# Set of vector fields with a particular form of commutator

I have a set of vector fields $$v_1, \dots, v_n$$. For each couple of indices $$i$$ and $$j$$, the commutator $$[v_i, v_j]$$ is linear in $$v_i$$ and $$v_j$$, i.e. there are two scalars $$a$$ and $$b$$ such that: $$[v_i, v_j] = a v_i + b v_j$$

Here, by "scalar", I mean that $$a$$ and $$b$$ are functions of the place, i.e. they depend on the position. They are not constant all over the manifold.

If I'm not wrong, this is not a general property of vector fields. Does it have a name? Are there known properties of such a set?

I can add that the $$v_i$$ are as many as the dimension of the space; moreover, they are linearly independent point by point. Are there further properties in this case?

• Are these arising in the context of invariant vector fields on a Lie group? Commented Mar 9, 2021 at 18:35
• Maybe it is connected, but not so directly, and, well, I do not know. Very roughly, the $v_i$ are the covariant Lyapunov vectors of a dynamic system. I noticed the relation among them, and now I'm wondering if it brings me some useful information. Commented Mar 9, 2021 at 21:18
• It tells you in particular that every pair generates an integral surface. I don't know the significance of the constants (rather than functions) other than suggesting some group invariance, Commented Mar 9, 2021 at 23:14
• Constants or functions? $a$ and $b$ are functions, i.e. they depend on the point. I clarified this in the question. I do not know much more on them. About the "group invariance" that you mention: do you have any additional hint? Commented Mar 10, 2021 at 9:11
• To be clear, the scalars $a$ and $b$ depend on the pair $(i, j)$, correct? Commented Mar 11, 2021 at 0:31

Assuming these are vector fields on some smooth manifold $$M$$, they span a finite-dimensional Lie subalgebra $$\mathfrak g\subseteq \mathfrak X(M)$$ (where $$\mathfrak X(M)$$ is the Lie algebra of all smooth vector fields on $$M$$). Under the condition that all of the vector fields in this subalgebra are complete (meaning their flows exist for all time), there is a simply connected Lie group $$G$$ whose Lie algebra is isomorphic to $$\mathfrak g$$, and a right action of $$G$$ on $$M$$ for which $$\mathfrak g$$ is the infinitesimal generator. This is essentially Theorem 20.16 in my Introduction to Smooth Manifolds (2nd ed.). (See also Problem 10-14.)
EDIT: The OP has now clarified that the coefficients $$a$$ and $$b$$ are meant to be scalar functions, not constants, so my answer is not relevant. Other than the point mentioned by Ted Shifrin that each pair of vector fields determines a foliation by surfaces, I don't know what else can be said in general.
• I'm not sure if I understood. The sub-space spanned by the $v_i$ does not include the commutators $[v_i, v_j]$, because the factors $a$ and $b$ are scalars, i.e. they are functions of the place. Thus I would say that the space spanned by the $v_i$ is not a sub-algebra. Am I wrong? On the other hand, the set obtained by combining the $v_i$ multiplying by any scalar is the set of all the vector fields, because, as highlighted in the final part of the question, the $v_i$ are linearly independent point by point. Commented Mar 10, 2021 at 8:54