Conference rusure::math

re: -.1  Info about Poincare's conjecture:

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Mark,

The Poincare conjecture conjectures that the only simply connected
three dimensional manifold is the three-sphere.  (A k-dimensional 
manifold is an object that looks locally like k-dimensional space; thus
the boundary of the unit ball in 3-d space is a 2-dimensional manifold.
The k-sphere is the boundary of a ball in (k+1) dimensions.)  So the
recent hoopla about the Poincare conjecture was that somebody last year
claimed that he had proved the Pwan-caray conjecture; however, when this 
guy came to Berkeley last year to give a seminar on his proof, it was 
found by the participants in the seminar to be a non-proof.  There 
was a big article in July 1987 Discover all about this.  By the way,
the analogue of the Poincare conjecture for dimensions higher than 
three is known to be true; the four-dimensional version was proved recently.

Ed

    re .0 .1
    Or simply stated:
         Let M be an n-manifold.  If M is homotopically equivalent to
    S^n then M is homeomorphic to M.
    
    The strange thing about this conjecture is that it had been proved
    to be true for n = 1, 2 and n > 4 a long time ago (50's?).  Yet
    the case for 3 and 4 are more difficult.  There was big excitement
    when M. Friedman of U.C Sandiego proved for n = 4 (Although I think
    he only proved it for the compact case).  Friedman won the Veblon
    and Fieltz prize for that year.  I guess there will be some application
    in physics since spacetime is a 4-manifold.
    Eugene
          

     typo alert
     
>>         Let M be an n-manifold.  If M is homotopically equivalent to
>>    S^n then M is homeomorphic to M.
                                    ^
                                    should be S^n
     
     (Prove: M is homeomophic to M.  State the homeomorphism
     explicitly. :-)

    re .3
    Thanks for the correction.
    Eugene
    

    It seems to me that the explanations in .1 and .2 are different. .1
    says "simply connected" which I understood to mean "path connected and
    all loops based at a given point are path homotopic." .2 says
    "homotopically equivalent to S^n" which seems to be both a stricter
    requirement, as well as one that depends upon n.
    
    Dan

    Simply connected does mean homotopy equivalence at least it is true
    on the complex plane.  I am not sure about what you mean by depends
    on n.
    Eugene

     Re "depends on n"
     
     I meant that, for example, unless S^3 is homotopy equivalent
     to S^4, which seems untrue, the condition depends is
     different for different n.
     
     The Discover magazine article said simply connected meant
     that [I can't remember exactly what they said] any sphere
     [?] within the space could be shrunk continuously to a
     point, always within the space.  It may not have been a
     sphere, but it definitely wasn't a "circle" (loop), which is
     what I thought it meant.
     
     Dan

    Re .7 .6
    Obviously S^3 cannot be possibly homeomorphic to S^4.  The definition
    of "simply connected" means exactly what you have defined in .7
    However, when someone say "simply connected n-manifold" he/she
    sometimes means "simply connected homology n-manifold".
    
    Obviously simply connected is not good enough.  R^3 is a 3-manifold,
    and is simply connected.  However, it is obviously not homeomorphic
    to S^3.  
    Eugene
    

    re .7 .8
    After listening to a talk, I realize that the condition of being
    simply connected might be sufficient for compact manifolds.  On
    the other hand, I think the standard way of stating the conjecture
    is as in .2 and .3.
    Eugene