What kind of 3-manifolds can arise as hypersurfaces $\{ f(x,y,z,w) = 0\} \subset \mathbb{R}^4$? Can they have nontrivial H1 or H2?
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asked Jul 3, 2010 at 18:18
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2$\begingroup$ What kind of function is $f$? For smooth $f$ you can get any closed subset of $\mathbb{R}^4$ as a zero set. $\endgroup$– Robin ChapmanCommented Jul 3, 2010 at 18:22
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3 Answers
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A simple construction that bears on the narrow version of John's question: If $M$ is a closed $n$-manifold that embeds in $\mathbb{R}^{n+1}$ (which can only happen if $M$ is orientable), then $M \times S^k$ embeds in $\mathbb{R}^{n+1+k}$. Thicken $M$ in $\mathbb{R}^{n+1}$, then cross with $I^k$ in the new dimensions, and then take the boundary of that. By induction, then, any product of spheres embeds in the next dimension.
On the other hand, Ryan Budney in arXiv:0810.2346 has both new results and a bibliography of the broad question of which closed 3-manifolds embed in $\mathbb{R}^4$. It is problem 3.20 in Kirby's problem list, it is an interesting open problem, and there have been several partial results.
answered Jul 3, 2010 at 18:40
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$S^1\times\mathbb R^2$ is an hypersurface in $\mathbb R^4$, and so is $S^2\times\mathbb R^1$.
answered Jul 3, 2010 at 18:29
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You can cut a two-torus out of $\mathbb{R}^3$ and I think that a simple modification will cut a three-torus out of $\mathbb{R}^4$. So the answer to your second question is "yes".
