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  1. Let GG be a Lie group and let BG\mathbf{B}G be the smooth stack of principal GG-bundles. Then we have a natural equivalence GΩBGG\simeq \Omega\mathbf{B}G. If we apply the internal hom [S 1,][S^1,-] to the homotopy pullback diagram defining ΩBG\Omega\mathbf{B}G (and if internal homs preserve homotopy pullbacks, something I need be reassured about) then we get a homotopy pullback diagram realizing the free loop space object G=[S 1,G]\mathcal{L}G=[S^1,G] as the based loop space of BG=[S 1,BG]\mathcal{L}\mathbf{B}G=[S^1,\mathbf{B}G]. On the other hand, since GG is a smooth group, then so is G\mathcal{L}G and we can consider the smooth stack BG\mathbf{B}\mathcal{L}G, whose based loop space is again G\mathcal{L}G.

    What is the relation between BG\mathbf{B}\mathcal{L}G and BG\mathcal{L}\mathbf{B}G? (if any)

    From what I read in the lines after equation (3.2) in Konrad Waldorf’s Spin structures on loop spaces that characterize string manifolds I guess that for a connected GG the two stacks BG\mathbf{B}\mathcal{L}G and BG\mathcal{L}\mathbf{B}G should actually be equivalent, but I don’t clearly see this yet.

    • CommentRowNumber2.
    • CommentAuthorDavidRoberts
    • CommentTimeJun 4th 2013

    I guess simple-connectedness would play a role here, no?

  2. Could be: Konrad Waldorf writes “Since Spin(n)Spin(n) is connected, LFMLFM is a principal LSpin(n)LSpin(n)-bundle over LMLM, see etc. etc.”, which is the sentence I derived my guess from (here MM is a spin manifold, FMFM is its frame bundle (i.e., the total space of the associated principal Spin-bundle) and LL stands for “loop space”). But since Spin(n)Spin(n) is actually simply connected (Konrad work with high enough nn), it may well be that the sentence should actually have read “Since Spin(n)Spin(n) is simply connected…”.

    • CommentRowNumber4.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 4th 2013

    I don’t know about the stacky-part, but in terms of Lie groups and their classifying spaces then LL and BB commute. Indeed, Map(X,)Map(X,-) and BB commute. For a Lie group GG, we have the fibration sequence GEGBGG \to E G \to B G with EGE G contractible. Then this maps to Map(X,G)Map(X,EG)Map(X,BG)Map(X,G) \to Map(X,E G) \to Map(X,B G) which is still a fibration sequence. Moreover, the contraction of EGE G maps to a contraction of Map(X,EG)Map(X, E G) whence the base of the fibration is BMap(X,G)B\Map(X,G). This argument works with C C^\infty-maps or with CWCW-complexes or whatever you require. There’s no requirement of connectivity or simple connectivity.

    • CommentRowNumber5.
    • CommentAuthorDavidRoberts
    • CommentTimeJun 5th 2013

    Hmm, and we should use the fact we can take EGEG to be a group, so that Map(X,EG)/Map(X,G)Map(X,EG)/Map(X,G) is perfectly well-defined as a homogeneous space and is a model for BMap(X,G)BMap(X,G). And shouldn’t this be just Map(X,BG)Map(X,BG)?

    • CommentRowNumber6.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 5th 2013

    You don’t need it to be a group, just a principle LGL G-bundle.

  3. Hi,

    I guess the connectivity assumption is needed in order for Map(S 1,BG)Map(S^1,BG) to be connected. If not, then it is only the conneceted component of the basepoint to be a BMap(S 1,G)BMap(S^1,G). But connected should be enough, with no need of simple connectedness.

    • CommentRowNumber8.
    • CommentAuthorUrs
    • CommentTimeJun 5th 2013
    • (edited Jun 5th 2013)

    Yes, that the two stacks (\infty-stacks) are equivalent follows directly from [S 1,][S^1, -] preserving all \infty-limits, which gives

    Ω[S 1,BG][S 1,ΩBG][S 1,G] \Omega [S^1, \mathbf{B}G] \simeq [S^1, \Omega \mathbf{B}G] \simeq [S^1, G]

    and hence by definition of B\mathbf{B} as the inverse equivalence in

    Grp(H)BΩH 1 */ Grp(\mathbf{H}) \stackrel{\overset{\Omega}{\leftarrow}}{\underoverset{\mathbf{B}}{\simeq}{\to}} \mathbf{H}^{\ast/}_{\geq 1}

    we have

    [S 1,BG]B[S 1,G] [S^1, \mathbf{B}G] \simeq \mathbf{B} [S^1, G]

    as soon as [S 1,BG][S^1, \mathbf{B}G] is connected, which it is precisely if for all nn \in \mathbb{N} there is an essentially unique GG-principal infinity-bundle on the circle times the germ of an nn-disk .

    (I am including some links just for the sake of eventual bystanders.)

    • CommentRowNumber9.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 5th 2013

    I’ve thought about this some more, and the connectivity assumptions are there to ensure that Map(X,BG)Map(X,B G) is connected since it is the identity component of this which is BMap(X,G)B Map(X,G).

    The reason being that a point of BMap(X,G)B Map(X,G) must have a lift to EMap(X,G)=Map(X,EG)E Map(X,G) = Map(X,E G). Now a point of Map(X,BG)Map(X, B G) classifies a GG-bundle over XX and a lift of this to EGE G is a section, whence a trivialisation. Thus a point of Map(X,BG)Map(X, B G) is in BMap(X,G)B Map(X, G) if (and only if) the associated bundle is trivial, which occurs if and only if it is in the identity component of Map(X,BG)Map(X, B G).

    Thus the general statement is that:

    BMap(X,G)=Map 0(X,BG) B Map(X,G) = Map_0(X, B G)