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The only exposition of de Rham homology I've found is an appendix to Uranga and Ibanezs book on String Phenomenology. It was brief and gave only basic outline of how to construct this homology.

Now de Rham's theorem asserts that there is an isomorphism between de Rham cohomology of smooth manifolds and that of singular cohomology; and so what appears to be an invariant of smooth structure, is actually an invariant of topological structure.

Is there a similar theorem showing an isomorphism between de Rham homology and singular homology?

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    $\begingroup$ What is deRham homology? $\endgroup$
    – Praphulla Koushik
    Commented Jan 17, 2019 at 4:02
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    $\begingroup$ I think one uses currents instead of differential forms... $\endgroup$
    – Francesco Polizzi
    Commented Jan 17, 2019 at 8:58
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    $\begingroup$ See Chapter IV of De Rham's book Differentiable manifolds. The result you want follows from Thm.16 in Sec. 21. $\endgroup$
    – Liviu Nicolaescu
    Commented Jan 17, 2019 at 9:28
  • $\begingroup$ I think the book of Breadon, Geometry and topology contains a proof (for cohomology) $\endgroup$
    – Ali Taghavi
    Commented Jan 17, 2019 at 10:42
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    $\begingroup$ @FrancescoPolizzi: There is a description using currents; but the book I've alluded to above uses submanifolds. I appreciate currents are more general, and subsume submanifolds by way of Stokes theorem; however, I find the description of homology via submanifolds more intuitive than the simplicial approach in Hatcher. To my mind it makes a better beginning. Though of course one needs to know what a manifold is - but intuitively we know what this is. $\endgroup$
    – Mozibur Ullah
    Commented Jan 17, 2019 at 17:20

2 Answers 2

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I guess that by de Rham homology you mean the homology groups $H_{k, \, \mathrm{dR}}(X)$ constructed on a closed manifold $X$ by using the complex of currents.

In that case, [1, Theorem 2 page 582] shows that there is an isomorphism between $H^{n-k}_{\mathrm{dR}}(X)$ and $H_{k, \, \mathrm{dR}}(X)$, where the cohomology is the usual one (constructed by using the complex of differential forms) and $n = \dim X$.

Now, using the standard De Rham isomorphism between $H^{n-k}_{\mathrm{dR}}(X)$ and the singular cohomology group $H^{n-k}_{\mathrm{sing}}(X, \, \mathbb{R})$, together with the Poincaré duality $H^{n-k}_{\mathrm{sing}}(X, \, \mathbb{R}) \simeq H_{k, \,\mathrm{sing}}(X, \, \mathbb{R})$, we deduce the desired isomorphism $$H_{k, \, \mathrm{dR}}(X) \simeq H_{k, \,\mathrm{sing}}(X, \, \mathbb{R}).$$

References

[1] Giaquinta, Mariano; Modica, Giuseppe; Souček, Jiří, Cartesian currents in the calculus of variations I. Cartesian currents, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 37. Berlin: Springer. xxiv, 711 p. (1998). ZBL0914.49001.

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I can't comment, so as an answer: The definition in your source via submanifolds does not work out the way the authors expect, I think. They essentially define an embedded cobordism group but they neglect to use oriented submanifolds, which would certainly be necessary for a theory with real coefficients.

More seriously though, even if one adds orientations, since already the usual oriented cobordism ring is larger than $\mathbb Z$, even $\mathbb R^n$ has nontrivial homology according to their definition (for $n>2$ or something).

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$
    – Alex M.
    Commented Sep 13 at 13:46
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    $\begingroup$ @AlexM. it does though -- for the precise notion of 'de Rham homology' referenced with source in the question (although it might not deserve the name), the answer is 'no, there is no such isomorphism'. I would have commented were there not a reputation requirement (now I can, yay!) $\endgroup$
    – Slown
    Commented Sep 13 at 13:57

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