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Jul 28, 2020 at 4:19 comment added Joe It's actually showed $RP^4 \# CP^2$ and $Q \# CP^2$ ($Q$ being the fake $RP^4$) are diffeomorphic in this article: [S. Akbulut, A fake 4-manifold, 4-Manifold Topology, Contemporary Math. 35 (1984), 75-141.]
Jun 18, 2020 at 15:53 comment added Arun Debray For these specific $M$ and $N$, I think $X := \mathbb{CP}^2 \# (S^2\times S^2)$ works: for some $i$ and $j$, $M \#_i X$ and $M \#_j X$ are diffeomorphic. This is because $M \# \mathbb{CP}^2$ and $N \#\mathbb{CP}^2$ are $S^2\times S^2$-stably diffeomorphic: one once again uses Kreck's modified surgery theory, but $M \# \mathbb{CP}^2$ and $N \# \mathbb{CP}^2$ aren't pin+ or pin-, so the calculation takes place in the 4th unoriented bordism group, where they are bordant.
Jun 18, 2020 at 15:47 comment added Arun Debray That's a great question, and I don't know what happens in general. The trick that makes this work is that $S^2\times S^2$ is null-bordant, allowing bordism to be used to check stable diffeomorphism. So something different would have to be done for $\mathbb{CP}^2$.
Jun 18, 2020 at 15:20 comment added Bruno Martelli Aha, I didn't know that. What happens if we replace $S^2 \times S^2$ with some other fixed 4-manifold $X$, like for instance the complex projective plane? Are the connected sums $M\#X$ and $N\# X$ always guaranteed to be non-diffeomorphic for these two specific 4-manifolds $M$ and $N$?
Jun 17, 2020 at 15:47 comment added Arun Debray It is often but not always true that homeomorphic $M$ and $N$ are $S^2\times S^2$-stably diffeomorphic. (It is true when $M$ and $N$ are oriented, by a theorem of Gompf.) One counterexample is Capell-Shaneson's fake $\mathbb{RP}^4$ and the standard $\mathbb{RP}^4$, in fact, which can be shown with Kreck's modified surgery theory, reducing the problem to the possible values they can take on in the 4th pin+ bordism group.
Jun 17, 2020 at 14:29 history answered Bruno Martelli CC BY-SA 4.0