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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJul 21st 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexI should have known, but I've actually had an intuition of $\mathbf{B}^n U(1)$ as an $(n+1)$-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 $U(1)$. Next, we consider $\mathcal{B}U(1)$, the classifying space for principal $U(1)$-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 $\mathbf{B}U(1)$ as the functor mapping a toplogical space $X$ to the 2-group of principal $U(1)$-bundles over $X$. So far we have gone from the Lie group $U(1)$ to the 2-group $\mathbf{B}U(1)$ of principal $U(1)$-bundles. Next step is going to (the 3-group of) principal $\mathbf{B}U(1)$-bundles. Therefore, on each open set $U_i$ of an open cover of a topological space $X$ we'll have the 2-group of line bundles on $U_i$. Transition functions will be given by line bundles on $U_{ij}$, 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 $\mathbf{B}U(1)$-gerbe, and so we are led to think of $\mathbf{B}^2U(1)$ as the functor mapping a topological space $X$ to the 2-category of $\mathbf{B}U(1)$-gerbes on $X$. 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 $\mathbf{B}U(1)$-gerbes (and to invert one), and this is accomplished by looking at transition functions, exactly as for principal $U(1)$-bundles: in the gerbe case, transition functions are principal $U(1)$-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 $\mathbf{B}^2U(1)$ and can consider principal $\mathbf{B}^2U(1)$-bundles; this will give the abelian 4-group $\mathbf{B}^3U(1)$, and so on.

   I should have known, but I’ve actually had an intuition of $\mathbf{B}^n U(1)$ as an $(n+1)$-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 $U(1)$.

   Next, we consider $\mathcal{B}U(1)$, the classifying space for principal $U(1)$-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 $\mathbf{B}U(1)$ as the functor mapping a toplogical space $X$ to the 2-group of principal $U(1)$-bundles over $X$.

   So far we have gone from the Lie group $U(1)$ to the 2-group $\mathbf{B}U(1)$ of principal $U(1)$-bundles. Next step is going to (the 3-group of) principal $\mathbf{B}U(1)$-bundles. Therefore, on each open set $U_i$ of an open cover of a topological space $X$ we’ll have the 2-group of line bundles on $U_i$. Transition functions will be given by line bundles on $U_{ij}$, 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 $\mathbf{B}U(1)$-gerbe, and so we are led to think of $\mathbf{B}^2U(1)$ as the functor mapping a topological space $X$ to the 2-category of $\mathbf{B}U(1)$-gerbes on $X$. 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 $\mathbf{B}U(1)$-gerbes (and to invert one), and this is accomplished by looking at transition functions, exactly as for principal $U(1)$-bundles: in the gerbe case, transition functions are principal $U(1)$-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 $\mathbf{B}^2U(1)$ and can consider principal $\mathbf{B}^2U(1)$-bundles; this will give the abelian 4-group $\mathbf{B}^3U(1)$, and so on.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJul 21st 2010
   • (edited Jul 21st 2010)
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   Author: Urs
   Format: MarkdownItexYes. A quick way to see the group structure on these $\infty$-stacks is to start with the $\infty$-prestacks defined as follows: let $U(1)$ denote the sheaf of $U(1)$-valued functions (continuous or smooth, depending on context) and $U(1)[n]$ the chain complex of sheaves concentrated with $U(1)$ in degree $n$. Let $\Xi : Ch_\bullet \to KanCplx$ be the Dold-Kan map,then $\Xi U(1)[n]$ 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 $\Xi U(1)[n]$ will in fact already be an $(\infty,1)$-sheaf. I like to write this guy $\mathbf{B}^n U(1)$ 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 $(\infty,1)$-sheaf we pass to the $\infty$-stackification $L \mathbf{B}^n U(1)$. Since $\infty$-stackification (by the very definition!) preserves finite products, it preserves group structures. That's one way to see these group structures on $\mathbf{B}^n U(1)$. More abstractly, one may realize that this makes $\mathbf{B}^n U(1)$ an [[Eilenberg-MacLane object]] in the $(\infty,1)$-sheaf topos. Then in a way I can go the reverse direction and deduce the group operations on $n$-gerbes from this: a $U(1)$ $n$-gerbe is the homotopy fiber of maps $X \to \mathbf{B}^{n+1} U(1)$.

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

   let $U(1)$ denote the sheaf of $U(1)$-valued functions (continuous or smooth, depending on context) and $U(1)[n]$ the chain complex of sheaves concentrated with $U(1)$ in degree $n$. Let $\Xi : Ch_\bullet \to KanCplx$ be the Dold-Kan map,then $\Xi U(1)[n]$ 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 $\Xi U(1)[n]$ will in fact already be an $(\infty,1)$-sheaf. I like to write this guy $\mathbf{B}^n U(1)$ 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 $(\infty,1)$-sheaf we pass to the $\infty$-stackification $L \mathbf{B}^n U(1)$. Since $\infty$-stackification (by the very definition!) preserves finite products, it preserves group structures.

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

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

3. • CommentRowNumber3.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJul 21st 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexThanks. I'm also thinking one should obtain the Lie $n$-algebra $\mathfrak{bu}_n$ from the Lie $n$-group $\mathbf{B}^n U(1)$ by looking at $U(1)$ $n$-gerbes on $Spec{\mathbf{K}[\epsilon]/(\epsilon^2)$, is this correct? A minor typo in what you write above: the chain complex of sheaves concentrated with $U(1)$ in degree $n$ should be $U(1)[-n]$ rather than $U(1)[n]$.

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

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

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeJul 21st 2010
   • (edited Jul 21st 2010)
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   Author: Urs
   Format: MarkdownItex> Thanks. I'm also thinking one should obtain the Lie $n$-algebra $\mathfrak{bu}_n$ from the Lie $n$-group $\mathbf{B}^n U(1)$ by looking at $U(1)$ $n$-gerbes on $Spec(\mathbf{K}[\epsilon]/(\epsilon^2))$, is this correct? Yes, in the sense that if you start with the simplicial sheaf $\mathbf{B}^n U(1)$, 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" $b^n \mathfrak{u}(1)$ 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 $b^n \mathfrak{u}(1)$ is $\mathbf{B}^n \mathbb{R}$. This follows from the fact that an $n$-form on the $n$-sphere may be extended to a closed $n$-form on the $(n+1)$-ball precisely if its integral over the $n$-sphere is 0. Now the $n$-morphisms in the Lie integration of $b^n \mathfrak{u}(1)$ are $n$-forms on $D^n$ modulo the relation that two are identified if there is a closed $n$-form on $D^{n+1}$ 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.

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

   Yes, in the sense that if you start with the simplicial sheaf $\mathbf{B}^n U(1)$, 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” $b^n \mathfrak{u}(1)$ 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 $b^n \mathfrak{u}(1)$ is $\mathbf{B}^n \mathbb{R}$.

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

   Now the $n$-morphisms in the Lie integration of $b^n \mathfrak{u}(1)$ are $n$-forms on $D^n$ modulo the relation that two are identified if there is a closed $n$-form on $D^{n+1}$ 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.

5. • CommentRowNumber5.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJul 21st 2010
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   Author: domenico_fiorenza
   Format: MarkdownItexFine, 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 <a href="http://ncatlab.org/nlab/show/2-group#weak_groups_5">weak <i>n</i>-group</a> 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 $n$-group way.

   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 $n$-group way.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeJul 21st 2010
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   Author: Urs
   Format: MarkdownItex> 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 $n$. 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 $F : Ab \to KanCplx$. Then the $\infty$-stack $\mathbf{B}^n U(1)$ is modeled as a group object by (the $\infty$-stackification of) the prestack $$ X \mapsto F(Hom(X,U(1))) \,. $$

   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 $n$. 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 $F : Ab \to KanCplx$. Then the $\infty$-stack $\mathbf{B}^n U(1)$ is modeled as a group object by (the $\infty$-stackification of) the prestack

   $$X \mapsto F(Hom(X,U(1))) \,.$$
7. • CommentRowNumber7.
   • CommentAuthordomenico_fiorenza
   • CommentTimeJul 23rd 2010
   • (edited Jul 23rd 2010)
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   Author: domenico_fiorenza
   Format: MarkdownItexso, here is a tentative naive picture of $String(n)$. the idea is that $String(n)$ is difficult, but $\mathbf{B}String(n)$ is easy, so let us look at the latter. Namely, as a topological space, $String(n)$ is a $\mathcal{B}U(1)$-bundle on $Spin(n)$ induced by a map $Spin(n)\to \mathcal{B}^2U(1)$. taking classifying spaces we obtain a map $\mathcal{B}Spin(n)\to \mathcal{B}^3U(1)$, whose homotopy fiber is $\mathcal{B}Spin(n)$. now, it is immediate to say explicitely which kind of functor is $\mathcal{B}String(n)$. Namely, the map $\mathcal{B}Spin(n)\to \mathcal{B}^3U(1)$ allow us to associate with every $Spin$-bundle an $U(1)$-2-gerbe. then by the universal property of homotopy fiber, the functor $\mathcal{B}String$ maps a topological space $X$ to the category of principal $Spin$-bundle over $X$ 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 $\mathcal{B}SO(n)$ classifies $O(n)$-bundles with a trivialization of the associated orientation bundle, which is fine. next, $\mathcal{B}Spin(n)$ should classify $SO(n)$-bundles with a trivialization of the associated $\mathbb{Z}/2\mathbb{Z}$-gerbe. mmm.. I have to convince me of this..

   so, here is a tentative naive picture of $String(n)$. the idea is that $String(n)$ is difficult, but $\mathbf{B}String(n)$ is easy, so let us look at the latter. Namely, as a topological space, $String(n)$ is a $\mathcal{B}U(1)$-bundle on $Spin(n)$ induced by a map $Spin(n)\to \mathcal{B}^2U(1)$. taking classifying spaces we obtain a map $\mathcal{B}Spin(n)\to \mathcal{B}^3U(1)$, whose homotopy fiber is $\mathcal{B}Spin(n)$. now, it is immediate to say explicitely which kind of functor is $\mathcal{B}String(n)$. Namely, the map $\mathcal{B}Spin(n)\to \mathcal{B}^3U(1)$ allow us to associate with every $Spin$-bundle an $U(1)$-2-gerbe. then by the universal property of homotopy fiber, the functor $\mathcal{B}String$ maps a topological space $X$ to the category of principal $Spin$-bundle over $X$ 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 $\mathcal{B}SO(n)$ classifies $O(n)$-bundles with a trivialization of the associated orientation bundle, which is fine. next, $\mathcal{B}Spin(n)$ should classify $SO(n)$-bundles with a trivialization of the associated $\mathbb{Z}/2\mathbb{Z}$-gerbe. mmm.. I have to convince me of this..

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeJul 23rd 2010
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   Author: Urs
   Format: MarkdownItexYes, 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 $(\infty,1)$-pullbacks $$ \array{ Spin &\to& P &\to& \mathbf{B}^2 U(1) &\to& {*} \\ \downarrow && \downarrow && \downarrow && \downarrow \\ * &\stackrel{x}{\to}& X &\to& \mathbf{B}String &\to& \mathbf{B}Spin \\ && && \downarrow && \downarrow \\ && && * &\to& \mathbf{B}^3 U(1) } $$ coming from a $Spin$-principal bundle $P \to X$ with String-structrure (witnessed by the factorization of its classifiyng map $X \to \mathbf{B}Spin$ through $\mathbf{B}String$). From that you read off that a String structure on $X$ is the same thing as a $U(1)$-gerbe on $P$ that restricts on each fiber to the canonical cocycle on $Spin$. 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).

   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 $(\infty,1)$-pullbacks

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

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

   From that you read off that a String structure on $X$ is the same thing as a $U(1)$-gerbe on $P$ that restricts on each fiber to the canonical cocycle on $Spin$.

   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).