Exercise 9 chapter 6

Vasco,

And finally, I have resumed the exercises with Ex. 9, though I must still spend some more time on Subsection 11.

nh.ch6.ex9.pdf (140.07 KB)

Gary

Gary

I have looked at your solution to exercise 9.

In part (i) the net mapping is a mapping from the Poincaré disc to the Klein disc via the hemisphere model, so I think that when you write "The semi-circle on the sphere appears as a straight line on the PD", it could be confusing, as this final mapping takes us to the Klein disc.

In part (ii) right at the end you write:

"Since $z$ and $z'$ fall on the real line,..."

But this is incorrect. In your Figure 1, and the figure (b) which accompanies the exercise, the complex plane is perpendicular to the plane of the paper and the "vertical" axis in your diagram is not the imaginary axis of the complex plane and the horizontal axis in your diagram is not the real axis.

Vasco

Gary

But z and z' don't lie on the real line. Look at your figure 2 on page 2.

Vasco

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Gary
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Gary

In my opinion in [21b] on p. 142 and [23a] on p. 147 there is no introduction of a second $\mathbb{C}$, and I think it's important to understand this. In [21b] for example, the two $z$ points on $\Sigma$ represent points on the sphere and are points in 3 dimensions, and the same is true for the $z$ point on $\Sigma$ in [23a]. They should not be thought of as complex numbers.

I know that the maths on page 146 introduces $X+iY$ as well as $x+iy$ but this is not the introduction of a second complex plane. Notice that points on the sphere require three coordinates to describe them, either $(X,Y,Z)$ or $(r,\phi,\theta)$. On the Riemann sphere $r=1$ and so we only need $(\phi,\theta)$.
For every point in the complex plane, requiring two numbers $(x,y)$ to specify them. The corresponding point under stereographic projection requires 3 numbers to specify it.

Vasco

Gary

I have to say that I disagree with you on this and especially when you say that Inversion cannot work unless the points $\widehat{\tilde{z}}$ and $\tilde{z}$ are in a second $\mathbb{C}$ plane. Inversion is a geometric concept and doesn't depend on the complex plane in any way. As I see it, the reason for the use of "$z$" for the points on the sphere is to remind us that given the points $z$ and $\widetilde{z}$ in the complex plane and using the tilde to denote stereographic projection, we obtain the points $\widehat{\tilde{z}}$ and $\tilde{z}$. The notation does not give any detail about the points themselves, but just tells us how they are obtained from the original points in the complex plane (i.e. by stereographic projection).

We may have to agree to differ on this one!

Vasco

Gary

If $K$ is a circle, then $\mathcal{I}_K$ means inversion in the circle $K$. The formula (4) on page 125 is a specific formula for inversion of the complex plane in the circle $K$, when the circle $K$ lies in the complex plane. The formula is only applicable in this scenario, whereas $\mathcal{I}_K$ makes no reference to the complex plane and just means inversion in the circle $K$.

Also, if you have applied $\mathcal{I}_K$ in the vertical complex plane, then multiplying the result by $e^{i\theta}$ rotates this vertical plane about an axis through the origin orthogonal to the plane, and so does not bring about any three dimensional effect.

As far as (21) is concerned on page 147, this is a formula for the $z$ in the complex plane which is the stereographic projection of $\widehat{z}$, the point $(\phi,\theta)$ on the sphere. It turns out from the maths that $|z|=\cot(\phi/2)$ and $\text{arg}z=\theta$, and so $z=|z|e^{i\text{arg}z}=\cot(\phi/2)e^{i\theta}$.

The quantity $e^{i\theta}$ does not rotate $z$ about the origin. To rotate $z$ about the origin you would have to write $ze^{i\theta}$.

If you wanted to project back to $\widehat{z}$ in [23b] you can find $\phi$ as $\phi=\cot^{-1}(|z|)$, and $\theta$ as $\theta=\tan^{-1}(y/x)$, where $y=\text{Im}{z}$ and $x=\text{Re}z$, and taking care to get the right value of $\theta$ from the signs of $x$ and $y$.

Vasco

Gary

I feel as though I am beginning to understand your thinking on this. Here are my thoughts:

It is clearly perfectly acceptable to have the complex plane oriented in any direction we like - horizontal, vertical, whatever. However if we have more than one complex plane existing at the same time, then we are in trouble, as there is no way to distinguish a complex variable in one plane from a complex variable in another plane, and immediately we do arithmetic with these complex numbers the 'action' can only take place in one plane, and will consequently produce erroneous results.

Subsection 6 on pages 290-293 on the subject of quaternions is an attempt to extend the ideas of the complex plane to more dimensions, and Needham does not use them outside this subsection.

Even when we are thinking about mappings and we have the z-plane being mapped to the w-plane, we only really have one complex plane. The plane is being mapped to itself.

To summarise my thoughts I would say that we can only ever have one complex plane existing at any given moment. Also, if we introduce one complex plane and use it to obtain a result about some three dimensional configuration of space, we cannot then forget about the first plane and introduce a second plane and use the result obtained from use of the first plane and carry on. If we do we will obtain erroneous results, because the results we obtained using the first plane will look to the second plane like its own results and we are in trouble.

Vasco