# A-paths on Lie algebroids?

In the paper integrability of Lie brackets M. Crainic and L. R. Fernandez define the notion of A-path as follows.

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

I'm trying to make sense of this definition. What does $\frac{d}{dt}\pi(a(t))$ mean? I don't understand for it should be a map from $I$ to $TM$.

Thanks.

• $TI \cong I \times I \hookrightarrow TM$, so just restrict to the fiber over $t$ and applies $\partial_t$. This is just differentiation of a curve on a manifold – user40276 Feb 29 '16 at 13:11
• This means that you have a curve on $M$ and on $A$ that are compatible in the sense that the vector field of tangent vectors to the curve in $TM$ is the anchor applied to the curve on $A$ – user40276 Feb 29 '16 at 13:14
• Do you mean I must use the embedding $I\hookrightarrow TI$? – PtF Feb 29 '16 at 13:16
• Not exactly. You just apply the vector field $\partial_t$ to $Da : TI \rightarrow TA$. – user40276 Feb 29 '16 at 13:18
• Ops! I mean $D(\pi \circ a) : TI \rightarrow TA$. – user40276 Feb 29 '16 at 13:23