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I am interested in seeing examples of two closed 4-manifolds $X_1,X_2$ such that $SW(X_i)=0$ and they are homeomorphic but not diffeomorphic.

So far in the literature I've only found examples which somehow used symplectic geometry and operations on the Seiberg-Witten invariant (knot surgery, rational blow down, star surgery, etc.) to distinguish one from the other. So I am interested in knowing some other (if known) techniques in this direction. The reason I've mentioned $SW(X_i)=0$ is to ensure that none of them admits symplectic structures (by work of Taubes). Trisection theory is potentially a way but I am not sure if anything is known in that direction or not. Thanks in advance.

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    $\begingroup$ These may not be the spirit of examples you're looking for, but you can produce examples using connected sums. See, for example, Ishida-Sasahira. The point is that Seiberg-Witten invariants vanish under connected sums of 4-manifolds with $b^+_2 \geq 1$, but this isn't true of the stronger Bauer-Furuta invariants. $\endgroup$ May 20, 2019 at 21:02

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$\mathbb CP^2\#\overline{K3}$ (trivial SW because it is a connected sum with $b^2_+>0$ for both pieces). This is an example without having to use the Bauer-Furuta invariants (contrast with Kyle's comment), but with the SW invariants of the 4-manifold's opposite (reverse $X$'s orientation into $\overline X$). I learned this example from Tim Perutz at some point: The symplectic manifold $X=K3\#\overline{\mathbb CP^2}$ has $b^2_+=3$ and $b^2_-=20$ and $b^1=0$ and nontrivial SW invariants (note that $b^2_++b^1+1$ is even, this quantity is the parity of the various SW moduli spaces and needs to be even for the ordinary SW invariants to possibly be nontrivial). By doing Fintushel-Stern knot surgery on K3 we get exotic copies $X_K$ (knot $K$) distinguished by the SW invariants. Then $SW(\overline X_K)=0$ because $b^2_+(\overline X_K)+b^1(\overline X_K)+1=21$ is odd, with each $\overline X_K$ homeomorphic to the non-symplectic $\overline{K3}\#\mathbb CP^2$. So the "technique" of orientation-reversal should be added to our arsenal -- related discussion is found in Draghici's paper "Seiberg-Witten invariants when reversing orientation" and its references, for example.

More interestingly (perhaps), Taubes studied a possibly infinite set (if not all diffeomorphic) of smooth 4-manifolds $\lbrace X_K\rbrace_K$ all homeomorphic to the non-symplectic $3\mathbb CP^2\#23\overline{\mathbb CP^2}$, obtained from knot surgery on K3 using different hyperbolic knots $K\subset S^3$. They all have vanishing SW invariants (and vanishing Bauer-Furuta invariants). But it is not known whether they are all diffeomorphic... if they are then there are some interesting properties of ASD Weyl curvature metrics on it!

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  • $\begingroup$ Interesting!! Thank you for sharing :) $\endgroup$ May 22, 2019 at 3:01
  • $\begingroup$ One thing is not clear to me, why $\bar{X_K}$ and $\bar{K3} # \mathbb CP^2$ are not diffeomorphic? i.e if X and Y are diffeomorphic does that mean $\bar{X}$ and $\bar{Y}$ are diffeomorphic? $\endgroup$ May 22, 2019 at 3:16
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    $\begingroup$ Yes, using the same diffeomorphism (pick your favorite definition of orientation and check). $\endgroup$ May 22, 2019 at 5:52
  • $\begingroup$ Do you know any example such that $X$ and $\bar{X}$ are not homeomorphic (or diffeomorphic) but both are symplectic? $\endgroup$ Mar 23, 2020 at 22:25
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Here is an example as Kyle suggested in his comment.

Claim is that $K3 \#K3 \# \bar{\mathbb {CP^2}}$ and $\mathbb{ \#_6 CP^2}\#_{39}\mathbb{ \bar{CP^2}}$ are homeomorphic but not diffeomorphic. Notice that they have same intersection form and since they are simply-connected, by Freedman, they are hoeomorphic. Observe that they have trivial Seiberg Witten invariant follows from Taubes result of vanishing Seiberg Witten invariant under connected sum.

In order to distinguish their diffeomorphism type, we are going to use stable cohomotopy Seiberg Witten invariat (ScSW).

Here are two important theorem in this context, one is corresponds to vanishing and the other one for non-vanishing

Theorem 1(Bauer) If a closed smooth connected 4 manifold satisfies one of the following 1) or 2) then they have non vanishing ScSW

1) If $X = X_1 \#X_2$ and $X_1$ has non vanishing ScSW and $b_2^+(X_2)=0$ 2) If $X= \#_n X_i$ such that $b_2^+(X_i)=3 (mod 4)$, $b_1(X_i)=0$. Each $X_i$ has a compatible spinc structure $s_i$ such that $SW_{s_i}(X_i)=1(mod 2)$ and $1<n<5$ and for $n=4$ we need $b_2^+(X)=4(mod 8)$

By this theorem $K3\#K3\#\mathbb {\bar{CP^2}}$ has non vanishing ScSW (since $K3$ is symplectic and by Taubes, the canonical spinc strure s corresponding to the symplectic structure has $SW_s(K3)=1$).

Now the vanishing part follows from the adjunction inequality

Theorem 2(Froyshov) Let $X$ be a smooth closed oriented smooth 4 manifold with $b_2^+>1$. Suppose $Y$ is a codim 1 embedded manifold which admists a positive scalar curvature. And the inclusion map induced non trivial map $H^2(X,\mathbb Q)\to H^2(Y,\mathbb Q)$, then ScSW vanishes.

Observe that $\mathbb {CP^2}$ sits in $\mathbb{ \#_6 CP^2}\#_{39}\mathbb{ \bar{CP^2}}$. There is a self intersection 1 sphere in $\mathbb CP^2$. If we blown up once, that will give us a self intersection 0 sphere in $\mathbb{ \#_6 CP^2}\#_{39}\mathbb{ \bar{CP^2}}$. Inface the boundary of such a neighbourhood is $S^1\times S^2$. And the copy of $S^2$ gives a non-torsion element in $H^2$. So by Theorem 2, ScSW=0.

And thus they are EXOTIC.

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  • $\begingroup$ In the spirit of both my answer and Kyle's comment, you can quickly do Fintushel-Stern knot surgery on K3 to get exotic copies of $K3\#K3$ visible with Bauer-Furuta's ScSW. $\endgroup$ Mar 24, 2020 at 2:27
  • $\begingroup$ @ChrisGerig How does knot surgery relate to Bauer-Furuta's ScSW? Is it the same way as with SW,i.e, multiplication by Alexander polynomial? I just started reading about ScSW today, so I do not have much deep knowledge. If possible refer me some paper. You already texted me one paper, is that a good source to learn? $\endgroup$ Mar 24, 2020 at 2:35

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