Timeline for Self-intersection of immersed surfaces and connected sum
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 20, 2013 at 15:56 | vote | accept | GFR | ||
| Jul 17, 2013 at 14:41 | answer | added | András Szűcs | timeline score: 3 | |
| Jul 17, 2013 at 14:21 | comment | added | András Szűcs | I doubt that $A = A#A$ unless $A = 0.$ I think $A#B = A + B$ always. | |
| Jul 2, 2013 at 21:40 | comment | added | JHM | My confusion is clarified: it is not unconditionally true that the connect sum $A \# B$ represents $[A] + [B]$ in $H_2$. | |
| Jul 2, 2013 at 19:57 | comment | added | JHM | But the formula seems to yield nonsense: if $M$ is a simply connected $4$-manifold, then Hurewicz tells us $H_2(M,\mathbb{Z})$ is generated by $2$-spheres (i.e. $\pi_2(M) \to H_2(M,\mathbb{Z})$ is iso). For an immersed 2-sphere $A$ having nontrivial self-intersection (which arises if the intersection form is nonzero) then we have $A=A \# A$ (w.r.t. integral homology if we orient the connect sum properly) and $A\cdot A = 4 A \cdot A$ -- !? | |
| Jul 2, 2013 at 19:35 | comment | added | JHM | @GFR I just meant a 2-sphere which is not a boundary of some 3-cycle. I am somewhat surprised that the formula $A\#B \cdot C = B \cdot C +A\cdot C$ is not unconditionally true (which would unconditionally confirm your above formula). | |
| Jul 2, 2013 at 18:42 | comment | added | GFR | @J.Martel what is a nonbounding 2-sphere? | |
| Jul 2, 2013 at 18:40 | comment | added | GFR | @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. | |
| Jul 2, 2013 at 18:02 | comment | added | Ben McKay | 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. | |
| Jul 2, 2013 at 18:02 | comment | added | JHM | 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. | |
| Jul 2, 2013 at 17:15 | comment | added | GFR | 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]. | |
| Jul 2, 2013 at 17:10 | comment | added | Ben McKay | 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? | |
| Jul 2, 2013 at 17:01 | history | asked | GFR | CC BY-SA 3.0 |