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I have been stuck for some time, thinking about the following question.

Let $G$ be a Lie group. Its classifying space $BG$ can be seen as the differentiable stack $[pt/G]$, which is of dimension $-dim(G)$.

Can something similar be said for the based loop group $\Omega G$ of pointed maps from $S^1$ to $G$ ?

I mean, is $\Omega G$ a differentiable stack and, if so, what is its dimension?

Best,

O.

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    $\begingroup$ Perhaps (I say this because I am not sure of what adjectives (like derived) to put everywhere) you can view it as a homotopy pullback (in an appropriate category of stacks) of the diagram $pt \rightarrow BG \leftarrow pt$. In this way you can calculate its tangent space (or, more appropriately, its cotangent complex) using the sequences induced by being a fibration. $\endgroup$
    – Elden Elmanto
    Commented Feb 24, 2015 at 16:38
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    $\begingroup$ In what useful sense does $BG$ have dimension $- \dim (G)$? This is certainly not its cohomological dimension, for example. Does "dimension" here mean something like "Euler characteristic of the tangent complex"? $\endgroup$
    – Qiaochu Yuan
    Commented Feb 24, 2015 at 19:05
  • $\begingroup$ @EldenElmanto: The homotopy pullback to which you are referring actually gives you $G$ back: $pt \to BG$ is a principal $G$-bundle in the world of stacks, and the pullback becomes a proncipal $G$-bundle over $pt$- and there's only one of these I know about ;-) $\endgroup$
    – David Carchedi
    Commented Feb 24, 2015 at 20:52
  • $\begingroup$ oh man I really wanted to say $G$ instead of $BG$! $\endgroup$
    – Elden Elmanto
    Commented Feb 24, 2015 at 21:22
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    $\begingroup$ @QiaochuYuan: My understanding is that the dimension of the differentiable stack $[M/G]$, where $G$ is a Lie group acting smoothly on a manifold $M$, is $\dim M - \dim G$. And $BG$ is $[pt/G]$. More generally, the dimension of an arbitrary differentiable stack is computed from any Lie groupoid presentation of it (webusers.imj-prg.fr/~gregory.ginot/papers/DiffStacksIGG2013.pdf). I'm afraid I do not know much beyond the definition. $\endgroup$
    – Oliver
    Commented Feb 25, 2015 at 10:29

2 Answers 2

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$\Omega G$ won't be a differentiable stack unless you are willing to go to infinite dimensions. Provided you are considering $S^1$ as a smooth manifold, Nerses answer above is correct- it gives you a nice sheaf on the site of manifolds (it's even concrete- a so-called diffeological space). It's also representable by an infinite dimensional manifold.

The other thing you could mean, is you might mean $S^1$ (or $B \mathbb{Z}$) as a homotopy type, or the "categorical circle." For this, you will have to go to derived manifolds (otherwise you will just recover $G$ back again), and then, $\Omega X = T[1] X$ is the shifted tangent bundle for any manifold $X,$ in particular, for $X=G$. $T[1] X$ is the graded manifold whose underlying space is $X$ and whose structure sheaf $\mathcal{O}$ is given by $\mathcal{O}^{-n}(U)=\Omega^{n}_{dR}(U).$

Edit: Actually, in the above, I am computing $\mathcal{L} X$ the free loops on $X$:

For $x \in X,$ the based loops at $X$ are the homotopy fibered product $* \times_{X} T[1]X$, but since evaluation at the base point of $S^1$ is just the vector bundle map $T[1]X \to X,$ which is a submersion, the ordinary pullback is a homotopy pullback, so we get simply $\Omega_x X=T_x[1] X=\left(T_x X\right)[1],$- the graded manifold associated to the graded vector space with $T_x X$ sitting in degree $-1.$ It's underlying manifold is just a point, but it's structure sheaf is given by the exterior algebra of the dual this vector space (but placed in non-positive degrees).

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    $\begingroup$ Isn't the shifted tangent bundle the free loop space, not the based loop space? $\endgroup$
    – Qiaochu Yuan
    Commented Feb 24, 2015 at 22:32
  • $\begingroup$ Do you recover $G$ in the non-derived setting because $\mathrm{Map}(X,G)$ is a set, so that the circle maps only trivially? $\endgroup$
    – Nerses Aramian
    Commented Feb 25, 2015 at 1:43
  • $\begingroup$ @QiaochuYuan: Good catch! I'll think about this and try to edit later. $\endgroup$
    – David Carchedi
    Commented Feb 25, 2015 at 3:14
  • $\begingroup$ @NersesAramian: Yes, that's right. Or more precisely, the representable sheaf defined by $G$ is $0$-truncated, and the $0$-truncation of the circle is the point. $\endgroup$
    – David Carchedi
    Commented Feb 25, 2015 at 3:15
  • $\begingroup$ In this paper, maths.qmul.ac.uk/~noohi/papers/MappingStacks.pdf, Noohi defines mapping stacks of topological stacks. In particular, there is a notion of free loop stack $L\chi:=Maps([S^1],\chi)$ and based loop stack $\Omega\chi$ of a stack $\chi$ ($\Omega\chi$ being by definition the fiber product of a point with the free loop stack $L\chi$ over $\chi$). For instance, it makes sense to speak of $\Omega[G]$ where $G$ is a Lie group and $[G]$ is the stack associated to the manifold $G$. $\endgroup$
    – Oliver
    Commented Feb 26, 2015 at 16:13
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The based loop-group $\Omega G$ can be thought of as a sheaf on the category of manifolds. You can simply declare $\Omega G(X)$ to be the subset of $C^\infty(X\times I, G)$ of functions that are constant on $X\times\{0,1\}$ with value at the unit of $G$.

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