group and product?

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    $\begingroup$ So you want to see a proof that the "chord multplication" agrees with the standard multiplication? $\endgroup$
    – Paul Frost
    Jan 26 at 10:41
  • $\begingroup$ @PaulFrost yes, I guess so. $\endgroup$ Jan 26 at 10:55

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Let $z_j = x_j + i y_j \in S^1$ ($j = 1,2$) be two elements on the unit circle.

In order to show that the two multiplications are the same it suffices to show that the slope of the chord between $z_1$ and $z_2$ coincides with the slope of the chord between $z_1 z_2 = (x_1 x_2 - y_1 y_2) + i (x_1 y_2 + x_2 y_1)$ and $1 + 0i$.

That means we have to show that $$ (x_1 - x_2) (x_1 y_2 + x_2 y_1) = (y_1 - y_2) (x_1 x_2 - y_1 y_2 - 1)$$ and with the help of $x_j^2 + y_j^2 = 1$ for $j = 1,2$, this is easily done.

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I know this already has an answer, but there is a more geometric way of thinking about it that you might find nice.

  • The points $1, x, y, z$ form a cyclic quadrilateral. To avoid confusing notation, call these points $A, B, C, D$ respectively. Assume $x$ comes before $y$ as you traverse the circle anticlockwise.
  • Since the line $\vec{AD}$ is parallel to $\vec{BC}$, this quadrilateral is a trapezium.
  • Since it's a cyclic quadrilateral, $A+C = B+D = \pi$. Since it's a trapezium, $A+B = C+D = \pi$.
  • It follows the whole diagram is symmetric under reflection in the perpendicular bisector of AD (which is also the perpendicular bisector of BC).
  • If you add O, the center of the circle, to the diagram, then the desired fact to prove is

$$\arg x + \arg y = \arg z$$ i.e. $$\angle AOB + \angle AOC = \angle AOD$$ But it's obvious (here we use the anticlockwise orientation of $ABCD$) that $\angle AOD = \angle AOC + \angle COD$, and by the symmetry $\angle COD = \angle AOB$. Hence,

$$\angle AOD = \angle AOC + \angle AOB$$

which is what we wanted to prove.

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  • $\begingroup$ Note that I made some assumptions about how nicely the trapezium could be laid out. It's not obvious to me this proof covers the case where $\arg x + arg y > 2\pi$. $\endgroup$ Jan 28 at 9:55
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Too long for a comment.

Consider the reducible degree three curve $${\cal C}: z(x^2+y^2-z^2)=0$$ in ${\Bbb P}^2_{\Bbb R}$ with $O$ the point $(x:y:z)=(1:0:1).$

Let $A,B,C,$ and $O=(1,0)$ be points on $x^2+y^2-1=0$ in ${\Bbb R}^2={\Bbb A}^2_{z=1}\subset{\Bbb P}^2.$

$A$ and $B$ in ${\Bbb P}^2$ define a line that intersects ${\cal C}$ in a point I call $\infty_{AB}.$ Then proceed as with the elliptic curve group law: $$A+B+\infty_{AB} =0$$ and similarly $$\infty_{OC} + O +C=0$$ remembering from projective geometry that $AB$ parallel to $OC$ means that $\infty_{AB}=\infty_{OC},$ we have $$A+B=-\infty_{AB}=-\infty_{OC}=O+C.$$ If we let $O$ be the identity for $+,$ we get $$A+B=C.$$

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