Gary
GaryVasco
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Post by Gary on May 29, 2016 23:04:05 GMT
Vasco,
Now that the mechanics of $f_\ddagger$ appear to be firm, I would like to discuss the meaning of it a bit more. I have a hard time connecting the plot, which I think accurately represents the symbols and functions, with Needham’s summary. Frankly, I am missing the whole point of the section!
Needham wrote on p. 257:
Given that f is defined in $P$, but not in $\tilde{P}$, it seems odd to see $f \circ \mathfrak{R}_L$, which applies f to a point in $\tilde{P}$. What is the thought behind this? Is it that we can apply f to a point in $\tilde{P}$ so long as the point in $\tilde{P}$ is a known Schwarzian reflection of a point in P?
Secondly, if the point is to extend f to $\tilde{P}$, why is $f^\ddagger$ a point in $\hat{P}$ rather than a point in $\tilde{\hat{P}}$? The appearance to me is that we have merely found an alternative path from a point in $P$ to a point in $\hat{P}$ rather than a function from $\tilde{P}$ to $\hat{\tilde{P}}$.
To put it another way, if we apply $f^\ddagger$ to a point a in $P$, we get a point that is on or near $f(a)$. Is the claim just that we can reach that same point by applying $\mathfrak{R}_\hat{L} \circ f$ to a point in $\tilde{P}$ that is a point reflected in L from $P$?
Can you explain how Needham accomplishes the continuation of f across the curve?
Gary
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Post by Admin on May 30, 2016 6:49:00 GMT
Vasco, Now that the mechanics of $f_\ddagger$ appear to be firm, I would like to discuss the meaning of it a bit more. I have a hard time connecting the plot, which I think accurately represents the symbols and functions, with Needham’s summary. Frankly, I am missing the whole point of the section! Needham wrote on p. 257: Given that f is defined in $P$, but not in $\tilde{P}$, it seems odd to see $f \circ \mathfrak{R}_L$, which applies f to a point in $\tilde{P}$. What is the thought behind this? Is it that we can apply f to a point in $\tilde{P}$ so long as the point in $\tilde{P}$ is a known Schwarzian reflection of a point in P? Gary I also have been struggling with all of this while trying to write my commentary. I think that I now understand it and I will write here my thoughts on your above questions. As soon as I have finished my commentary (not long I hope), I will send you a private message with a link to a draft version, or part thereof, so that you can give me some feedback if you will. Now to your specific question above: Don't forget that as Needham points out in the first paragraph of page 256 that $\mathfrak{R}_K\circ\mathfrak{R}_K$ is the identity mapping and so if $\widetilde{P}=\mathfrak{R}_L(P)$ then $\mathfrak{R}_L(\widetilde{P})=[\mathfrak{R}_L\circ\mathfrak{R}_L]P=P$. The regions $P$ and $\widetilde{P}$ are mapped to each other by $\mathfrak{R}_L$. I think from my first answer you can probably see that in a similar way to figure 28 on page 253 a point $a$ in $\widetilde{P}$ is mapped by $\mathfrak{R}_L$ to $\bar{a}$ in $P$ which is then mapped by $f$, which in its turn is mapped by $\mathfrak{R}_\hat{L}$ and so if we write $f^{\ddagger}=\mathfrak{R}_\hat{L}\circ f\circ\mathfrak{R}_L$ then $f^{\ddagger}$ is a continuation of $f$ across $L$ into $\widetilde{P}$. Hope this helps. Vasco
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Gary
GaryVasco
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Post by Gary on May 30, 2016 14:47:03 GMT
Vasco,
The discussion helps. It seems to be a matter of perspective. See if you think this metaphor is valid.
Suppose that Zeke, who lived in post-holocaust Minneapolis, Minnesota on the west side of the Mississippi River wanted to visit his aunt in neighboring St. Paul on the east side and give her a new hat for her birthday, but the bridge across the Mississippi River had collapsed (as it actually did in 2007, killing several people). Zeke found that he could reflect southward across the Minnesota boundary into Iowa, cross the river on a good bridge (analytic function) into Illinois at Rock Island, and then reflect northward to St. Paul, Minnesota (crossing two state boundaries, unfortunately for the metaphor). The trip was physically demanding, but he arrived in good shape.
Following the scheme on p. 257, we can label Zeke’s map “Mississippi $f^\ddagger$”. Minnesota is $P$, Iowa is $\tilde{P}$. But here is where the confusion enters! On p. 253, f operates on P, not on $\tilde{P}$ (which in [28] is the specific case $\overset{-}{P}$). These thoughts lead me to think that the last sentence of the initial half paragraph on p. 257 should read “the continuation $f^\ddagger$ of f to P, where $\tilde{P} = \mathfrak{R}_L(P)$”.
Gary
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Post by Admin on May 30, 2016 16:10:01 GMT
Gary
Love the metaphor, but my knowledge of geography in the US leaves a lot to be desired.
In figures 28 and 29 the function $f$ is defined on $P$ and the continuation is to $\overline{P}$ so that $f^*$ becomes the function defined on $\overline{P}$.
On page 257 Needham uses $\widetilde{P}$ instead of $\overline{P}$ and $f^{\ddagger}$ instead of $f^*$ because conjugation has been replaced by Schwarzian reflection.
So a point $a$ in $\overline{P}$ or $\widetilde{P}$ is reflected into $\overline{a}$ in $P$ or $\widetilde{a}$ in $P$ and then mapped to $f(\overline{a})$ or $f(\widetilde{a})$ and then reflected again to $\overline{f(\overline{a})}=f^*$ or $\widehat{f(\tilde{a})}=f^{\ddagger}$.
$f^*$ is then the analytic continuation of $f$ to $\overline{P}$ and $f^{\ddagger}$ is the analytic continuation of $f$ to $\widetilde{P}$.
I hope this helps but if not the diagrams in my commentary might.
Vasco
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Gary
GaryVasco
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Post by Gary on May 30, 2016 18:23:26 GMT
Vasco,
I looked at a map of England and nothing popped out at me as a rectangular pattern with smooth boundaries for reflection, so I opted for the Midwestern U.S., where our expansionist forefathers enthusiastically marked out large blocks of near rectangular and weakly guarded real estate.
I have to disagree with the following regarding p. 257:
On p. 257, there occurs the phrase “the continuation $f^\ddagger$ of $f$ to $\tilde{P} = \mathfrak{R}_L(P)$”. That means the first point to be reflected lies in $P$, not in $\tilde{P}$. But to make p. 253 consistent with p. 257, it should read as you have it. I think I understand this now. If so, then it would help to rewrite p. 257 to make it consistent with p. 253.
On a more philosophical level, we are trying to find the parallels between visual and symbolic representations. The visual metaphor and the symbolic structure of $f^\ddagger$ both follow a round about path to the end product. I find it misleading to include an arrow directly from the unreflected z to the final product and label it $f^*$, as in [28], or $f^\ddagger$. I can understand why it is done, and the shorthand of $f^\ddagger$ is useful, but perhaps there is a better way to present it. I understand that I might be an outlier in this respect. As I read p. 252, it seems that Needham wrote clearly and he points out that $f*$ is an entirely new function and not, in general, a continuation of $f$. But on p. 257, $f^\ddagger$ can not be a continuation of $f$ to $\tilde{P}$, because $f$ is defined on $\tilde{P}$, so there is no need for a continuation in that region. That is why I suggested the rewrite in my previous message.
Gary
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Post by Admin on May 30, 2016 18:50:21 GMT
Vasco, I looked at a map of England and nothing popped out at me as a rectangular pattern with smooth boundaries for reflection, so I opted for the Midwestern U.S., where our expansionist forefathers enthusiastically marked out large blocks of near rectangular and weakly guarded real estate. Gary I often wondered why the boundaries of states were mostly straight lines. Must have been problematic at times running into obstacles like rivers and mountains etc I disagree with you that the phrase you quote from the book means that the first point to be reflected lies in $P$. Maybe it's the way he's written it - it can be understood in two different ways: 1) which I think is your way - "the continuation $f^{\ddagger}$ of $f$ to $\widetilde{P}$, which (ie $f^{\ddagger}$) is $\mathfrak{R}_L(P)$. 2) my way and I think Needham's way - " the continuation $f^{\ddagger}$ of $f$ to $\widetilde{P}$, where $\widetilde{P}=\mathfrak{R}_L(P)$. Rather than being rewritten I think it needs clarifying. I disagree - $f$ is defined on $P$ if you accept what I have written above of course. Vasco
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Gary
GaryVasco
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Post by Gary on May 30, 2016 19:33:29 GMT
Vasco,
Just for the record, this is not at all my view and I don't regard it as Needham's view.
You write:
How does one reconcile that with Needham's definition of $f^\ddagger$?
$\hspace{5em}f^\ddagger = \mathfrak{R}_\hat{L} \circ f \circ \mathfrak{R}_L$
and the preceding line which contains
$\hspace{5em}\tilde{P} = \mathfrak{R}_L(P)$.
I read this as $f$ is defined on the domain $\mathfrak{R}_L$ and $\mathfrak{R}_L$ is defined on $P$. Since $\tilde{P} = \mathfrak{R}_L$, f must be defined on $\tilde{P}$. This is a reversal from the notation in [18], where it shows $f$ as defined on $P$. In general, I don't think there is anything mathematically wrong with the reversal, but it does lead to confusion, and since $f$ is defined on $\tilde{P}$ by the evidence of these two equations, it seems wrong to say "the continuation $f^\ddagger$ of $f$ to $\tilde{P}$", which implies that $f$ is not defined on $\tilde{P}$. If there is a continuation, it would be to P.
This is odd. Sentence 1 (first full sentence) on p. 257 reads "Let $f$ be an analytic mapping defined in a region P", so I see why you say $f$ is defined on $P$. I had gotten involved in the later sentences. Then, for the reasons given above, I think the page is incoherent.
Gary
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Post by Admin on May 30, 2016 19:57:03 GMT
Gary
To me all that $\widetilde{P}=\mathfrak{R}_L(P)$ means is that the region $\widetilde{P}$ is obtained by Schwarzian reflection of $P$ across $L$.
Choose a point $a$ in the region $\widetilde{P}$ and reflect it across $L$ by using the Schwarzian reflection $\mathfrak{R}_L$. This gives us the point $\widetilde{a}=\mathfrak{R}_L(a)$. This must be a point in $P$ since it has been reflected across $L$ from the region $\widetilde{P}$. So we can apply the mapping $f$ to this point $f\circ\mathfrak{R}_L(a)=f(\widetilde{a})$ We now reflect this point across $\widehat{L}$ by using the Schwarzian reflection $\mathfrak{R}_{\widehat{L}}$ $\mathfrak{R}_{\widehat{L}}\circ f(\widetilde{a})=\mathfrak{R}_{\widehat{L}}\circ f\circ\mathfrak{R}_L(a)=f^{\ddagger}$
That's it. I don't find any inconsistency between this and figure 28 in the book.
Vasco
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Gary
GaryVasco
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Post by Gary on May 30, 2016 20:38:16 GMT
Vasco,
Your interpretation is coherent and I have no argument with it, and I think it is probably what Needham intended. I find it very hard to interpret what he actually wrote in that way, but I think we will both get from Minneapolis to St. Paul.
Gary
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Post by Admin on May 30, 2016 20:52:13 GMT
Vasco, Your interpretation is coherent and I have no argument with it, and I think it is probably what Needham intended. I find it very hard to interpret what he actually wrote in that way, but I think we will both get from Minneapolis to St. Paul. Gary Gary I think you are right! Are they both places worthy of a visit? Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on May 30, 2016 21:23:18 GMT
Vasco,
Apologies for taking up math space with geography, but since you asked, I grew up in Minneapolis and went to school in both cities (known as the "Twin Cities"), so I have fond memories of the area. But I don't think I would like the job of head of the tourist bureau. It depends on what you like. Minneapolis has scenic lakes (one named Como!) and hundreds more are to be found in the surrounding countryside. The northern lakes of Minnesota are incredibly beautiful, at least to a native. But compared to more spectacular destinations such as the Grand Canyon, Yosemite, the Four Corners, Yellowstone Park, Glacier National Park and coastal California, or compared to cities like New York, New Orleans, San Francisco, and Seattle, the Twin Cities are a little bland. Compared to anyplace in Europe, there is a lack of architecture and history. For insight into the culture, you might try to stream the PBS program Prairie Home Companion, often broadcast from St. Paul. The Lake Wobegon segments are classic.
Gary
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Post by Admin on May 31, 2016 5:40:20 GMT
Gary
We could call it Topology! Thanks for the information.
Vasco
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