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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeJul 13th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexI am trying to understand if we can refine the following basic fact: **Basic fact**. _The geometric realization of the smooth stack $\mathbf{B}U(n)$ of $U(n)$-principal bundles is the classifying space $B U(n)$. Similarly for the smooth stack $\mathbf{B}U \times \mathbb{Z}$ of stable unitary bundles and the classifying space $B U \times \mathbb{Z}$ of topological K-theory._ This says that $\mathbf{B}U \times \mathbb{Z}$ is a "smooth/stacky refinement" of the _space_ $B U \times \mathbb{Z}$. **Goal.** I would like to have a smooth refinement, in this sense, of $B U \times \mathbb{Z}$ not just as a space, but as an $E_\infty$-space: I would like to equip the smooth stack $\mathbf{B}U \times \mathbb{Z} \in Sh_\infty(SmothMfd)$ with the structure of an abelian group object in $Sh_\infty(SmothMfd)$. I am not sure yet. What I am trying is to see is what happens when I apply a loop space machine objectwise and then $\infty$-stackify again. By which I mean this: **Possible strategy.** The smooth stack $\mathbf{B} U(n)$ is presented by the simplicial presheaf / $\infty$-presheaf $ *//C^\infty(-, U(n)) $, the presheaf that sends any smooth manifold $X$ to the groupoid with a single object and the (discrete) group $C^\infty(X,U(n))$ as morphisms. The coproduct $\coprod_n *//C^\infty(-, U(n))$ of all these is a presheaf of [[permutative categories]]. Hence we can objectwise construct the [[K-theory of permutative categories]] to obtain a presheaf of Segal-spectra $$ X \mapsto K^{Seg} ( \coprod_n *//C^\infty(- , U(n)) ) \,. $$ **Question**. Is this presheaf of algebraic K-theory spectra under $\infty$-stackification and geometric realization nicely related to the topological K-theory spectrum? For instance, can we usefully complete the following diagram somehow: $$ \array{ PSh_\infty(SmothMfd) &\hookrightarrow& Sh_\infty(SmthMfd) &\stackrel{{|-|}}{\to}& Top \\ \coprod_n *//C^\infty(-, U(n)) &\stackrel{stackify}{\mapsto}& \coprod_n \mathbf{B}U(n) & \stackrel{{geometric\, realization}}{\mapsto} & \coprod_n B U(n) \\ {}^{\mathllap{K^{Seg}(-)_1}}\downarrow && && \downarrow^{\mathrlap{(K^{Seg}(-))_1}} \\ K^Seg( \coprod_n *//C^\infty(-,U) )_1 && \stackrel{??}{\leftrightarrow} && B U \times \mathbb{Z} } $$ ?

   I am trying to understand if we can refine the following basic fact:

   Basic fact. The geometric realization of the smooth stack $\mathbf{B}U(n)$ of $U(n)$-principal bundles is the classifying space $B U(n)$. Similarly for the smooth stack $\mathbf{B}U \times \mathbb{Z}$ of stable unitary bundles and the classifying space $B U \times \mathbb{Z}$ of topological K-theory.

   This says that $\mathbf{B}U \times \mathbb{Z}$ is a “smooth/stacky refinement” of the space $B U \times \mathbb{Z}$.

   Goal. I would like to have a smooth refinement, in this sense, of $B U \times \mathbb{Z}$ not just as a space, but as an $E_\infty$-space: I would like to equip the smooth stack $\mathbf{B}U \times \mathbb{Z} \in Sh_\infty(SmothMfd)$ with the structure of an abelian group object in $Sh_\infty(SmothMfd)$.

   I am not sure yet. What I am trying is to see is what happens when I apply a loop space machine objectwise and then $\infty$-stackify again.

   By which I mean this:

   Possible strategy. The smooth stack $\mathbf{B} U(n)$ is presented by the simplicial presheaf / $\infty$-presheaf $*//C^\infty(-, U(n))$, the presheaf that sends any smooth manifold $X$ to the groupoid with a single object and the (discrete) group $C^\infty(X,U(n))$ as morphisms.

   The coproduct $\coprod_n *//C^\infty(-, U(n))$ of all these is a presheaf of permutative categories. Hence we can objectwise construct the K-theory of permutative categories to obtain a presheaf of Segal-spectra

   $$X \mapsto K^{Seg} ( \coprod_n *//C^\infty(- , U(n)) ) \,.$$

   Question. Is this presheaf of algebraic K-theory spectra under $\infty$-stackification and geometric realization nicely related to the topological K-theory spectrum?

   For instance, can we usefully complete the following diagram somehow:

   $$\array{ PSh_\infty(SmothMfd) &\hookrightarrow& Sh_\infty(SmthMfd) &\stackrel{{|-|}}{\to}& Top \\ \coprod_n *//C^\infty(-, U(n)) &\stackrel{stackify}{\mapsto}& \coprod_n \mathbf{B}U(n) & \stackrel{{geometric\, realization}}{\mapsto} & \coprod_n B U(n) \\ {}^{\mathllap{K^{Seg}(-)_1}}\downarrow && && \downarrow^{\mathrlap{(K^{Seg}(-))_1}} \\ K^Seg( \coprod_n *//C^\infty(-,U) )_1 && \stackrel{??}{\leftrightarrow} && B U \times \mathbb{Z} }$$

   ?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeJul 13th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexCan't we give $\mathbf{B} U\times\mathbb{Z}$ an abelian group structure representably, using the fact that we know what it classifies?

   Can’t we give $\mathbf{B} U\times\mathbb{Z}$ an abelian group structure representably, using the fact that we know what it classifies?

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJul 13th 2012
   • (edited Jul 13th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThanks for the reaction! > Can't we give $\mathbf{B}U \times \mathbb{Z}$ an abelian group structure representably, using the fact that we know what it classifies? Maybe, but I don't see this yet. It seems to me that this is at least not straightforward because classifying space theory says little about $\mathbf{H}(X,\mathbf{B}U \times \mathbb{Z})$ (by which I mean as usual the derived hom of $\infty$-stacks), but instead knows about $Top(X, B U \times \mathbb{Z})$; and these two in general only coincide on connected components: $$ \pi_0\mathbf{H}(X,\mathbf{B}U \times \mathbb{Z}) \simeq \pi_0 Top(X, B U \times \mathbb{Z}) \,. $$ Do you see what I mean? Isn't this a problem for the strategy that you suggest? On an unrelated note, I just see that around [p. 24](http://arxiv.org/pdf/math/9908097v2.pdf#page=24) of [Toën's thesis](http://arxiv.org/abs/math/9908097) is a comparison between "stacks of algebraic K-theory spectra" $\underline{\mathbf{K}}$ and "algebraic K-theory of homs of stacks" $\mathbf{K}$ that might be related to what I am after here.

   Thanks for the reaction!

   Can’t we give $\mathbf{B}U \times \mathbb{Z}$ an abelian group structure representably, using the fact that we know what it classifies?

   Maybe, but I don’t see this yet. It seems to me that this is at least not straightforward because classifying space theory says little about $\mathbf{H}(X,\mathbf{B}U \times \mathbb{Z})$ (by which I mean as usual the derived hom of $\infty$-stacks), but instead knows about $Top(X, B U \times \mathbb{Z})$; and these two in general only coincide on connected components:

   $$\pi_0\mathbf{H}(X,\mathbf{B}U \times \mathbb{Z}) \simeq \pi_0 Top(X, B U \times \mathbb{Z}) \,.$$

   Do you see what I mean? Isn’t this a problem for the strategy that you suggest?

   On an unrelated note, I just see that around p. 24 of Toën’s thesis is a comparison between “stacks of algebraic K-theory spectra” $\underline{\mathbf{K}}$ and “algebraic K-theory of homs of stacks” $\mathbf{K}$ that might be related to what I am after here.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeJul 13th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIs the problem the difference between $\mathbf{B} U = \mathbf{B} (\colim_n U(n))$ and $\colim_n (\mathbf{B} U(n))$ ?

   Is the problem the difference between $\mathbf{B} U = \mathbf{B} (\colim_n U(n))$ and $\colim_n (\mathbf{B} U(n))$ ?

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJul 13th 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexNo, that's okay. The problem that I see is that the homotopy type of stack maps into $\mathbf{B}U$ is very different from the homotopy type of continuos maps into $B U$. Only the connected components coincide. This is, I think, what prevents one from simply lifting the $\infty$-group structure in $Top$ to one in $Sh_\infty(SmthMfd)$ by a Yoneda-argument.

   No, that’s okay. The problem that I see is that the homotopy type of stack maps into $\mathbf{B}U$ is very different from the homotopy type of continuos maps into $B U$. Only the connected components coincide.

   This is, I think, what prevents one from simply lifting the $\infty$-group structure in $Top$ to one in $Sh_\infty(SmthMfd)$ by a Yoneda-argument.

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeJul 14th 2012
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   Author: Mike Shulman
   Format: MarkdownItexThen what exactly do you mean by saying that $\mathbf{B}U \times \mathbb{Z}$ is "the stack of stable unitary bundles"? That sounds to me like saying that it classifies stable unitary bundles; and so why can't we take the direct sum of stable unitary bundles to induce a monoidal structure?

   Then what exactly do you mean by saying that $\mathbf{B}U \times \mathbb{Z}$ is “the stack of stable unitary bundles”? That sounds to me like saying that it classifies stable unitary bundles; and so why can’t we take the direct sum of stable unitary bundles to induce a monoidal structure?

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJul 14th 2012
   • (edited Jul 14th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItex> That sounds to me like saying that it classifies stable unitary bundles; Oh sure, it does. But it does more. It's the "fine moduli space". That it classifies stable unitary bundles is the statement that $\pi_0$ of homs into is is the set of equivalence classes of stable bundles. But the thing is that $\pi_1$ of homs into it is very different from $\pi_1$ of homs into $B U \times \mathbb{Z}$: generally the object $\mathbf{B}U \times \mathbb{Z}$ is 1-truncated in $Sh_\infty(SmthMfd)$, while $B U \times \mathbb{Z}$ is untruncated in $Top$. > and so why can't we take the direct sum of stable unitary bundles to induce a monoidal structure? Oh, we can do that. This is what I expressed above as saying that $\coprod_n \mathbf{B}U(n)$ is a stack of permutative categories (under direct sum of bundles, yes). But for the construction of $K$-theory now, we need to $\infty$-groupify this monoidal structure, once in $Sh_\infty(SmthMfd)$, once in $Top$. And this is where I see what looks like a subtlety. How do I construct the _smooth_ $\infty$-group -- an object in $Grp(Sh_\infty(SmthMfd))$ -- generated by the stack of monoidal categories $\coprod_n \mathbf{B}U(n)$, such that it is usefully related to the $K$-theory spectrum $KU$ in $Top$ ? I admit that I may not be asking the right question yet. But let me know if you see now what I have in mind.

   That sounds to me like saying that it classifies stable unitary bundles;

   Oh sure, it does. But it does more. It’s the “fine moduli space”. That it classifies stable unitary bundles is the statement that $\pi_0$ of homs into is is the set of equivalence classes of stable bundles. But the thing is that $\pi_1$ of homs into it is very different from $\pi_1$ of homs into $B U \times \mathbb{Z}$: generally the object $\mathbf{B}U \times \mathbb{Z}$ is 1-truncated in $Sh_\infty(SmthMfd)$, while $B U \times \mathbb{Z}$ is untruncated in $Top$.

   and so why can’t we take the direct sum of stable unitary bundles to induce a monoidal structure?

   Oh, we can do that. This is what I expressed above as saying that $\coprod_n \mathbf{B}U(n)$ is a stack of permutative categories (under direct sum of bundles, yes).

   But for the construction of $K$-theory now, we need to $\infty$-groupify this monoidal structure, once in $Sh_\infty(SmthMfd)$, once in $Top$. And this is where I see what looks like a subtlety.

   How do I construct the smooth $\infty$-group – an object in $Grp(Sh_\infty(SmthMfd))$ – generated by the stack of monoidal categories $\coprod_n \mathbf{B}U(n)$, such that it is usefully related to the $K$-theory spectrum $KU$ in $Top$ ?

   I admit that I may not be asking the right question yet. But let me know if you see now what I have in mind.

8. • CommentRowNumber8.
   • CommentAuthorDavidRoberts
   • CommentTimeJul 15th 2012
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   Author: DavidRoberts
   Format: MarkdownItexThe phrase 'group completion' comes to mind...

   The phrase ’group completion’ comes to mind…

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeJul 16th 2012
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexSo is the question whether geometric realization $\Pi_\infty : Sh_\infty(SmthMfd) \to \infty Gpd$ preserves group completion?

   So is the question whether geometric realization $\Pi_\infty : Sh_\infty(SmthMfd) \to \infty Gpd$ preserves group completion?

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJul 16th 2012
    • PermaLink
    Author: Urs
    Format: MarkdownItexYes, or probably rather: if it doesn't give an equivalence, do we still have useful comparison maps?

    Yes, or probably rather: if it doesn’t give an equivalence, do we still have useful comparison maps?

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeJul 16th 2012
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexOkay, now I think I understand the question! Here's a try at an argument for why something analogous should be true in the 1-categorical case. The theories of groups and monoids are finite product-theories, which means that the free group and free monoid monads (certainly in any topos, or probably more generally a cartesian closed cocomplete category) can be built out of finite products and colimits. (The free-monoid monad is just $F(X) = \coprod_n X^n$, of course.) Thus, any functor between toposes that preserves finite products and colimits (such as $\Pi$ for a cohesive topos) will also commute with these monads. But by the [[adjoint lifting theorem]], for any $\mathbf{E}$, the left adjoint to the forgetful functor $Grp(\mathbf{E}) \to Mon(\mathbf{E})$ can be built out of these monads plus colimits (specifically, a reflexive coequalizer), and thus it will also be preserved by any finite-product-preserving and cocontinuous functor. It seems to me that this basic outline should also work in the $\infty$-case, modulo a suitable adjoint lifting theorem and a finite-product presentation of the theory of $\infty$-groups. However, I don't know to what extent it is known that the operation called "group completion" in classical algebraic topology is actually the left adjoint to $\infty Grp \to \infty Mon$.

    Okay, now I think I understand the question!

    Here’s a try at an argument for why something analogous should be true in the 1-categorical case. The theories of groups and monoids are finite product-theories, which means that the free group and free monoid monads (certainly in any topos, or probably more generally a cartesian closed cocomplete category) can be built out of finite products and colimits. (The free-monoid monad is just $F(X) = \coprod_n X^n$, of course.) Thus, any functor between toposes that preserves finite products and colimits (such as $\Pi$ for a cohesive topos) will also commute with these monads. But by the adjoint lifting theorem, for any $\mathbf{E}$, the left adjoint to the forgetful functor $Grp(\mathbf{E}) \to Mon(\mathbf{E})$ can be built out of these monads plus colimits (specifically, a reflexive coequalizer), and thus it will also be preserved by any finite-product-preserving and cocontinuous functor.

    It seems to me that this basic outline should also work in the $\infty$-case, modulo a suitable adjoint lifting theorem and a finite-product presentation of the theory of $\infty$-groups. However, I don’t know to what extent it is known that the operation called “group completion” in classical algebraic topology is actually the left adjoint to $\infty Grp \to \infty Mon$.

12. • CommentRowNumber12.
    • CommentAuthorMarc Hoyois
    • CommentTimeJul 16th 2012
    • (edited Jul 16th 2012)
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexFor the existence of group completions, Prop. 6.1.2.9 in [[Higher Topos Theory]] (and its proof) seem relevant. Regarding the abstract vs. classical group completion, they both satisfy the same higher universal property, according to [this MO answer](http://mathoverflow.net/questions/17128/group-completion-theorem#answer-17135). Added: actually the existence of group completions follows directly from Prop. 6.1.2.9 and 5.5.4.17, if you want to be lazy ;) Also an easier argument to show that any cocontinuous functor that preserves finite products commutes with group completions is that its right adjoint trivially commutes with the forgetful functors.

    For the existence of group completions, Prop. 6.1.2.9 in Higher Topos Theory (and its proof) seem relevant.

    Regarding the abstract vs. classical group completion, they both satisfy the same higher universal property, according to this MO answer.

    Added: actually the existence of group completions follows directly from Prop. 6.1.2.9 and 5.5.4.17, if you want to be lazy ;)

    Also an easier argument to show that any cocontinuous functor that preserves finite products commutes with group completions is that its right adjoint trivially commutes with the forgetful functors.

13. • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeJul 16th 2012
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    Author: Mike Shulman
    Format: MarkdownItexExcellent; so is all we need a higher adjoint lifting theorem and a presentation of the theory of $\infty$-groups?

    Excellent; so is all we need a higher adjoint lifting theorem and a presentation of the theory of $\infty$-groups?

14. • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeJul 16th 2012
    • PermaLink
    Author: Urs
    Format: MarkdownItexHm, nice. Thanks! I need to call it quits for today, but thanks, this should be useful. I see that a problem with how I put my question was that I started focusing on the construction instead of on the goal. But that is something I also still need to pinpoint explicitly: if or when the objectwise group completion of a stack of monoidal categories is the group completion of monoids in $\infty$-stacks. But not tonight.

    Hm, nice. Thanks! I need to call it quits for today, but thanks, this should be useful.

    I see that a problem with how I put my question was that I started focusing on the construction instead of on the goal. But that is something I also still need to pinpoint explicitly: if or when the objectwise group completion of a stack of monoidal categories is the group completion of monoids in $\infty$-stacks. But not tonight.

15. • CommentRowNumber15.
    • CommentAuthorMarc Hoyois
    • CommentTimeJul 16th 2012
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexIn the 1-categorical case the lifted functors are adjoint because the unit and counit lift as well (on a monoid object the (co)unit is necessarily a monoid homomorphism). This is still true in the $\infty$-case, but I'm not positive that this is enough for an adjunction, since we don't have any triangular identities (or do we?).

    In the 1-categorical case the lifted functors are adjoint because the unit and counit lift as well (on a monoid object the (co)unit is necessarily a monoid homomorphism). This is still true in the $\infty$-case, but I’m not positive that this is enough for an adjunction, since we don’t have any triangular identities (or do we?).