# 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: http://www.youtube.com/watch?feature=player_detailpage&list=PL6763F57A61FE6FE8&v=R_gDV17X7pc#t=2880s

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?

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Please add the problem itself to your post if you want help. (source is nice, but don't expect everybody to watch it just to help you) – example Apr 7 '12 at 15:43
Also, there's the option on Youtube to link to the exact time (48:15) you want to start at. – Tyler Apr 7 '12 at 16:11
@TylerBailey - i am not sure how to use it, if you want to edit it, please do so – Victor Apr 7 '12 at 16:14
The formulas resemble the addition law for the tangent function. Given any two real numbers $h_1$ and $h_2$, write $h_1 = \tan(x_1)$ and $h_2 = \tan(x_2)$. Then $h_3 = \tan(x_1+x_2)$. If $h = \tan(x)$ then $e(h) = (\cos(2x),\sin(2x))$. The graph of the parametrization is not the circle, as Tyler writes, but the circle with the point $(-1,0)$ missing. – KCd Apr 7 '12 at 16:41
@KCd Whoops, sorry! I'll delete my comment. I was being careless. – Tyler Apr 7 '12 at 16:58

Using some trigonometric identities 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 and the definition is after 17:15, corresponds to addition of angles. Just notice the if you denote the centre of the circle by $C$, angles $|\angle OCA|=\alpha$ and $|\angle OCB|=\beta$, and if $D=A*B$, the you have $|\angle BCD| = |\angle OCA|$, which gives you $|\angle OCD|=|\angle OCA|+|\angle ACB|+|\angle BCD|= \alpha+(\beta-\alpha)+\alpha=\alpha+\beta$.
(Try to draw a picture - it might help.)

Now to finish the problem, you only have to know that $$\tan (x+y)=\frac{\sin(x+y)}{\cos(x+y)}=\frac{\cos x\sin y+\sin x\cos y}{\cos x\cos y-\sin x\sin y}=\frac{\tan y+\tan x}{1-\tan x\tan y}.$$

Professor Wildberger mentions a paper by S. A. Shirali from Mathematical Gazette as a reference for this product. I did not find this paper online, but I found another paper by the same author, which mentions this topic.

• S. Shirali, Groups associated with conics, Math. Gaz. 93 (2009), 27-41
• Shailesh A. Shirali: Algebraic methods in plane geometry 1. The use of conic sections, DOI: 10.1007/s12045-008-0100-3
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Suppose $O = (1,0)$, $A = e(h_1)$, $B = e(h_2)$, and $e(h_1)*e(h_2)=e(h_3) = C$, where $OC$ is parallel to $AB$.

$$AB = e(h_2) - e(h_1) = \left (\dfrac{(1+h_1^2)(1-h_2^2) - (1-h_1^2)(1+h_2^2)}{(1+h_1^2)(1+h_2^2)}, \dfrac{2h_2(1+h_1^2)-2h_1(1+h_2^2)}{(1+h_1^2)(1+h_2^2)}\right) = \left( \dfrac{2(h_1^2-h_2^2)}{(1+h_1^2)(1+h_2^2)}, \dfrac{2(h_2-h_1)(1-h_1 h_2)}{(1+h_1^2)(1+h_2^2)} \right)$$

$$OC = \left( \dfrac{1-h_3^2}{1+h_3^2}, \dfrac{2h_3}{1+h_3^2} \right) - (1,0) = \left( \dfrac{-2h_3^2}{1+h_3^2}, \dfrac{2h_3}{1+h_3^2} \right)$$

$AB$ being parallel to $OC$ means that $$\dfrac{\frac{2(h_1^2-h_2^2)}{(1+h_1^2)(1+h_2^2)}}{\frac{-2h_3^2}{1+h_3^2}} = \dfrac{\frac{2(h_2-h_1)(1-h_1 h_2)}{(1+h_1^2)(1+h_2^2)}}{\frac{2h_3}{1+h_3^2}}$$ $$\dfrac{h_2+h_1}{h_3} = 1-h_1h_2$$ Rearrangement gives the desired result.

I think the interesting part is in the existence of group structures on different conics.

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