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This is probably a very simple question, but I have not found it addressed in the references that I know. Let $M$ be a closed and connected orientable $d$-dimensional manifold ($d\leq 8$) and let $[\alpha_{i}]\in H_{k}(M,\mathbb{Z})$, $k\leq 6$, $i=1,2,$ be $k$-homology classes of $M$ that admit representations $\iota_{i}\colon Z_{i}\hookrightarrow M$ in terms of closed oriented submanifolds $Z_{i}$, that is:

$\iota_{i \ast}[Z_{i}] = [\alpha_{i}]$

Since $k\leq 6$, every element in $H_{k}(M,\mathbb{Z})$ is representable by a closed oriented submanifold, so in particular $[\alpha] := [\alpha_{1}] + [\alpha_{2}]\in H_{k}(M,\mathbb{Z})$ is also representable by a closed oriented submanifold $\iota\colon Z\hookrightarrow M$. Is there any relation between $Z$ and $Z_{i}$? Can I "construct" a representative of $[Z]$ once I know $Z_{1}$ and $Z_{2}$?

Thanks.

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    $\begingroup$ Certainly not always by a connected sum; e.g., if $M$ is not connected then $\alpha_i$ could be supported in different components of $M$. $\endgroup$ Commented May 6, 2017 at 2:07
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    $\begingroup$ It's not clear to me that one can hope for something as trivial as a connected sum. Suppose that $k \ge n/2$; then it's quite possible that generic representatives of $M_1$ and $M_2$ do intersect one another and so you'd have to do a bit more than just connected sum. To see this concretely take $\alpha_i$ to be a basis for $H_1(T^2)$; then $\sum \alpha_i$ is representable but you have to resolve intersection points somehow. $\endgroup$
    – dvitek
    Commented May 6, 2017 at 2:34
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    $\begingroup$ @user43326: The problem is that there is no such thing as disjoint union of embedded submanifolds. $\endgroup$
    – Mark Grant
    Commented May 6, 2017 at 8:09
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    $\begingroup$ The most interesting case is codimension 2 (or codimension 1 with mod 2 coefficients) since then every class is representable by a submanifold. Csaba Nagy (currently a student of Diarmuid Crowley in Melbourne) wrote a nice construction of the embedding representing the sum in that case, but it doesn't seem to be online anywhere. $\endgroup$
    – Mark Grant
    Commented May 6, 2017 at 8:12
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    $\begingroup$ I think the "formal group law of complex cobordism" is an instructive example. The whole higher stuff in that group law seems to come from the fact that there is no nice way to represent the sum of two $\mathbb C P^{n-1}\subset\mathbb C P^n$ generating $H_{2n-2}(\mathbb C P^n)$ in general position by a homologous submanifold. $\endgroup$ Commented May 6, 2017 at 11:38

2 Answers 2

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There are a few well-known (at least to low-dimensional topologists) cases of this.

First, the obvious observation that if $k<d/2$, then one may take $Z_1\cap Z_2=\emptyset$ by a small isotopy, since this is the transversal case, so $Z=Z_1\cup Z_2$.

For $d=2, k=1$, this is well-known: we may assume $Z_1 \pitchfork Z_2$, so $Z_1\cap Z_2$ is a finite collection of points with transverse intersections modeled on $\{(x,y) \in \mathbb{R}^2 | xy=0\}$. We resolve the crossings by perturbing to $\{(x,y)\in \mathbb{R}^2 | xy = \epsilon\}$. The choice of the sign of $\epsilon$ depends on the local orientations at the points of $Z_1\cap Z_2$.

For $d=3, k=2$, this is known as double-curve sum to 3-manifold topologists. If $Z_1\pitchfork Z_2$, then $Z_1\cap Z_2$ is a collection of curves, locally modeled on $\{(x,y,z)\in \mathbb{R}^3 | xy=0\}$ (so the previous example crossed with $\mathbb{R}$). The same resolution of intersections works: $\{(x,y,z) \in \mathbb{R}^3 | xy=\epsilon\}$. (in the picture, $Z_1=R, Z_2=S$.)

MR0992331 ]

I believe that this pattern persists for codimension-one homology classes in all dimensions: when $k=d-1$ and $Z_1\pitchfork Z_2$, then $Z_1\cap Z_2$ should be locally modeled on $(Z_1\cap Z_2) \times \{xy=0\}$, and the ``double curve sum" should be $(Z_1\cap Z_2) \times \{xy=\epsilon\}$.

For $d=4, k=2$, when $Z_1\pitchfork Z_2$, then $Z_1\cap Z_2$ is finite, and locally modeled on $\{(x,y)\in \mathbb{C}^2 | xy=0\}$. Now the local orientations on $M, Z_1, Z_2$ are determined by the complex structure. Then the resolution is $\{(x,y)\in \mathbb{C}^2 | xy=\epsilon\}$ again. I don't know whether this resolution always works in codimension 2 for representable homology classes in general (crossed with $Z_1\cap Z_2$) - there may be a framing issue if the normal bundle of $Z_1\cap Z_2$ is non-trivial. I think this ought to also work in $d=8, k=4$, replacing $\mathbb{C}$ with the quaternions $\mathbb{H}$.

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  • $\begingroup$ I was implicitly assuming that the maps $\iota_i:Z_i\hookrightarrow M$ were smooth embeddings. I'm not sure what is known for locally flat embeddings in the topological category, and even less sure if the embeddings are not locally flat... $\endgroup$
    – Ian Agol
    Commented May 12, 2017 at 3:45
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My paper is now available at http://arxiv.org/abs/1705.03836. It describes a construction for the codimension-1 (mod 2 coefficients) and codimension-2 (twisted integer coefficients) case.

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