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Is there a simply-connected smooth closed 4-manifold with a characteristic class $x \in H_2(X; \mathbb{Z})$ such that $x$ can not be represented by a disjoint union of tori in $X$?

I would not know how to prove this without the characteristic hypothesis either so any thoughts on that would also be appreciated.

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In $H_2(CP^2)$, every class $nH$ where $H$ is a generator and n is odd is characteristic. However, if $n >3$, then such a class is not represented by a torus. It is not represented by a disjoint union of tori, either. For non-zero classes in $H_2(CP^2)$ have non-zero intersection numbers. So if you had a disjoint union of tori, then at most one would be non-trivial in homology (and that one would represent $nH$, contradicting the previous step.)

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  • $\begingroup$ Thanks for this example - this then shows that in $CP^2$ there are exactly two characteristic homology classes that can be represented by a disjoint union of tori. I wonder if there are any such 4-manifolds where no characteristic homology class can be represented by disjoint tori. $\endgroup$
    – user101010
    Commented Feb 21, 2021 at 16:39
  • $\begingroup$ Such a manifold might be hard to find; you can find characteristic tori (indeed spheres) in the only known examples of definite manifolds (sums of $CP^2$). But if your manifold is indefinite, then you will have characteristic classes of negative square, and obstructions to being represented by a torus are harder to come by. $\endgroup$
    – Danny Ruberman
    Commented Feb 21, 2021 at 18:54
  • $\begingroup$ @DannyRuberman Can you please expand the last implication: why must all but one be trivial? $\endgroup$
    – magicker72
    Commented Feb 22, 2021 at 3:43
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    $\begingroup$ @magicker72 If two where nontrivial, then they would need to intersect (since any two homologically nontrivial surfaces intersect), therefore you would not have disjoint surfaces. $\endgroup$
    – user101010
    Commented Feb 22, 2021 at 11:35

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