Slice knots and exotic $\mathbb R^4$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T05:03:58Z http://mathoverflow.net/feeds/question/42624 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/42624/slice-knots-and-exotic-mathbb-r4 Slice knots and exotic $\mathbb R^4$ Nikita Kalinin 2010-10-18T13:00:14Z 2011-03-09T08:29:13Z <p>In the <a href="http://arxiv.org/abs/math/0606464v1" rel="nofollow">http://arxiv.org/abs/math/0606464v1</a> I read </p> <p>"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)</p> <p>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.</p> <p>Added: <a href="http://arxiv.org/abs/math/0408379" rel="nofollow">http://arxiv.org/abs/math/0408379</a>. Does there exist some other construction?</p> http://mathoverflow.net/questions/42624/slice-knots-and-exotic-mathbb-r4/42649#42649 Answer by Tim Perutz for Slice knots and exotic $\mathbb R^4$ Tim Perutz 2010-10-18T15:01:23Z 2010-10-18T15:01:23Z <p>From Jacob Rasmussen's paper "Knot polynomials and knot homologies", arXiv:math/0504045, p.13 of ArXiv version:</p> <blockquote> 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. </blockquote> http://mathoverflow.net/questions/42624/slice-knots-and-exotic-mathbb-r4/57926#57926 Answer by Kelly Davis for Slice knots and exotic $\mathbb R^4$ Kelly Davis 2011-03-09T07:17:38Z 2011-03-09T08:29:13Z <p>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.</p> <p><strong>Definition:</strong> (<strong>Smoothly Slice Knot</strong>) 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 $K$.</p> <p><strong>Definition:</strong> (<strong>Flat Topological Embedding</strong>) Let $X$ be a topological manifold of dimension $n$ and $Y$ a topological manifold of dimension $m$ where $n &lt; m$. A topological embedding $\rho : X \rightarrow Y$ is flat if it extends to a topological embedding $\rho : X \times D^{m - n} \rightarrow Y$.</p> <p><strong>Definition:</strong> (<strong>Topologically Slice Knot</strong>) 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 $K$.</p> <p><strong>Lemma 1:</strong> 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.</p> <p><strong>Proof:</strong> First, let us assume there exists a smooth embedding $\rho: X_K \rightarrow \mathbb{R}^4$ and prove $K$ is smoothly slice. </p> <p>As $X_K$ is compact and $\rho$ a smooth embedding, $\rho(X_K)$ is compact. Thus, the Heine-â€“Borel Theorem (Theorem 2.41) <a href="http://tinyurl.com/28bthn" rel="nofollow">Rudin</a> 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.</p> <p>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) <a href="http://tinyurl.com/68d9oae" rel="nofollow">Kosinski</a> 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 $X$.</p> <p>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.</p> <p>Now, let us assume $K$ is smoothly slice and prove there exists a smooth embedding $\rho': X_K \rightarrow \mathbb{R}^4$.</p> <p>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) <a href="http://tinyurl.com/68d9oae" rel="nofollow">Kosinski</a> 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 $X_K$.)</p> <p>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 $\rho'$. </p> <p>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 $\rho': X_K \rightarrow \mathbb{R}^4$. <strong>QED</strong>.</p> <p><strong>Lemma 2:</strong> 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.</p> <p><strong>Proof:</strong> First, let us assume there exists a topological embedding $\rho: X_K \rightarrow \mathbb{R}^4$ and prove $K$ is topologically slice.</p> <p>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.</p> <p>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 $X_K$.</p> <p>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 $K$. </p> <p>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.</p> <p>Now, let us assume $K$ is topologically slice and prove there exists a topological embedding $\rho': X_K \rightarrow \mathbb{R}^4$.</p> <p>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 $X_K$.)</p> <p>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 $\rho'$. </p> <p>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 $\rho': X_K \rightarrow \mathbb{R}^4$. <strong>QED</strong>.</p> <p><strong>Definition:</strong> (<strong>Large Exotic $\mathbb{R}^4$</strong>) 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 $\mathbb{R}^4$.</p> <p><strong>Thorem:</strong> If there exists a knot $K$ that is topologically slice but not smoothly slice, then there exists a large exotic $\mathbb{R}^4$.</p> <p><strong>Proof:</strong> 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 $0$.</p> <p>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.</p> <p>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.</p> <p>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 <a href="http://tinyurl.com/68d9oae" rel="nofollow">Kosinski</a>, 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) <a href="http://tinyurl.com/4e6np84" rel="nofollow">Massey</a> 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) <a href="http://tinyurl.com/4e6np84" rel="nofollow">Massey</a> $\check{H}^3(\rho(X_K);\mathbb{Z})$ 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) <a href="http://tinyurl.com/68d9oae" rel="nofollow">Kosinski</a> 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 <a href="http://tinyurl.com/68d9oae" rel="nofollow">Kosinski</a> 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.</p> <p>So, as $\mathbb{R}^4 - int(\rho(X_K))$ is a non-compact connected topological four--manifold, Theorem 9.4.22 <a href="http://tinyurl.com/6xn23a8" rel="nofollow">Gompf and Stipsicz</a> implies $\mathbb{R}^4 - int(\rho(X_K))$ admits a smooth structure.</p> <p>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.</p> <p>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 <a href="http://tinyurl.com/6gkhwh8" rel="nofollow">Moise</a>. 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. </p> <p>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 $\mathbb{R}^4$.</p> <p>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 $\mathbb{R}^4$.</p> <p>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. <strong>QED</strong>.</p>