-1
$\begingroup$

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.27,Exer 3.28

(Exer 3.27) Consider the plane $H$ determined by the equation $x + y -z = 0$. What is a unit normal vector to $H$? Compute the image of $E:=H\cap \mathbb S^{2}$ under the stereographic projection $\Phi$.

Asked here and I answered agreeing with OP that the image is the circle $T: = \{|z-(1+i)|^2 = 3\}$, but now I ask:

Q1. Is $(2)$ below wrong?

$$\forall \ \text{planes} \ J, \exists \text{circle} \ C \in \mathbb S^2 : C = J \cap \mathbb S^2 \tag{2}$$

To clarify, compare that (2) is different from (1) below. We know that $$\forall \ \text{circles} \ C \in \mathbb S^2, \exists \ \text{plane} \ J : C = J \cap \mathbb S^2. \tag{1}$$

Q2. Do I go wrong in attempting to disprove (2) as follows?

I think $(2)$ contradicts Exer 3.27: I observe that $E$, whose image is the circle $T$, is both

If $(2)$ is true, then for $J=H, \exists C \in \mathbb S^2: C = H \cap \mathbb S^2 =: E ↯$.

Q3. How is Exer 3.27 consistent with the succeeding Exer 3.28?

-

(Exer 3.28) Prove that every circle in $\hat{\mathbb C}$ is the image of some circle in $\mathbb S^2$ under stereographic projection $\Phi$.

-

I rephrase: $$\forall \ R \in \hat{\mathbb C}, \exists C \in \mathbb S^2 : \Phi(C)=R$$

-

Now consider $T$, the circle in Exer 3.27 s.t. $\Phi(E)=T$.

It seems that by Exer 3.28, for $R= T, \exists C \in \mathbb S^2 : \Phi(C)=T$.

$$\therefore, \Phi(E) = \Phi(C) \ \text{while} \ E \ne C \ \text{, but} \ \Phi \ \text{is bijective ↯ ?}$$

$\endgroup$
0
+50
$\begingroup$

As already verified, all intersections of a circle and plane are circles either geodesic/great or small. I avoid symbols in the answer here.

If 2-parameter surface normal coincides with 1-parameter circle normal then the intersection is a great circle. The intersection is a geodesic.

From differential geometry standpoint $\psi$ is between arc and meridian. Cylindrical coordinates $(r,z)$ Clairaut constant derivative $\dfrac{d(r \sin \psi)}{dz}=0$ forms with respect to an arbitrary North-South polar axis.

If the 2-parameter surface normal does not coincide with 1-parameter circle normal, but makes an angle $\gamma$ of relative latitude then the intersection is a small circle. Intersection is like a parallel circle but not an equator.

Clairaut constant derivative $\dfrac{d(r \sin \psi)}{dz}$ is a constant $ =\tan \gamma$ forms with respect to an arbitrary North-South polar axis.

In stereographic projection all non-equatorial sections of a plane passing through North pole of the sphere at angle $\gamma$ to North-South polar plane are small circles.

$\endgroup$
  • $\begingroup$ Narasimham, is $E$ indeed an ellipse? $\endgroup$ – BCLC Aug 9 '18 at 5:23
  • $\begingroup$ Narasimham, I posted an answer. How is it please? $\endgroup$ – BCLC Aug 9 '18 at 6:13
0
$\begingroup$

(Q1) I suspect some relationship is being illustrated. The 1st 2 coordinates in $[1,1,-1]$ are the centre of the circle $\Phi(E)$ while the norm of $[1,1,-1]$ s the radius of $\Phi(E)$.

(Q2) Yes if the planes intersect $\mathbb S^2$ or $\emptyset$ is considered a circle in $\mathbb S^2$.

(Q3), (Q4) E is not an ellipse. It is a circle in 3D (see here too) parametrised as (WA)

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

Observe that while $x^2+y^2+xy=\frac12$ is an ellipse, $x^2+y^2+xy=\frac12 \wedge z= x+y$ is a circle.

$\endgroup$
  • 1
    $\begingroup$ I guess your major confusion was WA saying ellipse on the intersection. When you set $z=x+y$ into $x^2+y^2+z^2=1$ you get indeed an ellipse, but this is the projection of the intersection onto $xy$-plane, i.e. all the coordinates $(x,y)$ that satisfy both equations. $\endgroup$ – A.Γ. Aug 9 '18 at 9:16
  • $\begingroup$ @A.Γ. I was thinking projection but unable to express precisely. Thanks! Any idea for (Q1) please? $\endgroup$ – BCLC Aug 9 '18 at 9:18
  • 1
    $\begingroup$ To see that all intersections are circles: 1. Intersection of a plane $z=\text{const}$ and the sphere is a circle (or empty set). 2. Any plane can be rotated to become $z=\text{const}$, and sphere is rotation invariant. $\endgroup$ – A.Γ. Aug 9 '18 at 9:18
  • 1
    $\begingroup$ Is the image of $\Phi$ the plane $z=0$ or $z=-1$ (there are different conventions)? $\endgroup$ – A.Γ. Aug 9 '18 at 10:17
  • 1
    $\begingroup$ I do not see how knowing the normal may help finding the image. The exercise is pretty straightforward without the normal. $\endgroup$ – A.Γ. Aug 9 '18 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.