What's the Kirby Diagram of a universal $\mathbb{R}^4$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:36:42Z http://mathoverflow.net/feeds/question/54143 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54143/whats-the-kirby-diagram-of-a-universal-mathbbr4 What's the Kirby Diagram of a universal $\mathbb{R}^4$? Kelly Davis 2011-02-02T22:53:14Z 2011-07-27T06:19:07Z <blockquote> <p>What's the Kirby diagram of a universal $\mathbb{R}^4$?</p> </blockquote> <p><strong>Background</strong></p> <p>Define $\mathcal{R}$ as the set of smoothings of $\mathbb{R}^4$. For two oriented elements $R_1$, $R_2$ in $\mathcal{R}$ we can define their <em>end sum</em> $R_1 \natural R_2$ if we are given two proper embeddings $\gamma_i : [0, \infty) \rightarrow R_i$. </p> <p>We remove a tubular neighborhood of $\gamma_i((0, \infty))$ from each $R_i$ and glue the resulting $\mathbb{R}^3$ boundaries together respecting orientations. The result is the <em>end sum</em> $R_1 \natural R_2$ of $R_1$ and $R_2$. As $\gamma_i$ is unique up to ambient isotopy, $R_1 \natural R_2$ is well defined up to diffeomorphism.</p> <p>In <a href="http://projecteuclid.org/euclid.jdg/1214440258" rel="nofollow">"A universal smoothing of four-space"</a> Freedman and Taylor proved the existence of an element $U \in \mathcal{R}$ such that for any $R \in \mathcal{R}$ the <em>end sum</em> $U \natural R$ is diffeomorphic to $U$. This $U$ is the universal $\mathbb{R}^4$.</p> <p><strong>Foreground</strong></p> <p>In <a href="http://arxiv.org/abs/math/9712292" rel="nofollow">"An invariant of smooth 4-manifolds"</a> Taylor defines an invariant $\gamma(R) \in \{0,1,2,\ldots,\infty \}$ for $R \in \mathcal{R}$. Taylor defines $\gamma(R)$ to be $sup_K \{ min_X\{ \frac{1}{2} b_2(X) \} \}$, where $K$ ranges over compact $4$-manifolds smoothly embedding in $R$ and $X$ ranges over closed, spin $4$-manifolds with signature $0$ in which $K$ smoothly embeds. (Actually, Taylor defines $\gamma$ for all smooth $4$-manifolds, but we don't need this detail here.)</p> <p>Taylor goes on to prove that if $R \in \mathcal{R}$ and $\gamma(R) > 0$, then any handle decomposition of $R$ has infinitely many three handles.</p> <p>In <a href="http://tinyurl.com/6766kyq" rel="nofollow">"4-Manifolds and Kirby Calculus"</a> Stipsicz and Gompf prove, see page 376, that $\gamma(U) = \infty$. Thus, any Kirby Diagram of $U$ must have infinitely many three handles. </p> <p><strong>Question</strong></p> <p>What's the Kirby diagram of a such a $U$?</p> http://mathoverflow.net/questions/54143/whats-the-kirby-diagram-of-a-universal-mathbbr4/71352#71352 Answer by Bob Gompf for What's the Kirby Diagram of a universal $\mathbb{R}^4$? Bob Gompf 2011-07-26T20:31:16Z 2011-07-26T20:31:16Z <p>I would also like to know the answer to that. As far as I know, it is still a difficult, unsolved problem. The bit about 3-handles is a clue, but I haven't found any way to make use of it. </p>