Lubos Motl
Nov20-04, 04:16 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Xi mentioned Whitehead continuum, so let me start to learn and teach those\nwho did not know - much like I - what these things are.\n\nYou know what\'s a contractible manifold, right? Something that can\ncontinuously be shrunk to a point inside the manifold itself.\n\nFor example, a ball is a contractible manifold. All manifolds homeomorphic\nto the ball are contractible, too.\n\nAre *all* contractible manifolds homeomorphic to a ball? For dimensions\n1,2, the answer is classical and it is "yes".\n\nDimension 3 has its first counterexample - the Whitehead continuum.\n\nTake a S^3 - three-dimensional sphere.\n\nNow find a compact solid torus T_1 inside the sphere S^3. What is a solid\ntorus? I hope that it\'s just ordinary three-dimensional doughnut, i.e.\nfilled T^2, i.e. S^1 x disk.\n\nI believe that the complement of the solid torus inside S^3 is still a\nsolid torus.\n\nOK, now take another solid torus T_2 inside T_1, but in a very special\nway. It is *not* just inserting it in the topologically trivial way, and\nit is also not embedding it as a tube inside another tube (T_2 does not\ncontain the nontrivial cycle of T_1). I don\'t have the picture, but I\nguess that the figure on page 11 of\n\nhttp://www.arxiv.org/abs/quant-ph/0303089\n\nmay express how weird the embedding could be.\n\nWith this embedding, if you want to shrink T_2, you must either leave T_1,\nor you may stay inside T_1, but the T_2 must cross itself.\n\nYou embed T_3 inside T_2 in the same way as T_2 was inside T_1, and so on\n(to infinity).\n\nOK, now define W to be T_{\\infinity}, more precisely the intersection of\nall T_k for k=1,2,3, ...\n\nThe interesting space is S^3\\W which is a noncompact manifold without\na boundary. It\'s contractible, but not homeomorphic to S^3, and the reason\nis essentially that it is not simply connected at infinity.\n\nImprovements and simple proofs welcome.\n________________________________________ ______________________________________\nE-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/\neFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)\nWebs: http://schwinger.harvard.edu/~motl/ http://motls.blogspot.com/\n^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Xi mentioned Whitehead continuum, so let me start to learn and teach those
who did not know - much like I - what these things are.
You know what's a contractible manifold, right? Something that can
continuously be shrunk to a point inside the manifold itself.
For example, a ball is a contractible manifold. All manifolds homeomorphic
to the ball are contractible, too.
Are *all* contractible manifolds homeomorphic to a ball? For dimensions
1,2, the answer is classical and it is "yes".
Dimension 3 has its first counterexample - the Whitehead continuum.
Take a S^3 - three-dimensional sphere.
Now find a compact solid torus T_1 inside the sphere S^3. What is a solid
torus? I hope that it's just ordinary three-dimensional doughnut, i.e.
filled T^2, i.e. S^1 x disk.
I believe that the complement of the solid torus inside S^3 is still a
solid torus.
OK, now take another solid torus T_2 inside T_1, but in a very special
way. It is *not* just inserting it in the topologically trivial way, and
it is also not embedding it as a tube inside another tube (T_2 does not
contain the nontrivial cycle of T_1). I don't have the picture, but I
guess that the figure on page 11 of
http://www.arxiv.org/abs/http://www.arxiv.org/abs/quant-ph/0303089
may express how weird the embedding could be.
With this embedding, if you want to shrink T_2, you must either leave T_1,
or you may stay inside T_1, but the T_2 must cross itself.
You embed T_3 inside T_2 in the same way as T_2 was inside T_1, and so on
(to infinity).
OK, now define W to be T_{\infinity}, more precisely the intersection of
all T_k for k=1,2,3, ...
The interesting space is S^3\W which is a noncompact manifold without
a boundary. It's contractible, but not homeomorphic to S^3, and the reason
is essentially that it is not simply connected at infinity.
Improvements and simple proofs welcome.
__{_______________________________________________ _____________________________}
E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)
Webs: http://schwinger.harvard.edu/~motl/ http://motls.blogspot.com/
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
who did not know - much like I - what these things are.
You know what's a contractible manifold, right? Something that can
continuously be shrunk to a point inside the manifold itself.
For example, a ball is a contractible manifold. All manifolds homeomorphic
to the ball are contractible, too.
Are *all* contractible manifolds homeomorphic to a ball? For dimensions
1,2, the answer is classical and it is "yes".
Dimension 3 has its first counterexample - the Whitehead continuum.
Take a S^3 - three-dimensional sphere.
Now find a compact solid torus T_1 inside the sphere S^3. What is a solid
torus? I hope that it's just ordinary three-dimensional doughnut, i.e.
filled T^2, i.e. S^1 x disk.
I believe that the complement of the solid torus inside S^3 is still a
solid torus.
OK, now take another solid torus T_2 inside T_1, but in a very special
way. It is *not* just inserting it in the topologically trivial way, and
it is also not embedding it as a tube inside another tube (T_2 does not
contain the nontrivial cycle of T_1). I don't have the picture, but I
guess that the figure on page 11 of
http://www.arxiv.org/abs/http://www.arxiv.org/abs/quant-ph/0303089
may express how weird the embedding could be.
With this embedding, if you want to shrink T_2, you must either leave T_1,
or you may stay inside T_1, but the T_2 must cross itself.
You embed T_3 inside T_2 in the same way as T_2 was inside T_1, and so on
(to infinity).
OK, now define W to be T_{\infinity}, more precisely the intersection of
all T_k for k=1,2,3, ...
The interesting space is S^3\W which is a noncompact manifold without
a boundary. It's contractible, but not homeomorphic to S^3, and the reason
is essentially that it is not simply connected at infinity.
Improvements and simple proofs welcome.
__{_______________________________________________ _____________________________}
E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
eFax: +1-801/454-1858 work: +1-617/384-9488 home: +1-617/868-4487 (call)
Webs: http://schwinger.harvard.edu/~motl/ http://motls.blogspot.com/
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^