3
$\begingroup$

For a Lie group $G$, we can define a principal $G$ bundle as a submersion of manifolds $\pi:P \to X$ equipped with a free right $G$-action on $P$ that is transitive on the fibres over $X$.

What goes wrong with an analogous definition for 2-groups? For now, we can think of 2-groups as weak monoidal groupoid with monoidal inverses. The autoequivalences of a category $\operatorname{Aut}(C)$ form a 2-group. One can define a right action of a 2-group on a category $C$ as a monoidal functor from the 2-group to $\operatorname{Aut}(C)$. I will also consider only essentially finite two groups.

Now given a 2-group $G$, let $\pi:\mathfrak{P} \to \mathfrak{X}$ be a representable submersion of stacks over the category of manifolds (called $\mathrm{Man}$) equipped with the étale site. Suppose we define a "principal $G$ bundle" by an action of G on the category $\mathfrak{P}$ (on the right) such that the functor \begin{gather*} \mathfrak{P} \times G \to \mathfrak{P} \times_{\mathfrak{X}} \mathfrak{P} \\ (p,\gamma) \mapsto (p\cdot\gamma,p) \end{gather*} is an equivalence of categories over $\mathrm{Man}$. Note that the action is over $\mathrm{Man}$, i.e. $p$ and $p\cdot \gamma$ are over the same object.

I could not find such a definition in the literature. Does something go wrong? If such a definition is available, please share a reference.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ Both $\mathfrak P$ and $\mathfrak X$ are not mere categories, they are both fibred over your site. Don't you need some condition about compatibility of the action with this structure? Also, you should somehow ensure that $\mathfrak X$ is actually equivalent, via $\pi$, to the quotient of $\mathfrak P$ by the action of $G$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 12, 2023 at 5:44
  • $\begingroup$ The action is defined as a monoidal functor from the 2-group G to Aut($\mathfrak{P}$) where the automorphisms are over Man. So there is compatibility with that structure. Even the equivalence of categories is over Man. Let me add that clearly in the post. Let me think about the last point. $\endgroup$
    – Spai
    Commented Apr 12, 2023 at 7:34
  • $\begingroup$ Thanks, this is clearer to me now. Still, there is, I think, a subtlety there. On categorifying, properties turn into structures. So when you say $\operatorname{Aut}(\mathfrak P)$ is taken over Man, this actually means that objects of this category are pairs $(A,\alpha)$ where $A$ is an autoequivalence of $\mathfrak P$ and $\alpha$ is an isomorphism between $\pi_{\mathfrak P}$ and $\pi_{\mathfrak P}\circ A$, where $\pi_{\mathfrak P}$ is the fibration of $\mathfrak P$. Morphisms in Aut($\mathfrak P$) must respect those $\alpha$s. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 12, 2023 at 7:55
  • 1
    $\begingroup$ Furthermore these are not just any fibrations but the ones satisfying stack conditions, maybe these also have to be taken into account in determining Aut, I am not sure. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 12, 2023 at 7:56

3 Answers 3

Reset to default
8
$\begingroup$

The definition you are looking for is precisely Def. 6.1.5 in:

Nikolaus, Thomas; Waldorf, Konrad, Four equivalent versions of nonabelian gerbes, Pac. J. Math. 264, No. 2, 355-420 (2013). ZBL1286.55006.

This definition is based on the work of Toby Bartels (Higher gauge theory I: 2-bundles) and Christoph Wockel (Principal 2-bundles and their gauge 2-groups). The main advantage of the definition in the paper of Thomas Nikolaus and me is that it is as simple as possible: the action functor is a smooth functor (as opposed to a bibundle, or anafunctor), and the action is strict. Yet, principal 2-bundles in this sense are classified by Giraud's non-abelian cohomology, and are equivalent to bundle gerbes.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you! Indeed the definition in your article is most intuitive to me. I saw anafunctors in most definitions and wanted a "non-Morita style" bicategory of principal 2 bundles. In Definition 6.1.5, the morphisms are in terms of anafunctors. Is there a way to avoid them? $\endgroup$
    – Spai
    Commented Apr 13, 2023 at 13:56
  • $\begingroup$ I don't think so. 2-bundles related by invertible anafunctors have to be isomorphic, since they represent the same class in cohomology. Anafunctors are the correct notion of morphisms in the bicategory of Lie groupoids - it is simply natural that they appear here. $\endgroup$
    – Konrad Waldorf
    Commented Apr 14, 2023 at 14:07
  • $\begingroup$ Maybe it is helpful to remark that - under the equivalence with bundle gerbes - anafunctor-morphisms correspond to so-called "stable" morphisms. But stable morphisms are the true morphisms in the bicategory of bundle gerbes. $\endgroup$
    – Konrad Waldorf
    Commented Apr 14, 2023 at 14:14
6
$\begingroup$

A good place to start is :

  • Larry Breen. Notes on 1-and 2-gerbes. In J. Baez and J. May, editors, Towards Higher Categories, volume 152 of The IMA Volumes in Mathematics and its Applications, pages 193–235. Springer, 2009. arXiv:math/0611317

There are lots of other sources and various different approaches to the subject, but I have found Breen's notes very useful. Which approach suits you, of course, depends on your interests and background.

There are many other more recent treatments.

Remember that principal G bundles are also called G-torsors so search for 2-torsors. There are various papers with 2-torsors in their title, but note there are various levels of generalisation. For further information try the n-Lab page: principal 2-bundle and there you will find a reference to Toby Bartels, Higher gauge theory I: 2-Bundles, amongst lots of other useful sources. Which ones fit best with you is difficult to guess!!!

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you for Breen's article and the suggestion to look for 2-torsors. $\endgroup$
    – Spai
    Commented Apr 13, 2023 at 13:51
4
$\begingroup$

A modern presentation that fully covers the indicated cases can be found in the work of Nikolaus–Schreiber–Stevenson:

In particular, the proposed definition is true as stated, provided that an equivalence is understood as a (local) weak equivalence over Man.

Improve this answer
$\endgroup$

You must log in to answer this question.

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