14
$\begingroup$

In the http://arxiv.org/abs/math/0606464v1 I read

"If you want to prove existence of exotic smooth structure on $\mathbb R^4$ you can do this if you are in possession of a knot which is topologically slice but not smoothly slice (slice means zero slice genus). Freedman has a result stating that a knot with Alexander polynomial 1 is topologically slice. We now have an obstruction (s being non-zero) to being smoothly slice." (p.28)

What does it mean? Does anyone know the construction of exotic $\mathbb R^4$ using slice knots? Please, give me a references, if there are several constructions.

Added: http://arxiv.org/abs/math/0408379. Does there exist some other construction?

$\endgroup$
1

2 Answers 2

14
$\begingroup$

From Jacob Rasmussen's paper "Knot polynomials and knot homologies", arXiv:math/0504045, p.13 of ArXiv version:

Bob Gompf kindly pointed out another such application [of Rasmussen's $s$-invariant, a concordance invariant of knots extracted from Khovanov homology]. Namely, $s$ can also be used to give a gauge-theory free proof of the existence of an exotic $\mathbb{R}^4$. Indeed, Gompf has shown that to construct such a manifold, it suffices to exhibit a knot $K$ which is topologically but not smoothly slice (see Gompf and Stipsicz, "4-manifolds and Kirby calculus", p. 522 for a proof). By a theorem of Freedman, any knot with Alexander polynomial 1 is topologically slice, so we need only find a knot $K$ with $\Delta_K(t)=1$ and $s(K) \neq 1$. It is not difficult to provide such a knot - for example, the $(-3,5,7)$ pretzel knot will do.
$\endgroup$
0
25
$\begingroup$

I wrote this up for my own personal edification when I was going through Gompf and Stipsicz. Please, excuse the pedantry. I had to make sure even I understood it.

Definition: (Smoothly Slice Knot) A knot $K$ in $\partial D^4$, where $D^n$ is the standard closed $n$--dimensional disk, is smoothly slice if there exists a two-disk $D^2$ smoothly embedded in $D^4$ such that the image of $\partial D^2$ is $K$.

Definition: (Flat Topological Embedding) Let $X$ be a topological manifold of dimension $n$ and $Y$ a topological manifold of dimension $m$ where $n < m$. A topological embedding $\rho : X \rightarrow Y$ is flat if it extends to a topological embedding $\rho : X \times D^{m - n} \rightarrow Y$.

Definition: (Topologically Slice Knot) A knot $K$ in $\partial D^4$ is topologically slice if there exists a two-disk $D^2$ flatly topologically embedded in $D^4$ such that the image of $\partial D^2$ is $K$.

Lemma 1: Let $K$ be a knot in $\partial D^4$ and $X_K$ the two-handlebody obtained by attaching a two-handle to $D^4$ along $K$ with framing $0$. $X_K$ has a smooth embedding into $\mathbb{R}^4$ iff $K$ is smoothly slice.

Proof: First, let us assume there exists a smooth embedding $\rho: X_K \rightarrow \mathbb{R}^4$ and prove $K$ is smoothly slice.

As $X_K$ is compact and $\rho$ a smooth embedding, $\rho(X_K)$ is compact. Thus, the Heine-–Borel Theorem (Theorem 2.41) Rudin implies there exists an open ball $B^4 \subset \mathbb{R}^4$ such that $\rho(X_K) \subset B^4$. Therefore, $\rho$ can be used to create a smooth embedding $\rho: X_K \rightarrow S^4$, mapping $X_K$ to one hemisphere.

The zero-handle of $X_K$ is $D^4$. Thus, $\rho$ smoothly embeds this $D^4$ into $S^4$. The Disk Theorem (Corollary 4.7 Chapter III) Kosinski implies that any two orientation preserving smooth embeddings of $D^4$ into $S^4$ are isotopic. Thus, the closure of the complement of any open ball in $S^4$ is diffeomorphic to $D^4$. Hence, $S^4 - int(\rho(D^4))$ is diffeomorphic to $D^4$, where $int(X)$ denotes the interior of $X$.

By construction, $K$ is smoothly embedded in $\partial(S^4 - int(\rho(D^4)))$ and the $D^2$ core of the $X_K$ two-handle is smoothly embedded in $S^4 - int(\rho(D^4))$ such that the image of $\partial D^2$ is $K$. Thus, we conclude that if there exists a smooth embedding $\rho: X_K \rightarrow \mathbb{R}^4$, then $K$ is smoothly slice.

Now, let us assume $K$ is smoothly slice and prove there exists a smooth embedding $\rho': X_K \rightarrow \mathbb{R}^4$.

We now work backwards. As $K$ is smoothly slice, there exists a smooth embedding of $D^2$ into $D^4$ such that the image of $\partial D^2$ is $K$. As this $D^2$ is smoothly embedded, the Tubular Neighborhood Theorem (Theorem 4.2 Chapter III) Kosinski implies there exists a closed tubular neighborhood $D^2 \times D^2$ of $D^2$ smoothly embedded in $D^4$. (This closed tubular neighborhood will become the two-handle of $X_K$.)

Using the identity map glue a second $D^4$ to the above $D^4$ forming $S^4$ as $D^4 \cup_{\partial D^4} \bar{D}^4$, where $\bar{D}^4$ is $D^4$ with the opposite orientation. This new $D^4$ along with the closed tubular neighborhood above forms $X_K$. This two-handlebody $X_K$ is smoothly embedded into $S^4$. We notate this smooth embedding as $\rho'$.

We can modify the closed tubular neighborhood above, shrinking it if required, such that $S^4 - \rho'(X_K)$ is not empty. Thus, removing a point in $S^4 - \rho'(X_K)$ we obtain the desired smooth embedding $\rho': X_K \rightarrow \mathbb{R}^4$. QED.

Lemma 2: Let $K$ be a knot in $\partial D^4$ and $X_K$ the two-handlebody obtained by attaching a two-handle to $D^4$ along $K$ with framing $0$. $X_K$ has a topological embedding into $\mathbb{R}^4$ iff $K$ is topologically slice.

Proof: First, let us assume there exists a topological embedding $\rho: X_K \rightarrow \mathbb{R}^4$ and prove $K$ is topologically slice.

A slight variation of the logic in the previous proof implies $\rho$ can be used to create a topological embedding $\rho: X_K \rightarrow S^4$, mapping $X_K$ to one hemisphere.

Again, a variation of the logic in the previous proof implies that $D^4$ is diffeomorphic to $S^4 - int(\rho(D^4))$, where $\rho$ is acting on $D^4$ the zero-handle of $X_K$.

By construction, $K$ is topologically embedded in $\partial(S^4 - int(\rho(D^4)))$, and $D^2$, the core of the $X_K$ two-handle, is topologically embedded in $S^4 - int(\rho(D^4))$ such that the image of $\partial D^2$ is $K$.

The two-handle of $X_K$, which is homeomorphic to $D^2 \times D^2$, is topologically embedded into $S^4 - int(\rho(D^4))$ such that it extends the topological embedding of $D^2$, the core of the $X_K$ two-handle. Thus, we conclude that if there exists a topological embedding $\rho: X_K \rightarrow \mathbb{R}^4$, then $K$ is topologically slice.

Now, let us assume $K$ is topologically slice and prove there exists a topological embedding $\rho': X_K \rightarrow \mathbb{R}^4$.

Again, we work backwards. As $K$ is topologically slice, there exists a topological embedding of $D^2$ into $D^4$ such that the image of $\partial D^2$ is $K$ and there exists a topological embedding of $D^2 \times D^2$ into $D^4$ that extends the topological embedding of $D^2$. (This topological embedding of $D^2 \times D^2$ will become the two-handle of $X_K$.)

Using the identity map glue a second $D^4$ to the above $D^4$ forming $S^4$ as $D^4 \cup_{\partial D^4} \bar{D}^4$. This new $D^4$ along with the above topological embedding of $D^2 \times D^2$ forms $X_K$. This two-handlebody $X_K$ is topologically embedded into $S^4$. We notate this topological embedding as $\rho'$.

Using the same logic as in the previous proof, we can remove a point from $S^4 - \rho'(X_K)$ to construct the desired topological embedding $\rho': X_K \rightarrow \mathbb{R}^4$. QED.

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 that can not be smoothly embedded into $\mathbb{R}^4$.

Thorem: If there exists a knot $K$ that is topologically slice but not smoothly slice, then there exists a large exotic $\mathbb{R}^4$.

Proof: As $K$ is topologically slice, Lemma 2 implies there exists $\rho$ a topological embedding of $X_K$ into $\mathbb{R}^4$, where $X_K$ is the two-handlebody obtained by attaching a two-handle to $D^4$ along $K$ with framing $0$.

First let us prove $\mathbb{R}^4 - int(\rho(X_K))$ is a topological manifold. As $\rho$ is a topological embedding, $\rho(X_K)$ is a topological submanifold of $\mathbb{R}^4$. Hence, $int(\rho(X_K))$ is also a topological submanifold of $\mathbb{R}^4$. This implies $\mathbb{R}^4 - int(\rho(X_K))$ is a topological submanifold of $\mathbb{R}^4$. Thus, $\mathbb{R}^4 - int(\rho(X_K))$ is a topological manifold in its own right.

Next let us prove $\mathbb{R}^4 - int(\rho(X_K))$ is not compact. $X_K$, by construction, is compact. Thus, as $\rho$ is a topological embedding, $\rho(X_K)$ is a compact topological submanifold of $\mathbb{R}^4$. As $\mathbb{R}^4$ is not compact, it follows that $\mathbb{R}^4 - \rho(X_K)$ is not compact. Hence, as $\mathbb{R}^4 - \rho(X_K)$ has two ends, this implies $\mathbb{R}^4 - int(\rho(X_K))$ is not compact.

Now we will prove $\mathbb{R}^4 - int(\rho(X_K))$ is connected. In the standard handle presentation, $S^4$ has no one-handles. Thus, as a result of Theorem 3.4 Chapter VII Kosinski, we have $H_1(S^4;\mathbb{Z}) = 0$. As in Lemma 2, we can use $\rho$ to create a topological embedding $\rho: X_K \rightarrow S^4$. Hence, $\rho(X_K)$ is a closed, proper subset of the compact connected orientable manifold $S^4$. Thus, Alexander Duality (Theorem 6.6(a) Chapter XIV) Massey implies that the Cech cohomology group $\check{H}^3(\rho(X_K);\mathbb{Z})$ is isomorphic to reduced homology group $\tilde{H}_0(S^4 - \rho(X_K);\mathbb{Z})$. As $X_K$ is compact, $\rho(X_K)$ is compact. Thus, $\rho(X_K)$ is paracompact. Hence, (Page 372) Massey $\check{H}^3(\rho(X_K);\mathbb{Z})$ is isomorphic to $H^3(\rho(X_K);\mathbb{Z})$. As $\rho$ is a topological embedding, $H^3(\rho(X_K);\mathbb{Z})$ is isomorphic to $H^3(X_K;\mathbb{Z})$. The ``lower'' boundary $\partial_- X_K$ of $X_K$ is empty. Thus, $H^3(X_K;\mathbb{Z})$ is isomorphic to $H^3(X_K,\partial_- X_K;\mathbb{Z})$. Poincare duality (Theorem 5.1 Chapter VII) Kosinski then implies $H^3(X_K,\partial_- X_K;\mathbb{Z})$ is isomorphic to $H_1(X_K,\partial_+ X_K;\mathbb{Z})$. However, as $X_K$ has no three-handles, Theorem 3.4 Chapter VII Kosinski implies $H_1(X_K,\partial_+ X_K;\mathbb{Z}) = 0$. Thus, tracing isomorphisms, we have proven $\tilde{H}_0(S^4 - \rho(X_K);\mathbb{Z}) = 0$; in other words $S^4 - \rho(X_K)$ is connected. Removing a point from a four-manifold will not disconnect it; thus, $\mathbb{R}^4 - \rho(X_K)$ is connected. (Equivalently, one could prove this by applying Alexander Duality again.) Finally, as adding in boundary points also does not disconnect, $\mathbb{R}^4 - int(\rho(X_K))$ is also connected.

So, as $\mathbb{R}^4 - int(\rho(X_K))$ is a non-compact connected topological four--manifold, Theorem 9.4.22 Gompf and Stipsicz implies $\mathbb{R}^4 - int(\rho(X_K))$ admits a smooth structure.

As $\rho$ is a topological embedding, the restriction $\rho: \partial X_K \rightarrow \partial (\mathbb{R}^4 - int(\rho(X_K)))$ is a homeomorphism. Thus, $\partial X_K$ and $\partial (\mathbb{R}^4 - int(\rho(X_K)))$ are homeomorphic.

As $X_K$ is a smooth manifold, it induces a smooth structure on $\partial X_K$. Similarly, as $\mathbb{R}^4 - int(\rho(X_K))$ is a smooth manifold, it induces a smooth structure on $\partial (\mathbb{R}^4 - int(\rho(X_K)))$. A topological three-manifold admits a unique smooth structure Moise. Thus, the smooth structure on $\partial X_K$ is the same as that on $\partial (\mathbb{R}^4 - int(\rho(X_K)))$. Hence, $\partial X_K$ and $\partial (\mathbb{R}^4 - int(\rho(X_K)))$ are diffeomorphic.

The diffeomorphism of $\partial X_K$ and $\partial (\mathbb{R}^4 - int(\rho(X_K)))$ can be used to join $X_K$ and $\mathbb{R}^4 - int(\rho(X_K))$ along their respective boundaries forming a smooth manifold $R$. Obviously, $R$ is homeomorphic to $\mathbb{R}^4$.

We will now prove, by contradiction, that $R$ is exotic. Let us assume $R$ is diffeomorphic to $\mathbb{R}^4$. Thus, there exists a diffeomorphism $\varphi: R \rightarrow \mathbb{R}^4$. The restriction of this diffeomorphism to $X_K$, which is smoothly embedded in $R$, is a smooth embedding of $X_K$ into $\mathbb{R}^4$. However, Lemma 1 implies such a smooth embedding exists if and only if $K$ is smoothly slice. But, by hypothesis, $K$ is not smoothly slice. Thus, our assumption that there exists a diffeomorphism $\varphi: R \rightarrow \mathbb{R}^4$ leads to a contradiction. Therefore, no such diffeomorphism exists. Hence, $R$ is exotic, homeomorphic but not diffeomorphic to $\mathbb{R}^4$.

Finally, we prove $R$ is large. $X_K$, by construction, is compact. Also, by construction, $X_K$ is a smooth submanifold of $R$, an exotic $\mathbb{R}^4$. By hypothesis, $K$ is not smoothly slice. Thus, Lemma 1 implies $X_K$ can not smoothly embed in $\mathbb{R}^4$. Hence, we conclude, $R$ is large. QED.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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