Homeomorphism and the group structure on a circle | Algebraic Topology 2 | NJ Wildberger
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users cite it when asking about specific moments, such as the circle multiplication rule and points at infinity in projective geometry, and answerers point to it to explain how that product matches addition of angles.
Recommended by 1 people across 1 different months, 2012 to 2020.
What it is
The second lecture in N J Wildberger's beginner Algebraic Topology course at UNSW. It defines homeomorphism, shows that the line and circle are not homeomorphic, and introduces a group structure on a circle (or general conic) using Pascal's theorem, with some background on projective geometry.
What people said
Recommends Martin Sleziak · 2012-04 · Mathematics Stack Exchange · on “How do you do this example on a video lecture about algebraic topology?”
Using some trigonometric identities [LINK] you can get: $$\newcommand{\vp}{\varphi} \begin{align} \cos 2\vp &= \frac{\cos^2\vp-\sin^2\vp}{\cos^2\vp+\sin^2\vp}=\frac{1-\tan^2\vp}{1+\tan^2\vp}\\ \sin 2\vp &= \frac{2\sin\vp\cos\vp}{\cos^2\vp+\sin^2\vp}=\frac{2\tan\vp}{1+\tan^2\vp}\\ \end{align} $$ So $e(\tan\vp)=(\cos2\vp,\sin2\vp)$. If we let $\vp$ run through $\langle 0,\pi)$, we get all points on the circle. Also you can notice that the multiplication of the points on the circle defined in the video, starting cca around 15:15 [this video] and the definition is after 17:15 [this video] , corresponds to addition of angles. Just notice the if you denote the centre of the circle by $C$, angles $|\angle OC
Cites Victor · 2012-04 · Mathematics Stack Exchange · on “How do you do this example on a video lecture about algebraic topology?”
In 48:15 of this video, I don't understand what the product have to do with the problem 4 and what that product is meaning. Link: [this video] Show that with $e(h)=(\frac{1-h^2}{1+h^2}, \frac{2h}{1+h^2})$, $e(h_1)*e(h_2)=e(h_3)$ where $h_3=\frac{h_1 + h_2}{1-h_1h_2}$, also, any interesting background or properties about this problem or any of its significance?
Cites user123 · 2014-06 · Mathematics Stack Exchange · on “A question about three collinear points”
This video [this video] (at 44:00) says that in a projective space if three points are collinear, and two of those points lie at infinity, then the third point will also have to lie at infinity. I wonder why that has to be true. Why can I not have three collinear points such that two of them are points of intersection of parallel lines (and hence lie at infinity), and the third is, say, the origin? EDIT: One may think that the two points at infinity are the same point. However, what if we take one pair of parallel lines, and take points of intersection at both sides? I feel both these points are different, unless it is defined that both are the same.
Cites GENIVI-LEARNER · 2020-08 · Mathematics Stack Exchange · on “Intuitive Understanding of How Parallel Lines Meet in Projective Geometry”
I am watching a lecture on Topology, where it is mentioned that in Projective Geometry Parallel Lines Meet [this video] . I am interested in intuitive idea of how is that even possible. Is in projective geometry do we make an assumption that at infinity the parallel lines meet just like we make an assumption that any sequence diverging to infinity is same at the infinity?