6
$\begingroup$

According to the definition 1.1 of the paper Kan Replacement of simplicial manifolds by Chenchang Zhu https://arxiv.org/pdf/0812.4150.pdf,

A Kan simplicial manifold is a simplicial manifold $X$ such that for all $m \in \mathbb{N} \cup \lbrace 0 \rbrace $ and $0 \leq j \leq m$, the restriction map $Hom(\Delta^{m},X) \rightarrow Hom(\Lambda^{m}_{j}, X)$ is a surjective submersion.

I also encountered this notion in the definition 2.24 of the paper Higher Groupoid Bundles, Higher Spaces, and Self-Dual Tensor Field Equations by Branislav Jurco, Christian Samann, and Martin Wolf https://arxiv.org/pdf/1604.01639v2.pdf. and in the definition 1.2 of Integrating L∞-Algebras by Andre ́ Henriques (in the name of simplicial manifold satisfying Kan condition)https://arxiv.org/pdf/math/0603563.pdf.

But I could not find much good examples in each of the above 3 references and also anywhere else. I also could not construct one.

(Though I could find some examples like Cech $\infty$-groupoids and internal nerve of Lie groupoids in references Cech cocycles for differential characteristic classes – An ∞-Lie theoretic construction by Domenico Fiorenza, Urs Schreiber and Jim Stasheff and Kan Replacement of simplicial manifolds by Chenchang Zhu respectively. [Please check my 1st two comments for details]).

But it seems to me that, this notion is a very direct and natural generalisation of the notion Lie groupoid to Lie $\infty$-groupoid. (Though Lie $\infty$-groupoid is defined sometimes differently in some literatures). Though according to the discussion in https://ncatlab.org/nlab/show/Kan-fibrant+simplicial+manifold, it is not clear to me whether this notion is very useful or not from the perspective of homotopy theory, but the notion itself looks very elegant to me.

It would be very helpful for me if someone can suggest some interesting examples of Kan simplicial manifolds or suggest some literatures in this direction.

Thanks in advance.

$\endgroup$
2
  • 1
    $\begingroup$ Though I have not verified completely but I am expecting that the internal nerve of a Lie groupoid should be an example of Kan simplicial manifold. This fact is mentioned (without proof) just after the definition 1.1 of the paper Kan replacement of simplicial manifolds by Chenchang Zhu arxiv.org/pdf/0812.4150.pdf $\endgroup$ Commented Dec 24, 2020 at 14:18
  • $\begingroup$ In the definition 3.1.1 of Cech cocycles for differential characteristic classes – An ∞-Lie theoretic construction by Fiorenza, Schreiber and Stasheff arxiv.org/pdf/1011.4735.pdf, the notion of Smooth $\infty-$ groupoid is defined and it is mentioned that Kan simplicial manifolds is special kind of smooth $\infty-$ groupoids called representable smooth $\infty-$ groupoids. In Page 17 , in example 3.1.2 they mentioned that the Cech $\infty$-groupoid(defined in the same page) and nerves of Lie groupoids are examples of representable smooth $\infty-$ groupoids. $\endgroup$ Commented Dec 24, 2020 at 14:54

1 Answer 1

6
$\begingroup$

Kan simplicial manifolds are in the same relation to differentiable ∞-stacks (i.e., locally fibrant simplicial presheaves on the site of cartesian spaces and smooth maps) as smooth manifolds are to sheaves of sets on the same site. That is to say, Kan simplicial manifolds can be seen as the ∞-categorification of manifolds.

Some important examples include:

  1. Any ordinary manifold, interpreted as a constant simplicial object.

  2. The nerve of a Lie groupoid. In particular, the delooping of any Lie group, which represents principal bundles with this Lie group as a structure group.

  3. The Dold–Kan functor Γ applied to any nonnegatively graded chain complex of abelian Lie groups.

  4. In particular, applying Γ to the chain complexes U(1)[n], we get the Kan simplicial manifold representing bundle (n-1)-gerbes.

  5. The nonabelian analogue of Γ applied to any crossed module whose two constituent groups are Lie groups and the involved homomorphisms and actions are smooth.

  6. The nonabelian analogue of Γ applied to any (hyper)crossed complex whose constituent groupoids are Lie groupoids and the involved homomorphisms and actions are smooth.

  7. As a special case of the previous example, any simplicial Lie group is a Kan simplicial manifold.

$\endgroup$
11
  • $\begingroup$ Thank you very much Sir for the answer. Sir, does that mean that in each of these examples $Hom(\wedge_{j}^m,X)$ is a smooth manifold? (Since we need the restriction maps $Hom(\Delta^{m},X) \rightarrow Hom(\wedge^{m}_{j}, X)$ to be surjective submersions for all $m \in \mathbb{N} \cup \lbrace 0 \rbrace $ and $0 \leq j \leq m$ by definition of Kan simplicial manifolds. Is there any canonical smooth manifold structure on the Hom sets $Hom(\wedge_{j}^m,X)$? (Here $X$ is the Kan simplicial manifold) $\endgroup$ Commented Dec 24, 2020 at 19:26
  • 1
    $\begingroup$ @AdittyaChaudhuri: Yes, Hom(Λ^m_j, X) is a smooth manifold for any Kan simplicial manifold X, in particular, it is a smooth manifold in all examples given above. (Do note that Λ is the capital Greek letter lambda, not the wedge sign ∧.) This follows from the fact that Hom(Λ^m_j, X) can be presented as the domain of a finite sequence of base changes of surjective submersions. Since surjective submersions of smooth manifolds are closed under base changes along arbitrary smooth maps, this establishes the claim. $\endgroup$ Commented Dec 24, 2020 at 19:34
  • 1
    $\begingroup$ @AdittyaChaudhuri: This is Corollary 2.2 in arxiv.org/abs/math/0609420v6 $\endgroup$ Commented Dec 24, 2020 at 20:23
  • 1
    $\begingroup$ @AdittyaChaudhuri: Simplicial groups are a special case of Example 6. See Carrasco and Cegarro, Group-theoretic algebraic models for homotopy types. $\endgroup$ Commented Dec 24, 2020 at 20:28
  • 1
    $\begingroup$ @AdittyaChaudhuri: Yes. This also follows directly from the fact that surjective homomorphisms of Lie groups are submersions. Since any simplicial Lie group is a Kan complex, the matching maps are surjective homomorphisms of Lie groups, hence surjective submersions. $\endgroup$ Commented Dec 24, 2020 at 20:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .