Let $X$ be a contractible 2-dimensional simplicial complex. Are there nice necessary and sufficient conditions for $X$ to be embeddable in $\mathbb R^3$? Clearly it is necessary that the link of every vertex be a planar graph. Is this sufficient?
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asked Aug 13, 2011 at 18:50
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$\begingroup$ Its not necessary at all -- consider the space that's the product of an interval with the cone on a discrete space (say, with at least three elements). This embeds in $\mathbb R^3$ and is contractible, but does not satisfy your "necessary" condition. $\endgroup$– Ryan BudneyCommented Aug 13, 2011 at 20:04
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2$\begingroup$ @Ryan, it seems to me that no matter how you triangulate your example, all the links are suspensions of discrete spaces, which are planar graphs. $\endgroup$– Autumn KentCommented Aug 13, 2011 at 20:31
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$\begingroup$ (I guess at some points the links are cones on discrete spaces, but that's okay, too.) $\endgroup$– Autumn KentCommented Aug 13, 2011 at 20:32
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1$\begingroup$ Er, right. I had some examples in mind. Maybe they demonstrated something else. For example, think about the bundle over $S^1$ whose fibre is the cone on a three-point space, so the fibre is a "Y". If the monodromy of that bundle flips the two "arms" of the "Y", but keeps the base fixed, the total space would be a Moebius band with an annulus attached along the central circle. Give it a triangulation which is a refinement of the product CW-decomposition. The links of vertices are planar graphs but I believe it can't embed since if it did, the Moebius band would be orientable. $\endgroup$– Ryan BudneyCommented Aug 13, 2011 at 22:33
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$\begingroup$ So what this demonstrates is that the author's condition is very much just a local one. Perhaps you could view it as the 0-dimensional criterion in an obstruction theoretic approach. The above example demonstrates you need some kind of 1-dimensional obstruction, as well. $\endgroup$– Ryan BudneyCommented Aug 13, 2011 at 22:36
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1 Answer
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If your complex is finite, then figure out the possible ways of thickening it to a 3-manifold. The possible thickenings are determined by the various embeddings of the links of the vertices into $S^2$, then seeing if these induce compatible thickenings over the edges (determined by the same cyclic ordering over the link of the edge) and faces of the complex. If it can be thickened this way, then it must be a ball since it is a contractible 3-manifold.
answered Aug 13, 2011 at 21:09
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$\begingroup$ Very nice answre! This still leaves open the question of whether there exist 2-dimensional complexes all of whose links are planar, but that don't embed in $\mathbb R^3$. $\endgroup$ Commented Aug 13, 2011 at 22:15
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$\begingroup$ @Andre: I believe my 2nd example in the comments above address your question. $\endgroup$ Commented Aug 13, 2011 at 22:37
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7$\begingroup$ @Ryan: Your Y-shaped Moebius band example is not contractible. So, strictly speaking, it doesn't answer the question. However, a small variation does: replace "the total space would be a Moebius band with an annulus attached along the central circle" by "the total space would be a Moebius band with A DISK attached along the central circle". $\endgroup$ Commented Aug 13, 2011 at 22:48
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$\begingroup$ Ah, yes! My mind is all over the place today. $\endgroup$ Commented Aug 13, 2011 at 22:51