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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?

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    $\begingroup$ As to whether this is a general property of vector fields, certainly it isn't: for example, upon identifying $\mathfrak{sl}_2$ with left-invariant vector fields on $\operatorname{SL}_2$, we see that $E = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $F = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ have the property that $\{E, F, [E, F]\}$ is linearly independent. $\endgroup$
    – LSpice
    Commented Mar 10, 2021 at 17:11
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    $\begingroup$ By Frobenius theorem, your condition implies that any k vector fields in your family are tangent to a k-dimensional foliation. I can't detail that now but I think it is equivalent to saying that there are local coordinates $x_i$ and scalar functions $f_i$ in which your vector fields are the $f_i \frac{\partial}{\partial x_i}$ $\endgroup$
    – Nicolast
    Commented Mar 13, 2021 at 20:15
  • $\begingroup$ The hint sonds interesting! I started a bounty to encourage @Nicolas Tholozan to provide some more detail. $\endgroup$ Commented Mar 15, 2021 at 9:24
  • $\begingroup$ @NicolasTholozan OP says in the last part of the post that $F$ generate the whole tangent space therefore if we brutally apply Frobenius to the distribution generated by $F$ we don't get any information, the integral submanifold would be the whole space. $\endgroup$ Commented Mar 15, 2021 at 11:52
  • $\begingroup$ @DorianoBrogioli. Can you say something about the derivatives of $a$ and $b$? For example, can we assume that the only non-vanishing derivatives of $a$ and $b$ are the ones taken with respect to the vector fields $v$ and $u$? $\endgroup$ Commented Mar 15, 2021 at 11:54

2 Answers 2

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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 $v_1,\dots, v_k$.

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  • $\begingroup$ I should add a caveat: all the leaves obtained with Frobenius are in general only weak submanifolds, i.e. they come with a map $f_{\alpha}: N_\alpha \to X$, from a k-manifold $N_\alpha$ where $f(N_\alpha) = S_\alpha$, $f$ injective and $df$ is injective. This is not an embedding since $f_\alpha$ ma not be open on the image $S_\alpha$, so in general is not an homeomorphism onto the image (for example at limit points you can loose the injectivity or $S_\alpha$ may be dense in $X$). $\endgroup$ Commented Mar 16, 2021 at 11:11
  • $\begingroup$ A very simple example: let us take $F=\partial/\partial x$ and $G=\partial/\partial y$, in any $n$ dimensional space. We have $[F,G]=0$. The $S_{\alpha}$ should be the $x-y$ planes, aren't they? But there should be something wrong, since the planes have Euler characteristic 1, not 0. $\endgroup$ Commented Mar 25, 2021 at 18:40
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    $\begingroup$ You are right I've corrected it. Parallelizability $\Rightarrow$ existence of one non-vanishing vector field $\Rightarrow$ trivial Euler class $e(TS_\alpha)= 0 $. If $S_\alpha$ is closed then the Euler class can be identified with the Euler characteristic. If $S_\alpha$ is not-closed then the Euler class is always trivial. You can still get some constraints on the characteristic classes of $N_\alpha$ (which must vanish), this may tell you something about the topology in higher dimension. $\endgroup$ Commented Mar 25, 2021 at 19:26
  • $\begingroup$ Also note that the Euler class is defined only for orientable vector bundles thus when I speak about it I am already assuming orientability. $\endgroup$ Commented Mar 25, 2021 at 19:48
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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 !

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