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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.

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

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