In bordism theory and algebraic topology, 4d spin bordism group is generated by $K3$ surface, while 4d $SO$ bordism group generated by $\mathbf{CP}^2$.

$K3$'s 4-manifold signature is $- 16$ and $\mathbf{CP}^2$'s 4-manifold signature is $+1$.

  1. Are there ways that showing the 16 copies of $\mathbf{CP}^2$ is cobordant to the orientation reversing of $K3$? Namely, are there 5-manifold making these 17 disjoint pieces of manifolds (16 copies of $\mathbf{CP}^2$ and $\overline{K3}$) null bordant?

  2. Relatedly $\mathbf{CP}^2$ is a non-spin manifold. The $K3$ is a spin manifold. Can 16 copies of $\mathbf{CP}^2$ be a spin manifold in some modified way? So that the spin version of the 16 copies of $\mathbf{CP}^2$ becomes cobordant to $\overline{K3}$?


1 Answer 1


[Note: This answer was updated according to @MarcoGolla's comment to avoid a potential circularity.]

(1) The existence of such a bordism is already answered by your question: "$\mathbb{CP}^2$ generates $\Omega^{SO}_4$" literally means any closed oriented smooth 4-manifold is oriented cobordant to some copies of $\mathbb{CP}^2$ or $\overline{\mathbb{CP}^2}$. So I suppose that you are asking for an explicit bordism.

As @MarcoGolla pointed out, there is a diffeomorphism

$$K3\mathbin\#\mathbb{CP}^2\ \simeq\ 4\mathbb{CP}^2\mathbin\#19\overline{\mathbb{CP}^2}\,,$$

by Exercise 8.3.4(d) in Gompf and Stipsicz's book 4-manifolds and Kirby Calculus. Tracing the connected sums on both sides, we obtain a compact oriented smooth 5-manifold $W_1$ with boundary

$$\partial W_1\ \simeq\ K3\amalg 20\mathbb{CP}^2\amalg 4\overline{\mathbb{CP}^2}\,.$$

Gluing $W_1$ with 4 copies of $\mathbb{CP}^2\times I$ along the boundary, we obtain a compact oriented smooth 5-manifold $W_2$ with boundary

$$\partial W_2\ \simeq\ K3\amalg 16\mathbb{CP}^2\,.$$

Hence $W_2$ is what you want.

In addition to 4-manifolds and Kirby Calculus mentioned above, Scorpan's book The Wild World of 4-manifolds is also a good place to learn about 4-manifolds.

(2) I don't quite understand the question. If you have a connected spin manifold, it induces a spin structure on each connected component of its boundary.

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    $\begingroup$ $K3\#\mathbb{CP}^2$ is diffeomorphic to $4\mathbb{CP}^2\#19\overline{\mathbb{CP}}\vphantom{\mathbb{P}}^2$, by Exercise 8.3.4(d) in Gompf and Stipsicz's book 4-manifolds and Kirby calculus. $\endgroup$ Commented May 19 at 9:29
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    $\begingroup$ By the way, as far as I recall Wall's proof uses the existence of the cobordism, so the argument you're presenting might be a bit circular. (If you use the explicit diffeomorphism I suggested, then it isn't circular, though. Also, there might be other proofs of Wall's theorem that do not rely on the existence of the cobordism.) $\endgroup$ Commented May 19 at 14:47
  • $\begingroup$ TeX note: To get \# to space as a binary operator, you can use \mathbin\# rather than \ \#\ (or use something like \newcommand\csum{\mathbin\#} to avoid re-typing that repeatedly). The binary version of the sum-like operator \coprod is \amalg; compare, for example, $K3 \amalg 2\mathbb CP^2$ to $K3 \coprod 2\mathbb CP^2$. $\endgroup$
    – LSpice
    Commented May 19 at 17:13
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    $\begingroup$ @MarcoGolla Indeed, the construction you suggested is far better. I'm not aware of an alternative proof of Wall's theorem that does not rely on h-cobordism, so I admit my argument is a bit circular... Would you like to write an answer? Or would you mind if I modify my answer using the diffeomorphism you suggested? $\endgroup$
    – Leo
    Commented May 19 at 22:03
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    $\begingroup$ The main idea is yours---go ahead and modify your answer. $\endgroup$ Commented May 20 at 10:26

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