Let $E\rightarrow M$ be a vector bundle.

Kirill Mackenzie in the book General theory of Lie groupoids and Lie algebroids associates a Lie algebroid to $E\rightarrow M$ in the following steps:

  1. talk about zero-th and first order differential operators on $E\rightarrow M$, which are some nice maps of sections $\Gamma(M,E)\rightarrow \Gamma(M,E)$.
  2. realise these first order differential operators as sections of a vector bundle $\text{Diff}^1(M)\rightarrow M$.
  3. do some pullback along some morphism of vector bundles. Call the pullback as the Lie algebroid of derivations on $E\rightarrow M$.

The book also talks about Linear vector fields and says how these are related to the Lie algebroid of derivations.

I do not fully understand the idea of linear vector fields and first/zeroth order differential operators. But, that is not what I want to ask.

Given a principal bundle $P\rightarrow M$, one can consider the morphism of tangent bundles $TP\rightarrow TM$. As the action of $G$ on $TP$ is nice, this would induce a morphism of vector bundles $(TP)/G\rightarrow TM$. The Lie bracket on $\mathfrak{X}(P)=\Gamma(P,TP)$ is nice enough to induce a Lie bracket on $\Gamma(M,(TP)/G)$. Thus, we have a vector bundle over $M$, a morphism of vector bundles to the tangent bundle of $M$, a Lie bracket on sections of this vector bundle, and some more extra nice properties. This is a Lie algebroids. This vector bundle $(TP)/G\rightarrow M$ is called the Atiyah vector bundle and the Lie algebroid is called the Atiyah Lie algebroid.

Given a vector bundle $E\rightarrow M$, one can consider the associated principal bundle $GL(E)\rightarrow M$ and consider the construction mentioned above which would result in a Lie algebroid over $M$.

  1. Are these two procedures giving same Lie algebroid? I think the answer to this is positive, but I am not sure.

  2. Why would one want to involve differential operators and linear vector fields to associate a Lie algebroid when there is an easier way of considering the associated Atiyah Lie algebroid?

Even though I mentioned that I am not asking about the idea behind linear vector fields/differential operators, please see if you can say a few lines which might make my understand better.

  • $\begingroup$ As for the second question, on reading it I thought that Mackenzie was simply mimicking the construction of Lie algebra of module valued derivations, extended to the geometric setting of Lie algebroids. $\endgroup$ Dec 11, 2021 at 15:57
  • $\begingroup$ @NicolaCiccoli Hi. I am now aware of the notion of Lie algebra of module valued derivations... I only know about Lie algebra of derivations (of an $R$-algebra).. Can you please suggest some reference that discuss about this.. $\endgroup$ Dec 12, 2021 at 5:07

1 Answer 1


Question 1. The two procedures indeed give the same Lie algebroid. One possible way of seeing this is by considering the flows of vector fields: a section of $T(GL(E))/GL(n)$ is a vector field on the frame bundle that is $GL(n)$-invariant, and this means that its flow is by $GL(n)$-equivariant diffeomorphisms. But this is, in turn, equivalent to a family of vector bundle isomorphisms of $E$. When we differentiate this family, we obtain a linear vector field on $E$.

More conceptually, the categories of $GL(n)$-principal bundles and rank $n$-vector bundles (where we only consider automorphisms) are equivalent. Therefore the symmetries are the same, and in particular, the infinitesimal symmetries are the same. But these are precisely the $GL(n)$-invariant vector fields on the one hand, and linear vector fields on the other.

The link to differential operators is also not so difficult to see: a section of $E^{*}$ is the same as a linear function on $E$. So given a section of $E^*$, we may differentiate it by a vector field on $E$. In general this will not return a section of $E^*$, rather it will just be some function on $E$. Linear vector fields are precisely the vector fields on $E$ that send linear functions to linear functions. So we can think of linear vector fields as differential operators for $E^*$ (and by dualizing, for $E$ as well).

Question 2. The question of whether principal bundles or differential operators are 'easier' is probably a matter of one's background. But the Atiyah algebroid is useful for studying both.

I believe the reason Atiyah originally introduced the Atiyah algebroid was to study the question of the existence of holomorphic connections. Note that connections $\nabla$ are a type of differential operator: given a vector field $X$, then we get a first-order differential operator $\nabla_{X} : \Gamma(E) \to \Gamma(E)$, whose symbol is the vector field $X$.

The Atiyah algebroid sits in a nice short exact sequence of Lie algebroids:

$0 \to End(E) \to At(E) \to TM \to 0. $

The right-hand map is the anchor. From the perspective of principal bundles, this is the map you constructed in your question, and the kernel is the 'vertical bundle'. From the perspective of differential operators, where we view sections of the Atiyah algebroid as differential operators on $E$ of order $1$, the anchor map becomes the symbol. Hence, the kernel $End(E)$, corresponds to the differential operators of order $0$.

Looking back at the definition of a connection, you can see that it is precisely giving you a section of the symbol (i.e. anchor) map. So the existence of a connection on $E$ is exactly the same thing as an isomorphism $At(E) \cong End(E) \oplus TM$ (respecting the extension structure). In the world of smooth manifolds, short exact sequences of bundles always split: every bundle admits a connection. In the holomorphic world this is no longer the case: the existence of holomorphic connections is measured by the extension class of $At(E)$, which is a cohomology class in $H^{1}(T^{*}M \otimes End(E))$, called the Atiyah class of $E$.

  • $\begingroup$ Hi. Thanks for your answer.. I understand most of the first half of it.. Please see if you can suggest some reference to read more about what you said... For the second part, I am not sure how to see it... "this is the map you constructed in your question".. What are you referring to here? $\endgroup$ Dec 12, 2021 at 5:14
  • $\begingroup$ I'm not sure about what the best reference is. About the map in the second part: the anchor is the map $\rho: At(E) \to TM$. When we view $At(E) \cong T(GL(E))/GL(n)$, this is the derivative of the projection map $GL(E) \to M$ (this is what you wrote as $TP/G \to TM)$. When we view $At(E)$ as the space of first-order differential operators, then $\rho(D)$ is the symbol of $D$. In fact, $At(E)$ doesn't contain all first-order differential operators, only the ones whose symbol is a vector field. $\endgroup$ Dec 12, 2021 at 15:04

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