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I've read that a handle decomposition for 4-manifold determines a unique smooth structure, and I've also read that every 4-manifold admits a Kirby diagram. So when does a Kirby diagram fail to determine a handle decomposition?

This question occurred to me while looking at the standard Kirby diagram for the $E_8$-manifold.

Turion
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Kyle
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    I think I read somewhere in Gompf-Stipsicz that a 4-manifold has a handlebody decomposition iff it's smooth. The E8 manifold described by that diagram is the one that has the Poincare sphere as its boundary; the usual non-smoothable one is obtained by using Freedman's results to cap this off with a contractible 4-manifold and obtain a closed manifold with intersection form E8. –  Jul 22 '15 at 20:41
  • @MikeMiller: That makes a lot more sense. I was thinking the "cap" was somehow determined by the diagram. – Kyle Jul 22 '15 at 21:39
  • You might be thinking about the fact that if there's a way to add 3- and 4-handles and get a closed manifold, there's essentially a unique way of doing so. But this contractible cap is not smooth, much less a union of 3- and 4-handles. –  Jul 22 '15 at 21:40
  • @MikeMiller: I think that's exactly where I got tripped up. – Kyle Jul 22 '15 at 21:46
  • A similar question is here, even if I think there is something wrong, since the answer there seems to be in contrast with the one here (at least to my understanding, unless in that question they were assuming the 4-manifolds to be smooth) – Dario Jul 26 '15 at 16:43
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    @Dario Likely there 4-manifolds were assumed to be smooth by definition which is somewhat common (I am fairly certain B. Gompf did when answering the question). Also on that question Morse functions are not assumed to be proper which is strange to me. – PVAL-inactive Jul 26 '15 at 17:23

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It is not true that every topological $4$-manifold admits a Kirby diagram. In fact, a handle-decomposition of a $4$-manifold and a Kirby diagram for that $4$-manifold are equivalent notions. The Kirby diagram is exactly the attaching data for the handles. By a theorem of Morse, every smooth $n$-manifold admits a handle-decomposition so in particular every smooth $4$-manifold admits a Kirby diagram. Locally-flat embeddings can always be smoothed in dimensions $ \leq 3$, so as the attaching is being done in dimension $3$ one can always isotope the attaching maps to be smooth for each attached handle. In particular a topological handle-decomposition on a $4$-manifold can be refined to a smooth one, and hence a topological $4$-manifold with a handle-decomposition must admit a smooth structure coming from this decomposition. (As far as I know the source for this argument is due to Kirby. It appears here.)

However, it is not true that every topological $4$-manifold admits a smooth structure (i.e. Freedman's E8 manifold given by capping the E8 plumbing by a contractible 4-manifold in Mike Miller's comment). In particular, such a manifold cannot admit a topological handle-decomposition (or a Kirby diagram).

PVAL-inactive
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  • Ah, okay. I thought I misunderstood what constituted a Kirby diagram. The statement about 4-manifolds possessing Kirby diagrams must have said "smooth" somewhere... – Kyle Jul 22 '15 at 21:37