About solvable algebras and solvable structures

1. Lie algebras are defined like vector spaces with a binary operation $[-,-]$, bla, bla, bla. The result of the bracket has to be expressed as linear combinations with coefficients in $\mathbb{R}$, of course. Some Lie algebras are finite dimensional and other infinite dimensional.

2. Some Lie algebras are solvable, in the sense that the derived series arrives at 0 in a finite number of steps. As far as I know, they can be finite or infinite dimensional.

3. An example of an infinite dimensional Lie algebra: vector fields on a manifold $\mathfrak{X}(M)$. If we fix a distribution $\mathcal{Z}\subseteq \mathfrak{X}(M)$ in $M$, there is a (possibly infinite dimensional) Lie subalgebra consisting of _all the symmetries of the distribution_ $\mbox{Sym}(\mathcal{Z})$. Remember: they are vector fields $X$ such that

$$ [X,\mathcal{Z}]\subseteq \mathcal{Z} $$so for $Z\in \mathcal{Z}$, $[X,Z]$ is a linear combination with coefficients in $\mathcal{C}^{\infty}(M)$. The distribution $\mathcal{Z}$ will be, in the context of an ODE, the one generated by $A$ , and $Sym(\mathcal{S}(\{A\}))$ will consist of the generalized symmetries (after prolongation).

4. In [Lychagin 1991] it is said, loosely speaking, that if we focus on a finite dimensional Lie subalgebra $L$ of $Sym(\mathcal{Z})$ which is "pointwise independent of $\mathcal{Z}$", which "completes $\mathcal{Z}$ to $TM$" and which is solvable then we can find the integral manifolds of $\mathcal{Z}$ by quadratures. This is Lie theory, but with distributions.

For $X\in L,Y\in L$ we have that $[X,Y]$ has constant coefficients. But for $X\in L,Z\in \mathcal{Z }$ we have that $[X,Z]$ has "functional coefficients."

5. If above we consider $\mathcal{Z}=\mathcal{S}(A)$ and focus on the subalgebra of $Sym(\mathcal{S}(A))$ that corresponds to prolonged Lie point symmetries we get the Lie symmetry algebra. As far as I know, it is not necessarily finite-dimensional [Olver 1986] corollary 2.40.

6. On the other hand, solvable structures are made of vector fields in such a way that:

The first one is a symmetry of distribution $\mathcal{Z}$. The Lie bracket condition is, therefore, "with functional coefficients".

The following vector fields are symmetries of the previous distributions, so the Lie bracket condition is again "with functional coefficients".

7. In conclusion, the freedom obtained with solvable structures vs solvable algebras comes from two facts: (1) we can look for vector fields whose bracket depends not only on the other vector fields of the set but also on the vector fields of the distribution (the vector field of the equation if we are in the context of an ODE) and (2) the coefficients can be functions.