Exercise 14 chapter 6

Gary

I also found the same figure difficult to fathom. It took a long time to convince myself that I had it right. There's still a chance that I have it wrong, but the mathematics of the solution seemed almost trivial once I had understood Needham's diagram. I then drew a few of my own and it started to make sense.
I look forward to reading your interpretation of it. The holidays certainly are a distraction! I hope to be able to devote some of the time to the solution of the remaining exercises.

Vasco

Gary

I like your diagram 1a, especially as it makes it so much easier to understand Needham's diagram at the top of page 333.

On page 3 of your document you quote the text of part (i) of the exercise immediately below figure 2. The reference in the text of the exercise to "The figure above..." is immediately confusing because "The figure above" refers to Needhams diagram at the top of page 333 and not to your figure 2.

Also, the distance between $S$ and $z_p$ in figure 2 is not the real part of $z_p$, but the modulus $|z_p|$ of $z_p$.

At the end of paragraph 1 on page 3 you write "...which is perhaps parallel to the plane of the unit circle. I don't understand why you use the word "perhaps" since from page 284, referenced by Needham in the exercise, we know for certain that $\Pi_p$ is parallel to the unit circle.

In my opinion Needham's diagram on page 333 should be thought of as a diagram of the complex plane after the unit circle, the stereographic projection of the great circle, and the projective line L, have all been superimposed (collapsed down) onto the unit circle.

In the third paragraph of page 3 you refer to the fourth quadrant. I don't understand this, since your figure 2 could apply for any value of $\theta$ and therefore could be in any quadrant.

On page 4 paragraph 1 you again refer to the real part of $z_p$ when it should be $|z_p|$, and when you write $e^{i\phi}$ in this paragraph, it should be $e^{i\theta}$ and I don't agree with your specification of $z_p$ at the end of the paragraph, it should just be $z_p=[-\tan\phi]e^{i\theta}$.

Just after your definitiuon of equation (2) you say that $z_s$ is a 3-D point, but it isn't, it is a 2-d point lying in the complex plane.

Reading your solution I realised that I had omitted to write anything about the comparison with exercise 9 in my solution, so I will rectify that as soon as I can.

Vasco

Gary



I completely understand your point on this - I think, and it all depends on what we mean by something like $z_p$. Sometimes there is no confusion and $z_p$ can be used one minute as a label for a point and then a moment later as a complex number. Needham himself does this frequently. When we are working in the complex plane all this works well, however, when we start using (as in this exercise), more than one complex plane and we venture into three dimensions I think we (and that includes me, Needham and anybody else) need to be more careful to explain what we mean.
When we use stereographic projection from the sphere to the complex plane then we end up with a set of points $z_s$, as a set of complex numbers in the equatorial complex plane.
When we use central projection as defined on page 284 we also end up with a set of complex numbers $z_p$ in some other complex plane tangential to the sphere.
We now have two sets of complex numbers, derived in two different ways from the same set of points $\widehat{z}$ on the sphere. We then choose to think of these two sets of points as lying in the same complex plane, as in Needham's diagram on page 333.

I deliberately avoid thinking of this superimposition of the two planes as a "projection" to avoid leading myself down the garden path. The two complex planes in question need not be parallel for us to superimpose them. We can then think of the stereographic projection and central projection as inducing a transformation of the set of points $z_s$ to the set of points $z_p$, all taking place in the complex plane. Parts (ii)-(vi) of this exercise go on to explore the nature of this induced mapping on the complex plane. Retracted (see later post)

To answer your specific points above, think of $S$ as the origin of the complex plane containing the points $z_p$. Then $|z_p-S|=|z_p-0|=|z_p|$.



I don't think there is a contradiction in the problem. Following on from my comments above, the plane of projection IS a complex plane just as the equatorial plane of the sphere IS another complex plane. This is why I argue that $z_s$ and $z_p$ are points in two dimensions. We know that they both lie in their respective complex planes and so we only need two values to describe each point.
I now see from what you have written about layer-cake cosmologies, that my description of "collapsing down" the structure of figure 2 so that the planes are superimposed can lead to ways of thinking about the problem that I am not comfortable with, and so I withdraw it and replace it with what I have written below.

For me using $\widehat{z_p}$ is confusing because this is Needham's convention for points on the sphere and $z_p$ is not on the sphere.

I would say to you: try and get used to the idea of two complex planes; if it helps psychologically, then draw two separate diagrams, one for the stereographic projection, with the equatorial plane as the complex plane, and another for the central projection, with the plane of projection as the complex plane, and then at the end superimpose the two complex planes. I only put them on the same diagram for convenience. In my head I am thinking of them as completely separate.

Vasco

PS Why do you say in your comparison that "In both problems there is a stereographic projection from the south pole"? In Ex 14 there is no s-projection from the south pole.

Gary

I mean two diagrams like figure 2, not two 3 dimensional plots, so that each diagram has only one complex plane. One 3 dimensional seems sufficient.
I do not understand your vertical arrows except for the one from $z_p$.
You do not need another name for $z_p$ since it is still $z_p$.

Vasco

Gary

In a recent post above I wrote



This is not correct - I'm not sure why I wrote it - I may have been thinking of another exercise.

Anyway I have highlighted it in red in the original post.

Vasco

Gary

But those points don't go onto the collapsed $\mathbb{C}$ plane because they do not lie in the $\mathbb{C}$ plane, they just disappear from the scene.

Vasco