# Lie algebroid

In mathematics, Lie algebroids serve the same role in the theory of Lie groupoids that Lie algebras serve in the theory of Lie groups: reducing global problems to infinitesimal ones. Just as a Lie groupoid can be thought of as a "Lie group with many objects", a Lie algebroid is like a "Lie algebra with many objects".

More precisely, a Lie algebroid is a triple $(E, [\cdot,\cdot], \rho)$ consisting of a vector bundle $E$ over a manifold $M$, together with a Lie bracket $[\cdot,\cdot]$ on its module of sections $\Gamma (E)$ and a morphism of vector bundles $\rho: E\rightarrow TM$ called the anchor. Here $TM$ is the tangent bundle of $M$. The anchor and the bracket are to satisfy the Leibniz rule:

$[X,fY]=\rho(X)f\cdot Y + f[X,Y]$

where $X,Y \in \Gamma(E), f\in C^\infty(M)$ and $\rho(X)f$ is the derivative of $f$ along the vector field $\rho(X)$. It follows that

$\rho([X,Y])=[\rho(X),\rho(Y)]$

for all $X,Y \in \Gamma(E)$.

## Examples

• Every Lie algebra is a Lie algebroid over the one point manifold.
• The tangent bundle $TM$ of a manifold $M$ is a Lie algebroid for the Lie bracket of vector fields and the identity of $TM$ as an anchor.
• Every integrable subbundle of the tangent bundle — that is, one whose sections are closed under the Lie bracket — also defines a Lie algebroid.
• Every bundle of Lie algebras over a smooth manifold defines a Lie algebroid where the Lie bracket is defined pointwise and the anchor map is equal to zero.
• To every Lie groupoid is associated a Lie algebroid, generalizing how a Lie algebra is associated to a Lie group (see also below). For example, the Lie algebroid $TM$ comes from the pair groupoid whose objects are $M$, with one isomorphism between each pair of objects. Unfortunately, going back from a Lie algebroid to a Lie groupoid is not always possible,[1] but every Lie algebroid gives a stacky Lie groupoid.[2][3]
• Given the action of a Lie algebra g on a manifold M, the set of g -invariant vector fields on M is a Lie algebroid over the space of orbits of the action.
The space of sections of the Atiyah algebroid is the Lie algebra of G-invariant vector fields on P.

## Lie algebroid associated to a Lie groupoid

To describe the construction let us fix some notation. G is the space of morphisms of the Lie groupoid, M the space of objects, $e:M\to G$ the units and $t:G\to M$ the target map.

$T^tG=\bigcup_{p\in M}T(t^{-1}(p))\subset TG$ the t-fiber tangent space. The Lie algebroid is now the vector bundle $A:=e^*T^tG$. This inherits a bracket from G, because we can identify the M-sections into A with left-invariant vector fields on G. The anchor map then is obtained as the derivation of the source map $Ts:e^*T^tG \rightarrow TM$. Further these sections act on the smooth functions of M by identifying these with left-invariant functions on G.

As a more explicit example consider the Lie algebroid associated to the pair groupoid $G:=M\times M$. The target map is $t:G\to M: (p,q)\mapsto p$ and the units $e:M\to G: p\mapsto (p,p)$. The t-fibers are $p\times M$ and therefore $T^tG=\bigcup_{p\in M}p\times TM \subset TM\times TM$. So the Lie algebroid is the vector bundle $A:=e^*T^tG=\bigcup_{p\in M} T_pM=TM$. The extension of sections X into A to left-invariant vector fields on G is simply $\tilde X(p,q)=0\oplus X(q)$ and the extension of a smooth function f from M to a left-invariant function on G is $\tilde f(p,q)=f(q)$. Therefore the bracket on A is just the Lie bracket of tangent vector fields and the anchor map is just the identity.

Of course you could do an analog construction with the source map and right-invariant vector fields/ functions. However you get an isomorphic Lie algebroid, with the explicit isomorphism $i_*$, where $i:G\to G$ is the inverse map.