A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27

(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$.

  1. What is a unit normal vector to $H$?

  2. Compute the image of $X:=H\cap \mathbb S^{2}$ under the stereographic projection $\Phi$.


  • For 2, I computed $X$ to be a 3D circle, parametrised here, and its image to be $Y:= \Phi(X) = \{|z-(1+i)|^2 = 3\}$, but now I ask:


For 1, What's the relevance of asking about the unit normal vector?

I computed the unit normal vectors to be $[1,1,-1]\frac{\pm 1}{\sqrt{3}}$. I observe their terminal points to be on the unit sphere.

Here are the parametrisations:

$Y:= \Phi(X) = \{|z-(1+i)|^2 = 3\}$ is parametrised:

$$\begin{bmatrix} y_1(t)\\ y_2(t)\\ y_3(t) \end{bmatrix} = \begin{bmatrix} \sqrt{3}\cos(t) + 1\\ \sqrt{3}\sin(t) + 1\\ 0 \end{bmatrix} = \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\sqrt{3}\cos(t)+ \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}\sqrt{3}\sin(t)$$

$X$ is parametrised:

$$\begin{bmatrix} x_1(t)\\ x_2(t)\\ x_3(t) \end{bmatrix} = \begin{bmatrix} \sqrt{\frac 2 3} \cos[t]\\ -\sqrt{\frac 2 4} \sin[t] - \sqrt{\frac 2 {12}} \cos[t]\\ -\sqrt{\frac 2 4} \sin[t] + \sqrt{\frac 2 {12}} \cos[t] \end{bmatrix} = \begin{bmatrix} \sqrt{\frac 1 3}\\ -\sqrt{\frac 1 {12}}\\ \sqrt{\frac 1 {12}} \end{bmatrix}\sqrt{2}\cos(t)+ \begin{bmatrix} 0\\ -\sqrt{\frac 1 {4}}\\ -\sqrt{\frac 1 {4}} \end{bmatrix}\sqrt{2}\sin(t)$$

$$H = \{x + y -z = 0\} = \{[1,1,-1] \cdot [x,y,z]=0\} = \{[1,1,-1]\frac{1}{\sqrt{3}} \cdot [x,y,z]=0\}$$ is parametrised:

$$\begin{bmatrix} h_1(r,s)\\ h_2(r,s)\\ h_3(r,s) \end{bmatrix}=\begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}r + \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}s$$

  • 2
    $\begingroup$ Great circles on $\mathbb S^2$ are identified by such normal vectors. Since the stereographic projection is circle-preserving, my interpretation of this exercise is that it suggests exploring the correspondence between unit vectors and circles. $\endgroup$ – Giuseppe Negro Aug 10 '18 at 10:11
  • $\begingroup$ Thanks @GiuseppeNegro. I edited to include parametrisations. Soooo, $X$ is identified by unit normal vector of $H$? How exactly please? $\endgroup$ – BCLC Aug 10 '18 at 11:03
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    $\begingroup$ Any plane passing through the origin is characterised by its normal vectors (they are two, opposite in sign). Any great circle on the sphere is identified by an unique plane. Conclusion : any great circle is identified by a unit vector. The stereographic projection maps circles on the sphere to circles in the plane, defining a map that takes a unit vector, associates to it the corresponding great circle and then the corresponding circle in the plane. Maybe this exercise is meant to explore this map a little bit. $\endgroup$ – Giuseppe Negro Aug 10 '18 at 12:51
  • $\begingroup$ @GiuseppeNegro right yeah so what's the identification process exactly? How does unit normal of H lead to the image other than all the above? I'm thinking there's some shortcut pattern from the unit normal to the image $\endgroup$ – BCLC Aug 10 '18 at 13:12
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    $\begingroup$ I don't think so. In the end, you will have to perform that computation, maybe in disguise. I foresee no real shortcut, but I could be wrong. $\endgroup$ – Giuseppe Negro Aug 10 '18 at 13:13

Here is one possible explanation. We are to find the image of $H\cap{\Bbb S}^2$ under $\Phi$.

A solution may go like this: $$ Y\in\Phi(H\cap{\Bbb S}^2)\iff X=\Phi^{-1}(Y)\in H\cap{\Bbb S}^2 \subseteq H. $$ The latter can be described by $ax_1+bx_2+cx_3+d=0$, where $(a,b,c)$ is a normal vector, giving an equation for $Y$ once the mapping $\Phi^{-1}$ is explicitly calculated.

EDIT: What I mean here is that $\Phi$ is the mapping $$ \Phi\colon \underbrace{(x_1,x_2,x_3)}_{X}\in{\Bbb S}^2\to \underbrace{(y_1,y_2,y_3)}_{Y}=\left(\frac{x_1}{1-x_3},\frac{x_2}{1-x_3},0\right)\in{\Bbb R}^2. $$ It is a bijection and $$ \Phi^{-1}\colon(y_1,y_2,0)\to\Big(\underbrace{\frac{2y_1}{y_1^2+y_2^2+1}}_{x_1},\underbrace{\frac{2y_2}{y_1^2+y_2^2+1}}_{x_2},\underbrace{\frac{y_1^2+y_2^2-1}{y_1^2+y_2^2+1}}_{x_3}\Big). $$ The necessary and sufficient condition for $Y$ to belong to the image is that $X$ (on the sphere) belongs to the plane $H$ (if a point is on the sphere, but not on $H$, then it does not belong to $H\cap{\Bbb S}^2$, hence, not mapped to the image due to one-to-one mapping), i.e. orthogonal to the normal vector $$ x_1+x_2-x_3=0. $$ When we substitute (and get rid of the denominator) we get the circle $$ 2y_1+2y_2-(y_1^2+y_2^2-1)=0\iff (y_1-1)^2+(y_2-1)^2=3. $$

  • 1
    $\begingroup$ @BCLC I meant exactly $X=\Phi^{-1}(Y)\in H$, since $\Phi^{-1}$ will be on the sphere in any case, we do not need to borther about. $\endgroup$ – A.Γ. Aug 10 '18 at 9:23
  • $\begingroup$ Thanks A.Γ.! I added the parametrisations. May you please explain explicitly how we get an equation for $Y$ with $\Phi^{-1}(Y)$ and the unit normal vector? Also, how is $X \in H$ please? I actually don't quite see it from the parametrisations. $\endgroup$ – BCLC Aug 10 '18 at 11:02
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    $\begingroup$ @BCLC I've made some edit, but I am no longer sure that you actually need the normal vector. All you need is the equation for the plane $H$, which is given. $\endgroup$ – A.Γ. Aug 10 '18 at 11:52
  • $\begingroup$ Ayt thanks! ^-^ $\endgroup$ – BCLC Aug 10 '18 at 22:58

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