Does there exist a $4$-dimensional contractible open manifold that is not homeomorphic to $\Bbb{R}^4$?

This question might be answered on this platform already. However, I am lacking the expertise to make the connection myself.

I read about the Mazur manifolds which are contractible manifolds with boundary which fail to be diffeomorphic (how about homeomorphic?) to $\mathbb R^4$. How about their interior?

I also read that some results of Freedman prove that there exist contractible 4-manifolds that are not homeomorphic to the 4-ball but I couldn't find the exact Proposition or Theorem yet.

Solution 1:

Hint: Verify that interiors of Mazur manifolds are not simply-connected at infinity unlike $R^4$. use the fact that the homology spheres bounding Mazur manifolds are not simply-connected.

Here are some details. Let's start with the definition of (topological) Mazur manifolds. Let $W$ be a compact contractible 4-dimensional manifold $W$ (necessarily with nonempty boundary $Y=\partial W$). A manifold $W$ is called a topological Mazur manifold if the boundary $Y$ is not simply-connected. Equivalently (this part is the 3-dimensional Poincare conjecture), $Y$ is not homeomorphic to the sphere $S^3$.

Here is a bit you can ignore if you do not know what homology means:

The above definition can be "reversed": Let $Y$ be a closed (compact and with empty boundary) 3-dimensional homology sphere, i.e. a connected nonempty orientable 3-dimensional manifold such that $H_1(Y)=0$. Then it is proven by Michael Freedman that for every manifold $Y$ with this property there exists a contractible compact 4-dimensional manifold $W$ (with boundary) such that $\partial W$ is homeomorphic to $Y$. Moreover, topology of $Y$ uniquely determines topology of $W$ (this is due to Richard Stong, Freedman's student). Thus, whenever $Y$ is not simply-connected, the manifold $W$ is a topological Mazur manifold.

Now, you are saying:

I read about the Mazur manifolds which are contractible manifolds with boundary which fail to be diffeomorphic (how about homeomorphic?) to ${\mathbb R}^4$.

This part is not quite right: Additionally, a Mazur manifold is required to be compact. Also, the fact that such a manifold is not homeomorphic to $E^4={\mathbb R}^4$ is quite easy: A compact topological space cannot be homeomorphic to a noncompact one. Furthermore, a manifold with nonempty boundary cannot be homeomorphic to a manifold with empty boundary, such as ${\mathbb R}^4$.

I need one more definition:

Definition. A topological space $X$ is said to be simply-connected at infinity if the following two properties hold:

a. $X$ is 1-ended, meaning that for every compact $K\subset X$ there exists a compact $C\subset X$ whose interior contains $K$, such that any two points $x, y\notin C$ can be connected by a path in $X\setminus K$. (One can think of this property as "path-connected at infinity".)

b. For every compact $K\subset X$ there exists a compact $C\subset X$ whose interior contains $K$, such that every continuous map $f: S^1\to X\setminus C$ extends to a continuous map $F: D^2\to X\setminus K$.

Lemma 1. For each $n\ge 3$, the Euclidean space $E^n$ is simply-connected at infinity.

Proof. Each compact $K\subset E^n$ is contained in a round ball $B\subset E^n$. Take $C:= B$ for both parts (a) and (b) of the definition. Then $E^n\setminus B$ is homeomorphic to $S^{n-1}\times (0,\infty)$ (this is proven using spherical coordinates). Since $n\ge 2$, $S^{n-1}$ is simply-connected and, therefore, the product $S^{n-1}\times (0,\infty)$ is also simply-connected. hence, every map $S^1\to E^n\setminus B$ extends to a map $D^2\to E^n\setminus B$. Lemma follows. qed

Lemma 2. Suppose that $W$ is a compact manifold with connected nonempty boundary $Y=\partial W$ and $Y$ is not simply-connected. Then the manifold $X:= W\setminus Y$ is not simply-connected at infinity.

Proof. It is known that the boundary of $W$ has collar, which is an open neighborhood $U$ of $Y$ in $W$ homeomorphic to $Y\times [0,1)$.

See also Proposition 3.42 in Hatcher's "Algebraic Topology" for a self-contained proof.

In what follows, I will identity points of $U$ with their images in $Y\times [0,1)$.

Pick a homotopically-nontrivial loop $c_0: S^1\to Y$, i.e. one which does not extend to a map $D^2\to Y$. Then for every $t\in (0,1)$, the loop $$ c_t: S^1\to Y\times \{t\}\subset U, c_t(s)=(t, c_0(s)) $$ is homotopically nontrivial in $U$. (This is because the inclusion map $Y\to U$ is a homotopy-equivalence.)

Now, take the compact $K:= W\setminus U$. For every compact $C\subset X=int(W)= W\setminus Y$, there exists $t\in (0,1)$ such that $$ C\subset K\cup Y\times (0,t). $$ Therefore, there exists a loop $c_t$ as above, whose image is in $W\setminus C$ and such that $c_t$ is homotopically nontrivial in $U$. Hence, $X$ is not simply-connected at infinity.

Corollary. The interior of each topological Mazur manifold $W$ is not homeomorphic to $E^4$.

Proof. If $X=int(W)$ is homeomorphic to $E^4$, then $X$ is simply-connected at infinity, contradicting Lemma 2. qed