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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeJul 13th 2012

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

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

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

    Goal. I would like to have a smooth refinement, in this sense, of BU×B U \times \mathbb{Z} not just as a space, but as an E E_\infty-space: I would like to equip the smooth stack BU×Sh (SmothMfd)\mathbf{B}U \times \mathbb{Z} \in Sh_\infty(SmothMfd) with the structure of an abelian group object in Sh (SmothMfd)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 BU(n)\mathbf{B} U(n) is presented by the simplicial presheaf / \infty-presheaf *//C (,U(n)) *//C^\infty(-, U(n)) , the presheaf that sends any smooth manifold XX to the groupoid with a single object and the (discrete) group C (X,U(n))C^\infty(X,U(n)) as morphisms.

    The coproduct n*//C (,U(n))\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

    XK Seg( n*//C (,U(n))). 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:

    PSh (SmothMfd) Sh (SmthMfd) || Top n*//C (,U(n)) stackify nBU(n) geometricrealization nBU(n) K Seg() 1 (K Seg()) 1 K Seg( n*//C (,U)) 1 ?? BU× \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} }


    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeJul 13th 2012

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

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJul 13th 2012
    • (edited Jul 13th 2012)

    Thanks for the reaction!

    Can’t we give BU×\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 H(X,BU×)\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,BU×)Top(X, B U \times \mathbb{Z}); and these two in general only coincide on connected components:

    π 0H(X,BU×)π 0Top(X,BU×). \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” K̲\underline{\mathbf{K}} and “algebraic K-theory of homs of stacks” K\mathbf{K} that might be related to what I am after here.

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeJul 13th 2012

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

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJul 13th 2012

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

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

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeJul 14th 2012

    Then what exactly do you mean by saying that BU×\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?

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeJul 14th 2012
    • (edited Jul 14th 2012)

    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 π 0\pi_0 of homs into is is the set of equivalence classes of stable bundles. But the thing is that π 1\pi_1 of homs into it is very different from π 1\pi_1 of homs into BU×B U \times \mathbb{Z}: generally the object BU×\mathbf{B}U \times \mathbb{Z} is 1-truncated in Sh (SmthMfd)Sh_\infty(SmthMfd), while BU×B U \times \mathbb{Z} is untruncated in TopTop.

    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 nBU(n)\coprod_n \mathbf{B}U(n) is a stack of permutative categories (under direct sum of bundles, yes).

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

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

    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.

    • CommentRowNumber8.
    • CommentAuthorDavidRoberts
    • CommentTimeJul 15th 2012

    The phrase ’group completion’ comes to mind…

    • CommentRowNumber9.
    • CommentAuthorMike Shulman
    • CommentTimeJul 16th 2012

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

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJul 16th 2012

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

    • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeJul 16th 2012

    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)= nX nF(X) = \coprod_n X^n, of course.) Thus, any functor between toposes that preserves finite products and colimits (such as Π\Pi for a cohesive topos) will also commute with these monads. But by the adjoint lifting theorem, for any E\mathbf{E}, the left adjoint to the forgetful functor Grp(E)Mon(E)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 GrpMon\infty Grp \to \infty Mon.

    • CommentRowNumber12.
    • CommentAuthorMarc Hoyois
    • CommentTimeJul 16th 2012
    • (edited Jul 16th 2012)

    For the existence of group completions, Prop. 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. and, 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.

    • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeJul 16th 2012

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

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeJul 16th 2012

    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.

    • CommentRowNumber15.
    • CommentAuthorMarc Hoyois
    • CommentTimeJul 16th 2012

    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?).