For M a 4-mfld., every class in H_2(M;Z) can be represented by a Surface.

1. Apr 11, 2010

Bacle

Hi, I am trying to show that for M a 4-manifold,

and [a]_2 a class in H_2(M,Z) , there is always

a surface that represents [a]_2 , i.e., there

exists a surface S , and an embedding i of S into

M , with [ioS]_2 =[a]_2.

(Equiv.: there exists S, and an embedding i of S of M , so that a triangulation of S

induces the class [a]_2)

** What I have **

If M is simply-connected, so that Pi_1(M)=0

(Notation: Pi_1:=Fund. Grp.)

Then, by the Hurewicz Theorem (Hip, Hip Hurewicz!)

Pi_2(M) is actually Isomorphic to H_2(M;Z) , so that

every class in H_2(M;Z) can be represented as an

embedded sphere S^2 (possibly with self-intersections,

which can be smoothed away ).

**BUT** I can't think of what can be done if

M is not simply-connected.

Any Ideas.?

2. Apr 12, 2010

zhentil

Why can you smooth self-intersections? If I'm not wrong, this is precisely the reason why h-cobordism fails in 4 dimensions.

Here's an idea that works if your manifold has no two-torsion (also covers the simply connected case). Take a class, and take the poincare dual. Then identify the poincare dual with a line bundle whose generic section intersects the zero section in an embedded surface homologous to what you started with.

3. Apr 13, 2010

lavinia

how do you know that the map of the 2 sphere has no critical points?

4. Apr 13, 2010

zhentil

Maps from the two-sphere into a four-dimensional manifold are generically immersions, but not embeddings. There's no way to ensure that a given map from the two-sphere can be perturbed to be an embedding.

5. Apr 13, 2010

lavinia

I can construct ugly maps of the 2 sphere into R^4 that are not immersions. Are you saying that any such map is smoothly homotopic to an immersion?

6. Apr 13, 2010

zhentil

Yes.

7. Apr 14, 2010

lavinia

How does the proof go?

8. Apr 14, 2010

zhentil

It's standard transversality theory. I guess a good reference is Hirsch.

9. Apr 14, 2010

zhentil

But I guess in this case, you don't need any fancy stuff ;)

In your case, I would even go so far as to say that your ugly map is homotopic to the standard embedding of S^2 into R^4.

10. Apr 14, 2010

lavinia

what is a non-fancy proof?

11. Apr 15, 2010

zhentil

That R^4 is contractible :)

12. Apr 15, 2010

lavinia

right. So actually any homology class can be represented by an immersed sphere.

I wonder what sort of equivalence classes of immersed manifolds you get if you require the homotopies to be immersions for each time. So immersed M is equivalent to immersed N if they can be moved into each other through a 1 parameter family of immersions.

13. Apr 15, 2010

zhentil

Well again, in the case of R^4, there's not too much homology to worry about.

The second question is quite interesting. You might want to look into isotopy, which is the relevant idea in the case of embeddings. I'm not sure how much work has been done in terms of using isotopies to study homology. I can tell you this: if you go up to dimension six, you can guarantee that two-homology can be represented by embedded surfaces, and two homotopic embeddings are homotopic through immersions. If you go to dimension seven or higher, it would be through embeddings.

14. Apr 15, 2010

lavinia

these seem like wonderful theorems. reference or can you explain them? Are you referring to 2 dimensional homology classes only?