First examples of Lie-Rinehart algebras that are not coming from Lie algebroids

Question

I heard the idea of a Lie-Rinehart algebra first time from an algebraist.

I noticed there is a similarity between description of Lie algebroid on a manifold and the algebraic notion of Lie-Rinehart algebras.

Just for giving some context, I had to mention some background. This is only my interpretation. Any correct idea here is due to that algebraist and any wrong idea is due to me.

Please bear with me.

The question comes in the end.

Any notion of an "algebra" comes with two binary operations:

• $A\times A\rightarrow A$, called the addition map,
• $A\times A\rightarrow A$, called the multiplication map.

Two properties that are assumed for addition map are that of commutativity and associativity.

These two properties are however not assumed for the multiplicative map. That is where the notion of noncommutative algebras and non associative algebras comes into picture. Note that, "non commutative" means not necessarily commutative and "nonassociative" means not necessarily associative.

One interesting case of nonassociative algebras is that of a Lie algebra, where, the failure of the associativity is controlled by the Jacobi identity.

Now comes the notion of Lie-Rinehart algebra. There may be a very intuitive and algebraic way to think about Lie-Rinehart algebra, but, if you are comfortable with the notion of Lie algebroid over a manifold, I think it is better to approach Lie-Rinehart algebra from Lie algebroid point of view.

Recall that, given a manifold $M$, a Lie algebroid over $M$ consists of,

• a vector bundle $A\rightarrow M$,

• a morphism of vector bundles $\rho:A\rightarrow TM$,

• a Lie algebra structure on $\Gamma(M,A)$

satisfying the obvious conditions.

Let us seperate out the algebraic ideas here. We have

• a Lie algebra $L=\Gamma(M,A)$

• an associative algebra $\mathcal{O}=C^\infty(M)$

• $\mathcal{O}$-module structure on $L$ given by $(f,s)\mapsto fs$ where $fs:M\rightarrow A$ is given by $m\mapsto f(m)s(m)$ for $m\in M, f\in \mathcal{O}$, and $s\in L$,

• a map $L=\Gamma(M,A)\rightarrow Der(C^\infty(M))=\mathfrak{X}(M)$ induced from the anchor map $A\rightarrow TM$ by taking sections $\Gamma(M,A)\rightarrow \Gamma(M,TM)=\mathfrak{X}(M)=Der(C^\infty(M))$,

satisfying certain conditions.

With out mentioning the reference to manifold, writing down the above algebraic data, along with "certain conditions" gives the notion of Lie-Rinehart algebra.

A Lie-Rinehart algebra consists of,

• a Lie algebra $L$

• an associative (some people ask it to be commutative) algebra $A$,

• a map $\tau : A\times L\rightarrow L$ giving an $A$-module structure on $L$,

• a map $\rho : L\times A\rightarrow A$ giving an $L$-module structure on $A$,

such that the following conditions are satisfied:

• the map $\rho : L\times A\rightarrow A$ gives a morphism of Lie algebras $L\rightarrow Der(A)$

• the map $\tau : A\times L\rightarrow L$ gives a morphism of associative algebras $A\rightarrow End(L)$,

• the maps $ \rho,\tau$ are compatible, in the sense that, $[u,\tau(a,v)]=\tau(a,[u,v])+\tau(\rho(u,a),v)$ for all $u,v\in L$ and $a\in A$.

Given a smooth manifold $M$, and a Lie algebroid $A\rightarrow M$, we get a Lie-Rinehart algebra $(C^\infty(M), \Gamma(M,A)$.

Question : Given a manifold $M$, are there any (easy to understand) Lie-Rinehart algebras whose associative algebra part is $C^\infty(M)$ and the Lie algebra is not $\Gamma(M,A)$ for some Lie algebroid $A\rightarrow M$?

Answer 1

Yes, there are many such Lie-Rinehart algebras. To understand such examples we need to recall some classical geometric results. Note that a $C^\infty(M)$- module is the space of global sections of a smooth vector bundle over $M$ if and only if it is finitely generated (f.g.) projective. Thus, a Lie algebroid over a smooth manifold $M$ is equivalent to a f.g. projective Lie-Rinehart algebra over $C^\infty(M)$. Note that, involutive subbundle of the tangent bundle (Frobenius distribution) of a smooth manifold forms a Lie algebroid in a canonical way. More generally, we consider generalized involutive distributions or singular foliations (Stefan-Sussmann distributions) in differential geometry. These objects provides Lie-Rinehart subalgebras of $Der_{\mathbb{R}}(C^\infty(M))$, which are not necessarily projective but only finitely presented over $C^\infty(M)$. Thus, in general if we consider a Lie-Rinehart subalgebras of $Der_{\mathbb{R}}(C^\infty(M)) \cong \mathfrak{X}(M)$ whose underlying $C^\infty(M)$-module structure is not f.g. projective (for example take space of logarithmic derivations or logarithmic vector fields which is associate with a principal divisor but not a free divisor), it can't be viewed as the space of global sections of a smooth Lie algebroid over $M$. One more thing, there are many geometric objects which are not smooth but singular, appears as vanishing set or zero set of (smooth/ holomorphic/algebraic) functions. To study calculus on these objects we need to consider an algebro-geometric approach (since tangent bundles are not available) of Lie algebroids, which is quite different from the classical one (smooth context).

Answer 2

This point of view is based on the talk of Joel Villatoro titled paths in Lie-Rinehart algebras.

I may be misunderstanding what Joel Villatoro is mentioning. Correct me if I am saying something wrong.

The category of Lie algebroids is not closed under the direct/inverse limit. The infinite jet construction does not give a Lie algebroid structure on $M$, but, gives a Lie-Rinehart algebra over $C^\infty(M)$.

In this sense Lie-Rinehart algebras helps us to do more geometry.