Let $\mathcal{G} = G_1 \rightrightarrows G_0$ be a Lie Groupoid (although I am also interested in groupoids internal to other sites), the stack associated to $\mathcal{G}$, which is sometimes denoted $B \mathcal{G}$ is defined as the projection from the category of principal $\mathcal{G}$-bundles to $\mathsf{Man}$. However it is also well known that we can use the functor $r_1: \mathsf{LieGpd} \hookrightarrow \mathsf{Pre}(\mathsf{Man}, \mathsf{Gpd})$, defined as $$M \mapsto \left( C^\infty(M, G_1) \rightrightarrows C^\infty(M, G_0) \right)$$ and that the stackification of this presheaf of groupoids is equivalent to $B \mathcal{G}$.

However as Chris Schommer-Pries pointed out, $r_1(G \rightrightarrows *)$ is not a stack on $\mathsf{Man}$. It is however a stack on $\mathsf{Cart}$ the site of cartesian spaces. This can be seen by showing that if we let $\widehat{r}G = r_1(G \rightrightarrows *)$, then for $U \in \mathsf{Cart}$ equipped with a differentiably good open cover $\{ U_i \to U \}$, the canonical map

$$\widehat{r}G(U) \to \text{holim} \left( \prod_i \widehat{r}G(U_i) \rightrightarrows \prod_{i,j} \widehat{r}G(U_{ij}) \mathrel{\substack{\textstyle\rightarrow\\[-0.6ex] \textstyle\rightarrow \\[-0.6ex] \textstyle\rightarrow}} \prod_{i,j,k} \widehat{r}G(U_{ijk}) \right)$$ is an equivalence of groupoids. One can see this by using a specific model of $\text{holim}$ from Hollander's Homotopy Theory of Stacks Corollary 2.11, and then recognizing the right hand side as the groupoid of Principal $G$-bundles on $U$. Since $U$ is a Cartesian space, this groupoid will be connected and this will prove that the map is an equivalence.

My question then is the following:

Is there a necessary and sufficient condition that one could put on a Lie groupoid $\mathcal{G}$ such that $r_1(\mathcal{G})$ is already a stack on $\mathsf{Cart}$? I suspect that it should be any Lie groupoid, we would need to follow the same logic as above using principal groupoid bundles, i.e. building principal groupoid bundles from local data. Perhaps this has been done somewhere?

  • 1
    $\begingroup$ In the case $\mathcal{G} = (G \rightrightarrows *)$, the groupoid valued presheaf $r_1(\mathcal{G})$ is not a stack and is not $B\mathcal{G}$. You can see this because for any M, the value of the presheaf is a groupoid with only one object, while the groupoid $B\mathcal{G}(M)$ has distinct isomorphism classes of objects for each isomorphism class of $G$-bundle on $M$. In general you will always have to stackify $r_1(\mathcal{G})$ unless $\mathcal{G}$ is trivial (only identity morphisms). $\endgroup$ Sep 22, 2021 at 21:10
  • $\begingroup$ I apologize, I mean to set the underlying site to Cart, not Man. See the edit above. $\endgroup$ Sep 23, 2021 at 0:09

1 Answer 1


Note that $\hat r\mathcal{G}$ is always a prestack. This is basically equivalent to the fact that $C^\infty(-,G_1)$ is a sheaf, and it means that your functor $$ \hat r\mathcal{G}(U) \to \mathrm{holim}(...) $$ can only fail to be essentially surjective.

Essential surjectivity of this functor is equivalent to saying that every principal $\mathcal{G}$-bundle over a Cartesian space is trivializable.

A sufficient condition to garantee this is that your Lie groupoid $\mathcal{G}$ is an action groupoid (for an action of a Lie group $H$ on a smooth manifold $X$). In particular, this is true when your Lie groupoid is a Lie 2-group.

Namely, principal bundles for an action groupoid $X/\!/H$ are at the same time ordinary principal $H$-bundles (with additional structure). Any ordinary principal bundle has a global section over any Cartesian space. But a section is already enough to trivialize the $X/\!/H$-bundle.

You find these statements in Section 2.2. of:

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

  • $\begingroup$ Ah very nice! Thank you $\endgroup$ Sep 23, 2021 at 10:10

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