MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.? – Herb Apr 25 '10 at 1:35
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. – Yemon Choi Apr 25 '10 at 1:41
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? – Charlie Frohman Apr 25 '10 at 2:18
Two closely related results are and . – Emerton Apr 25 '10 at 5:55
up vote 11 down vote accepted

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.)

share|cite|improve this answer
I love these kinds of arguments! – Torsten Ekedahl Apr 25 '10 at 18:21
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. – Herb Apr 28 '10 at 5:34
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. – Dylan Thurston Apr 28 '10 at 6:52

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.

share|cite|improve this answer
Just beat me to it! :) – Somnath Basu Apr 25 '10 at 5:32
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 ;). – Lennart Meier Apr 25 '10 at 9:08
Note that this is analogous to what I did for $H^1(M)$ where one has $H^1(M)=[M,S^1]$. – Torsten Ekedahl Apr 25 '10 at 16:53

I do not have enough points to post a comment, so I will have to pose my question here. Could someone explain why the pre-image of $ \mathbb{CP}^{1} $ should represent the class? How does one ensure that the pushforward (with respect to the inclusion map into the original 4-manifold) of the top-homology class of the pre-image is equal to the given 2nd-homology class? It would be great if someone could show this in full detail.

share|cite|improve this answer
Leonard, please ask your question as an MO or MSE question as its own right. (I presume you are referring to Lennart Meier's comment on Torsten Ekedahl's answer.) For formatting and other reasons, having multiple questions in one thread can be very confusing on MO, and is discouraged. Brief answer: what is being used here is the Pontryagin-Thom construction for the group $SO(2)$. – Tim Perutz Nov 11 '12 at 14:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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