# Contractible manifolds of dimension $n\le 2$

I have to give a talk about the Whitehead manifold. I would like to stress the fact that it is the first given example of a contractible open (i.e. non compact) manifold which is not homeomorphic to $$\mathbb{R}^n$$ (when $$n=3$$). This happens because the Whitehead manifold is not simply-connected at infinity and as proved by Stallings this is equivalent to not be homeomorphic to $$\mathbb{R}^n$$. The first example of this kind occurs in dimension $$3$$ beacause as I read in many articles the only contractible manifold (up to homeomorphism) of dimension $$n\le 2$$ are: $$\begin{equation*}\{\text{pt.}\} \quad \mathbb{R} \quad \mathbb{R}^2 \end{equation*}$$ I'm thinking about a proof of this classification and I'm having problems with dimension $$2$$. So I have the following questions:

## Question 1

Can you suggest me any reference for the classification in dimension $$2$$ with a detailed proof?

## Question 2

Is the Stalling's result mentioned above valid for a general $$n$$-manifold or do we have to require that it is contractible?

• I am currently thinking of this. If $S$ is an open contractible surface, define $\Sigma$ to be its Alexandrov compactification (its one point compactification). If one can show $\Sigma$ is a topological sphere, then $S$ is homeomorphic to $\mathbb{S}^2\setminus \{pt\}$ which is $\mathbb{R}^2$. Showing $\Sigma$ is the sphere would be equivalent to show it is a simply-connected compact surface, and I think this is not a hard thing to do. Commented Nov 12, 2020 at 10:37
• Is there any relation between the fundamental group of $X$ and the fundamental group of its one point compactification? I know that the one point compactification is compact, so I should only have to prove that it is simply-connected. Commented Nov 12, 2020 at 11:12
• I'm not that sure in the general case. But here, suppose you have a loop in the compactification. If it is not passing through the infinity point, then it is a loop in $S$ and by contractibility, it is homotopic to a point. If the loop passes through the infinity point, I think (this is not a proof!) we can deform it locally to evitate the infinity point, and use the previous work. Commented Nov 12, 2020 at 11:15
• In fact, it seems the hard thing would be to show that the one point compactification will be a topological manifold Commented Nov 12, 2020 at 11:26
• Yes, I think is the hard part of the proof. Commented Nov 12, 2020 at 12:55

## 1 Answer

Forget the one-point compactification, it's a dead end. By the time you prove that a neighborhood of infinity looks like $$[0,\infty) \times S^1$$ (which is what you'd need to see the 1-pt compactification is a manifold) you're more or less already done.

Theorem: If $$\Sigma$$ is an open surface with $$H_1(\Sigma; \Bbb F_2) = 0$$, then $$\Sigma$$ is homeomorphic to $$\Bbb R^2$$.

Sketch of proof.

(1) First prove that $$\Sigma$$ has exactly one end; if $$K \subset \Sigma$$ is a compact subset, then $$\Sigma \setminus \text{int}(K)$$ may have many connected components, but only one of them is noncompact. The proof will be by contrapositive: if $$\Sigma$$ has two ends, show that $$H_1(\Sigma; \Bbb F_2) \neq 0$$. You will need to know that the inclusion $$\partial \Sigma \to \Sigma$$ is nonzero on first homology whenever $$\Sigma$$ is a noncompact surface with boundary.

(2) Use that $$\Sigma$$ has a compact exhaustion --- $$\Sigma = \bigcup \Sigma_n$$, where $$\Sigma_n$$ is a compact surface and $$\Sigma_n \subset \text{int}(\Sigma_{n+1})$$, so that $$\Sigma_{n+1} = \Sigma_n \cup S_n$$, where $$S_n$$ is also a compact surface. This can be justified using the fact that $$\Sigma$$ has a proper smooth function to $$\Bbb R$$, together with Sard's theorem.

(3) Using (1) and (2) together, observe that $$\Sigma \setminus \text{int}(\Sigma_n)$$ only has one noncompact piece. Modify your compact exhaustion so that $$\Sigma \setminus \Sigma_n$$ is connected and so that $$S_n$$ is connected.

(4) Prove that $$\partial \Sigma_n$$ is a single circle, as otherwise $$\Sigma$$ has positive genus (you'll glue on a pair of pants as $$S_n$$, because $$S_n$$ is connected), which would imply $$H_1(\Sigma;\Bbb F_2) \neq 0$$; again this involves some Mayer-Vietoris work.

(5) Prove that each $$\Sigma_1$$ is a disc and each $$S_n$$ is a cylinder. From here it follows that $$\Sigma \cong \Bbb R^2$$.

The details are not completely trivial, in particular on (3). I will not see or respond to comments, so please feel free to edit this answer as you desire. This gives a rough strategy for the general classification of noncompact surfaces without boundary, the ideas simplify when $$\Sigma$$ is contractible like this; similarly you can show that if $$H_1(\Sigma)$$ is finite, then $$\Sigma$$ is obtained by deleting finitely many points from a closed surface.

• Thanks very much for your answer! Could you please add the references for all of this? Commented Nov 12, 2020 at 14:47