October 2003 - mathfun@mailman.xmission.com

Re: [math-fun] Poincaré Conjecture
by asimovd＠aol.com 16 Oct '03

16 Oct '03

Gene asks: << What does "homotopy equivalent to S^n" mean? In general, two spaces X & Y are homotopy equivalent if there exist (continuous) maps f: X -> Y and g: Y -> X such that f o g is homotopic to the identity on Y and g o f is homotopic to the identity on X. But due a theorem of J.H.C. Whitehead and the simplicity of the homotopy groups of S^n (for any k with 0 < k < n, the continuous maps f: S^k -> S^n are all homotopic to each other, so the groups pi_k(S^n) for these k are all the 0 group), it readily follows that any connected n-manifold M with the same homotopy groups as S^n is automatically homotopy equivalent to S^n. In fact, to conclude that a connected n-manifold M is homotopy equivalent to S^n, it's sufficient -- by using Poincaré duality -- to assume only that pi_k(M) = 0 for k in the range 0 < k <= floor(n/2). * * * Dylan asks: << Does anybody know a minimal counterexample with compact manifolds? Not I. But I wonder if using (homeomorphic?) link complements in S^3 one could cook up an example by surgeries on the link, or gluing two of these complementds together in two different ways, so that non-homeomorphic 3-manifolds are obtained, but taking their product with the circle we get homeomorphic 4-manifolds. (A fertile place to look may be among pairs of non-homeomorphic but homotopy equivalent 3-manifolds.) --Dan

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[math-fun] ANSWER Insomniacs Problem Revisited:
by otto 16 Oct '03

16 Oct '03

>John Conway posted the original insomniacs problem below, which >assumes a 12 hour clock with no leading zeros. > >The following trickier problem also assumes a 12 digit clock. >""" >Here in washington state I am watching palindromic numbers >go by in local time on my on my computers 12-hour clock without >leading zeros. I count 281 seconds between two palindromes. How >long will it be before I count 281 seconds between two numbers again? The answer is approximately one year. Regards Otto John Conway's Original Problem: > There's a digital clock at my bedside that shows > hours minutes and seconds, and which I usually look > at the moment I wake up. A few days ago, I happened > to wake up at a palindromic time, and out of interest, > waited until the next palindromic time, and noticed the > interval between them. > > Last night exactly the same thing happened, but the > interval was exactly four times as long as in the previous > case. What did the clock read when I woke up last night? > > John Conway ------------------------------------------------------- -------------------------------------------------------

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[math-fun] ANSWER Insomniacs Problem Revisited:
by Otto Smith 16 Oct '03

16 Oct '03

>John Conway posted the original insomniacs problem below, which >assumes a 12 hour clock with no leading zeros. > >The following trickier problem also assumes a 12 digit clock. >""" >Here in washington state I am watching palindromic numbers >go by in local time on my on my computers 12-hour clock without >leading zeros. I count 281 seconds between two palindromes. How >long will it be before I count 281 seconds between two numbers again? The answer is approximately one year. Regards Otto John Conway's Original Problem: > There's a digital clock at my bedside that shows > hours minutes and seconds, and which I usually look > at the moment I wake up. A few days ago, I happened > to wake up at a palindromic time, and out of interest, > waited until the next palindromic time, and noticed the > interval between them. > > Last night exactly the same thing happened, but the > interval was exactly four times as long as in the previous > case. What did the clock read when I woke up last night? > > John Conway -------------------------------------------------------

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Re: [math-fun] Poincaré's Conjecture
by asimovd＠aol.com 16 Oct '03

16 Oct '03

<< However, is there anything wrong with this; PC is true because from the 3-sphere, a compact, simply connected manifold, through only operations which preserve homeomorphicity, we can attain every other compact, simply-connected 3-manifold. >> Jon, this is a nice idea. But you seem to be unaware of the most basic concept in mathematics -- proof. If you can PROVE this idea is true -- by defining which operations you mean precisely, and showing they preserve topological equivalence, and then showing that any other simply-connected manifold can be reached through these operations, THEN and only then have you proved this. Regards, Dan Asimov

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RE: [math-fun] The spatial universe is finite !!!???
by Meeussen Wouter (bkarnd) 16 Oct '03

16 Oct '03

when will we get some good data for the gravitational constant G ? CODATA's G= 6.67259 10^-11 m^3kg^-1 s^-2 is not exactly blowing my socks off. see http://www.iop.org/EJ/abstract/0957-0233/10/6/001 and especially Fig 1. Wouter. -----Original Message----- From: Michael Kleber [mailto:kleber@brandeis.edu] Sent: vrijdag 10 oktober 2003 15:37 To: math-fun Subject: Re: [math-fun] The spatial universe is finite !!!??? >I've heard a cosmologist say, approximately, that anyone who >thought anything about the universe, and hasn't updated that >thinking in light of the WMAP data, should now be considered a >historical relic. Confidence intervals for cosmological constants >have changed from "within an order of magniture" to "+- 2%" >in the past year. O brave new world... > >--Michael Kleber > kleber(a)brandeis.edu =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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RE: [math-fun] Poincaré's Conjecture
by Jon Perry 16 Oct '03

16 Oct '03

Thanks - I can see the weaknesses in my argument. However, is there anything wrong with this; PC is true because from the 3-sphere, a compact, simply connected manifold, through only operations which preserve homeomorphicity, we can attain every other compact, simply-connected 3-manifold. Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com -----Original Message----- From: Andy Latto [mailto:Andy.Latto@gensym.com] Sent: 14 October 2003 19:00 To: 'perry(a)globalnet.co.uk' Subject: RE: [math-fun] PoincarÃ©'s Conjecture > -----Original Message----- > From: Jon Perry [mailto:perry@globalnet.co.uk] > Sent: Tuesday, October 14, 2003 7:00 AM > To: math-fun > Subject: RE: [math-fun] PoincarÃ©'s Conjecture > > > is this clear, or do I require further explaning? Apparently, Dan Asimov's post, which I found completely clear, requires more explanation. Let me try to make it clearer. You think you have a new way of proving facts about topology. You are wrong. Your techniques claim to prove that all simply connected manifolds of a given dimension are homeomorphic. This is a false statement. Even if you discover that you have used compactness in your "proof", and weaken your claim to "all simply connected compact manifolds of a given dimension are homeomorphic", this is still false, since S2 x S2 and S4 are simply connected compact 4-manifolds that are not homemorphic. So I am absolutely certain, as is Dan, and everyone else on this mailing list with the slightest understanding of topology, is certain that your proof technique is seriously flawed. So the process of trying to understand it from your vague hints would not be a useful one, since it is certain to result, if you are able to make what you are saying sufficiently precise to be comprehensible, in discovering that what you are saying is in error. So what you should be saying isn't "I have a proof of a result that those who have studied topology intensely for decades have not been able to prove" (because you clearly don't), but rather "I must be confused in some way about the basic concepts of topology; I have what seems to me to be a simple argument, but it produces a contradiction; can you help me find the flaw?". You might get more useful feedback with that sort of attitude; otherwise, people will just think you're a crank, who is producing ravings, rather than mathematics. If you want to make your argument clear to others, the first step is to make it clear to yourself. Here are a few steps that may help in this process: 1. You use the term 'solid'. Mathematics deals only with precisely defined terms. This is not a standard technical term in topology. So you need to define it precisely. It would probably also be useful for you to prove some elementary theorems about "solid" spaces, using your definition. What are the properties of solid spaces that guarantee that the sections of a simply connected manifold are solid? What are the properties of solid spaces that guarantee that a space with "solid" projections must be homeomorphic to a sphere? Prove that all of these properties hold, directly from your definition of "solid". 2. Be careful with quantifiers. Are you saying that in every projection of a simply connected space to a lower dimension, every projection must be solid? Or are you just saying that every simply connected space has at least one projection in which every projection is solid? In either case, can you prove this? 3. Again, define your terms, and be careful when you use pre-existing terms. A projection, say from (x,y,z) space to (x,y) space, is a map that is many-1, that maps every point of the form (x,y,z) to the point (x,y). So the projection of a geomtric sphere, for example, would be a closed disk. I'm guessing that isn't what you mean when you say projection, that instead, you are looking at the manifolds (x,y,t) for each value of t. These would be circles (and notice that the circle is *not* simply connected, even though the sphere is), and in 2 cases consist not of a manifold 1 dimension lower at all, but of a single point. So if you are looking at these lower-dimensional manifolds, you need to deal with the fact that they need not be simply connected, and need not be manifolds at all. 4. The first step in proving something is to understand the question. The Poincare conjecture states that (A)Any compact, simply-connected 3-manifold is homeomorphic to the sphere. Since the statement (B)Any simply-connected 3-manifold is homeomorphic to the sphere is false (consider R3), any attempt to prove this statement is obviouly completely futile, and doomed to failure. So any proof of the Poincare conjecture that has a hope of being correct must make use of compactness. But you can't make use of compactness without knowing what compactness is. And I don't mean a vague idea; I mean a completely and absolutely precise understanding. So if you can't define compactness, you have no hope of proving the Poincare conjecture, and it would obviously be a complete waste of time for anyone to look at anything that you thought was a proof. Can you define compactness for me, and show me exactly where you use compactness in your proof? Andy Latto andy.latto(a)pobox.com > > Jon Perry > perry(a)globalnet.co.uk > http://www.users.globalnet.co.uk/~perry/maths/ > http://www.users.globalnet.co.uk/~perry/DIVMenu/ > BrainBench MVP for HTML and JavaScript > http://www.brainbench.com > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com > [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of Jon Perry > Sent: 13 October 2003 13:19 > To: math-fun > Subject: RE: [math-fun] PoincarÃ©'s Conjecture > > > Consider a circle that varies in radius from 0 to R over > time. Placing this circle at the start of a 3-d space and > extending over the z-axis produces a sphere. > > So, in 3d space we have 3 axis-dependent projections for a > space. Similarly in n-d space. > > Reversing the process and taking the projections for the > space, simply connected spaces consist only of 'solid' > projection, and hence are homeomorphic. Other spaces have > various but definitely different projective spaces, for > example the cylinder with a hole consists of a circle with a > hole when viewed down the tube, and a widening than shrinking > rectangle that splits when viewed from either perdendicular. > Jon Perry > perry(a)globalnet.co.uk > http://www.users.globalnet.co.uk/~perry/maths/ > http://www.users.globalnet.co.uk/~perry/DIVMenu/ > BrainBench MVP for HTML and JavaScript > http://www.brainbench.com > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of asimovd(a)aol.com Sent: 13 October 2003 08:31 To: math-fun(a)mailman.xmission.com Subject: Re: [math-fun] PoincarÃ©'s Conjecture Jon Perry writes: << ... I have developed a schema for characterizing ANY x-manifold, and in my scheme simply connected spaces fall into exactly one category, and hence are equivalent. >> It really would be better if you devoted some effort toward making your posts comprehensible. Here you don't say what kind of "equivalent" you are talking about, so your assertion conveys zero information to anyone reading it. If you mean topologically equivalent, then this assertion is false, since (e.g.) each of the 4-manifolds S^4 and S^2 x S^2 is simply-connected but they are not homeomorphic. You also don't say what kind of manifolds you are considering (e.g., compact). If noncompact manifolds are covered by your assertion, then R^2 and S^2 provide a simpler counter-example of two simply-connected manifolds of ther same dimension that are non-homeomorphic. --Dan Asimov

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[math-fun] a dismal arithmetic
by Marc LeBrun 15 Oct '03

15 Oct '03

Decimal hacks aren't usually my taste, but perhaps you'll find this trivia fun (inspired by the newest volume of "Winning Ways"...) To address the dismal state of math education, let's engineer a "dismal arithmetic" to be easier than decimal, so no student will be left behind. You'll still operate on pairs of digits, but you don't have to worry about carrying or in fact doing anything harder than comparing. Instead, for each pair of dismal digits, to Add, take the lArger, but to Multiply, take the sMaller That's it! For example: 169 + 248 ------ 269 and 169 x 248 ------ 168 144 + 122 -------- 12468 You can check that these associate, commute and that multiplication distributes over addition: 357 x (169 + 248) = 357 x 269 = 23567 = 13567 + 23457 = (357 x 169) + (357 x 248) Curiously, while 0 is the additive identity, 1 is *not* the multiplicative unity (Quiz: what *is* the unity?). In fact, 1 even has non-unit divisors, eg 1 x 2 = 1! (Quiz: what's the smallest dismal prime?) If we count the number of dismal partitions of n into distinct positive parts not greater than n we get the bizarre sequence 1 1 2 4 8 16 32 64 128 256 1 5 22 92 376 1520 6112 24512 98176 392960 2 22 200 1696... [not yet sloaned--superseeker thinks it might see a generating function, is it hallucinating?] (Quiz: notice that p(12) = 22 = p(21); does p(13) = 92 = p(n) for any other n?) Perhaps some funsters will enjoy developing this dismal enterprise further...

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Re: [math-fun] Poincaré's Conjecture
by asimovd＠aol.com 14 Oct '03

14 Oct '03

Jon Perry writes: << ... I have developed a schema for characterizing ANY x-manifold, and in my scheme simply connected spaces fall into exactly one category, and hence are equivalent. >> It really would be better if you devoted some effort toward making your posts comprehensible. Here you don't say what kind of "equivalent" you are talking about, so your assertion conveys zero information to anyone reading it. If you mean topologically equivalent, then this assertion is false, since (e.g.) each of the 4-manifolds S^4 and S^2 x S^2 is simply-connected but they are not homeomorphic. You also don't say what kind of manifolds you are considering (e.g., compact). If noncompact manifolds are covered by your assertion, then R^2 and S^2 provide a simpler counter-example of two simply-connected manifolds of ther same dimension that are non-homeomorphic. --Dan Asimov

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[math-fun] Is \sech necessary?
by Thane Plambeck 12 Oct '03

12 Oct '03

Apparently not---I just discovered that in LaTeX, there's no hyperbolic secant symbol built in. \sin, \cos, \tan, \sec, \csc, \cot, \arcsin, \arccos, \arctan, \sinh, \cosh, \tanh, \coth But no \sech? Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/ehome.htm

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Re: [math-fun] Poincaré's Conjecture
by asimovd＠aol.com 12 Oct '03

12 Oct '03

John writes: << The problem is a subtle one, and there have been quite a few experienced topologists who've thought they'd proved it, but had to retract (in some cases, quite embarrassingly). >> When I was a graduae student, I thought I might haved proved it until I mentioned this to Stephen Smale (who had won a Fields medal in 1966 for proving the n-dimensional Poincaré conjecture for n >= 5), who told me he, too, came up with the same erroneous proof as a grad student. (Alas, this did not foreshadow my ever winning a Fields medal.) The "proof" goes as follows: Suppose you have a simply-connected 3-manifold. RMove the interiors of two disjoint 3-balls, getting a manifold with two two-spheres as its boundary. A simple argument shows that it admits a nowhere-zero vector field that is inward on one boundary component and outward on the other one. Clearly the only possibility for any trajectory of the vector field is an interval connecting the two boundary components. A simple argument then shows this manifold has the structure of the cartesian product S^2 x [0,1] -- which shows the original manifold was in fact S^3. --Dan

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