It seems that usually smooth structures on compact 4-manifolds are distinguished by Seiberg-Witten/Donaldson invariants. And I don't know another direct way to do that. But in the case when $b_2^+$ is even the invariants vanish identically.

Does there exist an example of smooth compact exotic 4-manifold with $b_2^+$ even?


2 Answers 2


An often-overlooked invariant of a closed oriented 4-manifold $X$ is the closed oriented 4-manifold $\overline{X}$: the orientation-reversed manifold. Taking the Seiberg-Witten invariants of $\overline{X}$, one can distinguish smooth structures on manifolds $X$ with $b^->0$ odd, regardless of the parity of $b^+$. For instance, $\mathbb{C}P^2\# \overline{K3}$ has $b^+ = 20$ and infinitely many smooth structures (perform knot surgery on a K3 surface, blow up once, reverse orientation).

The real problem, then, is to understand what happens when both $b^+$ and $b^-$ are even.

I think one can distinguish infinitely many smooth structures on the connected sum of two K3 surfaces (so $b^+=6$, $b^-=38$) along the lines Tom Leness suggests in his answer.

The non-equivariant Bauer-Furuta invariant of a $\mathrm{Spin}^c$-structure $\mathfrak{s}$ on a closed oriented 4-manifold $X$ is a stable map $S^{(c_1(\mathfrak{s})^2−σ)/4}\to S^{b^+}$. When $X$ is a homotopy K3 surface, and $c_1(\mathfrak{s})^2=0$, we get the stable homotopy class of a map $S^4\to S^3$; this lives in $\mathbb{Z}/2$, and is actually in this case the mod 2 Seiberg-Witten invariant (I'm going on memory here, so this is a point to check). By Bauer's theorem, the invariant for a connected sum is the smash product of the invariants of the summands. Now, the smash-square of the Hopf map is the generator for $π^s_2(S^0)=\mathbb{Z}/2$. Hence, if $X_1$ and $X_2$ are homotopy K3's, and $s(X_i)$ is the number of $\mathrm{Spin}^c$-structures with $c_1(\mathfrak{s})^2=0$ and odd SW invariant, the product $s(X_1)s(X_2)$ is an invariant of $X_1\#X_2$. Via Fintushel-Stern knot surgery along on K3 along knots with suitable Alexander polynomials, one can make $s(X_i)$ arbitrarily large, and thereby distinguish infinitely many smooth structures on $K3 \# K3$.

As so often in this subject, one has the feeling that information is being lost: maybe $X_1\#X_2$ actually knows $X_1$ and $X_2$.

  • $\begingroup$ Thank you for the great answer! Do you know something about the case last case? $\endgroup$
    – Misha
    Aug 20, 2013 at 17:00
  • $\begingroup$ I've edited to include an example with $b^+$ and $b^-$ even (previously given in comments). $\endgroup$
    – Tim Perutz
    Aug 20, 2013 at 23:43

One source of such examples would, of course, be to take the connected sum of two manifolds with $b^+_2$ odd. The torsion component of the Bauer-Furuta invariants do not obviously vanish on such connected sums:



Bauer or Furuta could tell you whether this works quite quickly.

  • $\begingroup$ Many thanks for your nice answer. I am far from being an expert in the field, so I didn't know anything about these invariants. I looked through the papers and understood that there indeed is a formula for a connected sum. But I didn't found an explicit computation of an invariant. Do you know where I should look for it? $\endgroup$
    – Misha
    Aug 20, 2013 at 17:13

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