# 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 $$\theta$$.
• 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. – David Roberts Jan 19 at 9:26