$\DeclareMathOperator\At{At}\DeclareMathOperator\Obj{Obj}\DeclareMathOperator\Mor{Mor}$According to https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#idea the Atiyah Lie groupoid $\At(P)$ of a principal $G$ bundle $\pi:P \rightarrow X$ is a category for which $$\Obj(\At(P))=\lbrace \pi^{-1}(x): x \in X \rbrace$$ and $$\Mor(\At(P))=\big\lbrace f:\pi^{-1}(x)\rightarrow \pi^{-1}(y): \text{$f$ is a $G$ equivariant morphism}\big\rbrace.$$ Structure maps of this category are easy to guess. Now it is easy to see that $\At(P)$ is indeed a groupoid.

Although it is mentioned in https://ncatlab.org/nlab/show/Atiyah+Lie+groupoid#idea that the Atiyah Lie groupoid is indeed a Lie groupoid, I am not able to guess appropriate smooth structures on $\Obj(\At(P))$ and $\Mor(\At(P))$ such that the source and the target maps are surjective submersions and other structure maps are smooth.

Is there any natural choice of such smooth structures on both $\Obj(\At(P))$ and $\Mor(\At(P))$ such that $At(P)$ is a Lie groupoid so that if someone talks about the Atiyah Lie groupoid of a principal $G$ bundle then he/she is precisely assuming those natural choice of smooth structures on $\Obj(\At(P))$ and $\Mor(\At(P))$?

I would also be very grateful if someone point me to any literature in this direction.

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    $\begingroup$ @LSpice I was asking about the definition given in the section idea ncatlab.org/nlab/show/Atiyah+Lie+groupoid#idea not ncatlab.org/nlab/show/…. So are you saying both are actually same notion and the definition given in the idea section is just an informal notion? $\endgroup$ Jul 27, 2020 at 22:26
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    $\begingroup$ Yes, that is what I am saying. $\endgroup$
    – LSpice
    Jul 27, 2020 at 22:27
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    $\begingroup$ There is no definition in the 'idea' section; it is just an idea, and, as not rigorously defined, cannot be checked against a rigorous definition. $\endgroup$
    – LSpice
    Jul 27, 2020 at 22:38
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    $\begingroup$ @LSpice Ok I got your point. Thanks. $\endgroup$ Jul 27, 2020 at 22:47
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    $\begingroup$ The objects as defined on the nLab and on Wikipedia give isomorphic sets. That the morphism sets are the same is less obvious, but it boils down to knowing that a map between principal homogeneous $G$-spaces is determined entirely by its value at a single point. It's worth thinking about the case of a trivialisable bundle first. $\endgroup$
    – David Roberts
    Jul 27, 2020 at 23:42

1 Answer 1


Contrary to what is claimed in the comments, I would argue that the definition given in nLab's Idea section is rigorous enough to be an actual definition in a research-level paper, possibly with an additional phrase thrown in like “The sets of objects and morphisms are equipped with the obvious smooth structures that turn this groupoid into a Lie groupoid.”

Let's see how these smooth structures are constructed. Recall that the set of objects is $\{π^{−1}(x)\mid x∈X\}$, i.e., the set of fibers of $P$. Fibers are in a bijective correspondence with points in the base $X$, and the latter is a smooth manifold.

The set of morphisms is $\{f\colon π^{−1}(x)→π^{−1}(y)\mid \text{$f$ is a $G$-equivariant morphism}\}$. A morphism between two $G$-torsors $U→V$ is uniquely determined by its value $v∈V$ at some point $u∈U$. That is, for any pair $(u,v)∈U⨯V$ there is exactly one morphism that sends $u↦v$. The pair $(gu,gv)$ gives rise to the same morphism $U→V$ as $(u,v)$. It is also easy to see that the converse is true: $(u,v)$ and $(u',v')$ yield the same morphism if there is $g∈G$ such that $(u',v')=(gu,gv)$. Thus, the set of morphisms $U→V$ is $(U⨯V)/G$, where $G$ acts on $U⨯V$ via $g(u,v)=(gu,gv)$. The action of $G$ on $U⨯V$ is a smooth free proper action, so the quotient $(U⨯V)/G$ is a smooth manifold and the quotient map $U⨯V→(U⨯V)/G$ is a submersion.

From here, we see that the set of all morphisms is $(P⨯P)/G$ and therefore possesses a canonical smooth structure. The source and target maps are surjective submersions by the 2-out-of-3 property.

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    $\begingroup$ Thank you Sir very much for the answer! $\endgroup$ Jul 28, 2020 at 4:58
  • $\begingroup$ Calling a fibre $X$ if the base is already called $X$ is a bit awkward. I suggest to pick different letters in the middle paragraph, maybe $F_x$ and $F_y$. $\endgroup$ Jul 28, 2020 at 15:02

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