# When does a CW-complex of dimension 2 embedd in $R^4$ ?

Let X be a finite CW-complex of dimension two having just one 0-cell (+ finitely many 1-cells + finitely many 2-cells). Is it true that X can be embedded in $R^4$ ? If true, is it due to Stallings ?

More generally, let Y be a finite CW-complex of dimension two of which the 1-skeleton embeds in $R^2$; is is true that Y can be embedded in $R^4$ ?

(Reminder : any finite CW-complex of dimesnion two embeds in $R^5$.)

Any comment, or even better a reference, would help.

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Thanks for the help, from both inside and outside MathOverflow. Here is the fact: a finite 2-complex embeds in $R^4$ as soon as its 1-skeleton is planar. And the reference: it is a very particular case of a statement from a 1965 paper by John Stallings, \emph{Embedding homotopy types into manifolds} unpublished but available on his web page, math.berkeley.edu/~stall/embkloz.pdf –  Pierre de la Harpe Apr 5 '10 at 15:19

I think there is an obstruction called Van Kampen's obstruction to embedding an $n$-complex into $\mathbb{R}^{2n}$. If you could embed the $n$-complex into $\mathbb{R}^{2n}$ then it has a thickening to a manifold, and that manifold has an intersection pairing in dimension $n$. The obstruction is derived from this pairing. If the complex embeds the obstruction vanishes. For complexes of dimension greater than 2 this suffices, it is not a complete invariant for two complexes. This is worked out in a paper of Krushkal, Freedman, and Teichner and then an example of imcompleteness of the invariant is in a paper of Krushkal. The example given by FKT is of the two skeleton of the six simplex.

Of course that example has more than one vertex. If you could embed one vertex $2$-complexes in $\mathbb{R}^4$ then some weird stuff would be happening in that example as you collapse a maximal tree in the one skeleton.

The weird thing is that the one skeleton becomes planar. There is a folk theorem that the classification of four manifolds is undecidable because any finitely presented group is the fundamental group of a four manifold. The proof is to embed a two complex with one vertex, whose fundamental group is the chosen finitely presented group into $\mathbb{R}^4$ and take a regular neighborhood. Its gotta be Stallings. :)

I would guess that the two questions, one vertex, or planar $1$-skeleton are equivalent.

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One can see that the 2-skeleton of the 6-simplex can't embed from a theorem of Conway & Gordon, which says that for any embedding of the complete graph K_6 into S^3, there are a pair of disjoint 3-cycles (i.e. triangles) which have non-zero (odd) linking number. If one had a (locally flat) embedding of the 2-skeleton of the 6-simplex into R^4, then the link of a vertex would be an embedding of K_6 in S^3. Two linked triangles give rise to two spheres in R^4 with non-zero intersection, which is of course impossible. Maybe there's a way to generalize this argument to a one vertex complex? –  Ian Agol Mar 28 '10 at 16:41
Stallings's theorem is that a finite 2-complex embeds in R^4 up to homotopy. I.e. a finite 2-complex X is homotopy equivalent to a 2-complex Y such that Y embeds in R^4 –  Allan Edmonds Mar 28 '10 at 16:52

Shapiro's obstruction:

A. Shapriro, "Obstructions to the imbedding of a complex in Euclidean space, I. The first obstruction," Ann. of Math., 66 No. 2 (1957), 256--269.

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@Agol: In the one vertex case, how to prove there exist two 2-cells $A$ and $B$, of which the boundaries have non-zero linking number?