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  1. I should have known, but I’ve actually had an intuition of B nU(1)\mathbf{B}^n U(1) as an (n+1)(n+1)-group only a few minutes ago. I’ll share here this point of view for the two or three readers which should happen to be unaware of it (I’m sure this is widely known).

    The starting point is the Lie group U(1)U(1).

    Next, we consider U(1)\mathcal{B}U(1), the classifying space for principal U(1)U(1)-bundles, or equivalently of complex line bundles. And let us look to line bundles not up to isomorphism, but as a category. Tensor product and dual of line bundles make line bundles a group-like category, i.e., a 2-group. Moreover, since tensor product is symmetric, this is an abelian 2-group. More formally, we should think of BU(1)\mathbf{B}U(1) as the functor mapping a toplogical space XX to the 2-group of principal U(1)U(1)-bundles over XX.

    So far we have gone from the Lie group U(1)U(1) to the 2-group BU(1)\mathbf{B}U(1) of principal U(1)U(1)-bundles. Next step is going to (the 3-group of) principal BU(1)\mathbf{B}U(1)-bundles. Therefore, on each open set U iU_i of an open cover of a topological space XX we’ll have the 2-group of line bundles on U iU_i. Transition functions will be given by line bundles on U ijU_{ij}, acting by tensor product. On triple intersections we’ll have the tensor product of three line bundles which has to be trivialised in a coherent way. This should define a BU(1)\mathbf{B}U(1)-gerbe, and so we are led to think of B 2U(1)\mathbf{B}^2U(1) as the functor mapping a topological space XX to the 2-category of BU(1)\mathbf{B}U(1)-gerbes on XX. Now, we have to convince ourselves that this 2-category is a group-object and so a 3-group. But this reduces to saying how to multiply two BU(1)\mathbf{B}U(1)-gerbes (and to invert one), and this is accomplished by looking at transition functions, exactly as for principal U(1)U(1)-bundles: in the gerbe case, transition functions are principal U(1)U(1)-bundles and we use the abelian 2-group structure on these to multiply and invert them. In down to earth terms, we take the tensor product of the transition line bundles of the gerbe.

    One sees that this procedure iterates: we now have an abelian 3-group B 2U(1)\mathbf{B}^2U(1) and can consider principal B 2U(1)\mathbf{B}^2U(1)-bundles; this will give the abelian 4-group B 3U(1)\mathbf{B}^3U(1), and so on.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeJul 21st 2010
    • (edited Jul 21st 2010)

    Yes. A quick way to see the group structure on these \infty-stacks is to start with the \infty-prestacks defined as follows:

    let U(1)U(1) denote the sheaf of U(1)U(1)-valued functions (continuous or smooth, depending on context) and U(1)[n]U(1)[n] the chain complex of sheaves concentrated with U(1)U(1) in degree nn. Let Ξ:Ch KanCplx\Xi : Ch_\bullet \to KanCplx be the Dold-Kan map,then ΞU(1)[n]\Xi U(1)[n] is a simplicial sheaf, an \infty-prestack. By Dold-Kan it is an abelian group object.

    Actually, if we are on a small enough site such as CartSp then ΞU(1)[n]\Xi U(1)[n] will in fact already be an (,1)(\infty,1)-sheaf. I like to write this guy B nU(1)\mathbf{B}^n U(1) and its group structure is just that inherited from chain complexes of abelian groups.

    But on a larger site such as Diff to get an (,1)(\infty,1)-sheaf we pass to the \infty-stackification LB nU(1)L \mathbf{B}^n U(1). Since \infty-stackification (by the very definition!) preserves finite products, it preserves group structures.

    That’s one way to see these group structures on B nU(1)\mathbf{B}^n U(1). More abstractly, one may realize that this makes B nU(1)\mathbf{B}^n U(1) an Eilenberg-MacLane object in the (,1)(\infty,1)-sheaf topos.

    Then in a way I can go the reverse direction and deduce the group operations on nn-gerbes from this: a U(1)U(1) nn-gerbe is the homotopy fiber of maps XB n+1U(1)X \to \mathbf{B}^{n+1} U(1).

  2. Thanks. I’m also thinking one should obtain the Lie nn-algebra 𝔟𝔲 n\mathfrak{bu}_n from the Lie nn-group B nU(1)\mathbf{B}^n U(1) by looking at U(1)U(1) nn-gerbes on , is this correct?

    A minor typo in what you write above: the chain complex of sheaves concentrated with U(1)U(1) in degree nn should be U(1)[n]U(1)[-n] rather than U(1)[n]U(1)[n].

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeJul 21st 2010
    • (edited Jul 21st 2010)

    Thanks. I’m also thinking one should obtain the Lie nn-algebra 𝔟𝔲 n\mathfrak{bu}_n from the Lie nn-group B nU(1)\mathbf{B}^n U(1) by looking at U(1)U(1) nn-gerbes on Spec(K[ε]/(ε 2))Spec(\mathbf{K}[\epsilon]/(\epsilon^2)), is this correct?

    Yes, in the sense that if you start with the simplicial sheaf B nU(1)\mathbf{B}^n U(1), think of it as a simplicial sheaf on a site of smooth loci and then restrict to the simplicial sheaf of infinitesimal cells, then that “is” b n𝔲(1)b^n \mathfrak{u}(1) in its incarnation as an infinitesimal \infty-Lie groupoid.

    More elementary is to see it the other way round: convince yourself that the Sullivan-Hinich-Getzler-Henriques Lie integration of b n𝔲(1)b^n \mathfrak{u}(1) is B n\mathbf{B}^n \mathbb{R}.

    This follows from the fact that an nn-form on the nn-sphere may be extended to a closed nn-form on the (n+1)(n+1)-ball precisely if its integral over the nn-sphere is 0.

    Now the nn-morphisms in the Lie integration of b n𝔲(1)b^n \mathfrak{u}(1) are nn-forms on D nD^n modulo the relation that two are identified if there is a closed nn-form on D n+1D^{n+1} restricting to each on the upper and the lower hemisphere. With the above it follows that under this equivalence relation each form represents the number in \mathbb{R} that is its integral.

    This is the “integration without integration” thing.

  3. Fine, that’s the sense I had in mind, thanks!

    As far as concerns looing things the other way round, I agree that it is the best and most economical way to work out all details in a completely rigorous way. Yet, I’d like to stress the weak n-group point of view, which I find more suitable and intuitive for a first introduction to higher groups.

    Now I’ll try to think of String(n) in this naive weak nn-group way.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeJul 21st 2010

    Yet, I’d like to stress the weak n-group point of view, which I find more suitable and intuitive for a first introduction to higher groups.

    Yes, i see what you mean. The chain-complex argument may be thought of as a concrete model that exhibits concretely the n-group structure for all nn. Being a model, it involves choices which is a bit ugly. But it serves to exhibit the abstract construction.

    One could take any other funtorial construction of groupal Eilenberg-MacLane spaces F:AbKanCplxF : Ab \to KanCplx. Then the \infty-stack B nU(1)\mathbf{B}^n U(1) is modeled as a group object by (the \infty-stackification of) the prestack

    XF(Hom(X,U(1))). X \mapsto F(Hom(X,U(1))) \,.
    • CommentRowNumber7.
    • CommentAuthordomenico_fiorenza
    • CommentTimeJul 23rd 2010
    • (edited Jul 23rd 2010)

    so, here is a tentative naive picture of String(n)String(n). the idea is that String(n)String(n) is difficult, but BString(n)\mathbf{B}String(n) is easy, so let us look at the latter. Namely, as a topological space, String(n)String(n) is a U(1)\mathcal{B}U(1)-bundle on Spin(n)Spin(n) induced by a map Spin(n) 2U(1)Spin(n)\to \mathcal{B}^2U(1). taking classifying spaces we obtain a map Spin(n) 3U(1)\mathcal{B}Spin(n)\to \mathcal{B}^3U(1), whose homotopy fiber is Spin(n)\mathcal{B}Spin(n). now, it is immediate to say explicitely which kind of functor is String(n)\mathcal{B}String(n). Namely, the map Spin(n) 3U(1)\mathcal{B}Spin(n)\to \mathcal{B}^3U(1) allow us to associate with every SpinSpin-bundle an U(1)U(1)-2-gerbe. then by the universal property of homotopy fiber, the functor String\mathcal{B}String maps a topological space XX to the category of principal SpinSpin-bundle over XX together with a trivialization of the associated 2-gerbe.

    testing this point of view on the lower stages of the Withehead tower, one finds that SO(n)\mathcal{B}SO(n) classifies O(n)O(n)-bundles with a trivialization of the associated orientation bundle, which is fine. next, Spin(n)\mathcal{B}Spin(n) should classify SO(n)SO(n)-bundles with a trivialization of the associated /2\mathbb{Z}/2\mathbb{Z}-gerbe. mmm.. I have to convince me of this..

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeJul 23rd 2010

    Yes, exactly. I have typed this out at various places, I think. Maybe at string structure. (Also in my article with Zoran and in the last one with Stasheff and Sati).

    So we have a pasting diagram of (,1)(\infty,1)-pullbacks

    Spin P B 2U(1) * * x X BString BSpin * B 3U(1) \array{ Spin &\to& P &\to& \mathbf{B}^2 U(1) &\to& {*} \\ \downarrow && \downarrow && \downarrow && \downarrow \\ * &\stackrel{x}{\to}& X &\to& \mathbf{B}String &\to& \mathbf{B}Spin \\ && && \downarrow && \downarrow \\ && && * &\to& \mathbf{B}^3 U(1) }

    coming from a SpinSpin-principal bundle PXP \to X with String-structrure (witnessed by the factorization of its classifiyng map XBSpinX \to \mathbf{B}Spin through BString\mathbf{B}String).

    From that you read off that a String structure on XX is the same thing as a U(1)U(1)-gerbe on PP that restricts on each fiber to the canonical cocycle on SpinSpin.

    This is the way String structures are sometimes defined, for instance Konrad Waldorf defined it that way in his second-but-last article (or so).