∞-Lie theory (higher geometry)
Let $X$ by a differentiable manifold. The tangent Lie algebroid $T X$ of $X$ is the Lie algebroid that corresponds – in the sense of Lie theory – to
the codiscrete groupoid $X \times X$ of $X$.
the fundamental groupoid $\Pi_1(X)$ of $X$.
When the tangent Lie algebroid is regarded as a Lie ∞-algebroid it corresponds to
The space of objects of $T X$ is $X$ itself and its elements in degree 1 are the vectors on $X$, i.e. the elements in the tangent bundle of $X$. These are to be thouhght of as the infinitesimal paths in $X$.
So to some extent the tangent Lie algebroid is the tangent bundle $T X$ of $X$. More precisely, when using the definition of a Lie algebroid $E$ over $X$ as a diagram
where $\rho$ is the map called the anchor map, the tangent Lie algebroid is that whose anchor map is the identity map
Therefore the tangent Lie algebroid of $X$ is usually denoted $T X$, just as the tangent bundle itself.
The Chevalley-Eilenberg algebra of $T X$ is correspondingly fundamental: it is the deRham dg-algebra of differential forms on $X$:
So regarded as an NQ-supermanifold the tangent Lie algebroid is the shifted tangent bundle $\Pi T X$ equipped with its canonical odd vector field.
For $\mathfrak{a}$ any other Lie algebroid or Lie infinity-algebroid, a morphism of Lie infinity-algebroids
is flat $\mathfrak{a}$-valued differential form datum .
For instance if $\mathfrak{a} = \mathfrak{g}$ is an ordinary Lie algebra then a morphism $T X \to \mathfrak{g}$ is a flat Lie algebra valued differential form on $X$, i.e. an element $A \in \Omega^1(X) \otimes \mathfrak{g}$ such that $d A + [-,-](A \wedge A) = 0$.
If $\mathfrak{g} = \mathfrak{u}(1) = \mathbb{R}$ is the abelian 1-dimensional Lie algebra, then a morphism $T X \to \mathfrak{g}$ is just a closed differential 1-form on $X$.
More on $\infty$-Lie algebroid-valued differential forms is (eventually) at ∞-Lie algebroid valued differential forms.
The higher-order version of tangent Lie algebroids are jet bundle D-schemes.
One of the earliest reference seems to be