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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 4th 2013
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   Author: domenico_fiorenza
   Format: MarkdownItexLet $G$ be a Lie group and let $\mathbf{B}G$ be the smooth stack of principal $G$-bundles. Then we have a natural equivalence $G\simeq \Omega\mathbf{B}G$. If we apply the internal hom $[S^1,-]$ to the homotopy pullback diagram defining $\Omega\mathbf{B}G$ (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 $\mathcal{L}G=[S^1,G]$ as the based loop space of $\mathcal{L}\mathbf{B}G=[S^1,\mathbf{B}G]$. On the other hand, since $G$ is a smooth group, then so is $\mathcal{L}G$ and we can consider the smooth stack $\mathbf{B}\mathcal{L}G$, whose based loop space is again $\mathcal{L}G$. What is the relation between $\mathbf{B}\mathcal{L}G$ and $\mathcal{L}\mathbf{B}G$? (if any) From what I read in the lines after equation (3.2) in Konrad Waldorf's <a href="http://arxiv.org/abs/1209.1731">Spin structures on loop spaces that characterize string manifolds</a> I guess that for a connected $G$ the two stacks $\mathbf{B}\mathcal{L}G$ and $\mathcal{L}\mathbf{B}G$ should actually be equivalent, but I don't clearly see this yet.

   Let $G$ be a Lie group and let $\mathbf{B}G$ be the smooth stack of principal $G$-bundles. Then we have a natural equivalence $G\simeq \Omega\mathbf{B}G$. If we apply the internal hom $[S^1,-]$ to the homotopy pullback diagram defining $\Omega\mathbf{B}G$ (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 $\mathcal{L}G=[S^1,G]$ as the based loop space of $\mathcal{L}\mathbf{B}G=[S^1,\mathbf{B}G]$. On the other hand, since $G$ is a smooth group, then so is $\mathcal{L}G$ and we can consider the smooth stack $\mathbf{B}\mathcal{L}G$, whose based loop space is again $\mathcal{L}G$.

   What is the relation between $\mathbf{B}\mathcal{L}G$ and $\mathcal{L}\mathbf{B}G$? (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 $G$ the two stacks $\mathbf{B}\mathcal{L}G$ and $\mathcal{L}\mathbf{B}G$ should actually be equivalent, but I don’t clearly see this yet.

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeJun 4th 2013
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   Author: DavidRoberts
   Format: MarkdownItexI guess simple-connectedness would play a role here, no?

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

3. • CommentRowNumber3.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 4th 2013
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   Author: domenico_fiorenza
   Format: MarkdownItexCould be: Konrad Waldorf writes "Since $Spin(n)$ is connected, $LFM$ is a principal $LSpin(n)$-bundle over $LM$, see etc. etc.", which is the sentence I derived my guess from (here $M$ is a spin manifold, $FM$ is its frame bundle (i.e., the total space of the associated principal Spin-bundle) and $L$ stands for "loop space"). But since $Spin(n)$ is actually simply connected (Konrad work with high enough $n$), it may well be that the sentence should actually have read "Since $Spin(n)$ is simply connected...".

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

4. • CommentRowNumber4.
   • CommentAuthorAndrew Stacey
   • CommentTimeJun 4th 2013
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   Author: Andrew Stacey
   Format: MarkdownItexI don't know about the stacky-part, but in terms of Lie groups and their classifying spaces then $L$ and $B$ commute. Indeed, $Map(X,-)$ and $B$ commute. For a Lie group $G$, we have the fibration sequence $G \to E G \to B G$ with $E G$ contractible. Then this maps to $Map(X,G) \to Map(X,E G) \to Map(X,B G)$ which is still a fibration sequence. Moreover, the contraction of $E G$ maps to a contraction of $Map(X, E G)$ whence the base of the fibration is $B\Map(X,G)$. This argument works with $C^\infty$-maps or with $CW$-complexes or whatever you require. There's no requirement of connectivity or simple connectivity.

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

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeJun 5th 2013
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   Author: DavidRoberts
   Format: MarkdownItexHmm, and we should use the fact we can take $EG$ to be a group, so that $Map(X,EG)/Map(X,G)$ is perfectly well-defined as a homogeneous space and is a model for $BMap(X,G)$. And shouldn't this be just $Map(X,BG)$?

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

6. • CommentRowNumber6.
   • CommentAuthorAndrew Stacey
   • CommentTimeJun 5th 2013
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   Author: Andrew Stacey
   Format: MarkdownItexYou don't need it to be a group, just a principle $L G$-bundle.

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

7. • CommentRowNumber7.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJun 5th 2013
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   Author: domenico_fiorenza
   Format: MarkdownItexHi, I guess the connectivity assumption is needed in order for $Map(S^1,BG)$ to be connected. If not, then it is only the conneceted component of the basepoint to be a $BMap(S^1,G)$. But connected should be enough, with no need of simple connectedness.

   Hi,

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

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeJun 5th 2013
   • (edited Jun 5th 2013)
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   Author: Urs
   Format: MarkdownItexYes, that the two stacks ($\infty$-stacks) are equivalent follows directly from $[S^1, -]$ preserving all $\infty$-limits, which gives $$ \Omega [S^1, \mathbf{B}G] \simeq [S^1, \Omega \mathbf{B}G] \simeq [S^1, G] $$ and hence [[looping and delooping|by definition]] of $\mathbf{B}$ as the inverse equivalence in $$ Grp(\mathbf{H}) \stackrel{\overset{\Omega}{\leftarrow}}{\underoverset{\mathbf{B}}{\simeq}{\to}} \mathbf{H}^{\ast/}_{\geq 1} $$ we have $$ [S^1, \mathbf{B}G] \simeq \mathbf{B} [S^1, G] $$ as soon as $[S^1, \mathbf{B}G]$ is connected, which it is precisely if for all $n \in \mathbb{N}$ there is an essentially unique $G$-[[principal infinity-bundle]] on the circle times the germ of an $n$-disk . (I am including some links just for the sake of eventual bystanders.)

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

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

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

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

   we have

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

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

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

9. • CommentRowNumber9.
   • CommentAuthorAndrew Stacey
   • CommentTimeJun 5th 2013
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   Author: Andrew Stacey
   Format: MarkdownItexI've thought about this some more, and the connectivity assumptions are there to ensure that $Map(X,B G)$ is connected since it is the identity component of this which is $B Map(X,G)$. The reason being that a point of $B Map(X,G)$ must have a lift to $E Map(X,G) = Map(X,E G)$. Now a point of $Map(X, B G)$ classifies a $G$-bundle over $X$ and a lift of this to $E G$ is a section, whence a trivialisation. Thus a point of $Map(X, B G)$ is in $B Map(X, G)$ if (and only if) the associated bundle is trivial, which occurs if and only if it is in the identity component of $Map(X, B G)$. Thus the general statement is that: $$ B Map(X,G) = Map_0(X, B G) $$

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

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

   Thus the general statement is that:

   $$B Map(X,G) = Map_0(X, B G)$$