# Non embedding of $Y\times Y$ into $\mathbb{R}^3$

I know that this is a well known result, but where can I find a proof? I am also interested to see more general non-embedding results of this type.

Theorem. Let $$Y$$ be the union of two segments with one segment being attached in the middle of the other one. Then there is no topological embedding of the space $$Y\times Y$$ into $$\mathbb{R}^3$$.

It was a homework problem in my undergraduate (sophomore) topology class taught by Professor Karol Sieklucki at University of Warsaw!

• @AntonPetrunin This was a standard course of topology of metric spaces, but there were 4 IMO students in the class and the problems were far beyond of what was covered in lectures. Another problem was the Moore-Young theorem: mathoverflow.net/a/27248/121665 Apr 13, 2018 at 17:43
• Okey, did you get to homology groups (maybe I am wrong, but it looks like a senseless pain without them). Apr 13, 2018 at 17:58
• No idea. For me this is a folklore result and I do not know any references. This is why I ask. Apr 13, 2018 at 20:50
• Piotr, the question can be essentially equivalently asked about embeddings into $S^3$. Years ago, Professor Karol Borsuk asked about the non-embeddability of nice products $\ X\times Y$ into $S^n$. As I remember, He got the first result for starters, then others followed (not too many of them), including one or both parents of Greg K., then I presented a strong result during a conference in Auburn (AL) but didn't publish it; then someone from Poland published a similar but weaker result -- that's all I remember at this moment; I will not recall much extra later after all these years. Jan 17, 2019 at 4:22
• "two segments with one segment being attached in the middle of the other one" is not the most elegant way to describe this space. (:-) Jan 17, 2019 at 4:26

If it did embed as $Y \times Y \subset \mathbb{R}^3$, then you could choose a sufficiently small sphere $S$ in $\mathbb{R}^3$ centered at the point $(y_0,y_0)$, where $y_0$ is the point in where the two segments are glued. Then $S \cap Y \times Y \subset S$ is an embedding of the complete bipartite graph $K_{3,3}$ into a two sphere. But this graph is not planar, which is a contradiction. This argument seems to assume the embedding is sufficiently nice, e.g. simplicial with respect to some triangulations of $Y \times Y$ and $\mathbb{R}^3$. Maybe others can share a different argument that is valid for any topological embedding.

• Could you elaborate why the embedding is $K_{3,3}$. I don't quite "see" this. Jan 17, 2019 at 1:31
• Think of $Y$ as consisting of three edges $E, F, G$ glued at a vertex $v$. A small sphere in $Y$ centered at $v$ consists of three points, one for each of $E,F,G$. A small sphere in $Y \times E$ centered at $(v,v)$ consists of four points, $E,F,G,E'$, where $E'$ lies in $\{v\} \times E$. All of $E,F,G$ are joined to $E'$ by arcs, one for each of the squares $E \times E$, $F \times E$, $G \times E$. The same is true in $Y \times F$ and $Y \times G$. Now glue these three spaces together (along $Y \times \{v\}$) to form $Y \times Y$. The spheres fit together to form $K_{3,3}$. Jan 17, 2019 at 12:51

Theorem 10 of "An Alternative Proof that 3-Manifolds Can be Triangulated" by R.H. Bing [Ann. of Math. Vol. 69, (1959), pp. 37-65] states that any topologically embedded simplicial 2-complex $K$ in a triangulated 3-manifold $M$ is near a PL embedding of $K$ into $M$.

In particular, a $2$-dimensional finite simplicial complex has a topological embedding in $\mathbb R^3$ if and only if it has a PL embedding in $\mathbb R^3$. Together with Robert Bell's answer this resolves the question.

I will show that $$Y^k$$ cannot embed in $$\mathbb{R}^p$$ when $$p < 2k$$.

Write $$\mathrm{Conf}(2,X) = \{(x,x') \in X^2 \, | \; x \neq x'\}$$ for the deleted diagonal of $$X$$, and note that any injection $$Y \hookrightarrow X$$ induces an $$S_2$$-equivariant map $$\mathrm{Conf}(2,Y) \to \mathrm{Conf}(2, X)$$.

We need the following corollary of Theorem 1.8 from my paper "Configuration space in a product".

If an embedding $$A \subseteq B$$ induces a homotopy equivalence of pairs $$(A^2, \mathrm{Conf}(2,A)) \simeq (B^2, \mathrm{Conf}(2,B))$$, then it also induces a homotopy equivalence $$\mathrm{Conf}(2,A^k) \simeq \mathrm{Conf}(2,B^k)$$ for all $$k \geq 0$$.

Here is a short proof of the corollary in the case $$k=2$$.

Two points in $$A^2$$ are distinct if and only if they are distinct in their first coordinate, or their second coordinate, or both. In other words, $$\mathrm{Conf}(2,A^2)$$ is covered by the two open subsets $$A \times \mathrm{Conf}(2,A)$$ and $$\mathrm{Conf}(2,A) \times A$$, and the intersection of these sets is $$\mathrm{Conf}(2,A) \times \mathrm{Conf}(2,A)$$. As a consequence, $$\mathrm{Conf}(2,A^2)$$ is the homotopy pushout of a diagram that depends only on the pair $$(A^2, \mathrm{Conf}(2,A))$$. By our assumption, the inclusion $$A \subseteq B$$ induces a pointwise homotopy equivalence on these pushout diagrams, and hence on homotopy pushouts. This concludes the proof for $$k=2$$; the case of general $$k$$ requires a larger homotopy colimit, but is otherwise similar.

"Reordering the cars in the driveway twice" gives an equivalence $$\mathrm{Conf}(2,Y) \simeq \mathrm{Conf}(2, \mathbb{R}^2)$$, (even when restricted to the driveway and a little part of the street, two cars may wind around each other), leading to an equivalence of pairs $$(Y^2, \mathrm{Conf}(2,Y)) \simeq (\mathbb{R}^4, \mathrm{Conf}(2,\mathbb{R}^2))$$ induced by the usual inclusion $$Y \subset \mathbb{R}^2$$. By the corollary, $$\mathrm{Conf}(2, Y^k)$$ is homotopy equivalent to $$\mathrm{Conf}(2, \mathbb{R}^{2k})$$, and moreover, this map is $$S_2$$ equivariant.

Since $$\mathrm{Conf}(2, \mathbb{R}^{p}) \simeq_{S_2} (S^{p-1}, \tau)$$, where $$\tau$$ denotes the antipodal action, and similarly $$\mathrm{Conf}(2,Y) \simeq_{S_2} \mathrm{Conf}(2, \mathbb{R}^{2k}) \simeq_{S_2} (S^{2k-1}, \tau)$$, any embedding $$Y^{k} \subseteq \mathbb{R}^p$$ induces an $$S_2$$-map $$(S^{2k-1}, \tau) \to (S^{p-1}, \tau),$$ which is impossible for $$p<2k$$ by the Borsuk-Ulam theorem.

• Thank you for your proof. I will read it carefully, but I need time (busy). Jan 30, 2019 at 17:53