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1. • CommentRowNumber1.
   • CommentAuthorDavidRoberts
   • CommentTimeDec 9th 2010
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   Author: DavidRoberts
   Format: MarkdownItexQuestion left at [[cohomology]] by Marc Hoyois.

   Question left at cohomology by Marc Hoyois.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeDec 9th 2010
   • (edited Dec 9th 2010)
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   Author: Urs
   Format: MarkdownItexThanks for alerting me. That warning there was a reminder that some discussion about the bigrading is missing, but I had never gotten back to it. I'll try to do so later.

   Thanks for alerting me. That warning there was a reminder that some discussion about the bigrading is missing, but I had never gotten back to it. I’ll try to do so later.

3. • CommentRowNumber3.
   • CommentAuthorMarc Hoyois
   • CommentTimeDec 20th 2011
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   Author: Marc Hoyois
   Format: TextI suddenly remembered about that old question. I can now answer it with more confidence: the description of the bigrading in the article is indeed wrong (both spheres are the same). The motivic bigrading is really only relevant in the stable theory, where P^1 has to be inverted so that smooth projective varieties become dualizable (just like with Grothendieck's pure motives, one had to invert the Lefschetz motive to get cohomology theories satisfying Poincare duality). So this is very specific to the motivic world.

   I suddenly remembered about that old question.
   I can now answer it with more confidence: the description of the bigrading in the article is indeed wrong (both spheres are the same).
   The motivic bigrading is really only relevant in the stable theory, where P^1 has to be inverted so that smooth projective varieties become dualizable (just like with Grothendieck's pure motives, one had to invert the Lefschetz motive to get cohomology theories satisfying Poincare duality). So this is very specific to the motivic world.
4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeDec 20th 2011
   • (edited Dec 20th 2011)
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   Author: Urs
   Format: MarkdownItexI guess that in the context of $\mathbb{A}^1$-homotopy my description of the geometric 1-sphere as a quotient of the interval was wrong, not the fact that there is a geometric sphere in the game. I am looking at [Morel-Voevodsky, page 78-79](http://www.math.uiuc.edu/K-theory/0305/nowmovo.pdf#page=79) where the categorical and then the geometric 1-sphere are defined, and then the two induced suspensions noticed, which give the bigrading.

   I guess that in the context of $\mathbb{A}^1$-homotopy my description of the geometric 1-sphere as a quotient of the interval was wrong, not the fact that there is a geometric sphere in the game.

   I am looking at Morel-Voevodsky, page 78-79 where the categorical and then the geometric 1-sphere are defined, and then the two induced suspensions noticed, which give the bigrading.

5. • CommentRowNumber5.
   • CommentAuthorMarc Hoyois
   • CommentTimeDec 20th 2011
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   Author: Marc Hoyois
   Format: TextHi Urs, I mean to say there is nothing canonical a priori about the geometric circle: it is just a random variety. The motivic analogue of the exotic spheres from equivariant homotopy theory would be all (categorical) suspensions of smooth schemes (as these form a set of generators of the A^1-homotopy category). In the stable homotopy category you need to add G_m-desuspensions to get a set of generators, which is why things start looking bigraded. It happens that many interesting motivic spectra (including cobordim, K-theory, and motivic cohomology) belong to the localizing subcategory generated by the bigraded spheres, which means that the bigraded homotopy groups detect equivalences between them. But the two families of spheres described in the article should still be the same over any "site with interval": this is Lemma 2.3.12 in Morel-Voevodsky, with X=S^n.

   Hi Urs,
   I mean to say there is nothing canonical a priori about the geometric circle: it is just a random variety. The motivic analogue of the exotic spheres from equivariant homotopy theory would be all (categorical) suspensions of smooth schemes (as these form a set of generators of the A^1-homotopy category). In the stable homotopy category you need to add G_m-desuspensions to get a set of generators, which is why things start looking bigraded. It happens that many interesting motivic spectra (including cobordim, K-theory, and motivic cohomology) belong to the localizing subcategory generated by the bigraded spheres, which means that the bigraded homotopy groups detect equivalences between them.
   
   But the two families of spheres described in the article should still be the same over any "site with interval": this is Lemma 2.3.12 in Morel-Voevodsky, with X=S^n.
6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeDec 20th 2011
   • (edited Dec 20th 2011)
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   Author: Urs
   Format: MarkdownItexVoevodsky supposedly "constructs a bigraded motivic cohomology theory". Could you state the precise definition of that?

   Voevodsky supposedly “constructs a bigraded motivic cohomology theory”. Could you state the precise definition of that?

7. • CommentRowNumber7.
   • CommentAuthorMarc Hoyois
   • CommentTimeDec 20th 2011
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   Author: Marc Hoyois
   Format: MarkdownItexIn the spirit of the article, there are several $\infty$-categories in which motivic cohomology can be described as a mapping space, depending on your taste. Voevodsky's original construction takes place in the stable category $DM^{eff}_-$ of complexes of presheaves with transfers. He defines the "motivic complexes" $\mathbb{Z}(q)$ for every $q\geq 0$, and proves that the motivic cohomology groups $H^{\ast,q}(X,\mathbb{Z})$ are the homotopy groups of the mapping spectrum from "chains" on X to $\mathbb{Z}(q)$. One can also define, "upstream" in the A^1-homotopy category, the Eilenberg-Mac Lane spaces $K(\mathbb{Z}(q),p)$ for $p\geq q\geq 0$ which represent $H^{p,q}$. These can be assembled into the Eilenberg Mac-Lane spectrum HZ, which happens to be an $\Omega$-spectrum (this is Voevodsky's cancellation theorem), so its (p,q)-suspension also represents $H^{p,q}$ (now for all $p,q\in\mathbb{Z}$). The whole picture can be summarized in a diagram with 6 categories: the three nonlinear categories of spaces, S^1-spectra (invert S^1), and T-spectra (invert G_m also), and their three linearizations: spaces with transfers, S^1-spectra with transfers, and T-spectra with transfers. $DM^{eff}_-$ is the same as S^1-spectra with transfers (through a Dold-Kan equivalence). T-spectra with transfers are the same as $DM_-$, and also the same as modules over HZ. There are suspension/loop and free/forgetful adjunctions relating all 6 categories, and motivic cohomology is representable in any of them. The Eilenberg-Mac Lane space $K(\mathbb{Z}(q),p)$ is just the free space with transfers associated to the (p,q)-sphere, which probably gives the shortest possible definition of motivic cohomology.

   In the spirit of the article, there are several $\infty$-categories in which motivic cohomology can be described as a mapping space, depending on your taste. Voevodsky’s original construction takes place in the stable category $DM^{eff}_-$ of complexes of presheaves with transfers. He defines the "motivic complexes" $\mathbb{Z}(q)$ for every $q\geq 0$, and proves that the motivic cohomology groups $H^{\ast,q}(X,\mathbb{Z})$ are the homotopy groups of the mapping spectrum from "chains" on X to $\mathbb{Z}(q)$.

   One can also define, "upstream" in the A^1-homotopy category, the Eilenberg-Mac Lane spaces $K(\mathbb{Z}(q),p)$ for $p\geq q\geq 0$ which represent $H^{p,q}$. These can be assembled into the Eilenberg Mac-Lane spectrum HZ, which happens to be an $\Omega$-spectrum (this is Voevodsky’s cancellation theorem), so its (p,q)-suspension also represents $H^{p,q}$ (now for all $p,q\in\mathbb{Z}$).

   The whole picture can be summarized in a diagram with 6 categories: the three nonlinear categories of spaces, S^1-spectra (invert S^1), and T-spectra (invert G_m also), and their three linearizations: spaces with transfers, S^1-spectra with transfers, and T-spectra with transfers. $DM^{eff}_-$ is the same as S^1-spectra with transfers (through a Dold-Kan equivalence). T-spectra with transfers are the same as $DM_-$, and also the same as modules over HZ. There are suspension/loop and free/forgetful adjunctions relating all 6 categories, and motivic cohomology is representable in any of them. The Eilenberg-Mac Lane space $K(\mathbb{Z}(q),p)$ is just the free space with transfers associated to the (p,q)-sphere, which probably gives the shortest possible definition of motivic cohomology.

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeDec 20th 2011
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   Author: Urs
   Format: MarkdownItexThanks for taking the time to explain. I don't know how much more time you have to do so. Is there a writeup that collects this big picture coherently, say that discusses that diagram of 6 categories? Here my next question: you talk about "$(p,q)$-suspension", "$(p,q)$-spheres", etc. What is the definition of this? From your previous comments I take it that it is not given by the categorical and the geometric 1-sphere. So what is it instead? Thanks.

   Thanks for taking the time to explain. I don’t know how much more time you have to do so. Is there a writeup that collects this big picture coherently, say that discusses that diagram of 6 categories?

   Here my next question:

   you talk about “$(p,q)$-suspension”, “$(p,q)$-spheres”, etc. What is the definition of this? From your previous comments I take it that it is not given by the categorical and the geometric 1-sphere. So what is it instead?

   Thanks.

9. • CommentRowNumber9.
   • CommentAuthorMarc Hoyois
   • CommentTimeDec 20th 2011
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   Author: Marc Hoyois
   Format: MarkdownItexNo, the (p,q)-sphere is the same as the one defined in Morel-Voevodsky, p.79. The complete picture is unfortunately scattered through several articles of Voevodsky. I wrote a very quick summary with references in http://math.northwestern.edu/~hoyois/papers/hopkinsmorel.pdf, section 3. Theorem 3.10 has precise representability statements.

   No, the (p,q)-sphere is the same as the one defined in Morel-Voevodsky, p.79.

   The complete picture is unfortunately scattered through several articles of Voevodsky. I wrote a very quick summary with references in http://math.northwestern.edu/~hoyois/papers/hopkinsmorel.pdf, section 3. Theorem 3.10 has precise representability statements.

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeDec 20th 2011
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    Author: Urs
    Format: MarkdownItex> No, the (p,q)-sphere is the same as the one defined in Morel-Voevodsky, p.79. Then I am confused about what you are trying to tell me, sorry. How is that not what the $n$Lab entry said, about which you commented > the description of the bigrading in the article is indeed wrong (both spheres are the same). ? > I wrote a very quick summary with references in [http://math.northwestern.edu/~hoyois/papers/hopkinsmorel.pdf](http://math.northwestern.edu/~hoyois/papers/hopkinsmorel.pdf), section 3. Thanks, I'll try to find a second to look at it...

    No, the (p,q)-sphere is the same as the one defined in Morel-Voevodsky, p.79.

    Then I am confused about what you are trying to tell me, sorry. How is that not what the $n$Lab entry said, about which you commented

    the description of the bigrading in the article is indeed wrong (both spheres are the same).

    ?

    I wrote a very quick summary with references in http://math.northwestern.edu/~hoyois/papers/hopkinsmorel.pdf, section 3.

    Thanks, I’ll try to find a second to look at it…

11. • CommentRowNumber11.
    • CommentAuthorMarc Hoyois
    • CommentTimeDec 20th 2011
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    Author: Marc Hoyois
    Format: MarkdownItexWhat was wrong in the article is that it claimed that the geometric circle is $\mathbb{A}^1/\{0,1\}$, while in fact this is the categorical circle (aka S^1, aka the nodal curve). The real geometric circle is $\mathbb{G}_m$, but I was pointing out that it does not actually play any special role in unstable motivic homotopy theory (and it is not something that is generally associated to a lined topos, as the article was implying). I see now how my "both sphere are the same" comment could be interpreted to mean the opposite of what I meant…

    What was wrong in the article is that it claimed that the geometric circle is $\mathbb{A}^1/\{0,1\}$, while in fact this is the categorical circle (aka S^1, aka the nodal curve). The real geometric circle is $\mathbb{G}_m$, but I was pointing out that it does not actually play any special role in unstable motivic homotopy theory (and it is not something that is generally associated to a lined topos, as the article was implying). I see now how my “both sphere are the same” comment could be interpreted to mean the opposite of what I meant…

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeDec 20th 2011
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    Author: Urs
    Format: MarkdownItexOkay, thanks, so this seems to be what I said in #4 > I guess that [...] my description of the geometric 1-sphere as a quotient of the interval was wrong, not the fact that there is a geometric sphere in the game.

    Okay, thanks, so this seems to be what I said in #4

    I guess that […] my description of the geometric 1-sphere as a quotient of the interval was wrong, not the fact that there is a geometric sphere in the game.

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeDec 20th 2011
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    Author: Urs
    Format: MarkdownItex> it is not something that is generally associated to a lined topos, as the article was implying Let's see, while my original statement was wrong, is there not a correct statement to be isolated after all? Namely: for any line object we can ask for its corresponding multiplicative group, the subobject of multiplicatively invertible elements. That seems to be something one can consider for any lined topos.

    it is not something that is generally associated to a lined topos, as the article was implying

    Let’s see, while my original statement was wrong, is there not a correct statement to be isolated after all?

    Namely: for any line object we can ask for its corresponding multiplicative group, the subobject of multiplicatively invertible elements. That seems to be something one can consider for any lined topos.

14. • CommentRowNumber14.
    • CommentAuthorMarc Hoyois
    • CommentTimeDec 20th 2011
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    Author: Marc Hoyois
    Format: MarkdownItexOh, in my head I was thinking of site with interval (as in Morel-Voevodsky) when I said lined topos, but the definition of lined topos is more specific. Still, it seems to me that this relation between G_m and A^1 is coincidental and is not why we care about G_m in motivic homotopy theory (see my first post).

    Oh, in my head I was thinking of site with interval (as in Morel-Voevodsky) when I said lined topos, but the definition of lined topos is more specific. Still, it seems to me that this relation between G_m and A^1 is coincidental and is not why we care about G_m in motivic homotopy theory (see my first post).

15. • CommentRowNumber15.
    • CommentAuthorMarc Hoyois
    • CommentTimeAug 6th 2013
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    Author: Marc Hoyois
    Format: MarkdownItex> Namely: for any line object we can ask for its corresponding multiplicative group, the subobject of multiplicatively invertible elements. That seems to be something one can consider for any lined topos. Hmm… but A1-homotopy theory does not use the structure of lined topos on $((Sm/S)_{Nis}, \mathbb{A}^1)$ (i.e., no ring structure is used). You could equivalently use the interval object $\mathbb{A}^n$ for any $n\geq 2$, and then the explanation breaks down. Is there any other [[lined topos]] where the object $\mathbb{G}_m$ is considered a "sphere"?

    Namely: for any line object we can ask for its corresponding multiplicative group, the subobject of multiplicatively invertible elements. That seems to be something one can consider for any lined topos.

    Hmm… but A1-homotopy theory does not use the structure of lined topos on $((Sm/S)_{Nis}, \mathbb{A}^1)$ (i.e., no ring structure is used). You could equivalently use the interval object $\mathbb{A}^n$ for any $n\geq 2$, and then the explanation breaks down.

    Is there any other lined topos where the object $\mathbb{G}_m$ is considered a “sphere”?

16. • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeAug 6th 2013
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    Author: Urs
    Format: MarkdownItex> but A1-homotopy theory does not use the structure of lined topos The [universal characterization](http://ncatlab.org/nlab/show/motivic+homotopy+theory#SymmetricMonoidalStructureAndUniversalProperty) that you added recently, has $\mathbb{A}^1$ and $\mathbb{G}_m$ right there in the axioms. But I may well be missing something.

    but A1-homotopy theory does not use the structure of lined topos

    The universal characterization that you added recently, has $\mathbb{A}^1$ and $\mathbb{G}_m$ right there in the axioms.

    But I may well be missing something.

17. • CommentRowNumber17.
    • CommentAuthorMarc Hoyois
    • CommentTimeAug 6th 2013
    • (edited Aug 6th 2013)
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    Author: Marc Hoyois
    Format: MarkdownItexMy point is that to define the object of invertible elements, one needs to consider $\mathbb{A}^1$ as a ring object, which motivic homotopy does not do. But I'm wondering if this explanation of the origin of a bigrading still holds water for some actual lined topoi.

    My point is that to define the object of invertible elements, one needs to consider $\mathbb{A}^1$ as a ring object, which motivic homotopy does not do. But I’m wondering if this explanation of the origin of a bigrading still holds water for some actual lined topoi.

18. • CommentRowNumber18.
    • CommentAuthorMarc Hoyois
    • CommentTimeAug 6th 2013
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    Author: Marc Hoyois
    Format: MarkdownItexBy the way, I think a better explanation for exotic gradings is that there often is a symmetric monoidal stable ∞-category around, which one is interested in for external reasons, and its [[Picard group]] is then a natural grading group for cohomology (it is the maximal grading monoid for which one has suspension isomorphisms). The motivic bigrading and the $RO(G)$-grading in equivariant homotopy are examples of this, and in $K(n)$-local homotopy theory one sometimes explicitly considers Picard-graded cohomology. In the motivic and equivariant case the exotic stabilization is in some sense the universal way to force Atiyah duality (for smooth proper schemes and compact $G$-manifolds, respectively).

    By the way, I think a better explanation for exotic gradings is that there often is a symmetric monoidal stable ∞-category around, which one is interested in for external reasons, and its Picard group is then a natural grading group for cohomology (it is the maximal grading monoid for which one has suspension isomorphisms). The motivic bigrading and the $RO(G)$-grading in equivariant homotopy are examples of this, and in $K(n)$-local homotopy theory one sometimes explicitly considers Picard-graded cohomology.

    In the motivic and equivariant case the exotic stabilization is in some sense the universal way to force Atiyah duality (for smooth proper schemes and compact $G$-manifolds, respectively).

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeAug 6th 2013
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    Author: Urs
    Format: MarkdownItexHm, that's interesting. Could you expand on that?

    Hm, that’s interesting. Could you expand on that?

20. • CommentRowNumber20.
    • CommentAuthorUrs
    • CommentTimeAug 6th 2013
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    Author: Urs
    Format: MarkdownItexMaybe I should be more specific: * could you indicate a little bit more in which way the motivic and the equivariant bigrading come from Picard groups? * could you indicate a little bit more in which way this "forces [[Atiyah duality]]"? Thanks!

    Maybe I should be more specific:

    • could you indicate a little bit more in which way the motivic and the equivariant bigrading come from Picard groups?

    • could you indicate a little bit more in which way this “forces Atiyah duality”?

    Thanks!

21. • CommentRowNumber21.
    • CommentAuthorMarc Hoyois
    • CommentTimeAug 6th 2013
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    Author: Marc Hoyois
    Format: MarkdownItexIn any symmetric monoidal ∞-category $C$, you could define $$ H^\star(X,A) = \pi_0 C(X, \star\otimes A), \quad \star\in Pic(C). $$ The restriction to invertible objects gives suspension isomorphisms (there might be more to it though). If $C$ is stable you always have $\mathbb{Z}\subset Pic(C)$ which gives the standard integer grading. In the motivic case the Picard group of $SH(S)$ contains $\{S^{p,q}\}\cong \mathbb{Z}\times\mathbb{Z}$ as a subgroup, which gives the usual bigrading. It even contains $\mathbb{Z}\times K_0(S)$, and even more exotic elements besides, but it seems that $\mathbb{Z}\times\mathbb{Z}$ is the largest subgroup that exists over any base. In equivariant stable homotopy theory the Picard group has the subgroup $\{S^V , V \in K_0(BG)\}\cong RO(G)$ (but I believe there are more exotic elements as well). I'm no expert in equivariant homotopy, but apparently enlarging the grading helps to prove theorems (see page 12 of [http://www.math.uchicago.edu/~may/PAPERS/Tate.pdf](http://www.math.uchicago.edu/~may/PAPERS/Tate.pdf)), and $RO(G)$ is just a convenient explicit subgroup of Pic. The Picard-grading in $K(n)$-local homotopy theory is considered in section 14 of [Morava K-theories and localization](http://www.math.rochester.edu/people/faculty/doug/otherpapers/kn.pdf). By Atiyah duality I mean that smooth proper schemes (resp. compact G-manifolds) become dualizable in the stable category. So certainly the stabilizations achieve this, and I'm pretty sure it's exactly this property that motivates the constructions in both cases. For equivariant homotopy this is mentioned in May's book above (same page) as a crucial difference between naive and genuine spectra, and in the motivic world the idea actually goes back to Grothendieck's construction of pure motives where one inverts the Lefschetz motive to make the monoidal category rigid. I don't know a precise universality property, but here's some evidence. In the pointed homotopy category $Ho(Top_*)$, compact manifolds are not quite dualizable but they are $S^n$-dualizable for some n, which means they have evaluation and coevaluation maps to and from $S^n$ that satisfy the triangle identities. Thus, inverting $S^n$ is exactly what is needed to get a true duality. Similarly, in motivic homotopy theory, smooth proper schemes are $T^{\wedge n}$-dualizable for some $n$, and in equivariant homotopy, compact G-manifolds are $S^V$-dualizable for some $V$ (not completely sure about this last one, I may be extrapolating).

    In any symmetric monoidal ∞-category $C$, you could define

    $$H^\star(X,A) = \pi_0 C(X, \star\otimes A), \quad \star\in Pic(C).$$

    The restriction to invertible objects gives suspension isomorphisms (there might be more to it though). If $C$ is stable you always have $\mathbb{Z}\subset Pic(C)$ which gives the standard integer grading.

    In the motivic case the Picard group of $SH(S)$ contains $\{S^{p,q}\}\cong \mathbb{Z}\times\mathbb{Z}$ as a subgroup, which gives the usual bigrading. It even contains $\mathbb{Z}\times K_0(S)$, and even more exotic elements besides, but it seems that $\mathbb{Z}\times\mathbb{Z}$ is the largest subgroup that exists over any base. In equivariant stable homotopy theory the Picard group has the subgroup $\{S^V , V \in K_0(BG)\}\cong RO(G)$ (but I believe there are more exotic elements as well). I’m no expert in equivariant homotopy, but apparently enlarging the grading helps to prove theorems (see page 12 of http://www.math.uchicago.edu/~may/PAPERS/Tate.pdf), and $RO(G)$ is just a convenient explicit subgroup of Pic.

    The Picard-grading in $K(n)$-local homotopy theory is considered in section 14 of Morava K-theories and localization.

    By Atiyah duality I mean that smooth proper schemes (resp. compact G-manifolds) become dualizable in the stable category. So certainly the stabilizations achieve this, and I’m pretty sure it’s exactly this property that motivates the constructions in both cases. For equivariant homotopy this is mentioned in May’s book above (same page) as a crucial difference between naive and genuine spectra, and in the motivic world the idea actually goes back to Grothendieck’s construction of pure motives where one inverts the Lefschetz motive to make the monoidal category rigid.

    I don’t know a precise universality property, but here’s some evidence. In the pointed homotopy category $Ho(Top_*)$, compact manifolds are not quite dualizable but they are $S^n$-dualizable for some n, which means they have evaluation and coevaluation maps to and from $S^n$ that satisfy the triangle identities. Thus, inverting $S^n$ is exactly what is needed to get a true duality. Similarly, in motivic homotopy theory, smooth proper schemes are $T^{\wedge n}$-dualizable for some $n$, and in equivariant homotopy, compact G-manifolds are $S^V$-dualizable for some $V$ (not completely sure about this last one, I may be extrapolating).

22. • CommentRowNumber22.
    • CommentAuthorUrs
    • CommentTimeAug 6th 2013
    • (edited Aug 6th 2013)
    • PermaLink
    Author: Urs
    Format: MarkdownItex> In any symmetric monoidal ∞-category $C$, you could define > $$ H^\star(X,A) = \pi_0 C(X, \star\otimes A), \quad \star\in Pic(C). $$ Ah, I see. Just for the record, let's say how this relates to the usual notion of [[twisted cohomology]] of Eilenberg-Steenrod-type generalized cohomology in terms of [[(infinity,1)-module bundles]]: For $E$ an $E_\infty$-ring and $X$ a homotopy type, let $$ C \coloneqq [X, E Mod] \,. $$ Write $\mathbb{I}_X^E \colon X \to \ast \to E Mod$ for the tensor unit. We have $Pic([X,E Mod]) \simeq [X, Pic(E Mod)]$ is the sub-$\infty$-category of $E$-line bundles. Given such a line bundle $\chi$, the corresponding $\chi$-twisted $E$-cohomology of $X$ is the sections of $\chi$. This is indeed $$ Hom( \mathbb{I}_X^E, \chi \otimes E ) \,. $$ > The restriction to invertible objects gives suspension isomorphisms (there might be more to it though). In the above case at least the restriction to invertible objects in the image of $[X,B GL_1(E)] \hookrightarrow Pic([X,E Mod])$ is what qualifies the twists as still being about $E$-cohomology (because for the invertible objects the fibers all look like $E$). Sections of more general objects are still something twisted, but no longer twisted $E$-cohomology. For what it's worth. More later, am forced offline right now.

    In any symmetric monoidal ∞-category $C$, you could define

    $$H^\star(X,A) = \pi_0 C(X, \star\otimes A), \quad \star\in Pic(C).$$

    Ah, I see.

    Just for the record, let’s say how this relates to the usual notion of twisted cohomology of Eilenberg-Steenrod-type generalized cohomology in terms of (infinity,1)-module bundles:

    For $E$ an $E_\infty$-ring and $X$ a homotopy type, let

    $$C \coloneqq [X, E Mod] \,.$$

    Write $\mathbb{I}_X^E \colon X \to \ast \to E Mod$ for the tensor unit. We have $Pic([X,E Mod]) \simeq [X, Pic(E Mod)]$ is the sub-$\infty$-category of $E$-line bundles. Given such a line bundle $\chi$, the corresponding $\chi$-twisted $E$-cohomology of $X$ is the sections of $\chi$. This is indeed

    $$Hom( \mathbb{I}_X^E, \chi \otimes E ) \,.$$

    The restriction to invertible objects gives suspension isomorphisms (there might be more to it though).

    In the above case at least the restriction to invertible objects in the image of $[X,B GL_1(E)] \hookrightarrow Pic([X,E Mod])$ is what qualifies the twists as still being about $E$-cohomology (because for the invertible objects the fibers all look like $E$). Sections of more general objects are still something twisted, but no longer twisted $E$-cohomology. For what it’s worth.

    More later, am forced offline right now.

23. • CommentRowNumber23.
    • CommentAuthorMarc Hoyois
    • CommentTimeAug 6th 2013
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    Author: Marc Hoyois
    Format: MarkdownItex> In the above case at least the restriction to invertible objects in the image of $[X,B GL_1(E)] \hookrightarrow Pic([X,E Mod])$ is what qualifies the twists as still being about $E$-cohomology (because for the invertible objects the fibers all look like $E$). Sections of more general objects are still something twisted, but no longer twisted $E$-cohomology. For what it's worth. I'm confused. Did you mean $[X,B GL_1(E)] \hookrightarrow [X,E Mod]$ or are you emphasizing the difference between $Pic(E Mod)$ and $E Line$? In Remark 1.3 in [http://arxiv.org/pdf/1002.3004v2.pdf](http://arxiv.org/pdf/1002.3004v2.pdf) they say that they could have defined twists using $Pic(E Mod)$ rather than just $E Line$. In any case, this looks like further evidence that Picard-graded cohomology is interesting in a wide variety of contexts.

    In the above case at least the restriction to invertible objects in the image of $[X,B GL_1(E)] \hookrightarrow Pic([X,E Mod])$ is what qualifies the twists as still being about $E$-cohomology (because for the invertible objects the fibers all look like $E$). Sections of more general objects are still something twisted, but no longer twisted $E$-cohomology. For what it’s worth.

    I’m confused. Did you mean $[X,B GL_1(E)] \hookrightarrow [X,E Mod]$ or are you emphasizing the difference between $Pic(E Mod)$ and $E Line$? In Remark 1.3 in http://arxiv.org/pdf/1002.3004v2.pdf they say that they could have defined twists using $Pic(E Mod)$ rather than just $E Line$.

    In any case, this looks like further evidence that Picard-graded cohomology is interesting in a wide variety of contexts.

24. • CommentRowNumber24.
    • CommentAuthorUrs
    • CommentTimeAug 6th 2013
    • (edited Aug 6th 2013)
    • PermaLink
    Author: Urs
    Format: MarkdownItexI was just using the sequence of maps $$ \mathbf{B} GL_1(E) \simeq E Line \hookrightarrow Pic(E Mod) \hookrightarrow E Mod $$ and pointed out that among $Pic(E Mod) \hookrightarrow E Mod$ those objects which genuinely deserve to be called twists for $E$-cohomology in that _locally_ they classify plain $E$-cohomology are those in the inclusion $\mathbf{B} GL_1(E) \simeq E Line \hookrightarrow Pic(E Mod)$. As you and these authors note, judging just from the formalism one can play with more general "twists", but in this sense only the twists in this inclusion are genuinely "twists of plain $E$-cohomology". This was just meant as a (naive) argument to restrict _at least_ to tensoring with objects in $Pic(E Mod)$.

    I was just using the sequence of maps

    $$\mathbf{B} GL_1(E) \simeq E Line \hookrightarrow Pic(E Mod) \hookrightarrow E Mod$$

    and pointed out that among $Pic(E Mod) \hookrightarrow E Mod$ those objects which genuinely deserve to be called twists for $E$-cohomology in that locally they classify plain $E$-cohomology are those in the inclusion $\mathbf{B} GL_1(E) \simeq E Line \hookrightarrow Pic(E Mod)$.

    As you and these authors note, judging just from the formalism one can play with more general “twists”, but in this sense only the twists in this inclusion are genuinely “twists of plain $E$-cohomology”.

    This was just meant as a (naive) argument to restrict at least to tensoring with objects in $Pic(E Mod)$.

25. • CommentRowNumber25.
    • CommentAuthorUrs
    • CommentTimeAug 6th 2013
    • PermaLink
    Author: Urs
    Format: MarkdownItex> this looks like further evidence that Picard-graded cohomology is interesting in a wide variety of contexts. Yes, certainly! I am glad you amplified this point of view and its examples in motivic and equivariant cohomology. I didn't appreciate this before and am happy to know about it now. In fact this inspired me to finally go and brush-up the entry _[[twisted cohomology]]_ a little. I have now added a brief mentioning of this definition in symmetric monoidal categories there. More could be done, but I am out of steam now. But if you feel energetic to add some of your above examples to the entry, I'd greatly appreciate it. See [another thread](http://nforum.mathforge.org/discussion/905/twisted-cohomology/?Focus=41072#Comment_41072) about edits to the "twisted cohomology"-entry.

    this looks like further evidence that Picard-graded cohomology is interesting in a wide variety of contexts.

    Yes, certainly! I am glad you amplified this point of view and its examples in motivic and equivariant cohomology. I didn’t appreciate this before and am happy to know about it now.

    In fact this inspired me to finally go and brush-up the entry twisted cohomology a little. I have now added a brief mentioning of this definition in symmetric monoidal categories there. More could be done, but I am out of steam now. But if you feel energetic to add some of your above examples to the entry, I’d greatly appreciate it.

    See another thread about edits to the “twisted cohomology”-entry.

26. • CommentRowNumber26.
    • CommentAuthorMarc Hoyois
    • CommentTimeAug 6th 2013
    • (edited Aug 6th 2013)
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexRe #24: I see. Maybe "locally" can be interpreted in a more algebro-geometric sense for general invertible modules? Don't they look locally like $E$ in the Zariski topology on $Spec(E)$?

    Re #24:

    I see. Maybe “locally” can be interpreted in a more algebro-geometric sense for general invertible modules? Don’t they look locally like $E$ in the Zariski topology on $Spec(E)$?

27. • CommentRowNumber27.
    • CommentAuthorUrs
    • CommentTimeAug 6th 2013
    • (edited Aug 6th 2013)
    • PermaLink
    Author: Urs
    Format: MarkdownItexHm [silly comment removed, let me think...]

    Hm [silly comment removed, let me think…]