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What is the definition of straight line in $\hat{\mathbb{C}}$?

Is it defined as $\{x\in\mathbb{C}: \frac{Re(x-a)}{Re(b)} = \frac{Im(x-a)}{Im(b)}\}\cup \{\infty\}$?

($a,b$ are complex numbers and $b\neq 0$)

I'm currently reading "Ahlfors - Complex analysis", and this terminology 'line' appears in p.79 Thm. 13:

The cross ratio $(z_1,z_2,z_3,z_4)$ is real iff $z_1,z_2,z_3,z_4$ lie on a circle or on a straight line.

What's the definition of line?

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    $\begingroup$ It's not unreasonable to think of a line as a circle that passes through $\infty$. $\endgroup$ – Hurkyl Apr 10 '14 at 17:50
  • $\begingroup$ @Hurkyl What is your definition of 'circle'? Regarding the Riemann sphere as $S^2$, yes, lines are latitudes and altitudes of $S^2$. Do you mean this? Then again, how do you define latitudes and altitudes? $\endgroup$ – user140374 Apr 10 '14 at 17:51
  • $\begingroup$ @Hurkyl So is there a way to define a line in a form regarding it as a circle passes through $\infty$ formally? $\endgroup$ – user140374 Apr 10 '14 at 17:59
  • $\begingroup$ Admittedly, what I really do is see that fractional linear transformations map lines and circles to lines and circles (given their ordinary plane geometry interpretations), and so I form a concept that encapsulates both (this concept usually gets named "circle"). A visualization that might help is to construct the circle that passes through the origin and has center on the real axis, and then taking the "limit" as the center slides off to infinity results in the imaginary axis. $\endgroup$ – Hurkyl Apr 10 '14 at 18:00
  • $\begingroup$ I actually learned about this stuff long before I encountered the complex numbers, though, through studying inversive geometry. $\endgroup$ – Hurkyl Apr 10 '14 at 18:03
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(a) In ${\mathbb C}\sim{\mathbb R}^2$ a line is a line and is given by an equation of the form $ax+by=c$ with $a^2+b^2\ne0$. A parametric representation of such a line would be $$\ell:\quad t\mapsto z(t)= p t+ q\qquad(-\infty<t<\infty)$$ with $p\in{\mathbb C}\setminus\{0\}$.

(b) When ${\mathbb C}$ is extended to the Riemann sphere $\hat{\mathbb C}$ by adjunction of the point $\infty$ then it make sense to let the lines from (a) to go through this point $\infty$ as well.

(c) The "symmetry group" of $\hat{\mathbb C}$ is no longer the group of euclidean similarities $z\mapsto c\>z+ d$, $\>c\ne0$, but is the larger group ${\cal M}$ of Moebius transformations. In the world of ${\cal M}$ euclidean lines are no longer recognizable, but "circles" are. Any three different points $z_k\in\hat{\mathbb C}$ determine a unique "circle" $$C:=\{z\in\hat{\mathbb C}\>|\>{\rm cross\ ratio}(z_1,z_2,z_3,z)\in{\mathbb R}\cup\{\infty\}\}\ .$$ The stereographic image of $C$ will be either a circle or straight line in ${\mathbb C}$. In other words: When restricting such a "circle" to the subset ${\mathbb C}\subset\hat{\mathbb C}$ that is modeled by a sheet of paper we see either an ordinary euclidean circle in ${\mathbb C}$, or if the given "circle" happens to pass through the point $\infty$, we see a euclidean line.

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  • $\begingroup$ Thank you very much! Could you please give more details for (c)?? (fyi, I'm familiar with Mobius group) What do you mean : Any three distinct points $z_k$ determine a unique circle? Do you mean any of "circle" in complex plane and "straight line" in the Riemann sphere? $\endgroup$ – user140374 Apr 10 '14 at 19:01
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It is 1-point compactification of a Euclidean straight line, as you suspected.

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