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Hi, everyone:

For the sake of context, I am a graduate student, and I have taken classes in algebraic topology and differential geometry. Still, the 2 proofs I have found are a little too terse for me; they are both around 10 lines long, and each line seems to pack around 10 pages of results. Of course, I am considering cases for "reasonable" spaces, being the beginner I am at this point.

It would also be great if someone knew of similar results for H_1 (equiv. H_3).

Thanks in Advance.

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  • $\begingroup$ Thanks, but this is a different question from the one you are referring/linking to. The only thing common to both is that both questions are on, or relate-to, 4-manifolds. AFAIK, the intersection form assumes the existence of representative surfaces, but does not prove their existence. An etiquette question: I was able to answer my own previous question, the one you linked to. But since I saw it was of such low interest in this site, I thought I would not comment on this. Is that O.K in this site.? $\endgroup$
    – Herb
    Apr 25, 2010 at 1:35
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    $\begingroup$ Herb: my mistake regarding the question -- I read too hastily, and will delete my comments. As for the etiquette question: if you were able to answer a question you posed then I think it is preferred that you add a comment to that effect; you're even allowed to post an answer yourself and accept it. This is because open questions get sporadically and automatically bumped to the top by the software, so it's still worth updating or correcting them. $\endgroup$
    – Yemon Choi
    Apr 25, 2010 at 1:41
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    $\begingroup$ Given an element of the $\mathbb{Z}$ homology of your four manifold $M$, call it $[z]$ and a map of an oriented surface $h:F\rightarrow M$, so that the map induced on homology $h_*$ takes the fundamental class of $[F]$ to $[z]$? Take a singular chain representing the class [z]$, identify the interior of edges of copies of the $2$-simplex in pairs, corresponding to the boundary of the chain being zero. Take the closure of the relation to get an oriented surface, and the map by pasting. You want more? $\endgroup$ Apr 25, 2010 at 2:18
  • $\begingroup$ Two closely related results are mathoverflow.net/questions/1489 and mathoverflow.net/questions/21171 . $\endgroup$
    – Emerton
    Apr 25, 2010 at 5:55

2 Answers 2

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Torsten's answer was good, but there are also more elementary answers. Here's one, which is essentially a big transversality argument, followed by a mild de-singularization.

Let me consider $M$ to be a triangulated 4-manifold, and represent a class in $H_2(M)$ as a sum of 2-faces of the triangulation. For each 2-face appearing with multiplicity $n$ in the sum, take $n$ copies of the face, pushed off of each other slightly and meeting only at the edges.

Now, along each edge (1-face) of the triangulation, since we started with a 2-cycle the total number of triangles meeting there is $0$. (This is a signed count if we are working in $H_2(M; \mathbb{Z})$, and means that there are an even number if we are in $H_2(M; \mathbb{Z}/2)$.) Along each edge, pair up the incident triangles in an arbitrary way that's compatible with the orientations, and resolve the intersections along the interior of the edge according to that pairing. (A neighborhood of the edge looks like $D^3 \times I$; the incident triangles are coming in from fixed directions, i.e., at fixed points in $S^2 \times I$; so given any pairing of the points on $S^2$, we can just join them up and avoid the edge altogether. It's easier to think about what happens in a 3-manifold, where it's very similar, you just have to be more picky about how you pair the incident triangles.)

After the last step, we have a surface $S$ with codimension-2 singularities at points in $M$. For each such singularity, consider a small ball $B$ in $M$ and consider $S \cap \partial B$. This is a link $L$ in a 3-manifold (oriented or not, depending on which homology we look at). Replace $S \cap B$ with a Seifert surface for $L$, and we're done. (You could also use any surface with boundary $L$ inside the 4-ball $B$ instead of the Seifert surface, of course. Frequently you can get lower genus that way.)

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    $\begingroup$ I love these kinds of arguments! $\endgroup$ Apr 25, 2010 at 18:21
  • $\begingroup$ Thank you both. Dylan's answer is nice , in that one can see the actual geometry. Torsten's answer is a bit out-of-my-league at this point, but I will still try to understand it. $\endgroup$
    – Herb
    Apr 28, 2010 at 5:34
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    $\begingroup$ Note that these "actual geometry" arguments get stuck at some point, before the result is false. Specifically, $H_6$ is always represented by surfaces, but local techniques like I sketch above get stuck already for $H_4$ (independent of the ambient space), since not every 4-manifold is the boundary of a 5-manifold. $\endgroup$ Apr 28, 2010 at 6:52
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I do not know if this provides more details:

I assume the closed ($4$-)manifold $M$ to be oriented (we need the submanifold to be oriented to have an integral homology class and the construction I am about to give will make the normal bundle oriented) and as transversality arguments are a little bit tricky in the purely topological case I shall also assume $M$ is smooth. If $b\in H_2(M,\mathbb Z)$ we let $c\in H^2(M,\mathbb Z)$ be its Poincaré dual. The exponential sequence of sheaveas $0\rightarrow\mathbb Z\rightarrow\mathcal C_M\rightarrow\mathcal C_M^*\rightarrow0$ of sheaves on $M$ (where $\mathcal C_M$ is the sheaf of continuous complex-valued functions) and the fact that the cohomology of $\mathcal C_M$ is trivial (because of existence of partitions of unity) shows that $c$ may be represented as the first Chern class of a complex line bundle $L$. A general section of this line bundle is transversal to the zero section and hence its zero set is a $2$-submanifold of $M$. Its normal bundle is the restriction of $L$ to $S$ and is therefore orientable and hence so is $S$. Its class is then equal to $b$.

Essentially the same argument (but with a slightly different exponential sequence) shows that every class $w$ in $H^1(M,\mathbb Z/2)$ is the Stiefel-Whitney class of a real line bundle and the zero set of a general section will have its class equal to the Poincaré dual of $w$ in $H_3(M,\mathbb Z/2)$ and hence any such class can be represented by a non-orientable $3$-dimensional submanifold. As for a class in $H_3(M,\mathbb Z)$ its Poincaré dual in $H^1(M,\mathbb Z)$ is the inverse image of the fundamental class in $H^1(S^1,\mathbb Z)$ of a map $M \rightarrow S^1$ (this is because every element of $H^1(X,\mathbb Z)$ for a nice enough $X$ is represented in that way). The inverse image of a regular value of this map (which exists by Sard's lemma) will then be a $3$-dimensional orientable (in fact with trivial normal bundle) submanifold of $M$ representing the given class in $H_3(M,\mathbb Z)$. Finally, elements of $H_1(M,\mathbb Z)$ are easy to deal with. By the Hurewicz theorem any class of $H_1(M,\mathbb Z)$ is the image of the fundamental class of $S^1$ under a map $S^1 \rightarrow M$. As the dimension ($1$) of $S^1$ is smaller than half ($2=4/2$) the dimension of $M$ a general such function will be an embedding.

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    $\begingroup$ Just beat me to it! :) $\endgroup$ Apr 25, 2010 at 5:32
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    $\begingroup$ If one does not like sheaf cohomology, one can circumvent its use like follows: it holds that $H^2(M) = [M, \mathbb{CP}^\infty]$ since $\mathbb{CP}^\infty$ is an Eilenberg-MacLane space. Since $M$ is four-dimensional, you can homotope it via cellular approximation into $\mathbb{CP}^2$ and you can make it transverse to $\mathbb{CP}^1\subset \mathbb{CP}^2$. Take the preimage and this is then your wished-for surface. - But this may just be one of the terse proofs Herb has found ;). $\endgroup$ Apr 25, 2010 at 9:08
  • $\begingroup$ Note that this is analogous to what I did for $H^1(M)$ where one has $H^1(M)=[M,S^1]$. $\endgroup$ Apr 25, 2010 at 16:53

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