One can define twisted cohomology theories via bundles of classifying spaces. In particular, given a cohomology theory $h^{*}$ and a corresponding $\Omega$-spectrum $E_{n}, \varepsilon_{n}$, we can consider on a space $X$ a bundle with fiber $E_{n}$, and define a twisted version of $h^{n}$ as the set of homotopy classes of sections of such a bundle. If the bundle is trivial, we recover the ordinary cohomology theory.

Let us consider a manifold $X$ and its de-Rham cohomology. I suppose there are no problems in defining the Eilenberg-MacLane spaces $K(\mathbb{R}, n)$, even if $\mathbb{R}$ is not countable. For a fixed odd-dimensional form $H$, we can define the (even and odd) twisted de-Rham cohomology groups, via the twisted coboundary $d + H\wedge$, for $d$ the de-Rham differential.

Is there a way to relate the two previous approaches? In particular, given an odd-form $H$, is there a suitable bundle of real Eilenberg-MacLane spaces, whose homotopy classes of sections correspond to the twisted cohomology classes defined via $d + H\wedge$?

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    $\begingroup$ I believe the answer to this question (and probably far more than you ever wanted to know) is contained in the paper "Periodic twisted cohomology and T-duality" by Bunke, Schick and Spitzweck, in Asterisque 337 (2011) - arxiv.org/abs/0805.1459 $\endgroup$ Commented Dec 1, 2011 at 16:13
  • $\begingroup$ In general Brown's representability theorem expresses which cohomology-esque theories can be represented by Eilenberg Mac-Lane-esque spaces. The answer, roughly, is "all of them". Have you looked into that? $\endgroup$
    – Will Sawin
    Commented Dec 1, 2011 at 21:29

2 Answers 2


You cannot compare both cohomologies unless $H$ has degree $1$, because otherwise the cohomology of $d+H\wedge$ is not $\mathbb{Z}$-graded.

If $H$ has degree $1$ then the cohomology of $d+H\wedge$ is the cohomology of $X$ with local coefficients corresponding to the flat line bundle with $1$-form $H$.

The only twisted coefficients associated to singular cohomology with real coefficients are the usual local coefficiens. This follows from the fact that, for $E_n=K(\mathbb{R},n)$, the bundles over $X$ with fiber $E_n$ are classified by $B(\operatorname{Aut}^h(E_n))=K(\operatorname{Aut}\mathbb{R},1)$. Here $\operatorname{Aut}^h$ is the topological group (or $A$-infinity space) of self-homotopy equivalences, and $\operatorname{Aut}$ is just an automorphism group. Therefore, bundles over $X$ with fiber $E_n$ are classified by homomorphisms $\pi_1(X)\rightarrow \operatorname{Aut}(\mathbb{R})$, which correspond to local systems.

  • $\begingroup$ Thank you for your answer. If $H$ has not degree 1, of course the theory is not $\mathbb{Z}$-graded, but I thought it was possible to use bundles of cartesian products of Eilenberg-MacLane spaces (of odd or even degrees) in some way. Your answer seems to exclude this. $\endgroup$
    – Fabio
    Commented Dec 1, 2011 at 16:55
  • $\begingroup$ Fabio, you are on the right. Fernando is only talking about Z-graded cohomology. But if you instead ask about 2-periodic cohomology (Z/2-graded) then I believe you can have more interesting twists. $\endgroup$ Commented Dec 1, 2011 at 21:33
  • $\begingroup$ Sure, what I meant is that, if $H$ doesn't have degree $1$, then the associated twisted cohomology (which is very interesting and studied in the literature) cannot correspond to any twisted generalized cohomology in the sense Fabio first defined, as the former is not $\mathbb{Z}$-graded, but the later is. $\endgroup$ Commented Dec 1, 2011 at 22:26
  • $\begingroup$ That's true, of course a $\mathbb{Z}$-graded theory cannot be equal to a $\mathbb{Z}_{2}$-graded one, but the doubt I have is the following. Let us consider, for example, the even cohomology: in order to twist it, can we consider a bundle whose fiber is $K(\mathbb{R}, 0) \times K(\mathbb{R}, 2) \times \cdots$? In this way we must consider $Aut(K(\mathbb{R}, 0) \times K(\mathbb{R}, 2) \times \cdots)$, which maybe does not split as the direct sum of the single automorphism groups in each degree. Can we obtain in this way the twisted cohomology as defined via $d + H \wedge$? $\endgroup$
    – Fabio
    Commented Dec 2, 2011 at 0:00
  • $\begingroup$ That's true, it doesn't split because real Eilenberg-MacLane spaces do have real cohomology, and it's easy enough (free on one generator at the indicated dimension) so that the homotopy automorphism group can be computed. There's a challenge here! $\endgroup$ Commented Dec 2, 2011 at 8:37

You should turn your d+H into a flat superconnection of degree one. Here is the example which works for K-theory. You consider the $\mathbb{Z}$-graded complex $\Omega(M)[b,b^{-1}]$ with $b$ of degree $-2$ and define the superconnection by $d+bH$ for the closed $3$-form $H$. There is an equivalence between the $\infty$-categories of such superconnections and bundles of chain complexes (representations of the singular complex of your underlying manifold), see Block-Smith. If you then apply the Eilenberg-MacLane equivalence between the categories of chain complexes and $H\mathbb{Z}$-modules, then you get a bundle of $H\mathbb{Z}$-modules. This (or the bundle of its $\infty$-loop spaces) is what you are lokking for. Actually, all these steps are equivalences so that you can go backwards.


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