Gary
GaryVasco
Posts: 3,352 
|
Post by Gary on May 2, 2016 0:53:57 GMT
Vasco,
Regarding Section XI.2 Rigidity, p. 249, para (-1), ln 3: “By choosing to look along s we find that amplification vanishes at each point of s.”
Would you agree that “along” should be interpreted as a viewer looking at the segment from the side and looking at the length from one end to another, so that the line of sight is perpendicular to the segment?
Gary
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 2, 2016 5:03:14 GMT
Vasco, This question is related to the previous. In the middle of p. 249, we find $f(z) \tilde{} z^m <==> \{ L_1,..., L_{m-1}$ show nothing, but image visible with $L_m$ $\hspace{6em} f'(p) = 0, f''(p) = 0, ..., f^{(m-1)}(p) = 0$, but $f^m(p) \neq 0$ I get the idea that he is drawing an analogy between the derivatives and lenses, and I get the idea that "the higher the derivative that vanishes at p, the greater the degree of crushing at p." But I can't relate that to the two lines following the brackets. I made an image of $z^4$ with derivatives up to the 5th plotted at 0, thinking that might help, but it didn't. Can you elucidate this? nh.ch5.crushing.p.249.pdf (60.95 KB) Gary |
|
|
|
Post by Admin on May 2, 2016 6:08:06 GMT
Vasco, Regarding Section XI.2 Rigidity, p. 249, para (-1), ln 3: “By choosing to look along s we find that amplification vanishes at each point of s.” Would you agree that “along” should be interpreted as a viewer looking at the segment from the side and looking at the length from one end to another, so that the line of sight is perpendicular to the segment? Gary Gary I don't see it in that way, no. Because the mapping $f$ is analytic then as Needham writes: "The amplification of $f$ at a point of $s$ may be read off by looking in any direction". (my italics) This is because all arrows from the point on $s$ are amplified by the same amount (see figure 10 on page 197). Needham then goes on to write: "By choosing to look along $s$ we find that the amplification vanishes at each point of $s$" Here he means choosing the arrow which is tangent to the point on $s$, so that we are looking along $s$ as we travel its length. Because at the top of page 249 we have assumed that $s$ is crushed to a point, then the amplification along $s$ must be zero. Vasco |
|
|
|
Post by Admin on May 2, 2016 8:10:28 GMT
Vasco, This question is related to the previous. In the middle of p. 249, we find $f(z) \tilde{} z^m <==> \{ L_1,..., L_{m-1}$ show nothing, but image visible with $L_m$ $\hspace{6em} f'(p) = 0, f''(p) = 0, ..., f^{(m-1)}(p) = 0$, but $f^m(p) \neq 0$ I get the idea that he is drawing an analogy between the derivatives and lenses, and I get the idea that "the higher the derivative that vanishes at p, the greater the degree of crushing at p." But I can't relate that to the two lines following the brackets. I made an image of $z^4$ with derivatives up to the 5th plotted at 0, thinking that might help, but it didn't. Can you elucidate this? View AttachmentGary Gary On page 205 about halfway through the first paragraph of subsection 2 Breakdown of Conformality Needham writes: "In fact we will show later that the behaviour of a mapping very near to a critical point is essentially given by $z^m, m\geq 2$" I can't find this 'later' anywhere in the book, but it will probably be something like this: Think of the Taylor series at the critical point p of order $m-1$: Lets look at a small circle about p of radius $\epsilon$, and let $z=p+\epsilon$. The Taylor series is $f(z)=f(p)+(z-p)f^{(1)}(p)+(z-p)^2f^{(2)}(p)/2!+...+(z-p)^{m-1}f^{(m-1)}(p)/(m-1)!+(z-p)^mf^{(m)}(p)/m!+...$ If all the derivatives are zero up to and including $f^{(m-1)}$, then $f(z)=(z-p)^mf^{(m)}(p)/m!+...$ or $f(z)\sim z^m$ for $z$ close to $p$. You might see this more easily if you think of the critical point $p$ being at the origin, in which case we can write $f(z)=z^mf^{(m)}(0)/m!+...$ or $f(z)\sim z^m$ for $z$ close to $0$. So the maths bang in the middle of page 249 means that if (and only if) $f(z)\sim z^m$ then what is to the right of the big curly bracket is the case. Each of the two lines is an alternative way of saying the same thing. The symbol $\Longleftrightarrow$ means if and only if. Hope this helps Vasco |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 2, 2016 17:20:53 GMT
Vasco,
Thank you for the answer to the question regarding the sentence “By choosing to look along s we find that amplification vanishes at each point of s.” It sounds right to me and I accept it. But could we not reach the same conclusion by moving along (not looking along) s one point at a time and looking in any direction for the edge of an infinitesimal disk? If we saw nothing, then there is no amplification.
The explanation of the "if and only if" statement seems well constructed and relevant, so I am a little embarrassed to admit that I still can't fit it into Needham's essay.
In the case of $f(z) = z^4$, with critical point p = 0, $f(0) = f'(0) = f''(0) = f^{3}(0)$, but $f^{4}(0) = 24$. Is this what is meant by
$\hspace{3em}"L_1, ..., L_{m-1}$ show nothing, but image visible with $L_m"$?
Then $f^{5}(0) = 0$, and all higher order derivatives are again zero (or are they undefined?).
Considering the concrete example again, an infinitesimal vector at 0 has length $\epsilon$, our function $f(z) = z^4$ crushes it to length $\epsilon^4$ and $f^{4}(0)$ amplifies it to $(24 \epsilon^4)$ (or is it $(24 \epsilon)$?. How do we know that will be enough amplification to make the image visible, assuming it was visible before crushing? Or are we only interested in the fact that amplification is non-zero?
In the second to last paragraph, Needham states that "all the derivatives must vanish". Doesn't this contradict his previous statement that $f^{m}(p) \neq 0$ and the fact that $f^{4}(0) = 24$.
By the way, is Needham using $\tilde{}$ to indicate same order of magnitude?
Gary
|
|
|
|
Post by Admin on May 3, 2016 15:04:27 GMT
Thank you for the answer to the question regarding the sentence “By choosing to look along s we find that amplification vanishes at each point of s.” It sounds right to me and I accept it. But could we not reach the same conclusion by moving along (not looking along) s one point at a time and looking in any direction for the edge of an infinitesimal disk? If we saw nothing, then there is no amplification. Gary At a point $p$ of $s$ if we look in any direction other than along $s$ we are in the dark, we do not know what the amplification is. However if we look along $s$ then we know that it is zero and because $f$ is analytic it must be zero in all directions emanating from $p$. The reason we know that the amplification is zero along $s$ is because we start with the assumption that we know that $f$ crushes $s$ to a point. (top of page 249) I will post an answer to your other questions soon. Vasco |
|
|
|
Post by Admin on May 3, 2016 15:17:10 GMT
Gary This is my second post about your last post - don't miss my first post.In the case of $f(z) = z^4$, with critical point p = 0, $f(0) = f'(0) = f''(0) = f^{3}(0)$, but $f^{4}(0) = 24$. Is this what is meant by $\hspace{3em}"L_1, ..., L_{m-1}$ show nothing, but image visible with $L_m"$? For an explanation of this read subsection 1 Degrees of Crushing on pages 204-5 of chapter 4. Yes they are all zero from 5 onwards. Think of the Taylor series for your function: $z^4f^{(4)}(0)/4!=z^4\cdot 24/24=z^4$ as expected! No, in your concrete example the only thing that happens is that the origin-centred circle of radius $\epsilon$ is crushed by the mapping to a circle of radius $\epsilon^4$, and so you would need lens $L_4$ to see it. No, because in the first instance he is talking about the general case of how a function behaves at a critical point, but in the second case, the penultimate paragraph on page 249, he is arguing that in the case of a function which behaves as described in the italicised statement at the top of page 249, every derivative at every point on $s$ is zero, and so in this special case the entire domain is crushed to a point In this case I think a better interpretation would be "behaves like" (see paragraph 1 of subsection 2 Breakdown of Conformality on page 205) Vasco |
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 4, 2016 4:40:54 GMT
Vasco,
Thank you for all the good answers. In the first question, I think I was mentally attaching an $\epsilon$ disc to each point of s before applying the function.
I won't reply point by point, but I it seems from your answers and another reading that Needham is setting up a distinction between two situations.
The First Situation, Capturing the Essential Character of Analytic Rigidity
The first situation is described in the result paragraph beginning on line 2, p. 249 of XI, 2.
Quote: "If even an arbitrarily small segment of a curve is crushed to a point by an analytic mapping, then its entire domain will be collapsed down to that point."
He returns to this in the penultimate paragraph, finding that the segment s consists entirely of critical points for which f' = 0. Then, considering f' as a mapping, he wrote "We have just seen that this mapping possesses the same property as f did: it crushes s to a point." I surmise that the reason that f' possesses the same property as f of crushing s to a point (not necessarily the same point) is because f' = 0 for all p in s. From there, its turtles all the way down. "All the derivatives of f must vanish" and infinitesimal disks "centred on s" get "totally crushed." Needham didn't give an example function, but perhaps f(z) = const. would quality. Are others possible? From the next page (250), we deduce that this function is unique for the specified constant, so perhaps no others are possible, only other ways of expressing f.
The Second Situation: Critical Points Conferring Insight Into the Truth of the Quote
The second situation is when the amplification vanishes at a critical point p "leading to the [false] impression that an infinitesimal disc centred there is crushed down to a point." Here I would emphasize the words "impression" and "centred there," because it is "merely a 'trick of the light'. This is where the lens of m power comes into play. In this situation, derivatives up to and including (m-1) vanish at p, but $f^{m}(p)$ somewhat miraculously reveals the shrunken disk (for f(z) that behave like $z^{m}$). Derivatives of order greater than m also vanish at p. With the given example of $z^{m}$, 0 is a critical point. Now this is surprising. The derivative f'(0) vanishes, so the amplification must vanish at 0. Yet there is still a hidden image.
What is the point that Needham is making: Is it that the first situation illustrates analytic rigidity by using the special case of a function that crushes any curve in its domain to a point? Presumably, other analytic functions demonstrate rigidity by preserving infinitesimal triangles and squares in their domains.
I see that the second situation (critical points) is different and interesting, but I don't quite see what it has to do with rigidity. Both situations deal with analytic functions. Do both situations illustrate rigidity, or only the first?
I see that I have the first and second designations reversed from your answer. I just took them in order of their first appearance on the page.
Gary
|
|
|
|
Post by Admin on May 4, 2016 14:57:58 GMT
Gary
The "First Situation", as you called it in your post, is in italics at the top of page 249 and the way I see things, it requires to be proved.
This is what the rest of subsection 2 on the same page is all about.
The important thing to bear in mind for the proof is that the analytic mapping is assumed to be constant on $s$.
I think it is important to realise that in chapter 4 when Needham introduces the idea of the amplitwist he is choosing the infinitesimal $\epsilon$ such that under the magnification of $L_1$, the image plane (e.g. the RHS of figure 10 on page 197) appears to be circular. It is very important to read and understand footnote 3 at the bottom of page 205.
If $f'$ is zero then $f\sim \epsilon^2f^{(2)}$ and this is why we cannot see the disk unless we use $L_2$ and so on.
In the penultimate paragraph we apply this idea to each derivative and find that any infinitesimal disc centred on $s$ must be totally crushed. This would not be the case if $f$ were not constant on $s$, because then the amplification of points on $s$ would not be zero.
The final paragraph on page 249 argues that it follows from this that the entire domain is crushed to a point, thus proving the statement at the top of page 249. Clearly the mapping is a constant mapping.
On page 250 subsection 3 Uniqueness, in the first paragraph, Needham shows that the mapping $(A-B)\equiv 0$ over its complete domain by using the result just found above under Rigidity, and this means that $A\equiv B$, or in other words the mapping found in the first paragraph of subsection 3 is unique. I think Needham has now arrived at the point where he wanted to be, right from the beginning of section XI.
In the "Second Situation" I hope that most of your questions are explained by the above. The reason that in the general case there is still an image when the amplification is zero is because $f$ is non-zero, and although it cannot be seen under $L_1$, it is still there behaving like $\epsilon^2$!
To summarise I think that Needham first shows that as more and more of the higher order derivatives vanish, the greater the degree of crushing. He then applies this to the situation described at the top of page 249 and finds that in this particular situation all the derivatives vanish and so the crushing is total.
This result is then used in the next subsection, 3 Uniqueness, to prove a uniqueness theorem.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on May 4, 2016 15:57:17 GMT
To summarise I think that Needham first shows that as more and more of the higher order derivatives vanish, the greater the degree of crushing. He then applies this to the situation described at the top of page 249 and finds that in this particular situation all the derivatives vanish and so the crushing is total. This result is then used in the next subsection, 3 Uniqueness, to prove a uniqueness theorem. Vasco, The quoted paragraph helps to explain XI,2. I will also try to remember that most of the diagrams we have seen were viewed through $L_1$. I hope that Needham won't apply either function to the state of California, where I now reside, but I am probably safe, as the functions would also crush the University of San Francisco where he holds a position. Gary |
|
|
|
Post by Admin on May 4, 2016 16:54:46 GMT
Gary
Highly amusing! We could do with a few laughs in chapter 5, especially after all that curvature stuff.
Vasco
|
|
