# Morphism of Lie groups $\theta:G\rightarrow H$ giving an equivalence of categories $BG\rightarrow BH$?

Given a morphism of Lie groups $$\theta:G\rightarrow H$$  and a principal $$G$$ bundle $$\pi:P\rightarrow M$$ there are (at least) two ways to assign a principal $$H$$ bundle.

1. See that the morphism of Lie groups $$\theta:G\rightarrow H$$ gives an action of $$G$$ on $$H$$ by $$g.h=\theta(g).h$$. Given an action of $$G$$ on manifold (Lie group in this case) $$H$$ there is an associated fibre bundle $$P\times_G H\rightarrow M$$ with fibre $$H$$. This gives a principal $$H$$ bundle.
2. For principal bundle $$\pi:P\rightarrow M$$, we can find an open cover $$\{U_\alpha\}$$ of $$M$$ and  (transition) maps $$g_\alpha g_\beta:U_{\alpha\beta}\rightarrow G$$ satifsying the cocycle condition $$g_{\alpha\beta}g_{\beta\gamma}=g_{\alpha\gamma}$$ on $$U_\alpha\cap U_\beta\cap U_\gamma$$. Then the compositions $$\tau_{\alpha\beta}=\theta\circ g_{\alpha\beta}:U_{\alpha\beta}\rightarrow G\rightarrow H$$ also satifies the cocycle condition $$\tau_{\alpha\beta}\tau_{\beta\gamma}=\tau_{\alpha\gamma}$$ on $$U_\alpha\cap U_\beta\cap U_\gamma$$. One can then produce a principal $$H$$ bundle over $$M$$ given this open cover $$\{U_\alpha\}$$ of $$M$$ and smooth maps $$\tau_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow H$$ satisfying the cocycle condition. This gives a principal $$H$$ bundle.

It is a good exercise (that I have not tried) to check that principal $$H$$ bundles obtained from above two methods are (naturally) isomorphic.

Given a Lie group $$G$$, let $$BG$$ denote the category of principal $$G$$ bundles. Objects are principal $$G$$ bundles and morphisms are $$G$$-equivariant morphisms.

Given a morphism of Lie groups $$\theta:G\rightarrow H$$, above construction gives a functor (at the level of objects) $$B\theta:BG\rightarrow BH$$. It is not difficult to see that, a $$G$$-equivarint map induce a $$H$$-equivariant map. This gives a functor.

I am trying to understand what can we say about $$\theta:G\rightarrow H$$ if we know that $$B\theta:BG\rightarrow BH$$ is an equivalence of categories? Does it have to be a diffeomorphism? Any comments are welcome.

It is a diffeomorphism, since there is an equivalence of bicategories $$DifferentiableStacks \simeq LieGroupoids[W^{-1}]$$ where the RHS is the bicategorical localisation of the usual 2-category of Lie groupoids a la Pronk. Because of the special nature of the domain of your $$\theta$$, namely it is a Lie groupoid with one object (call it $$\mathbf{B}G$$; note the boldface B!), then in fact $$LieGroupoids[W^{-1}](\mathbf{B}G,\mathbf{B}H) \simeq LieGroupoids(\mathbf{B}G,\mathbf{B}H)$$ is an equivalence of categories. The latter category is isomorphic to the category whose objects are homomorphisms $$G\to H$$ and whose 2-arrows are elements of $$H$$, acting by conjugation on homomorphisms. Tracing what happens to the quasi-inverse $$\phi\colon BH\to BG$$ to $$B\theta$$ through these equivalences of categories one gets a homomorphism $$\psi\colon H\to G$$ such that $$\psi\circ \theta$$ is conjugate to the identity map on $$G$$, and $$\theta\circ \psi$$ is conjugate to the identity map on $$H$$. This is enough to know that $$\theta$$ is a diffeomorphism, since we can pre- and post-compose $$\psi$$ with the inner automorphisms to get an inverse for $$\theta$$.

• a localisation of bicategories as defined by Pronk, and as explained at the linked page. Jan 19, 2019 at 8:36
• Thanks. Can this be done with out using "bicategorical localisation"? I thought this is some simple question which has very little to do with stacks and all that... Is the question that serious that it has to be treated with localization and all that? Jan 19, 2019 at 9:23
• Sure, you can probably prove it other ways, but this is the quickest way I know with the tools I have that I can prove $\theta$ is a diffeomorphism. $BG$ is a stack, after all, so you might as well use that. YMMV. Jan 19, 2019 at 9:26
• Oh. Ok Ok. :) I am not comfortable with Localization and all that.. I will read it :) You are also teaching me English... Google says YMMV means "your mileage may vary" :D Thanks Jan 19, 2019 at 9:37