# A-Paths as morphisms of Lie Algebroids $TI\longrightarrow A$?

In the paper Integrability of Lie Brackets Marius Crainic and Rui Fernandes describe obstructions to integrate a Lie algebroid to a Lie groupoid. The process of integration relies on the construction of the Weinstein groupoid which is the quotient of the space of A-paths by a suitable equivalence relation.

Given a Lie algebroid $A\stackrel{\pi}{\longrightarrow} M$, an $A$-path is a path $a:I\longrightarrow A$ of class $C^1$ such that $$\sharp a(t)=\frac{d}{dt}\pi(a(t))$$ where $\sharp:A\longrightarrow TM$ is the anchor map.

The RHS of the above equality (I believe) stands for the map $$t\longmapsto ((\pi\circ a)(t), (\pi\circ a)^\prime(t)).$$

They state:

An $A$-path can be seen as a morphism of vector bundles $a\ dt:TI\longrightarrow A$ covering $\pi\circ a:I\longrightarrow M$ and this gives a Lie algebroid morphism $TI\longrightarrow A$.

I don't understand this statement.

Firstly, what does $dt$ in $a\ dt$ mean? Is it just a notation or what?

Furthermore, can anyone explain-me with details the equivalence between A-paths and Lie algebroid morphisms $TI\longrightarrow A$?

• I think this question is not suitable for this site. Anyway $a dt$ sends $(t, \partial_t)$ to $a(t)$. Mar 30, 2016 at 12:12
• Thanks. Could you give me an explicit description of this $\partial_t$ and this $dt$? I'm a bit confused with those notations.
– PtF
Mar 30, 2016 at 12:13
• $t$ is the coordinate of the real line, $\partial_t$ is the usual differentiation as a derivation and $dt$ its dual. Maybe you should study basic differential geometry before trying Lie algebroids/grupoids. Mar 30, 2016 at 12:18
• Thanks again. Indeed, I've never had a course on differential geometry although I've already studied vector and principal bundles, so most of time I can follow all the arguments, but sometimes I get stuck on some basic stuffs. Thanks for the tip anyway.
– PtF
Mar 30, 2016 at 12:21
• @user40276 Any book recommendation for those basic parts of DG I should know?
– PtF
Mar 30, 2016 at 12:23

As for the question, from the definition of a Lie algebroid morphism, check that a vector bundle map $$F:TI \to A$$ covering $$\gamma: I \to M$$ is equivalent to a path in the algebroid $$\widetilde{\gamma}: I \to A$$. Where $$\widetilde{\gamma}(t_0)= F(d/dt|_{t_0})$$. Think of $$d/dt$$ as the cannonical section of the vector bundle $$TI \to I$$.
In order to be an algebroid morphism, we have two compatibility conditions. Compatibility with the anchor will imply that $$\widetilde{\gamma}$$ is an $$A$$ path. Compatibility with the Lie bracket will hold trivially (check).
• I know this answer is old, but for future readers: I would not say that compatibility with the Lie bracket is a trivial thing to check here. For non-base-preserving morphisms of Lie algebroids, according to the paper you linked, the definition of Lie bracket compatibility is surprisingly involved, and only becomes simple if one knows that there is a pushforward of vector fields along $\gamma$, which will not be the case in general, not even if $\gamma$ is injective. To me it seems there are a few more steps one has to consider. Oct 25, 2018 at 12:31
• Lukas, you are right about the computating being a little more delicate than I let on. My suggestion on how to do this calculation (in a way that is useful even for the more complicated examples) is to show that the failure of a anchor preserving vector bundle map $F \colon A \to B$ covering $f \colon M \to N$ to be a Lie algebroid homomorphism can be interpreted as a section of $\wedge^2 A^* \otimes f^* B$. Showing this basically amounts to showing that the complicated looking equation defining bracket compatibility is $C^\infty(M)$-linear. Jun 25, 2020 at 14:32