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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?

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  • $\begingroup$ Are these arising in the context of invariant vector fields on a Lie group? $\endgroup$
    – Ted Shifrin
    Commented Mar 9, 2021 at 18:35
  • $\begingroup$ 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. $\endgroup$
    – Doriano Brogioli
    Commented Mar 9, 2021 at 21:18
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    $\begingroup$ 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, $\endgroup$
    – Ted Shifrin
    Commented Mar 9, 2021 at 23:14
  • $\begingroup$ 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? $\endgroup$
    – Doriano Brogioli
    Commented Mar 10, 2021 at 9:11
  • $\begingroup$ To be clear, the scalars $a$ and $b$ depend on the pair $(i, j)$, correct? $\endgroup$
    – Travis Willse
    Commented Mar 11, 2021 at 0:31

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Doriano Brogioli
    Commented Mar 10, 2021 at 8:54
  • $\begingroup$ I edited the question to clarify this point. $\endgroup$
    – Doriano Brogioli
    Commented Mar 10, 2021 at 9:01
  • $\begingroup$ @DorianoBrogioli: OK, this makes my answer more or less irrelevant. $\endgroup$
    – Jack Lee
    Commented Mar 11, 2021 at 1:48
  • $\begingroup$ However, this highlights an important difference of notation between mathematics and physics. In physics, a "scalar" is any number which changes with the position. A single number, uniform for all the universe, is called a "constant" or "universal constant", e.g. F, the Farady constant, or c, the speed of light in vacuum. Instead, I realize now from the answer that, in mathematics, "scalar" (without specifying "function") means what in physics is a "constant". It's a subtlety, but it is worth knowing it. $\endgroup$
    – Doriano Brogioli
    Commented Mar 11, 2021 at 8:49
  • $\begingroup$ @JackLee Hi professor, could I ask your assistance here, please? $\endgroup$
    – Antonio Maria Di Mauro
    Commented Mar 11, 2021 at 21:18

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