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1. • CommentRowNumber1.
   • CommentAuthornilay
   • CommentTimeMar 1st 2018
   • PermaLink
   Author: nilay
   Format: MarkdownItexHi all -- I have some basic questions about the cup product on the smooth Deligne complexes as defined at [[Beilinson–Deligne cup product]]: 1. the product is a bit odd in the sense that it's written $\mathbb{Z}[i]_D^\bullet \otimes\mathbb{Z}[j]_D^\bullet \to \mathbb{Z}[i+j]_D^\bullet$, i.e. it looks like a graded algebra in the category of chain complexes (of sheaves). Is this the best way to think about it? 2. where does this multiplication come from, abstractly? For instance we can produce an $E_\infty$-algebra if we take a suitable homotopy construction on commutative algebras -- can we write the Deligne complexes as such? Beilinson, in the 1985 paper linked on the page above, makes a remark (Remark 1.2.6) about the cup product coming from such an inverse homotopy limit but that seems to be in the wrong direction, i.e. along a given $k$th Deligne complex as opposed to the collection of complexes... what am I missing? Or maybe more generally can we show that this graded algebra in chain complexes is some sort of unit in a symmetric monoidal $\infty$-category? Thanks! Nilay

   Hi all –

   I have some basic questions about the cup product on the smooth Deligne complexes as defined at Beilinson–Deligne cup product:

   1. the product is a bit odd in the sense that it’s written $\mathbb{Z}[i]_D^\bullet \otimes\mathbb{Z}[j]_D^\bullet \to \mathbb{Z}[i+j]_D^\bullet$, i.e. it looks like a graded algebra in the category of chain complexes (of sheaves). Is this the best way to think about it?

   2. where does this multiplication come from, abstractly? For instance we can produce an $E_\infty$-algebra if we take a suitable homotopy construction on commutative algebras – can we write the Deligne complexes as such? Beilinson, in the 1985 paper linked on the page above, makes a remark (Remark 1.2.6) about the cup product coming from such an inverse homotopy limit but that seems to be in the wrong direction, i.e. along a given $k$th Deligne complex as opposed to the collection of complexes… what am I missing? Or maybe more generally can we show that this graded algebra in chain complexes is some sort of unit in a symmetric monoidal $\infty$-category?

   Thanks! Nilay

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMar 2nd 2018
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   Author: Urs
   Format: MarkdownItexThe idea is that 1. differential cohomology is a combination of a cohomology theory with differential form representatives of its realified image; 1. the cup product on differential cohomology is the corresponding combination of the cup product in the cohomology theory with the wedge product of differential forms. At the very least, this immediately explains those shifts in degree that you are pointing to. These are just those of the cup prouct in ordinary cohomology and of the wedge product of differential forms.

   The idea is that

   1. differential cohomology is a combination of a cohomology theory with differential form representatives of its realified image;

   2. the cup product on differential cohomology is the corresponding combination of the cup product in the cohomology theory with the wedge product of differential forms.

   At the very least, this immediately explains those shifts in degree that you are pointing to. These are just those of the cup prouct in ordinary cohomology and of the wedge product of differential forms.

3. • CommentRowNumber3.
   • CommentAuthorDmitri Pavlov
   • CommentTimeMar 2nd 2018
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   Author: Dmitri Pavlov
   Format: MarkdownItexRe #1.1: The differential cohomology groups are bigraded (even though most of them coincide), so a graded algebra in (co)chain complexes is indeed a good way to think about it. (And a filtered algebra in cochain complexes is even better.)

   Re #1.1: The differential cohomology groups are bigraded (even though most of them coincide), so a graded algebra in (co)chain complexes is indeed a good way to think about it. (And a filtered algebra in cochain complexes is even better.)

4. • CommentRowNumber4.
   • CommentAuthornilay
   • CommentTimeMar 4th 2018
   • (edited Mar 4th 2018)
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   Author: nilay
   Format: MarkdownItexThanks Urs, Dmitri. That makes sense. To maybe rephrase my question #2: I've seen it noted in a few papers that the DB cup product is commutative up to coherent homotopy. Is there a proof written down somewhere of this? One thing that I have seen explained is that we can write $\mathbb{Z}[k]_D^\bullet=\text{cone}(\underline{\mathbb{Z}} \hookrightarrow \Omega^{\lt k})[1]$. The shifted cone of a map of cdgas has at least one multiplicative structure that is commutative up to homotopy. If you write out the multiplication in this case, it's kind of a stupid product but it's of a similar form to the DB cup product. Of course, this only defines a product on the $k$th Deligne complex, for each $k$, so that can't be quite what's going on.

   Thanks Urs, Dmitri.

   That makes sense. To maybe rephrase my question #2: I’ve seen it noted in a few papers that the DB cup product is commutative up to coherent homotopy. Is there a proof written down somewhere of this?

   One thing that I have seen explained is that we can write $\mathbb{Z}[k]_D^\bullet=\text{cone}(\underline{\mathbb{Z}} \hookrightarrow \Omega^{\lt k})[1]$. The shifted cone of a map of cdgas has at least one multiplicative structure that is commutative up to homotopy. If you write out the multiplication in this case, it’s kind of a stupid product but it’s of a similar form to the DB cup product. Of course, this only defines a product on the $k$th Deligne complex, for each $k$, so that can’t be quite what’s going on.