# Spectral realization of the natural transformation from bordism to singular homology

There is a geometric version of the definition of singular homology $H_n$ in terms of continuous maps of "pseudomanifolds" ($n$-dimensional simplicial complexes such that every $(n-1)$-simplex is contained in exactly two $n$-simplices with opposite induced orientations and every simplex is contained in some $n$-simplex) and a notion of cobordism between such maps. Maps from smooth manifolds are a special case of maps from pseudomanifolds, so there is induced a natural map

$MSO_k X \longrightarrow H_k X.$

(This question phrases it better: Is there a initial "bordism-like" homology theory?)

These theories are of representable, in the sense that

$MSO_k X = \pi_k(MSO \wedge X)$ and $H_k X = \pi_k(H\mathbb{Z} \wedge X)$.

Is there a map $MSO \to H\mathbb Z$ of spectra inducing this natural homomorphism? If so, what can one say about it?

• Would this not just be the zeroth Postnikov section of $MSO$? – JHF Jul 2 '17 at 20:00
• @JHF: I guess so. I see that this map exists and is the natural candidate, but it's not obvious from me from this description, that (1), this is actually the thing that considers a manifold as a pseudomanifold and thus induces the map motivating the question, and (2), well, literally anything else about this map. – jdc Jul 3 '17 at 21:17
• I should add that you can see this from coherence of a system of Thom classes of MSO(k), if one picks the orientations consistently, and I think this is the same thing you are describing, but I don't feel I actually know anything about these classes except that they must exist because the Serre spectral sequence of the fiberwise-compactified sphere bundle of the tautological k-plane bundle over BSO(k) collapses. So I have a mysterious system of suspension-coherent classes that I think gives you the Postnikov section you talk about simply because the lower-degree reduced homology vanishes ... – jdc Jul 3 '17 at 21:24
• ... but this all seems very inexplicit and unsatisfying. – jdc Jul 3 '17 at 21:24
• The map $MSO \to H\mathbb{Z}$ induces a map $\pi_0 MSO \to \pi_0 H\mathbb{Z} \cong \mathbb{Z}$, whose source $\pi_0 MSO \cong MSO_0(\Sigma^\infty_+ *)$ are maps of $0$-manifolds into a point, which is just the data of the $0$-manifold up to oriented bordism, which is classified by their signed count of the points, which one can think of as the Euler characteristic. In particular, it sends a generator to a generator. One way to lift this map on $\pi_0$ to spectra is to use the Postnikov section I mentioned in my previous comment. Now, I think $H^0(MSO; \mathbb{Z}) \cong \mathbb{Z})$... – JHF Jul 3 '17 at 21:49

We seek a map $MSO \to H\mathbb{Z}$. It's not hard to see that any such map factors through $MSO \to H\pi_0 MSO$, and homotopy classes of maps from $H\pi_0 MSO \to H\mathbb{Z}$ are in bijection with $\pi_0 MSO \to \mathbb{Z}$, so we just need to understand the map $MSO \to H\mathbb{Z}$ in $\pi_0$.
Now we can think of $\pi_0 MSO$ as $0$-dimensional manifolds up to cobordism, and we can think of $\pi_0 H\mathbb{Z} \cong \mathbb{Z}$ as $0$-dimensional "pseudomanifolds" up to cobordism. But $0$-manifolds are the same as $0$-pseudomanifolds, so the map $\pi_0 MSO \cong \mathbb{Z} \to \mathbb{Z}$ is the identity.
This shows that the map $H\pi_0 MSO \xrightarrow{\sim} H\mathbb{Z}$ is an equivalence, so the map $MSO \to H\mathbb{Z}$ is just the zeroth Postnikov stage $$MSO \to H\pi_0 MSO \simeq H\mathbb{Z}.$$