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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.

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See also mathoverflow.net/questions/22473/… –  Torsten Ekedahl Oct 25 '10 at 7:15
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2 Answers 2

up vote 7 down vote accepted

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).

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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.

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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. –  Allan Edmonds Oct 23 '10 at 12:25
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