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Re: [math-fun] looking for reference to a problem
by asimovd＠aol.com 17 Oct '03

17 Oct '03

Nick Baxter asks for references to the Colllege Board question: << Problem: Given an equilateral pyramid (square base), glue two regular tetrahedrons (same edge length) to two non-adjacent triangular faces. How many faces does the resulting polyhedron have? >> I may be wrong, but as I recall it, the problem was to glue a regular tetrahedron to a regular octahedron along one common face. --Dan

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[math-fun] Late Show physics
by Shel Kaphan 16 Oct '03

16 Oct '03

Last night David Letterman had physicist Brian Greene as a guest. The topics were the origins of the universe, inflation, and even a little about string theory. At one point in the conversation they talked about what might have existed "prior" to the big bang. A short excerpt: Letterman: But even in nothingness, nothingness in and of itself is something. Greene: Yes, it does wind up taxing language to describe these ideas using common words, but the notion would be that the familiar things we know about like space and time and matter really would not have existed. It would have been either a different realm, or a realm literally not composed of anything we know of or could describe. Letterman: Like Trenton.

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

16 Oct '03

Gene asked: << > Can we say that two compact n-dimensional manifolds M1 and M2 are > homotopy equivalent if for all k, 1<=k<=n the homotopy groups pi[k](M1) > and pi[k](M2) are isomorphic? Dylan answered: << No, that's not true. What is true is that if there is a map between M1 and M2 which induces the identity on all homotopy groups (not just in this range) then M1 and M2 are homotopy equivalent. >> A simple example of spaces that have the same homotopy groups but are not homotopy equivalent is P^k x S^m and P^m x S^k, where say m > k > 1. (Note to Dylan: The condition you mention -- usually called "weak homotopy equivalence" -- is the same as homotopy equivalence for "nice" spaces like manifolds, and more generally simplicial or CW complexes, but there are stranger spaces that can be weakly homotopy equivalent without being homotopy equivalent.) --Dan --Dan

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[math-fun] Re: Poincare Conjecture
by Dan Hoey 16 Oct '03

16 Oct '03

"Jon Perry" <perry(a)globalnet.co.uk> writes: > Thanks - I can see the weaknesses in my argument. Very good, if true. > 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. Yes, there's something wrong with that. Any statement of the form "PC is true because <sentence>" is hogwash. If you were going to prove PC, you would need more than a sentence to express a convincing proof. It might be accurate to say "If I could prove <sentence>, that would imply PC". That depends on the sentence, of course. If you ever convinced someone that you might have a proof of PC, then it might be appropriate to say something of the form, "I have proved <sentence>, and that implies PC." But you made the statement in the hogwash form. Moreover, you made that statement in reply to a message that shows that you not only did not present a realistic approach to a proof of PC, but were apparently unclear on the concept of what a proof of PC would be, and that you were beginning to sound like a crank. I hope your statement is not a proof, because it certainly isn't a proof of the Poincare conjecture. Dan Hoey Hoey(a)AIC.NRL.Navy.Mil

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[math-fun] special characters in subject lines
by Richard Schroeppel 16 Oct '03

16 Oct '03

Some mailers bounce messages that have special characters in the Subject line. Examples are the final vowel in Poincare, and the final consonant in Gauss. If you want your Math-Fun message to go to everyone in the group, avoid the fancy non-ASCII letters in your Subject lines. Rich rcs(a)cs.arizona.edu

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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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