Can you flip the end of a large exotic $\mathbb{R}^4$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:58:44Z http://mathoverflow.net/feeds/question/59046 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59046/can-you-flip-the-end-of-a-large-exotic-mathbbr4 Can you flip the end of a large exotic $\mathbb{R}^4$ Kelly Davis 2011-03-21T08:32:40Z 2011-04-02T08:22:36Z <blockquote> <p>Can you flip the end of a large exotic $\mathbb{R}^4$</p> </blockquote> <h2>Background</h2> <p><strong>Definition</strong> (<strong>Exotic $\mathbb{R}^4$</strong>): 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.</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 $K'$ that can not be smoothly embedded into $\mathbb{R}^4$.</p> <p><strong>Definition</strong> (<strong>End of a large exotic $\mathbb{R}^4$</strong>): 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 $R$.</p> <p><strong>Remark</strong>: The previous definition varies slightly from the standard definition of "end", however it will be used for the remainder of this question. (See <a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> Exercise 9.4.11 for the standard definition.)</p> <p><strong>Remark</strong>: 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 $R$.</p> <p><strong>Definition</strong> (<strong>Flip of the end of a large exotic $\mathbb{R}^4$</strong>): 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.</p> <p><strong>Remark</strong>: The previous definition is also non-standard. I am not aware of any standard definitions that carry, more-or-less, the same meaning.</p> <p>So, at this stage the meaning of the question is hopefully clear.</p> <h2>Foreground</h2> <p>In our attempt to flip the end of a large exotic $\mathbb{R}^4$ an inconvenient truth stands in our way:</p> <p><strong>Theorem 1</strong> (<strong>Uncountably many flips fail</strong>): There are uncountably many large exotic $\mathbb{R}^4$'s that one can not flip the end of.</p> <p>Quickly, let us see why this is true. Lemma 9.4.2 along with Addendum 9.4.4 of <a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> state:</p> <p><strong>Lemma 1</strong>: 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 <code>$n\langle-1\rangle$</code>, such that $X - int(Y)$ and $L - int(K)$ are orientation-preserving diffeomorphic.</p> <p>Theorem 9.4.3 of <a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> states:</p> <p><strong>Theorem</strong>: Any $L$ as appears in Lemma 1 is a large exotic $\mathbb{R}^4$. </p> <p>The two statements above lead to:</p> <p><strong>Lemma</strong>: One can not flip the end of any $L$ as appears in Lemma 1.</p> <p><strong>Proof</strong>: 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 <code>$n\langle-1\rangle$</code>. However, according to Donaldson's Theorem (<a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip. <strong>QED</strong></p> <p>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:</p> <p><strong>Definition</strong> (<strong>Radial Family</strong>): 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 $\{R_t | 0 &lt; t \le \infty \}$.</p> <p><strong>Remark</strong>: If $R_t$ is a member of a radial family, then $R_t$ is a smooth manifold as it inherits a smooth structure from $R$</p> <p>Theorem 9.4.10 of <a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> states:</p> <p><strong>Theorem</strong>: If $\{L_t | 0 &lt; t \le \infty \}$ 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.</p> <p>This leads directly to a proof of Theorem 1.</p> <p><strong>Proof</strong>: 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 <code>$n\langle-1\rangle$</code>. Again, according to Donaldson's Theorem (<a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip.<strong>QED</strong></p> <p>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:</p> <p><strong>Definition</strong> (<strong>Simply Connected at Infinity</strong>): 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 $\pi_1(Z - C') \rightarrow \pi_1(Z - C)$.</p> <p><strong>Definition</strong> (<strong>End Sum</strong>): 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 $Z_2$.</p> <p><strong>Remark</strong>: 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 (<a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> Definition 9.4.6).</p> <p><strong>Remark</strong>: 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 $\mathbb{R}^4$.</p> <p><strong>Remark</strong>: $X$ of Lemma 1 is simply connected at infinity.</p> <p><strong>Theorem 2</strong>: If $\{L_t | 0 &lt; t \le \infty \}$ 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 $R \natural L_t$.</p> <p><strong>Proof</strong>: 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 <code>$n\langle-1\rangle$</code>. Again, according to Donaldson's Theorem (<a href="http://www.amazon.com/4-Manifolds-Calculus-Graduate-Studies-Mathematics/dp/0821809946/ref=sr_1_1?s=books&amp;ie=UTF8&amp;qid=1300693632&amp;sr=1-1" rel="nofollow">Gompf and Stipsicz</a> Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip.<strong>QED</strong></p> <p>Now we can revel in this despair! </p> <p>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, <a href="http://mathoverflow.net/questions/42624/slice-knots-and-exotic-mathbb-r4/57926#57926" rel="nofollow">Davis</a>.) 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.</p> <h2>Question</h2> <blockquote> <p>Can you flip the end of a large exotic $\mathbb{R}^4$?</p> </blockquote> http://mathoverflow.net/questions/59046/can-you-flip-the-end-of-a-large-exotic-mathbbr4/59922#59922 Answer by Agol for Can you flip the end of a large exotic $\mathbb{R}^4$ Agol 2011-03-29T02:25:21Z 2011-03-29T14:26:19Z <p>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 $\mathbb{R}^4_{std}$). </p> <p>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 $\mathbb{R}^4_{std}$. </p> <p>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. </p> <p>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). </p> <p>I realize that this does not answer the spirit of your question, which is whether there is a large exotic $\mathbb{R}^4$ <em>which does not have a standard product end</em> and which admits a flip. </p>