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Let $Y$ denote the space homeomorphic to the (sans serif) letter $${\huge\mathsf Y}$$ or, equivalently, the space of three closed intervals glued together at one endpoint. Consider the space $Y\times Y$. Here is my attempt at a drawing:

$Y\times Y$

What I drew is not embedded in $\mathbb R^3$ -- it intersects itself.

  • Is there any topological embedding $Y\times Y\to\mathbb R^3$?
  • If not, why not?
  • What is the strategy for attacking problems like this?
Caleb Stanford
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user134824
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    What context did you encounter the problem in? –  May 04 '16 at 15:09
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    I think I can prove that there are no PL embeddings (and probably hence that there are no tame embeddings), but the proof is not pretty. There is no single method to approaching embedding problems, which are in general quite hard. The most accessible and easiest to use tool is Alexander duality, but that is not accessible here since your space is contractible. –  May 04 '16 at 15:17
  • It seems like the tricky part is at the obvious "center point" of the product. – Thomas Andrews May 04 '16 at 15:18
  • Basically, I can intuitively see how it is impossible - if you have a line segment times $Y$ and then try to attach another $Y$ times a line at the middle of the first, then you are stuck on one of the three "sides" of the original $Y$, so you can't get the third segment "over". Not sure how to prove that rigorously. – Thomas Andrews May 04 '16 at 15:21
  • @MikeMiller In a geometry/topology class we were given the problem of drawing this space along with several others. It was the only one for which I could not find a non-self-intersecting drawing. – user134824 May 04 '16 at 16:14

1 Answers1

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The product space consists of 9 rectangular sheets, glued together along two of the edges of each sheet. There are 6 glue-edges where three sheets meet, and all of these edges meet at a center point.

Assume we have an embedding and intersect it with a sphere around the center point that is small enough not to contain any points on the boundary of the embedding.

There are 6 points on the sphere that represents glue-edges (just pick an arbitrary one if the glue-edge crosses the sphere multiple times), and the sheets themselves cross the sphere in curves between these points.

Together, the whole figure drawn on the sphere is the complete bipartite graph $K_{3,3}$ -- but that is well known not to be planar, so it can't be drawn on a sphere without intersections!

user21820
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hmakholm left over Monica
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    It seems to me you're assuming a sort of tameness/transversality here. I see no reason that this intersection should be a graph as opposed to some horrible subspace for a small enough sphere. For the PL case, though, this is very clean. –  May 04 '16 at 15:42
  • It is possible that you could do this in the wild case using results from this article of Bing (who else?), but I haven't thought about it. –  May 04 '16 at 16:17
  • @MikeMiller: What I'm depending on is that the intersection will contain a graph as a (proper or not) subset. I thought I had an argument that this will always be the case no matter how wild the embedding is, but I can't quite get all the i's dotted and t's crossed. I'll keep thinking. – hmakholm left over Monica May 04 '16 at 16:18
  • I agree. Good luck with it; wild topology makes my bones rattle. I'm going to walk away happy with the PL result. :) –  May 04 '16 at 16:19