There is a nice theorem that for every simply connected, closed, smooth, oriented manifold $M$, we have for some $m \in \mathbb{N}$: $$M \# \left(\mathop{\#}^m \left(\mathbb{C}\mathbb{P}^2 \# \overline{\mathbb{C}\mathbb{P}^2} \right) \right) \cong \left(\mathop{\#}^{m+b^+_2(M)} \mathbb{C}\mathbb{P}^2 \right) \# \left(\mathop{\#}^{m+b^-_2(M)} \overline{\mathbb{C}\mathbb{P}^2} \right)$$ The $b^\pm_2(M)$ are the positive and negative eigenvalues of the intersection form. This has far-reaching consequences, for example any 4-manifold invariant that is multiplicative under direct sum $\#$ will only measure the intersection form on simply connected manifolds.

Is there a similar statement for non-simply-connected 4-manifolds? What is the closest we can get?

  • 5
    $\begingroup$ Given that blow up/downs won't change the fundamental group, I'm not sure what kind of answer you're hoping for. $\endgroup$ – Donu Arapura Sep 12 '14 at 17:17
  • $\begingroup$ @DonuArapura, Ideally, a connected sum of $S^1 \times S^3$, $\mathbb{CP}^2$ and $\overline{\mathbb{CP}^2}$, but that's probably to much to expect. $\endgroup$ – Manuel Bärenz Sep 12 '14 at 17:19
  • $\begingroup$ But the fundamental group would have to be free for that to work, wouldn't it? $\endgroup$ – Donu Arapura Sep 12 '14 at 17:23
  • $\begingroup$ @DonuArapura, right. So an interesting, more realistic, similar statement could be that such a sum is possible for free $\pi_1$. Or maybe it would be possible to have a connected sum with $\mathbb{CP}^2$'s and a manifold with trivial intersection form. I know, maybe it's a soft question, but maybe there is a sensible generalisation. $\endgroup$ – Manuel Bärenz Sep 12 '14 at 17:27
  • 4
    $\begingroup$ Wall's theorem tells you that if $X$ and $X'$ are $h$-cobordant 4-manifolds, then $X\#k(S^2\times S^2)$ and $X'\#k(S^2\times S^2)$ are diffeomorphic for some $k$ (conjecturally, $k=1$ is enough). This is somehow related (just blow up once and you get a similar statement). Somehow it looks like you're looking for a "minimal model" of a 4-manifold -- much like in the minimal model program in algebraic geometry. $\endgroup$ – Marco Golla Sep 12 '14 at 20:14

Two smooth oriented 4-manifolds $M,N$ with fundamental group $\pi$ admit maps to $B\pi$, the classifying maps of their universal covers. The manifolds $M$ and $N$ become diffeomorphic after connect summing with copies of complex projective space if and only if the images of the fundamental classes $[M],[N]$ in $H_4(B\pi)$ under the classifying maps are equal, for some choice of maps to $B\pi$, i.e. if we mod out $H_4(B\pi)$ by the action of outer automorphisms of $\pi$. This follows from Kreck's surgery theory with normal 1-type $B\pi \times BSO$. This theory really tells us about connect summing with $S^2 \times S^2$s, but $\#^2(\mathbb{CP}^2 \# \overline{\mathbb{CP}^2}) \cong S^2 \times S^2 \# \mathbb{CP}^2 \# \overline{\mathbb{CP}^2}$. It turns out that the class in $H_4$ is determined by the stable isomorphism classes of the $\mathbb{Z}\pi$ modules $\pi_2(M)$, $\pi_2(N)$.

Thus for general groups you need many minimal models. For groups with vanishing $H_4$ the situation is rather similar to the trivial group situation; you just need to construct some 4-manifold with that group as its fundamental group, for the base manifold. (Analogous to a hidden $S^4$ on the right hand side of your equation.)

| cite | improve this answer | |
  • $\begingroup$ Wow, that's really interesting. I'm actually constructing a TQFT that can be twisted by a 4-cocycle of a group, so thanks! $\endgroup$ – Manuel Bärenz Oct 27 '15 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.