MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $U$?

share|cite|improve this question
Gompf probably knows the answer to this. – Jim Conant Feb 3 '11 at 2:52
Waiting for G̶o̶d̶o̶t̶ Gompf. – Kelly Davis Feb 3 '11 at 18:08
Godot never showed up, but Gompf did, and it seems as if this is an open problem. – Kelly Davis Jul 27 '11 at 6:21

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.

share|cite|improve this answer
Welcome to MO!! – Zarathustra Jul 26 '11 at 21:07
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 $\mathbb{R}^4$. – Kelly Davis Jul 27 '11 at 6:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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