[math-fun] Re: Poincare Conjecture
"Jon Perry" <perry@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@AIC.NRL.Navy.Mil
Well is anthing wrong with this one then; Let P_n be a compact, simply connected n-mainfold. It is known that P_n is homeomorphic to S_n for n>=4. As P_4 for example is homeomorphic to S_4, and P_7 is homeomorphic to S_7, then P_4*P_3=P_7 must be homeomorphic to S_7=S_4*S_3, so P_3 is homeomorphic to S_3. Jon Perry perry@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
On Thu, 16 Oct 2003, Jon Perry wrote:
Well is anthing wrong with this one then;
Let P_n be a compact, simply connected n-mainfold. It is known that P_n is homeomorphic to S_n for n>=4. As P_4 for example is homeomorphic to S_4, and P_7 is homeomorphic to S_7, then P_4*P_3=P_7 must be homeomorphic to S_7=S_4*S_3, so P_3 is homeomorphic to S_3.
Quite a lot. You don't say what "*" means, so one thing might be the assumption that Pm*Pn is again a simply-connected manifold, which is false if * means direct product. Another is the assumption that we can "cancel" from a homeomorphism between A*B and A*C to deduce the existence of one between B and C. This again is false if * means direct product. John Conway
On Thu, Oct 16, 2003 at 12:43:45PM -0400, John Conway wrote:
On Thu, 16 Oct 2003, Jon Perry wrote:
Well is anthing wrong with this one then;
Let P_n be a compact, simply connected n-mainfold. It is known that P_n is homeomorphic to S_n for n>=4. As P_4 for example is homeomorphic to S_4, and P_7 is homeomorphic to S_7, then P_4*P_3=P_7 must be homeomorphic to S_7=S_4*S_3, so P_3 is homeomorphic to S_3.
Quite a lot. You don't say what "*" means, so one thing might be the assumption that Pm*Pn is again a simply-connected manifold, which is false if * means direct product.
No, actually, that's right: the fundamental group of a direct product is the product of the fundamental groups. The error is, instead, the assertion that any simply connected n manifold, n>=4, is a sphere: there are many 4-manifolds with trivial fundamental group, for instance. (There is some confusion in terminology, between the concepts of "trivial fundamental group" and "homotopy equivalent to S^n". Among topologists, "simply connected" always means the former, as far as I know. The two conditions are equivalent for 3-manifolds.)
Another is the assumption that we can "cancel" from a homeomorphism between A*B and A*C to deduce the existence of one between B and C. This again is false if * means direct product.
Yes, theorems like this are very unlikely to be true. In this case, for instance, a thrice perforated sphere and a once perforated torus are not homeomorphic, but their products with an interval are homeomorphic. Does anybody know a minimal counterexample with compact manifolds? Peace, Dylan
On Thu, 16 Oct 2003, Dylan Thurston wrote:
On Thu, Oct 16, 2003 at 12:43:45PM -0400, John Conway wrote:
Quite a lot. You don't say what "*" means, so one thing might be the assumption that Pm*Pn is again a simply-connected manifold, which is false if * means direct product.
No, actually, that's right: the fundamental group of a direct product is the product of the fundamental groups.
Oops! What a blunder! [The reason I made it is that I was thinking of circle x circle; but of course the circle is simply connected.]
The error is, instead, the assertion that any simply connected n manifold, n>=4, is a sphere: there are many 4-manifolds with trivial fundamental group, for instance.
I actually noticed this, but thought the above nonsense was an easier "kill"! Sorry, JHC
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
(There is some confusion in terminology, between the concepts of "trivial fundamental group" and "homotopy equivalent to S^n". Among topologists, "simply connected" always means the former, as far as I know. The two conditions are equivalent for 3-manifolds.)
What does "homotopy equivalent to S^n" mean? __________________________________ Do you Yahoo!? The New Yahoo! Shopping - with improved product search http://shopping.yahoo.com
On Thu, Oct 16, 2003 at 12:27:44PM -0700, Eugene Salamin wrote:
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
(There is some confusion in terminology, between the concepts of "trivial fundamental group" and "homotopy equivalent to S^n". Among topologists, "simply connected" always means the former, as far as I know. The two conditions are equivalent for 3-manifolds.)
What does "homotopy equivalent to S^n" mean?
There are a number of ways to say it. Here's one: A compact n-dimensional manifold M is "simply connected" if every map of a circle into M can be contracted to a point. It is "homotopy equivalent to S^n" if in addition, every map of a k-sphere, 1<=k<=n, can be contracted to a point. Alternatively (theorem), M is homotopy equivalent to S^n if M is simply connected and the homology groups of M, 1<=k<=n, vanish. (There is a more general notion of homotopy equivalence, which is a little harder to state.) Peace, Dylan
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
On Thu, Oct 16, 2003 at 12:27:44PM -0700, Eugene Salamin wrote:
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
(There is some confusion in terminology, between the concepts of "trivial fundamental group" and "homotopy equivalent to S^n". Among topologists, "simply connected" always means the former, as far as I know. The two conditions are equivalent for 3-manifolds.)
What does "homotopy equivalent to S^n" mean?
There are a number of ways to say it. Here's one:
A compact n-dimensional manifold M is "simply connected" if every map of a circle into M can be contracted to a point.
It is "homotopy equivalent to S^n" if in addition, every map of a k-sphere, 1<=k<=n, can be contracted to a point.
But the identity map of the n-sphere into itself is not contractable. 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?
Alternatively (theorem), M is homotopy equivalent to S^n if M is simply connected and the homology groups of M, 1<=k<=n, vanish.
(There is a more general notion of homotopy equivalence, which is a little harder to state.)
Peace, Dylan
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On Thu, Oct 16, 2003 at 02:51:14PM -0700, Eugene Salamin wrote:
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
On Thu, Oct 16, 2003 at 12:27:44PM -0700, Eugene Salamin wrote:
--- Dylan Thurston <dpt@exoskeleton.math.harvard.edu> wrote:
(There is some confusion in terminology, between the concepts of "trivial fundamental group" and "homotopy equivalent to S^n". Among topologists, "simply connected" always means the former, as far as I know. The two conditions are equivalent for 3-manifolds.)
What does "homotopy equivalent to S^n" mean?
There are a number of ways to say it. Here's one:
A compact n-dimensional manifold M is "simply connected" if every map of a circle into M can be contracted to a point.
It is "homotopy equivalent to S^n" if in addition, every map of a k-sphere, 1<=k<=n, can be contracted to a point.
But the identity map of the n-sphere into itself is not contractable.
Yes, there's a typo above: I should have written 1<=k<n.
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?
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. (Alternatively, if the M1 and M2 are simply connected and there is a map that induces an isomorphism on homology, then M1 and M2 are homotopy equivalent. Since homology groups H^k(M) vanish for k > n, this is close to your statement above.) (The case of spheres is special, since there is some special behaviour when the homotopy/homology groups are 0.)
(There is a more general notion of homotopy equivalence, which is a little harder to state.)
In fact, it's not so hard to state: M1 and M2 are homotopy equivalent if there are maps f: M1 -> M2 g: M2 -> M1 so that fg and gf can both be deformed to get the identity map. (Apologies for the excessive seriousness of this math...) Peace, Dylan
participants (5)
-
Dan Hoey -
dpt＠exoskeleton.math.harvard.edu -
Eugene Salamin -
John Conway -
Jon Perry