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Given a connected CW-complex $X$ I'm interested in if a given homology class $\sigma \in H_n(X)$ can be represented by a manifold meaning if there is a map $f : M^n \to X$ from a oriented manifold $M$ for which $f_*([M^n]) = \sigma$. Obviously this is always true for $n = 1$ and I could prove it for $n = 2$, but it seems this doesn't hold for any $n$.

For example I found this answer which talks about the case where $X$ is itself a manifold. It says there are cases where $\sigma$ is not represented by a manifold for $n = 7$. Are there similar results for $X$ that are not necessary manifolds?

I'm especially interested in the simpler case where $H_i(X) = 0$ for $1 < i < n$

Matthias Deiml
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2 Answers2

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The question in the title differs from the question spelled out in the post: In the title, you ask for embedded manifolds, in the post you ask for just maps from manifolds. I think the version of the question asking for embedded manifolds but $X$ an arbitrary CW-complex is not very well-behaved, so let me answer the question in the post.

One way to think about this is that there is also a homology theory based on oriented manifolds mapping to $X$, called oriented bordism, $\operatorname{MSO}_*(X)$. The construction which assigns to a class represented by an oriented manifold with map to $X$ the image of its fundamental class in $H_*(X)$ comes as a natural transformation $$ \operatorname{MSO}_*(X) \to H_*(X) $$ of homology theories. In fact, it lifts to a map of spectra, $\operatorname{MSO}\to H\mathbb{Z}$, and this is the bottom map in the Postnikov tower for $\operatorname{MSO}$. This way, the question of when homology classes of $X$ are in the image of this natural transformation relates this to differentials in the Atiyah-Hirzebruch spectral sequence for $\operatorname{MSO}_*(X)$. The existence of homology classes which are not in the image corresponds to the fact that the map $\operatorname{MSO}\to H\mathbb{Z}$ does not split, but one can in fact work out explicit obstructions which lead to the examples you referred to. All this only depends on the homotopy type of $X$ (as opposed to the embedding question).

Achim Krause
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  • @AchimKrause I can see why a splitting of $MSO\to H\mathbb{Z}$ would make the homology map surjective, but why does the failure of the splitting imply the failure of homology surjectivity? – Greg Friedman Oct 27 '20 at 18:12
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Turns out (as @archipelago noted) this is called the Steenrod Problem. The answer I linked also holds for $X$ that are not manifolds. In particular every class can be represented for $n \leq 6$. There are example for classes that cannot be represented for $n = 7$

Matthias Deiml
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