What's the Kirby diagram of a universal $\mathbb{R}^4$?


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 end sum $R_1 \natural R_2$ if we are given two proper embeddings $\gamma_i : [0, \infty) \rightarrow R_i$.

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 end sum $R_1 \natural R_2$ 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.

In "A universal smoothing of four-space" Freedman and Taylor proved the existence of an element $U \in \mathcal{R}$ such that for any $R \in \mathcal{R}$ the end sum $U \natural R$ is diffeomorphic to $U$. This $U$ is the universal $\mathbb{R}^4$.


In "An invariant of smooth 4-manifolds" 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.)

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.

In "4-Manifolds and Kirby Calculus" Stipsicz and Gompf prove, see page 376, that $\gamma(U) = \infty$. Thus, any Kirby Diagram of $U$ must have infinitely many three handles.


What's the Kirby diagram of a such a $U$?

  • $\begingroup$ Gompf probably knows the answer to this. $\endgroup$
    – Jim Conant
    Feb 3, 2011 at 2:52
  • 4
    $\begingroup$ Waiting for G̶o̶d̶o̶t̶ Gompf. $\endgroup$ Feb 3, 2011 at 18:08
  • 1
    $\begingroup$ Godot never showed up, but Gompf did, and it seems as if this is an open problem. $\endgroup$ Jul 27, 2011 at 6:21

1 Answer 1


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.

  • 3
    $\begingroup$ Welcome to MO!! $\endgroup$ Jul 26, 2011 at 21:07
  • $\begingroup$ Wow, welcome to MO, and thanks for the answer. I wasn't aware that there was still no progress on the Kirby diagram of universal $\mathbb{R}^4$. $\endgroup$ Jul 27, 2011 at 6:18

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