Lie algebroid associated to a vector bundle

Question

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.

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 $E$.

Answer 2

I am trying to write my understanding of "Lie algebroid of derivations $\mathcal{D}(E)\rightarrow M$ associated to a vector bundle $E\rightarrow M$".

Given a vector bundle $E\rightarrow M$, we want to associate a Lie algebroid over $M$. The first thing that comes to mind about Lie algebroid (even before a smooth map to $M$) is that there is a Lie algebra.

Given a vector bundle $E\rightarrow M$, we have a collection of vector spaces $\{E_m\}_{m\in M}$ indexed by elements of $M$. Every vector space gives a Lie algebra $\rm{End}(V)$. This collection of vector spaces $\{E_m\}_{m\in M}$ would also give a collection of Lie algebras $\{\rm{End}(E_m)\}_{m\in }$ indexed by elements of $M$. Even though fibers of vector bundles are vector spaces, it is usually not the case that the fiber of Lie algebroid is a Lie algebra. So, it is unlikely that this collection $\{\rm{End}(E_m)\}_{m\in M}$ of Lie algebras indexed over elements of $M$ would give an(y) interesting Lie algebroid over $M$. I would like to know if this is of any interest, any comments are welcome.

Given a vector bundle $E\rightarrow M$, there is another vector space associated to it, namely, the global sections $\Gamma(M,E)$. This would also give a Lie algebra $\rm{End}(\Gamma(M,E))$. But, there are two issues here:

1. Is there a vector bundle $P\rightarrow M$ for which $\Gamma(M,P)=\rm{End}(\Gamma(M,E))$.
2. Is there any nice vector bundle map $P\rightarrow TM$ that fits in the definition of Lie algebroid?

If we focus on subsets of ${\rm End}(\Gamma(M,E))$ that are $C^\infty(M)$-linear then there is the endomorphism bundle. Otherwise, it is not clear what could be a choice. Note that we should also be able to assign a map from that subspace of sections to the sections of the tangent bundle. It is not clear how do you assign a vector field on $M$ for a $C^\infty(M)$-linear map of sections.

The terms "vector field on $M$" and "linear maps $\Gamma(M,E)\rightarrow \Gamma(M,E)$ (satisfying some conditions)" together should remind the notion of connection on the vector bundle $E\rightarrow M$. Given a connection $\nabla:\Gamma(M,TM)\times \Gamma(M,E)\rightarrow \Gamma(M,E)$, for each vector field $X\in \Gamma(M,TM)$ we have the associated linear map $\nabla(X,-):\Gamma(M,E)\rightarrow \Gamma(M,E)$. (Un)forturnately, given a connection, the maps $\nabla(X,-)$ are not $C^\infty(M)$-linear. In fact, we have the property $$\nabla(X,-) (f\alpha)=f\nabla(X,-)(s)+X(f)s.$$

One can extract this property and look at maps $\Gamma(M,E)\rightarrow \Gamma(M,E)$ satisfying the property mentioned above. To be more precise, we are looking at maps $D:\Gamma(M,E)\rightarrow \Gamma(M,E)$ for which there exists a vector field $D_X\in \Gamma(M,TM)$ satisfying the condition $$D(f\alpha)=fD(\alpha)+D_X(f)\alpha.$$ These maps are called as derivative endomorphism in Mackenzie's General theory of Lie groupoids and Lie algebroids

If we denote this subset of derivative endomorphisms by $S$, we have a map $S\rightarrow \Gamma(M,TM)$, sending $D$ to $D_X$ (at this point, I do not know if for a given $D$, there can be more than one vector fields $D_X$ associated to $D$). Now, we need to know if this $S$ is the space of sections of some vector bundle.

This construction of vector bundles is done in two steps. Firstly, we look at a more general notion of first order differential operators (see regarding first order differential operator and derivative endomorphism). It is "understood" that, this first order differential operators are sections of vector bundle ${\rm Diff}^1(E)\rightarrow M$ (if you know any place that gives some details about this bundle, please leave references in comments).

This set of derivative endomorphisms is a subset of first order differential operators that has something to do with vector fields, as in the following diagram,

To "construct" the vector bundle associated to the vector space on the top left corner, an obvious first step to do is to take "pullback" with appropriate maps from ${\rm Diff}^1(E)$ and $TM$ to an appropriate manifold (that is also a vector bundle over $M$).

Whenever we want something common between two properties, we take "intersection". An appropriate notion of "intersection" is the pullback. The question is pullback with respect to which manifolds and which maps. The candidate for that is ${\rm{Hom}}(T^*M, {\rm End}(E))$. They define some bundle morphisms ${\rm Diff}^1(E)\rightarrow {\rm{Hom}}(T^*M, {\rm End}(E))$ and $TM\rightarrow {\rm{Hom}}(T^*M, {\rm End}(E))$ and declare the pullback to be the Lie algebroid of derivations $\mathcal{D}(E)\rightarrow M$ as in the following diagram,

This is the construction of Lie algebroid of derivations for a vector bundle $E\rightarrow M$.