# 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 $a(t)$. – user40276 Mar 30 '16 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 '16 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. – user40276 Mar 30 '16 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 '16 at 12:21
• @user40276 Any book recommendation for those basic parts of DG I should know? – PtF Mar 30 '16 at 12:23

## 1 Answer

You may find it useful to look at Lectures on the Integrability of Lie Brackets by the same authors. I believe it takes a slower approach to the subject you seem to be reading about.

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).

This question is probably better suited for math stackexchange, however. If you have more questions about learning this topic it may be better if you ask them there.

• 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. – Lukas Miaskiwskyi Oct 25 '18 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. – Joel Villatoro Jun 25 '20 at 14:32