Consider a map $f:M\to N$. We say that a vector field $X\in \mathfrak{X}(M)$ projects on $Y\in \mathfrak{X}(N)$ if they are $f$-related vector fields. We will say that $X\in \mathfrak{X}(M)$ is projectable if it projects to some vector field on $N$. See @saunders1989geometry page 67.
By the way, if $f$ is a smooth submersion, giving a vector field $Y\in \mathfrak{X}(N)$ there is at least one projectable vector field $X$ which is $f$-related to $Y$. See this answer in math.stackexchange
Given an involutive distribution $\mathcal{X}$ (i.e., a foliation) on a manifold $M$, a vector field $Y$ is said to be projectable with respect to $\mathcal{X}$ (in the sense of @paola.frobenius) if for every $X\in \Gamma(M,\mathcal{X})$ we have
$$ [Y,X]\in \Gamma(M,\mathcal{X}) $$They are also known as foliate vector fields, basic vector fields, base-like vector fields, foliated vector fields,... (see [Molino 1988] page 35)
In particular, it may happen that $\mathcal{X}=\mathcal{S}(\{X\})$, and the requirement becomes
$$ [Y,X]=\rho X $$We can say that $Y$ is a $X$-projectable vector field.
But, what is happening when we reduce an ODE with a symmetry? I.e., when we have a vector field $A$ associated to an ODE and
$$ [A,X]=\xi A $$The point here is that the vector field is not $X$-projectable, but the distribution $\{A\}$ is. The projection converts the vector field $A$ in a nonlocal vector field!?
But we would have had the same phenomenon if $[A,X]=\xi A+\rho X$, i.e., if $X$ were a cinf-symmetry.
This open the door to: a distribution $\mathcal{A}=\mathcal{S}(\{A_1,A_2\ldots\})$ is $X$-projectable if $\mathcal{S}(\{A_1,A_2\ldots,X\})$ is involutive, that is, $X$ is a cinf-symmetry of distribution $\mathcal{A}$.
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Author of the notes: Antonio J. Pan-Collantes
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