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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeAug 30th 2011
   • PermaLink
   Author: Urs
   Format: MarkdownItexadded the definition to [[Beilinson-Deligne cup product]]. Also expanded the list of references here and at [[Deligne cohomology]].

   added the definition to Beilinson-Deligne cup product.

   Also expanded the list of references here and at Deligne cohomology.

2. • CommentRowNumber2.
   • CommentAuthordomenico_fiorenza
   • CommentTimeFeb 14th 2012
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexat [[Beilinson-Deligne cup product]] I'd rescale the degrees so to have $\mathbb{Z}$ in degree $n+1$ and $\Omega^n$ in degree $0$, so to make it a [[chain complex]] instead of a cochain complex. This will require a minimal (but straightforward) redefinition of the cup product dropping a few $(-1)^{\alpha\beta}$ here and there. I have already worked out the correct signs, but I'll wait feedback on this grading shift before implementing it.

   at Beilinson-Deligne cup product I’d rescale the degrees so to have $\mathbb{Z}$ in degree $n+1$ and $\Omega^n$ in degree $0$, so to make it a chain complex instead of a cochain complex. This will require a minimal (but straightforward) redefinition of the cup product dropping a few $(-1)^{\alpha\beta}$ here and there. I have already worked out the correct signs, but I’ll wait feedback on this grading shift before implementing it.

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 14th 2012
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   Author: Harry Gindi
   Format: MarkdownItexDomenico, I think there might be some problems..

   Domenico, I think there might be some problems..

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeFeb 14th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexHi Domenico, yes, certainly we should change that. Did I type that? Looks like I did. Yes, let's change to the canonical chain complex degrees. (I can't do it right now...)

   Hi Domenico,

   yes, certainly we should change that. Did I type that? Looks like I did. Yes, let’s change to the canonical chain complex degrees.

   (I can’t do it right now…)

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeFeb 14th 2012
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   Author: domenico_fiorenza
   Format: MarkdownItexHi Harry, which problems do you see?

   Hi Harry,

   which problems do you see?

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 15th 2012
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   Author: Harry Gindi
   Format: MarkdownItexI thought that the cup product has some subtleties that make it work only with the cohomological grading (although I may be mistaken).

   I thought that the cup product has some subtleties that make it work only with the cohomological grading (although I may be mistaken).

7. • CommentRowNumber7.
   • CommentAuthordomenico_fiorenza
   • CommentTimeFeb 15th 2012
   • (edited Feb 15th 2012)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexyou're right: with the cohomological grading the product is graded commutative, with the shifted homological grading it is not. or better, it is shifted graded commutative, which is just a way of restating the fact that the original one (the cohomological one) was graded commutative. this is, however, precisely what has to happen. think to the basic example of the cup product $H^i(X,\mathbb{Z})\otimes H^j(X,\mathbb{Z})\to H^{i+j}(X,\mathbb{Z})$: this is a manifestation of a morphism $K(\mathbb{Z},i)\times K(\mathbb{Z},j)\to K(\mathbb{Z},i+j)$ which is in turn an incarnation of the product $\mathbb{Z}\otimes \mathbb{Z}\to \mathbb{Z}$ translated into a morphism of complexes $\mathbb{Z}[i]\otimes \mathbb{Z}[j]\to \mathbb{Z}[i+j]$.and this latter operation is not graded commutative, but it is shifted graded commutative; I mean, the composition $\mathbb{Z}[i]\otimes \mathbb{Z}[j]\stackrel{\sigma}{\to}\mathbb{Z}[j]\otimes \mathbb{Z}[i]\stackrel{\cdot}\mathbb{Z}[i+j]$ is $(-1)^{ij}$ times the product $\mathbb{Z}[i]\otimes \mathbb{Z}[j]\to \mathbb{Z}[i+j]$. now that I'm writing this I notice it seems that in nLab we are missing a neat description of the cup product $H^i(X,A)\otimes H^j(X,B)\to H^{i+j}(X,A\otimes B)$ in abelian cohomology in terms of the morphism $\mathbf{B}^i A\times \mathbf{B}^j B\to \mathbf{B}^{i+j}(A\otimes B)$ induced by the natural isomorphism of compelxes $A[i]\otimes B[j]\to (A\otimes B)[i+j]$

   you’re right: with the cohomological grading the product is graded commutative, with the shifted homological grading it is not. or better, it is shifted graded commutative, which is just a way of restating the fact that the original one (the cohomological one) was graded commutative. this is, however, precisely what has to happen. think to the basic example of the cup product $H^i(X,\mathbb{Z})\otimes H^j(X,\mathbb{Z})\to H^{i+j}(X,\mathbb{Z})$: this is a manifestation of a morphism $K(\mathbb{Z},i)\times K(\mathbb{Z},j)\to K(\mathbb{Z},i+j)$ which is in turn an incarnation of the product $\mathbb{Z}\otimes \mathbb{Z}\to \mathbb{Z}$ translated into a morphism of complexes $\mathbb{Z}[i]\otimes \mathbb{Z}[j]\to \mathbb{Z}[i+j]$.and this latter operation is not graded commutative, but it is shifted graded commutative; I mean, the composition $\mathbb{Z}[i]\otimes \mathbb{Z}[j]\stackrel{\sigma}{\to}\mathbb{Z}[j]\otimes \mathbb{Z}[i]\stackrel{\cdot}\mathbb{Z}[i+j]$ is $(-1)^{ij}$ times the product $\mathbb{Z}[i]\otimes \mathbb{Z}[j]\to \mathbb{Z}[i+j]$.

   now that I’m writing this I notice it seems that in nLab we are missing a neat description of the cup product $H^i(X,A)\otimes H^j(X,B)\to H^{i+j}(X,A\otimes B)$ in abelian cohomology in terms of the morphism $\mathbf{B}^i A\times \mathbf{B}^j B\to \mathbf{B}^{i+j}(A\otimes B)$ induced by the natural isomorphism of compelxes $A[i]\otimes B[j]\to (A\otimes B)[i+j]$

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeFeb 15th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItex> it seems that in nLab we are missing a neat description of the cup product $H^i(X,A)\otimes H^j(X,B)\to H^{i+j}(X,A\otimes B)$ in abelian cohomology in terms of the morphism $\mathbf{B}^i A\times \mathbf{B}^j B\to B^{i+j}(A\otimes B)$ induced by the natural isomorphism of compelxes $A[i]\otimes B[j]\to (A\otimes B)[i+j]$ I think you are right that this is missing. Do you have the time to add it? (Myself, I am all absorbed by operads right now... :-)

   it seems that in nLab we are missing a neat description of the cup product $H^i(X,A)\otimes H^j(X,B)\to H^{i+j}(X,A\otimes B)$ in abelian cohomology in terms of the morphism $\mathbf{B}^i A\times \mathbf{B}^j B\to B^{i+j}(A\otimes B)$ induced by the natural isomorphism of compelxes $A[i]\otimes B[j]\to (A\otimes B)[i+j]$

   I think you are right that this is missing. Do you have the time to add it?

   (Myself, I am all absorbed by operads right now… :-)

9. • CommentRowNumber9.
   • CommentAuthordomenico_fiorenza
   • CommentTimeFeb 15th 2012
   • (edited Feb 15th 2012)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexwe actually have something at [[monoidal Dold-Kan correspondence]], but there it is in terms of cochain complexes rather than in terms of chain complexes. which is obviously fine but I'm getting the impression that if we want to systematically adopt the $H^n(X,A)=\pi_0\mathbf{H}(X,\mathbf{B}^n A)$ point of view throughout nLab, some consistent clean up and harmonization work needs to be done at the various abelian cohomology entries. In the spring semester I'm not teaching so I'll gladly devote a part of my time to trying to do a few steps in this direction (cleaning up the nLab a bit is one of my recurring dream projects :) ). I will be quite busy till march 9, after that I'll be at work on this. given this premise I find it is convenient to pospone to then also the revision of the Beilinson-Deligne cup product: thanks, Harry, for having prevented me from making an edit towards a direction we have not yet supplied a fairly complete background for :)

   we actually have something at monoidal Dold-Kan correspondence, but there it is in terms of cochain complexes rather than in terms of chain complexes. which is obviously fine but I’m getting the impression that if we want to systematically adopt the $H^n(X,A)=\pi_0\mathbf{H}(X,\mathbf{B}^n A)$ point of view throughout nLab, some consistent clean up and harmonization work needs to be done at the various abelian cohomology entries. In the spring semester I’m not teaching so I’ll gladly devote a part of my time to trying to do a few steps in this direction (cleaning up the nLab a bit is one of my recurring dream projects :) ). I will be quite busy till march 9, after that I’ll be at work on this. given this premise I find it is convenient to pospone to then also the revision of the Beilinson-Deligne cup product: thanks, Harry, for having prevented me from making an edit towards a direction we have not yet supplied a fairly complete background for :)