- mathfun@mailman.xmission.com - mailman.xmission.com

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

1 0

[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...

2 3

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

3 3

[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

3 2

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

4 5

[math-fun] Mrs Perkins' Quilt; new Fermat Composite
by Richard Schroeppel 12 Oct '03

12 Oct '03

I'd like to encourage posters to present a little bit more background when they mention a problem. In the case of Mrs. Perkins' Quilt, the puzzle is to dissect an integer sized square into smaller integer squares, with non-integer-square pieces forbidden. (And the gcd of the piece sizes must be 1, to exclude dissecting a 6^2 into four 3^2.) In the case of "Lens Spaces", there's probably no short definition, so we're stuck. But the recreational problems are usually easy to state. It's easy for those of us with high-speed net connections to regard giving a URL as equivalent, even better, than summarizing background material. But to some folks, those URLs represent minutes of download time each, often loading useless decorations of no information value. (I've been on both sides of this fence, so I have some sympathy for the bandwidth challenged.) Rich rcs(a)cs.arizona.edu --- In other news, John Cosgrave &al have a new Fermat Composite. -- Rich Date: Sun, 12 Oct 2003 11:59:12 -0400 Sender: Number Theory List <NMBRTHRY(a)LISTSERV.NODAK.EDU> From: John Cosgrave <john.cosgrave(a)spd.dcu.ie> Subject: F[2478782], a new largest known composite Fermat number To: NMBRTHRY(a)LISTSERV.NODAK.EDU Dear colleagues, In Ferbuary this year I announced the discovery of a new largest known composite Fermat number: the 645817-digit prime (3*2^2145353 + 1) is a factor of F[2145352] = 2^(2^2145352) + 1, only [then] the second prime factor - with more than 100,000 digits - of a Fermat number. See www.spd.dcu.ie/johnbcos for complete details and further links. Now - in a slight state of shock, and delight - I wish to announce a further computational advance: the 746190-digit prime (3*2^2478785 + 1) is a factor of F[2478782] = 2^(2^2478782) + 1, and is now the fourth more-than-100,000 digit factor of a Fermat number (a 'smaller' one - now the third in ranking - was found by Takahiro Nohara on August 4th this year; for details see Wilfrid Keller's Fermat numbers site at http://www.prothsearch.net/fermat.html) There will most likely be some GFN results following, which I will put up at my web site over the next several days. (3*2^2478785 + 1) was discovered on M?ire N? Bhaoill's computer (a 2.6GHz Pentium IV) - pre-sieved with Paul Jobling's newpgen and PRP-ed with George Woltman's program - and then subjected to Yves Gallot's Proth (primality testing) and ProthF (generalised Fermat number testing). She (M?ire) is one of my supportive St Patricks College Proth-Gallot group members (http://www.spd.dcu.ie/johnbcos/proth-gallot_group_(spd).htm) and the new number actually appeared as a PRP-probable prime on her machine on June 22nd of this year. I didn't notice it until Tues 7th October... The Fermat divisibility result - "3*2^2478785+1 Divides GF(2478782, 2)" - appeared on my office screen at 11:02 today (the independent primality test was done at home), minutes before my double 2nd year number theory lecture was due to begin. Of course I had to tell my students, and one of them suggested we should all go across the road for coffee and cake, which is exactly what we did. We were there for almost two and a half hours, during which time they came to know of Euclid (and perfect numbers), ... , Mersenne, Fermat, Euler, Landry, ... , Brillhart-Morrison, CRAY, Wilfrid Keller, Jeffrey Young, Richard Crandall, Richard Brent, Proth, Yves Gallot, George Woltman, Paul Jobling, ... I will put up digital photos of computer screens sometime next week on my home page. John Cosgrave

1 0

[math-fun] Re: Can you help to me
by Richard Guy 12 Oct '03

12 Oct '03

In this connexion, has anyone discovered any more spoof odd perfect numbers, such as Descartes's 2 2 2 2 3 7 11 13 22021 which is perfect if you pretend that 22021 is prime ? R. On Thu, 9 Oct 2003, Jud McCranie wrote: > At 07:20 AM 10/9/2003, f.firoozbakht(a)sci.ui.ac.ir wrote: > > >a(n) is the n-th odd number that sigma(n)=2n+6. > > > >a(1)=8925, a(2)=32445, a(3)=442365, a(4)=? > > > >I know a(4) is greater than 2*10^9 . > > There are no terms between 2*10^9 and 6.5*10^9. > >

1 0

[math-fun] Mrs. Perkins Quilt -- Orders 89, 90 improved over UPIG
by Ed Pegg Jr 12 Oct '03

12 Oct '03

Lance Gay has cracked a problem in Unsolved Problems in Geometry. Quoting RK Guy -- known records with "some slightly increasing lack of confidence" are as follows: {1 >> 1}, {4 >> 2}, {6 >> 3}, {7 >> 4}, {8 >> 5}, {9 >> 6,7}, {10 >> 8,9}, {11 >> 10-13}, {12 >> 14-17}, {13 >> 18-23}, {14 >> 24-29}, {15 >> 30-39,41}, {16 >> 40,42-50}, {17 >> 51-66,68,70}, {18 >> 67,69,71-87}, {19 >> 88-100} RK can pencil in a correction for that last bit. Lance Gay (Lance.Gay(a)ngc.com) found order-18 solutions for side 89 and 90 Mrs. Perkins Quilts. These appear to be better than the current known values. Side=89 Order=18 - [48 41][5 5 11 20][2 8][41 9][2 9][3 7][12][8 28][20] Side=90 Order=18 - [49 41][8 12 21][41 9 7][3 9][2 8][11][2 28][5 5][21] Lance also asks for a source for the best know solutions for side 100-250. The only source I knew of was Duijvestijn, A. J. W. ftp://ftp.cs.utwente.nl/pub/doc/dvs/TableI. -- but this seems to be down. Other pertinent refs: http://mathworld.wolfram.com/MrsPerkinsQuilt.html http://mathworld.wolfram.com/PerfectSquareDissection.html I must humbly take blame for botching a diagram on Eric's site -- the order 7 square there is wrong (but the sequence right under is correct). Nick Gardner pointed out the mistake to me. For an exercise, try to dissect the order 7 square into 9 smaller squares. An Ascii solution is below, along with the Duijvestijn code. --Ed Pegg Jr, www.mathpuzzle.com (Fixed width Ascii art) ___ ___ ___ ___ ___ ___ ___ | | | | | | | | | | | | | |___ ___|___ ___| | | | | | |___ ___ ___|___|___| | | | | | | |___|___ ___| | | | | | | | | | | | | | | | |___ ___ ___ ___|___ ___ ___| [3 2 2][1 1 2][4 1][3]

4 3

Re: [math-fun] The spatial universe is finite !!!???
by asimovd＠aol.com 11 Oct '03

11 Oct '03

John asks: << Is there a shape for space-time that makes space compact but that doesn't have an invariant factorization as space x time? >> The lens spaces L(7,1) and L(7,2) (compact 3-manifolds) are known to be topologically distinct but homotopy equivalent. I wonder if L(7,1) x R is homeomorphic to L(7,2) x R. IF this is the case, then L(7,1) x R = L(7,2) x R is a topological 4-manifold with a non-unique factorization as (compact 3-manifold) x R. (Is this what you meant, John?) --Dan

2 1

[math-fun] Hodge conjecture
by Jon Perry 10 Oct '03

10 Oct '03

The Hodge conjecture, right, is nothingmore than visualising infinite space as a finite product of our imagination, given the neccessary and sufficient conditions that all is well and good with the owrld. In other words, every that will be will be, and nothing that won't be can't be. Or, we exit to think and thinking is existence. Personally, I see the Hodge conjecture as the rationalization of the human problem into reality, with the human influence removed. 7/10 this is correct, but it is the 3/10 times that this is incorrect that really buggers things up. Hopefully the second world in better than the first world. Peace to all hippies. 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

2 2