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?

  • $\begingroup$ 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. $\endgroup$ – Didier Nov 12 '20 at 10:37
  • $\begingroup$ 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. $\endgroup$ – John117 Nov 12 '20 at 11:12
  • $\begingroup$ 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. $\endgroup$ – Didier Nov 12 '20 at 11:15
  • 1
    $\begingroup$ In fact, it seems the hard thing would be to show that the one point compactification will be a topological manifold $\endgroup$ – Didier Nov 12 '20 at 11:26
  • $\begingroup$ Yes, I think is the hard part of the proof. $\endgroup$ – John117 Nov 12 '20 at 12:55

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.

  • $\begingroup$ Thanks very much for your answer! Could you please add the references for all of this? $\endgroup$ – John117 Nov 12 '20 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.