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Let $A\longrightarrow M$ and $B\longrightarrow N$ be two Lie algebroides and $\Phi:A\longrightarrow B$ a morphism of Lie algebroids covering $\phi:M\longrightarrow N$. Let $\alpha, \beta\in \Gamma(A)$. If there are $\alpha^\prime, \beta^\prime\in \Gamma(B)$ such that $$\Phi\circ \alpha=\alpha^\prime\circ \phi\quad \textrm{e}\quad \Phi\circ \beta=\beta^\prime\circ \phi,\quad\quad (1)$$ then $$\Phi\circ [\alpha, \beta]=[\alpha^\prime, \beta^\prime]\circ \phi.$$ This just follows from the definition of Lie algebroid morphism.

Now, fix a point $x\in M$. Suppose there are $\alpha^\prime, \beta^\prime\in \Gamma(B)$, which may vary with $x$, such that $$\Phi(\alpha(x))=\alpha^\prime(\phi(x))\quad \textrm{and}\quad \Phi(\beta(x))=\beta^\prime(\phi(x)).\quad\quad (2)$$ Is it true that $$\Phi([\alpha, \beta](x))=[\alpha^\prime, \beta^\prime](\phi(x))?\quad\quad (3)$$

Notice that, differently from $(1)$, $(2)$ is a equality in a single point as well as $(3)$.

I do believe this is true but I don't have any idea how to show it.

Thanks

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  • $\begingroup$ what is the definition of morphism of Lie algberoid? $\endgroup$
    – Dennise
    Commented Mar 30, 2017 at 19:49
  • $\begingroup$ The Lie bracket $[-,-]$ on the sections of a Lie algebroid is a first order differential operator in each argument. So I don't think your condition (2) is sufficient to imply (3). Suppose there are $\alpha'', \beta'' \in \Gamma(B)$ that do intertwine with $\alpha, \beta$, not just at $x$. Then you are asking whether $[\alpha',\beta'](\phi(x)) = [\alpha'',\beta''](\phi(x))$. In general, this equality holds when $\alpha'$ and $\alpha''$ agree both in their values and first derivatives at $x$, same for $\beta'$ and $\beta''$. But your condition (2) is not sufficient to guarantee that. $\endgroup$
    – Igor Khavkine
    Commented Mar 30, 2017 at 21:33
  • $\begingroup$ @Dennise, a morphism of Lie algebroids is defined to preserve the "anchor" and induce a Lie algebra homomorphism on the sections. A quick definition is here. Further basic queries should probably be put in their own question, perhaps over at math.SE. $\endgroup$
    – Igor Khavkine
    Commented Mar 31, 2017 at 9:15
  • $\begingroup$ Just stumbled over these comments, and wanted to mention that the definition of a Lie algebroid morphism is more difficult if one considers Lie algebroids over different base spaces, as in general one cannot pushforward sections. This is no problem in this question, where we assume being able to pushforward things properly, but it is if one wants a general definition. For future adventurers looking for the proper definition, for example Fernandes and Crainic mention one in arxiv.org/abs/math/0611259 (Igor's source does not give one, it seems to me) $\endgroup$
    – user126256
    Commented Oct 25, 2018 at 11:55

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