Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can you flip the end of a large exotic $\mathbb{R}^4$

Background

Definition (Exotic $\mathbb{R}^4$): An exotic $\mathbb{R}^4$ is a smooth manifold $R$ homeomorphic but not diffeomorphic to $\mathbb{R}^4$, where $\mathbb{R}^4$ is equipped with its standard smooth structure.

Definition (Large exotic $\mathbb{R}^4$): A large exotic $\mathbb{R}^4$ is an exotic $\mathbb{R}^4$ containing a four-dimensional compact smooth submanifold $K'$ that can not be smoothly embedded into $\mathbb{R}^4$.

Definition (End of a large exotic $\mathbb{R}^4$): If $R$ is a large exotic $\mathbb{R}^4$ and $D^4$ is a four-dimensional disk topologically embedded into $R$ such that $K' \subset D^4$, then $R - D^4$ is an end of $R$.

Remark: The previous definition varies slightly from the standard definition of "end", however it will be used for the remainder of this question. (See Gompf and Stipsicz Exercise 9.4.11 for the standard definition.)

Remark: If $R - D^4$ is the end of a large exotic $\mathbb{R}^4$, then $R - D^4$ is a smooth manifold that inherits a smooth structure from $R$.

Definition (Flip of the end of a large exotic $\mathbb{R}^4$): Given $R - D^4$, the end of a large exotic $\mathbb{R}^4$, a flip of $R - D^4$ is a diffeomorphism $f: R - D^4 \rightarrow R - D^4$ that maps the "inner region" of $R - D^4$, that "near" the removed $D^4$, to the "outter region", that "near infinity", and vica-versa.

Remark: The previous definition is also non-standard. I am not aware of any standard definitions that carry, more-or-less, the same meaning.

So, at this stage the meaning of the question is hopefully clear.

Foreground

In our attempt to flip the end of a large exotic $\mathbb{R}^4$ an inconvenient truth stands in our way:

Theorem 1 (Uncountably many flips fail): There are uncountably many large exotic $\mathbb{R}^4$'s that one can not flip the end of.

Quickly, let us see why this is true. Lemma 9.4.2 along with Addendum 9.4.4 of Gompf and Stipsicz state:

Lemma 1: There exist pairs $(X,Y)$ and $(L,K)$ of smooth, oriented four-manifolds with $X$ simply connected, $Y$ and $K$ compact, $X$ and $L$ open (i.e. noncompact and boundaryless), $L$ homeomorphic to $\mathbb{R}^4$, and $X$ with negative definite intersection form not isomorphic to $n\langle-1\rangle$, such that $X - int(Y)$ and $L - int(K)$ are orientation-preserving diffeomorphic.

Theorem 9.4.3 of Gompf and Stipsicz states:

Theorem: Any $L$ as appears in Lemma 1 is a large exotic $\mathbb{R}^4$.

The two statements above lead to:

Lemma: One can not flip the end of any $L$ as appears in Lemma 1.

Proof: Assume one could flip the end of $L$. Thus, one could use this flip to glue $L$ to the "end" of $X$ and obtain a simply connected closed smooth four-manifold with negative definite intersection form not isomorphic to $n\langle-1\rangle$. However, according to Donaldson's Theorem (Gompf and Stipsicz Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip. QED

Now we have shown that $L$ can not be flipped. Before we show how uncountably many large exotic $\mathbb{R}^4$'s can not be flipped, we need the definition:

Definition (Radial Family): Let $R$ be an exotic $\mathbb{R}^4$. Thus, there exists a homeomorphism $h:\mathbb{R}^4 \rightarrow R$. Define $R_t$ as the image under $h$ of the open ball of radius $t$ centered at $0$ in $\mathbb{R}^4$. A radial family is a set of the form $\{R_t | 0 < t \le \infty \}$.

Remark: If $R_t$ is a member of a radial family, then $R_t$ is a smooth manifold as it inherits a smooth structure from $R$

Theorem 9.4.10 of Gompf and Stipsicz states:

Theorem: If $\{L_t | 0 < t \le \infty \}$ is a radial family for an $L$ as appears in Lemma 1 and $r$ is such that $K \subset L_r$, then $\{L_t | r \le t \le \infty \}$ is an uncountable family of non-diffeomorphic large exotic $\mathbb{R}^4$'s.

This leads directly to a proof of Theorem 1.

Proof: Assume one could flip the end of $L_t$ for $r \le t \le \infty$, where all notation is as in the previous theorem. Thus, one could use this flip to glue $L_t$ to the "end" of $X$ less the image of $L - L_t$ and obtain a simply connected closed smooth four-manifold with negative definite intersection form not isomorphic to $n\langle-1\rangle$. Again, according to Donaldson's Theorem (Gompf and Stipsicz Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip.QED

Things are seeming rather hopeless at this point. In fact, things are worse than they seem! But, before we can revel in this despair, we must introduce two definitions:

Definition (Simply Connected at Infinity): Let $Z$ be a topological manifold. $Z$ is simply connected at infinity if for any compact subset $C$ of $Z$ there exists a compact subset $C'$ of $Z$ that contains $C$ and is such that the inclusion $Z - C' \rightarrow Z - C$ induces the trivial map $\pi_1(Z - C') \rightarrow \pi_1(Z - C)$.

Definition (End Sum): Let $Z_1$ and $Z_2$ be non-compact oriented smooth four-manifolds that are simply connected at infinity. Choose two proper smooth embeddings $\gamma_i : [0, \infty) \rightarrow Z_i$. Remove a tubular neighborhood of $\gamma_i((0, \infty))$ from each $Z_i$ and glue the resulting $\mathbb{R}^3$ boundaries together respecting orientations. The result is the end sum $Z_1 \natural Z_2$ of $Z_1$ and $Z_2$.

Remark: The requirement that $Z_i$ is simply connected at infinity guarantees that $\gamma_i$ is unique up to ambient isotopy and thus $Z_1 \natural Z_2$ is unique up to diffeomorphism (Gompf and Stipsicz Definition 9.4.6).

Remark: If $R_1$ and $R_2$ are exotic $\mathbb{R}^4$, then they are non-compact oriented smooth four-manifolds that are simply connected at infinity and $R_1 \natural R_2$ is a smooth manifold homeomorphic to $\mathbb{R}^4$.

Remark: $X$ of Lemma 1 is simply connected at infinity.

Theorem 2: If $\{L_t | 0 < t \le \infty \}$ is a radial family for an $L$ as appears in Lemma 1 and $r$ is such that $K \subset L_r$, where $K$ is as in Lemma 1, then for $R$ an exotic $\mathbb{R}^4$ and $t$ such that $r \le t \le \infty$ there exists no flip of $R \natural L_t$.

Proof: The proof is basically a slight variation on the above theme. Assume one could flip the end of $R \natural L_t$ for $r \le t \le \infty$. Thus, one could use this flip to glue $R \natural L_t$ to the "end" of $X$ less the image of $L - L_t$ end summed with $R$, in other words with the flip glue $R \natural L_t$ to $R \natural (X - (L - L_t))$, and obtain a simply connected closed smooth four-manifold with negative definite intersection form not isomorphic to $n\langle-1\rangle$. Again, according to Donaldson's Theorem (Gompf and Stipsicz Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip.QED

Now we can revel in this despair!

However, other ways of creating large exotic $\mathbb{R}^4$'s exist. For example, given a topologically slice knot that is not smoothly slice one can create a large exotic $\mathbb{R}^4$. (See, for example, Davis.) Such a large exotic $\mathbb{R}^4$, as far as I can see, might admit an end flip. But, I'm not sure. Thus, we end where we began.

Question

Can you flip the end of a large exotic $\mathbb{R}^4$?

share|improve this question
    
I wonder if one can prove all large exotic $\mathbb{R}^4$ are of the form $R \natural L_t$, where the notation is as in Theorem 2? One could then use Theorem 2 to answer the original question with a "No." However, I've no idea how to prove all large exotic R4 are of the form $R \natural L_t$. –  Kelly Davis Apr 2 '11 at 8:19
    
Another idea would be to use $U$ universal $\mathbb{R}^4$. Theorem 2 implies $U \natural L_t$ does not flip. As $U \natural L_t = U = U \natural R$ for $R$ any large exotic $\mathbb{R}^4$, we know $U$ and $U \natural R$ do not flip. Maybe this could be used to show $R$ doesn't flip? –  Kelly Davis Apr 3 '11 at 10:08

1 Answer 1

As stated, your question is equivalent to the existence of a large exotic 4-ball (a smooth $D^4$ which cannot be smoothly embedded into $\mathbb{R}^4_{std}$).

The existence of a flip would give rise to an exotic $S^4$, by gluing the $D^4$ at infinity using the flip diffeomorphism. Removing a (small) standard ball from this $S^4$ gives a large exotic $D^4$, since it contains a smooth submanifold $K'$ which cannot embed in $\mathbb{R}^4_{std}$.

Conversely, if you had a large exotic $D^4$, then you could adjoin a collar neighborhood of $S^3\times \mathbb{R}$ to get an exotic $\mathbb{R}^4$ which has a standard end (diffeomorphic to $S^3\times \mathbb{R}$), and therefore admits a flip.

Although I'm not an expert, I'm certain that the existence of a large exotic 4-ball is open (otherwise, the 4D smooth Schoenflies conjecture would imply the 4D smooth Poincare conjecture).

I realize that this does not answer the spirit of your question, which is whether there is a large exotic $\mathbb{R}^4$ which does not have a standard product end and which admits a flip.

share|improve this answer
    
You're playin' fast-and-loose with the boundaries there, but I see what you mean. However, as you mention, this is not a proof that no flip exists; it's only a list of consequences of the existence of a flip. –  Kelly Davis Mar 29 '11 at 19:49
6  
Right Kelly, but part of the function of mathoverflow answers is to determine if a problem is open. –  Ian Agol Mar 30 '11 at 4:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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