Set of vector fields with a particular expression of the commutator I have a set $F$ of vector fields. The commutator $[v, u]$ is linear in $v$ and $u$ point by point, i.e., for each couple of vector fields $v\in F$ and $u\in F$, there are two scalars $a$ and $b$ such that:
$$
[v, u] = a v + b u
$$
I remark that, 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. Moreover, I remark that $a$ and $b$ can be different for each couple of vector fields.
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 $F$ contains as many elements as the dimension of the space; moreover, they are linearly independent point by point. This means that any vector field $u$ can be written as the sum of the $v_i\in F$
multiplied by suitable scalars (functions of place) $a_i$. Are there further properties in this case?
 A: Let me develop my remark. It is actually a very nice problem in ``elementary'' differential geometry.
Here is my claim:
Proposition: Let $(v_1,\ldots, v_n)$ be pointwise linearly independent vector fields on an $n$-dimensional manifold $M$. Then the following are equivalent:
$(i)$ For all $i,j$, $$[v_i,v_j] \in \mathcal C^\infty(M) v_i\oplus \mathcal C^{\infty}(M) v_j$$
$(ii)$ Locally, there exist non-vanishing functions $(f_i)_{1\leq i \leq n}$ such that the vector fields $\hat v_i = f_i v_i$ satisfy $[v_i, v_j] = 0$ for all $i,j$.
Remark first that condition $(i)$ is invariant under multiplying each vector field by a function, which shows that $(ii) \Longrightarrow (i)$. For the other direction let us first prove two lemmas:
Lemma 1: Locally, there exists a non-zero function $f_n$ and functions $(b_i)_{1\leq i\leq n-1}$ such that
$$[f_n v_n, v_i] = b_i v_i~.$$
Proof: By property $(i)$, the distribution spanned by $(v_1,\ldots, v_{n-1})$ is integrable. Locally, it is thus given by the kernel of a closed $1$-form $\alpha$.
Since $(v_1,\ldots, v_n)$ are pointwise linearly independent, the function $\alpha(v_n)$ does not vanish. Let us set $f_n = \frac{1}{\alpha(v_n)}$ and $\hat v_n = f_n v_n$. By construction $\alpha(\hat v_n) \equiv 1$.
For all $1\leq i \leq n-1$, write
$$[\hat v_n, v_i] = a_i \hat v_n + b_i v_i.$$
We have
$$a_i = \alpha([\hat v_n, v_i]) = \hat v_n \cdot (\alpha(v_i)) - v_i \cdot (\alpha(\hat v_n)) - \mathrm d \alpha(\hat v_n, v_i) =0$$
since $\alpha(\hat v_n) \equiv 1$, $\alpha(v_i)\equiv 0$ and $\alpha$ is closed. We conclude that $[\hat v_n, v_i] = b_i v_i$. QED
Lemma 2: Locally, there exist positive functions $(f_i)_{1\leq i \leq n-1}$ such that
$$[\hat v_n, f_i v_i] = 0$$
for all $i$.
Proof: This equation is equivalent to
$$\hat v_n \cdot \log (f_i) = -b_i.$$
Let $\phi_t$ denote the flow of $\hat v_n$. Locally, let $L$ be some leaf of the foliation tangent to $(v_1,\ldots, v_{n-1})$. Then every point $y$ in some neighbourhood of $L$ can be written as $\phi_t(x)$ for a unique $t\in \mathbb R$ and $x\in V$. Set
$$f_i(\phi_t(x)) = \exp\left(-\int_{s=0}^t b_i(\phi_s(x))\right)~.$$
Then $f_i$ satisfies the above equation. QED
Proof of the proposition:
We have modified $(v_1,\ldots, v_n)$ into $(\hat v_1, \ldots, \hat v_n)$ such that $[\hat v_n, \hat v_i] = 0$ for all $i$. Now, by induction, we can find local functions $(f_i)_{1\leq i \leq n-1}$ on the leaf $L$ such $[f_i \hat v_i, f_j \hat v_j] = 0$. We extend them to $\phi_t$-invariant functions in a neighbourhood of $V$. Then we still have
$$[\hat v_n, f_i \hat v_i] = 0.$$
Thus
$${\phi_t}_* (f_i \hat v_i) = f_i v_i~,$$
hence
$${\phi_t}_* [f_i \hat v_i, f_j \hat v_j] = [f_i \hat v_i, f_j \hat v_j].$$
Since $[f_i \hat v_i, f_j \hat v_j]$ vanishes on $L$ and is $\phi_t$-invariant, it vanishes in a neighbourhood of $L$. QED
Remark 1: I assumed $n= \dim (M)$ for simplicity, but now, if $n< \dim(M)$ then $v_1,\ldots, v_n$ span the tangent distribution to some foliation, and locally one can modify $v_1,\ldots, v_n$ independently on each leaf so that they commute.
Remark 2: I don't know what happens near points where $v_1,\ldots, v_n$ are not linearly independent anymore. It is quite an interesting question !
A: Extended comment as asked by OP.
Let's call your manifold $X$, $\dim X = n$ .
Your assumption is that there are $v_1,\dots, v_n\in  \mathcal{T}(X)$ (vector fields) that are pointwise linearly independent and moreover
$[v_i,v_j] = a_{i,j} v_i + b_{i,j} v_j$ for $a_{i,j}, b_{i,j} \in C^{\infty}(X)$.
Remark 1. since $v_1,\dots, v_n$ are linearly independent $X$ is parallelizable. It means that its tangent bundle $TX \simeq X\times \mathbb{R}^n$. This is a global property and is not true in general, for example the even-dimensional spheres $\mathbb{S}^{2k}$ do not have this property because you cannot even find a single never vanishing vector field over $\mathbb{S}^{2k}$.
Now, consider a pair $v_1,v_2$. At each $x\in X$, consider $D_x = Span(v_1(x), v_2(x))$ this is a 2-dimensional subspace of the tangent space $T_x X$ which varies smoothly within $x$, in other words a distribution.
The condition $[v_1,v_2] = a_{1,2} v_1 + b_{1,2} v_2$  now says that $[v_1,v_2](x) \in D_x$. Therefore Frobenius theorem applies telling us that $D$ is an integrable distribution.
Being integrable, it means in particular that $X = \bigsqcup_\alpha S_\alpha$ where $S_\alpha$ are disjoint 2-dimensional maximally integrable (immersed) submanifolds for $D$.
In other words, for each $x\in X$ there is an $S_\alpha$ s.t. $T_x S_\alpha = D_x =Span (v_1,v_2)$, the tangent space coincides with the distribution.
Notice that I am considering maximal integrable submanifolds, this is the same thing as when you solve an ODE and you look for maximal solutions.
For the definition of immersed submanifold see my comment below. Keep in mind that $S_\alpha $ is image of a 2-manifold $N_\alpha$ under an immersion map $f_\alpha$. $S_\alpha$ may not be a manifold with the topology inherited by $X$ hence calling them surfaces is an abuse of language. When I say $T S_\alpha$  I mean $d f_\alpha (T N)$ (think of $f_\alpha$ as a parametrization).
The surfaces $S_\alpha$ are invariant under the flow of $v_1,v_2$ since they are now tangent vector fields to $S_\alpha$ hence you can solve the flow equations over $S_\alpha$.
We can also say something about the topology of the surfaces $S_\alpha$ (to be precise about the topology of $N_\alpha$).
Indeed since $v_1,v_2$ are linearly independent, their restriction to $S_\alpha$ provides a global framing for $S_\alpha$ since $\forall s \in S_\alpha, Span(v_1(s), v_2(s)) = T_s S_\alpha$ and $v_1,v_2$ are globally defined over $S_\alpha$. This shows that $S_\alpha$ (read $N_\alpha$) is parallelizable.
The only closed, parallelizable 2-manifolds are tori $\mathbb{S}^1\times \mathbb{S}^1$ hence if one of the $S_\alpha$ is closed, it must be a torus.
Note that the topology of the $S_\alpha$ may change with $\alpha$.
If $S_\alpha$ is not closed, then parallelizability tells us that $S_\alpha$ must be orientable (hence no Mobius bands). Also all characteristic classes of $T N_\alpha$ must vanish but in dimension two for non-closed surfaces this doesn't say anything more than orientability (for higher dimensions you get more constraints).
Final note. Consider now $v_1,\dots, v_k$, $k<n$. They induce a distribution of $k$-planes, $D_x = Span (v_1,\dots, v_k)$.
As above this distribution is involutive: $[\xi,\zeta](x) \in D_x $ for any $\xi,\zeta$ sections of $D$. Hence we can apply Frobenius and find that $X$ is foliated $X = \bigsqcup_\alpha S_\alpha$ with $\dim S_\alpha=k$. The  $k$-manifolds $N_\alpha$ will be parallelizable as above and invariant under the flow of $v_1,\dots, v_k$.
