# 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.

## Description

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

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

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

${\displaystyle \rho ([X,Y])=[\rho (X),\rho (Y)]}$

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

## Examples

• Every Lie algebra is a Lie algebroid over the one point manifold.
• The tangent bundle ${\displaystyle TM}$  of a manifold ${\displaystyle M}$  is a Lie algebroid for the Lie bracket of vector fields and the identity of ${\displaystyle TM}$  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 ${\displaystyle TM}$  comes from the pair groupoid whose objects are ${\displaystyle M}$ , 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 Atiyah algebroid of a principal G-bundle P over a manifold M is a Lie algebroid with short exact sequence:
${\displaystyle 0\to P\times _{G}{\mathfrak {g}}\to TP/G{\xrightarrow {\rho }}TM\to 0.}$
The space of sections of the Atiyah algebroid is the Lie algebra of G-invariant vector fields on P.
• A Poisson Lie algebroid is associated to a Poisson manifold by taking E to be the cotangent bundle. The anchor map is given by the Poisson bivector. This can be seen in a Lie bialgebroid.

## 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, ${\displaystyle e:M\to G}$  the units and ${\displaystyle t:G\to M}$  the target map.

${\displaystyle T^{t}G=\bigcup _{p\in M}T(t^{-1}(p))\subset TG}$  the t-fiber tangent space. The Lie algebroid is now the vector bundle ${\displaystyle A:=e^{*}T^{t}G}$ . 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 ${\displaystyle Ts:e^{*}T^{t}G\rightarrow TM}$ . 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 ${\displaystyle G:=M\times M}$ . The target map is ${\displaystyle t:G\to M:(p,q)\mapsto p}$  and the units ${\displaystyle e:M\to G:p\mapsto (p,p)}$ . The t-fibers are ${\displaystyle p\times M}$  and therefore ${\displaystyle T^{t}G=\bigcup _{p\in M}p\times TM\subset TM\times TM}$ . So the Lie algebroid is the vector bundle ${\displaystyle A:=e^{*}T^{t}G=\bigcup _{p\in M}T_{p}M=TM}$ . The extension of sections X into A to left-invariant vector fields on G is simply ${\displaystyle {\tilde {X}}(p,q)=0\oplus X(q)}$  and the extension of a smooth function f from M to a left-invariant function on G is ${\displaystyle {\tilde {f}}(p,q)=f(q)}$ . 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 ${\displaystyle i_{*}}$ , where ${\displaystyle i:G\to G}$  is the inverse map.

### Example

Consider the Lie groupoid

${\displaystyle \mathbb {R} ^{2}\times U(1)\rightrightarrows \mathbb {R} ^{2}}$

where the target map sends

${\displaystyle ((x,y),e^{i\theta })\mapsto {\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}}$

Notice that there are two cases for the fibers of ${\displaystyle T^{t}(\mathbb {R} ^{2}\times U(1))}$ :

{\displaystyle {\begin{aligned}t^{-1}(0)\cong &U(1)\\t^{-1}(p)\cong &\{(a,u)\in \mathbb {R} ^{2}\times U(1):ua=p\}\end{aligned}}}

This demonstrating that there is a stabilizer of ${\displaystyle U(1)}$  over the origin and stabilizer-free ${\displaystyle U(1)}$ -orbits everywhere else. The tangent bundle over every ${\displaystyle t^{-1}(p)}$  is then trivial, hence the pullback ${\displaystyle e^{*}(T^{t}(\mathbb {R} ^{2}\times U(1)))}$  is a trivial line bundle.