2
$\begingroup$

Can anyone give me the reference for this statement?:

Let $M$ be a closed oriented smooth 4-manifold. Any element $a\in H_2(M)$ can be represented by a smoothly embedded, oriented surface.

I found this statement and the proof at Saveliev's book, Lectures on the topology of 3-manifold, but I think it is not a complete proof and I couldn't fill the gap. Let me know other reference.

$\endgroup$
1

2 Answers 2

7
$\begingroup$

This result actually holds in all dimensions.

Let $M^n$ be a closed smooth manifold $M^n$ of any dimension $n\geqslant 3$. Every element $\alpha \in H_2(M,\mathbb Z)$ is represented by a smoothly embedded closed surface.

You can prove it as follows:

  1. Take a cycle $a_1\sigma_1 + \ldots + a_n\sigma_n$ representing $\alpha$. The coefficients $a_i$ are integers: by writing $a\sigma$ as $\pm(\sigma +\ldots +\sigma)$ you can suppose they are all $\pm 1$.
  2. Since it is a cycle, restrictions on edges must cancel in pairs. You can glue correspondingly the triangles along these edges and get a map $f:S\to M^n$ from some (possibly disconnected) 2-dimensional complex $S$.
  3. This complex $S$ is obtained from finitely many oriented triangles by gluing the edges in pairs (with orientation-reversing maps): it is necessarily a closed oriented surface (note: this is not true for higher-dimensional cycles where you only get onlu a kind of "pseudo-manifold" which might have singular codimension-2 stratum).
  4. Up to homotopy you can take $f$ smooth. You can then put the map $f$ in general position. If the dimension $n$ of $M^n$ is $n\geqslant 5$ then $f$ is necessarily injective and you are done.
  5. IF $n=4$ you may have isolated double points. These can be removed via some surgery which replaces locally the two intersecting transverse 2-discs with an annulus. The surgery modifies $S$ (genus increases by one) but not the class $\alpha$.
  6. If $n=3$ you may have double and triple points. These can also be removed via some similar surgery.

For a reference, I suggest you the nice and readable book "The wild world of 4-manifolds" from A. Scorpan which treats the 4-dimensional case and also the general $n$-dimensional one (with further discussion and references inside concerning the general problem of realizing an integral class by a manifold).

$\endgroup$
11
$\begingroup$

By Poincaré duality, there is an isomorphism

$H_2(M, \mathbb{Z}) \cong H^2(M, \mathbb{Z})$.

Now let $PD(a) \in H^2(M, \mathbb{Z})$ be the Poincaré dual of $a$. Since $H^2(M, \mathbb{Z})$ classifies line bundles on $M$, there exists a line bundle $L$ such that $c_1(L)=PD(a)$. Take a general smooth section of $L$. Then its zero set is a smoothly embedded oriented surface $\Sigma \subset M$ such that its fundamental class $[\Sigma]$ is equal to $a$.

See Donaldson-Kronheimer ["The geometry of 4-manifolds", Chapter 1] for more details.

$\endgroup$
1
  • 4
    $\begingroup$ Note that this argument shows that for any closed, orientable, smooth n-manifold, any homology class in dimension (n-2) is represented by a smooth submanifold. Similarly, an (n-1)-dimensional homology class can be represented by the inverse image of a regular value of a map to the circle representing its Poincaré dual. $\endgroup$ Oct 23, 2010 at 12:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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