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• CommentRowNumber1.
• CommentAuthorJon Beardsley
• CommentTimeSep 26th 2017

Hi all,

I suspect this is basic stuff, and could be sussed out from something that you all have written (perhaps Urs’ big book), but if I have a manifold $M$ is there an object, I’ll call it $X$ for now, such that (some version of) maps from $M$ to $X$ classify twists of de Rham cohomology on $M$? In other words, the trivial map should just give us regular de Rham cohomology, and non-trivial maps should give us twists. I’m thinking in analogy with a map $Y\to Pic(R)$, where $R$ is a ring spectrum, giving me twists of $R$-homology of $Y$.

Apologies if this is very basic, or perhaps nonsensical.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeSep 27th 2017
• (edited Sep 27th 2017)

Which twists are you thinking of, besides those classified by $B (\mathbb{Z}/2)$? Do you want to include twists by closed 3-forms, which twist not the $\mathbb{Z}$-graded but the $\mathbb{Z}/2$-graded de Rham complex?

• CommentRowNumber3.
• CommentAuthorJon Beardsley
• CommentTimeSep 27th 2017
• (edited Sep 27th 2017)

Well I don’t really know this stuff very well, I thought you could generally twist the differential of the de Rham complex by a 1-form. I’m not sure where this fits into what you’re saying. I guess it must be the twists classified by B(Z/2) that you mention?

• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeSep 27th 2017
• (edited Sep 27th 2017)

The Z/2 twists would correspond to a two-fold covering space. Presumably one could take a flat connection on that, i.e. some closed 1-form to make some sort of twisting.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeSep 28th 2017
• (edited Sep 28th 2017)

Okay, I see. Right, so as David says, the $\mathbb{Z}/2$-twisting is for the unoriented case (“pseudo-forms”). Consider the sheaf on CartSp of crossed complexes which to test space $U$ assigns the crossed complex which is the de Rham complex on $U$ (shifted to some degree $n$) with the groupoid $B \mathbb{Z}/2$ acting on any $k$-forms by sending them to their negative. Write $\flat \mathbf{B}^n \mathbb{R}/(\mathbb{Z}/2) \in Smooth \infty Grpd = Sh_\infty(CartSp)$ for the image of this under the Dold-Kan correspondene for crossed complexes. By construction this comes with a canonical forgetful morphism

$\array{ \flat \mathbf{B}^n \mathbb{R}/(\mathbb{Z}/2) \\ \downarrow \\ B \mathbb{Z}/2 }$

This morphism, regarded as an object in the slice $\infty$-topos over $B \mathbb{Z}/2$, is the coefficients for $\mathbb{Z}/2$-twisted de Rham cohomology in degree $n$. (We could instead use sheaves of spectra on $CartSpace$ instead, to unify the varying degrees more elegantly, but this is a minor technical point.)

For the twists by closed differential 1-forms, consider the homomorphism of semifree dgc-algebras

$\array{ \left( d h_1 = 0 \right) &\longrightarrow& \left( \array{ d h_1 = 0 \\ d \omega_n = h_1 \wedge \omega_n } \right) \\ h_1 &\mapsto& h_1 }$

Under Lie integration this again yields a morphism in $Sh_\infty(CartSp)$, and, regarded as an object in the slice over its codomain, this is the coefficient for twisted de Rham cohomology with twists by closed 1-forms, I think.