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I was reading the paper Non abelian Differentiable gerbes (page 24) and came across notion of Lie algebra bundles associated to a Lie group bundle.

I am not comfortable with these notions and google gave http://www.pphmj.com/Images/PT11.pdf which says

In 1966, A. Douady and M. Lazard constructed a Lie group bundle $\Gamma$ (not necessarily Hausdorff) for a given Lie algebra bundle $\xi$ such that the Lie algebra bundle of a Lie group bundle $\Gamma$ is isomorphic to a given Lie algebra bundle $\xi$ in their remarkable paper.

The paper they have mentioned is

A. Douady and M. Lazard, Espaces fibres en algebre de Lie et en groupes, Invent. Math. 1 (1966), 133-151. (digizeitschriften)

I do not read that language.

Can some one give a reference in English where this result is discussed, briefly atleast.

I am guessing that Lie algebra bundle over a manifold $M$ associated to a Lie group bundle over a manifold $M$ should be something where each fibre of $x\in M$ is Lie algebra $\mathfrak{g}_x$ of Lie group $G_x$ which is fibre of $x$ of Lie group bundle.

Any comments are welcome.

Edit : I assume the notion of group bundle is same as that of the paper Notes on 1-gerbes and 2-gerbes (because this paper is closely related to the paper Non abelian differentiable gerbes mentioned above). According to that paper the following is the definition of group bundle :

the standard notion of an $X$-group scheme $G$ will correspond in a topological context to that of a bundle of groups on a space $X$. By this we mean a total space $G$ above a space $X$ that is a group in the cartesian monoidal category of spaces over $X$. In particular, the fibers $G_x$ of $G $ at points $x \in X$ are topological groups, whose group laws vary continuously with $x$.

Thus, I consider the possibility that the fibers are not necessarily isomorphic.

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    $\begingroup$ If you mean that $P \to M$ is a bundle of groups (so equipped with a fiber map $m: P \times_M P \to P$ and $i: P \to P$, satisfying fiberwise the group axioms) whose fiber is isomorphic to the Lie group $G$, then it has a section $s: M \to P$ given by sending each point to the identity in the fiber. Pulling back the fiberwise tangent bundle along $s$ gives you the vector bundle you want; taking the fiberwise differential of $m$ gives you the Lie bracket, so you have a bundle of Lie algebras, fiberwise isomorphic to $\mathfrak g$. $\endgroup$
    – mme
    Commented Jul 26, 2018 at 13:27
  • $\begingroup$ @MikeMiller bundle I am not aure if they mean there is just one Lie group such that every fibre is isomorphic to that Lie group... I guess they mean each fibre is a different Lie group, that is why the name bundle of groups I think. Assuming that is the case, you ar saying consider the section $s:M\rightarrow P$ that sends each element to identity of fibre, consider tangent bundle $TP\rightarrow P$ and pull back of this $TP\rightarrow P$ along $s:M\rightarrow P$ and then call it Lie algebra bundle associated to $P\rightarrow M$.. is this what you are saying?? $\endgroup$
    – Praphulla Koushik
    Commented Jul 26, 2018 at 14:54
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    $\begingroup$ First, the paper of Douady and Lazard is really great. Reading mathematical French is not that hard, the overlap between advanced vocabulary in the two languages is very significant. Second, why are you insisting in your comment that each fibre is a different Lie group? The notion of a fibre bundle relies on local triviality. Perhaps the paper kchmackenzie.staff.shef.ac.uk/publications/cpblgpsg.pdf (especially Def. 2.8 there) can be useful. (For what it's worth, I found that paper by looking at English-language paper citing Douady and Lazard, this is something that you can also do.) $\endgroup$
    – Vladimir Dotsenko
    Commented Mar 12, 2020 at 9:35
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    $\begingroup$ Your question was about references in English where work of Douady and Lazard is discussed. Did you look at the paper of Mackenzie I linked? Did it help to clarify your concerns? You seem to insist not just on getting help but getting help in exactly a very specific way of your choosing. That's not how one learns maths. $\endgroup$
    – Vladimir Dotsenko
    Commented Mar 12, 2020 at 12:29
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    $\begingroup$ @VladimirDotsenko I forgot to tell, checking for books/papers written in English that cited a non-English language paper is a nice way to get some idea of what is there in that paper... Thanks for the tip :) $\endgroup$
    – Praphulla Koushik
    Commented Mar 13, 2020 at 3:39

3 Answers 3

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Let me first discuss Lie algebra bundles. If I understand the question right, then we are given a smooth vector bundle $E\to M$ together with a fiber respecting smooth fiberwise bilinear mapping $[\;,\;]:E\times_M E \to E$ which is fiberwise a Lie algebra structure. The question now is: when is the Lie algebra isomorphism class of the Lie algebras $E_x$ locally constant on $M$? This is the case if the Lie algebra is rigid: This means that the orbit through the Lie algebra structure on $\mathbb R^n$ (the typical fiber) of $\operatorname{GL}(n)$ in $L^2_{\text{skew}}(\mathbb R^n\times \mathbb R^n,\mathbb R^n)$ under the action $(A,F)\mapsto A\circ F\circ (A^{-1}\times A^{-1})$ is open in the real subvariety of Lie algebra structures (i.e., also satisfying the Jacobi identity). Semisimple Lie algebras are rigid. Nilpotent and solvable ones are not, in general: These have real moduli which may change in a smooth way with $x\in M$. So the isomorphism class of the Lie algebra $E_x$ is locally constant in $M$ around a rigid Lie algebra $E_x$.

By exponentiation this result carries over to fiber bundles with a smooth Lie group structure.

For information on rigid Lie algebras see papers by Michel Goze.
One recent paper is:

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  • $\begingroup$ Sir, thank you for the answer and the link... I will see that paper.. Can you please give give an outline of associating Lie algebra bundles for Lie group bundles.. $\endgroup$
    – Praphulla Koushik
    Commented Mar 14, 2020 at 4:01
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Refer the paper by Basavavannappa S KIRANAGI(B S KIRANAGI) on Lie Algebra Bundles, Bull. Sci. Math.2 series, 102(1978), 57-62. there are many papers by him.

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    $\begingroup$ Welcome to MathOverflow! "By him" - your username implies you are the author, and you should say so explicitly. $\endgroup$
    – Glorfindel
    Commented Oct 17, 2022 at 16:35
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    $\begingroup$ Not to mention, that paper seems rather hard to find, so while it's a reasonable reference, perhaps either a link or a more accessible paper would be helpful in addition. $\endgroup$
    – David Roberts ♦
    Commented Oct 18, 2022 at 1:42
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Are you familiar with the Lie algebroid associated to a Lie groupoid? A Lie group bundle is in particular a Lie groupoid $\Gamma \rightrightarrows M$ where the source and target maps are equal, and the associated Lie algebra bundle is the Lie algebroid of this Lie groupoid.

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