Can the Poincaré $3$-homology sphere be smoothly embeded in $\mathbb R^4$? If it is not the case, how can we fix it?
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3$\begingroup$ 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"? $\endgroup$– j.c.Commented Nov 14, 2012 at 15:48
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4$\begingroup$ 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? $\endgroup$– Mark GrantCommented Nov 14, 2012 at 16:01
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$\begingroup$ I wonder if someone would add here the reference for the Freedman result. Thanks! $\endgroup$– tweetie-birdCommented Nov 14, 2012 at 21:15
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2$\begingroup$ why is this question closed? It seems to be related to several interesting things. $\endgroup$– Daniel PomerleanoCommented Nov 15, 2012 at 5:21
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$\begingroup$ I didn't vote to close but the question needs a fair bit of work following the guidelines of mathoverflow.net/howtoask $\endgroup$– j.c.Commented Nov 15, 2012 at 13:13
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1 Answer
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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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2$\begingroup$ In the book "4-Manifolds and Kirby Calculus" by Robert E. Gompf,András I. Stipsicz, pag.345. $\endgroup$– FluxCommented Nov 14, 2012 at 21:42
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$\begingroup$ They admit topological embeddings but not necessarily smooth embeddings; the page you cite is specific about that point. $\endgroup$– j.c.Commented Nov 15, 2012 at 13:16
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$\begingroup$ ...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? $\endgroup$– FluxCommented Nov 15, 2012 at 15:34
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1$\begingroup$ 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.) $\endgroup$ Commented Nov 17, 2012 at 14:20