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May 8, 2011 at 9:27 comment added mikhail skopenkov Simply connectedness is not important: a punctured manifold S^3 x S^3 does not embed in R^6, because otherwise the restrictions to (* x S^3) and (S^3 x *) have a unique transversal intersection point.
Jul 13, 2010 at 17:13 comment added BS. A punctured 2-torus is an open parallelizable manifold, and it does not embed in $\mathbb{R}^2$, because of non-trivial intersection on $H_1$. Maybe adding "simply connected" works.
Mar 29, 2010 at 3:19 comment added Anton Petrunin @Igor, I guess that the result on parallelizable n-manifold which you mention is proved by h-principle --- one simply deforms a frame into a coordinate frame. I just want to say that the same thing can not work here in principle. (In other words: It should be a counterexample and if NOT then proof will be VERY hard.)
Mar 16, 2010 at 19:49 comment added Igor Belegradek @Anton, this is interesting but I am not sure it is relevant to what I am asking.
Mar 16, 2010 at 17:56 comment added Anton Petrunin If one takes figure-eight-immersion of cylinder $S^1\times(0,1)$ then you can not deform it into an embedding.
Mar 15, 2010 at 3:32 history edited Igor Belegradek CC BY-SA 2.5
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Mar 15, 2010 at 3:26 history edited Igor Belegradek CC BY-SA 2.5
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Mar 15, 2010 at 2:35 comment added Igor Belegradek Here is the point I was trying to make: suppose you have an open parallelizable $n$-manifold; how does one decide whether it embeds into $\mathbb R^n$? The original question describes a formally more special situation but it is unclear to me whether it is really more special.
Mar 15, 2010 at 2:22 comment added Petya Universal cover of a subset of ${\mathbb R}^n$ could be immersed to ${\mathbb R}^n$ by trivial reason (just by the projection).
Mar 15, 2010 at 2:16 history answered Igor Belegradek CC BY-SA 2.5