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Let me first say that my background is theoretical physics so I find it hard to look at some of the mathematical literature.

The kind of problem I am interested in is the following one. Consider a 4-dimensional smooth, simply connected orientable manifold M and suppose A and B are immersed surfaces. Denote with $A\#B$ the connected sum of A and B (since M is simply connected I think there is only one way of doing this). Denote with $A\cdot B $ the intersection number of A and B in M. Is it true that the self-intersection number of $A\#B $ in M is given by $(A \# B) \cdot (A \# B) = A \cdot A + 2 A \cdot B + B \cdot B $?

If A, B are 2-spheres, looking at Freedman and Quinn, Topology of 4-Manifolds page 24 I would think so, but I am not sure I am understanding the book well. What could be said, perhaps making additional assumptions, if A and B are surfaces of more complicated topology?

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  • $\begingroup$ Do you mean the sum as cycles, rather than the connected sum? Or by $A#B$ do you mean to glue $A$ and $B$ together to make a smooth surface near each point where they intersect? $\endgroup$
    – Ben McKay
    Commented Jul 2, 2013 at 17:10
  • $\begingroup$ In the question above I meant the usual connected sum - remove a disk and glue along the boundary. However I think that, if what I wrote is correct, one could think of the connected sum of A and B as a representative of the homology class [A+B]. $\endgroup$
    – GFR
    Commented Jul 2, 2013 at 17:15
  • $\begingroup$ Bearing in mind the familiar fact that $A#S^2,A$ are homologous, i would test your proposed formula by first asking for a simply connected 4-manifold which has a nonbounding 2-sphere with nonzero selfintersection number. $\endgroup$
    – JHM
    Commented Jul 2, 2013 at 18:02
  • $\begingroup$ If they intersect twic, do you glue each time? If they intersect nontransversally, do you perturb them first to get transverse intersections? I suppose the question then only depends on their homology classes and then you didn't really need to glue them; you could have just added the cycles they represent. $\endgroup$
    – Ben McKay
    Commented Jul 2, 2013 at 18:02
  • $\begingroup$ @BenMcKay: I am not sure. One probably needs additional hypotheses on how A and B intersect. Here is what I found on Freedman and Quinn: "Intersection number behave nicely with respect to sums: suppose A is an immersed sphere, B,C disks or spheres. [...] Then $A\#B \cdot C= B \cdot C + A \cdot C$". Note that I have omitted the dependence on the chosen arc between A and B as I assume the ambient manifold is simply-connected. If A and B do not intersect I think that, as you say, it is enough to consider just the cycles, but from the book it seems that more general configurations are allowed. $\endgroup$
    – GFR
    Commented Jul 2, 2013 at 18:40

1 Answer 1

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The answer is: yes, the formula holds.

For any $4n$-dimensional manifold and $2n$-dimensional immersed submanifolds $A$ and $B$. (Everything must be oriented if you want consider an integer formula, intersections with signs. If you drop the condition on orientability, then you still can count the points of intersections mod 2.) Now the formula holds, because intersection is dual to the cohomological multiplication, and $A\#B$ is equal to $A+B$.

The parity of the dimensions is necessary, because cohomological multiplication (as well as the intersection number) is not commutative, but anticommutative.

Actually one can generalize the statement further (the dimensions of the immersed submanifolds might be greater than the half of the dimension of the ambient manifold). In this case the intersection number one can be replaced by the homology class represented by the transversal intersection of the immersed submanifolds.

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  • $\begingroup$ I see an "answer" but not a proof. How can you convince me that the connect sum of two $2$-spheres is not homotopic (actually homeomorphic) to the single $2$-sphere? $\endgroup$
    – JHM
    Commented Jul 18, 2013 at 4:23
  • $\begingroup$ Of course the connected sum of two spheres is homeomorphic to one sphere, But we are considering the represented homology classes. $A$ and $B$ denote the represented homology classes , not just the manifolds themselves. $\endgroup$
    – András Szűcs
    Commented Jul 18, 2013 at 6:00
  • $\begingroup$ Dear Martel, I think the proof is that "intersection (of homology classes) is dual to the cohomological multiplication, and A#B is equal to A+B", Andras $\endgroup$
    – András Szűcs
    Commented Jul 18, 2013 at 9:19

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