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Can the Poincare'3-homology sphere be smoothly embeded in R^4? If it is not the case, how can we fix it?

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closed as off topic by Dan Petersen, David White, Ian Agol, Oscar Randal-Williams, Chris Gerig Nov 14 '12 at 21:07

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The answer is no, as a google search found for me: en.wikipedia.org/wiki/Rokhlin%27s_theorem#The_Rokhlin_invariant . What do you mean by "how can we fix it"? –  j.c. Nov 14 '12 at 15:48
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Any homology $3$-sphere admits a locally flat topological embedding in $4$-space, by a result of Freedman. Is this the kind of fix you were after? –  Mark Grant Nov 14 '12 at 16:01
    
I wonder if someone would add here the reference for the Freedman result. Thanks! –  tweetie-bird Nov 14 '12 at 21:15
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why is this question closed? It seems to be related to several interesting things. –  Daniel Pomerleano Nov 15 '12 at 5:21
    
I didn't vote to close but the question needs a fair bit of work following the guidelines of mathoverflow.net/howtoask –  j.c. Nov 15 '12 at 13:13

1 Answer 1

If I remember correctly, given M any homology 3-sphere then M x [0,1] can be embeded in R^4. Is this right?

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In the book "4-Manifolds and Kirby Calculus" by Robert E. Gompf,András I. Stipsicz, pag.345. –  Flux Nov 14 '12 at 21:42
    
They admit topological embeddings but not necessarily smooth embeddings; the page you cite is specific about that point. –  j.c. Nov 15 '12 at 13:16
    
...but I cannot understand why in this case the Whitney embedding result doesn't work...it hold for a general smooth manifold... is a homology 3-sphere not smooth? –  Flux Nov 15 '12 at 15:34
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Wouldn't Whitney embedding theorem need 6 dimensions? Am I missing something? Wikipedia says you start with an immersion with transverse self-intersections, then you fix this. (I paraphrase.) –  tweetie-bird Nov 17 '12 at 14:20

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