Say we are working in the following toy situation. Let $X$ and $Y$ be vector fields over a smooth manifold $M$, and say we have some equation involving the Lie bracket, e.g. $[X,Y] = X+Y$. Say $Y$ is fixed, and we wish to solve for all $X$ that satisfies this equation.

This becomes a PDE in the following sense, as far as I understand. First we work locally and write $$ X = \sum_{i=1}^n X^i \partial_i, ~~~~ Y = \sum_{i=1}^n Y^i \partial_i, ~~~~ [X,Y] = \sum_{i=1}^n \sum_{j=1}^n (X^j \partial_j Y^i - Y^j \partial_j X^i)\partial_i $$ over some chart $U$, where $X^i, Y^i$ are functions and $\partial_i = \frac{\partial}{\partial x^i}$ is the associated local basis of the tangent bundle.

Then, plugging the above in the equation, we isolate the functions in front of each $\partial_i$ that appears and we get for fixed $i$, $$ \sum_{j=1}^n (X^j \partial_j Y^i - Y^j \partial_j X^i) = X^i + Y^i. $$ So in particular, this is a system of PDEs where we wish to solve for the functions $X^j$ over a domain that is diffeomorphic to $\mathbb{R}^n$. Thus locally we "forget" about the manifold structure and just work in the flat case (if I understand this part correctly).

But because we are forgetting about the global structure, working locally like this seems very insufficient to finding a solution $X$ over all of $M$. How do we go from local to global to solve for the vector field $X$?

EDIT: I'm more interested in the question of the existence of a solution $X$, rather than an explicit solution.

  • $\begingroup$ Not an expert hence I will just comment. The idea is that you just "cut and paste" the solution via partition of unity, If the atlas is given (which I think it's your case) calculation should be possible (but I suspect they will usually be really bad looking). Usual calculus theorem gives you unicity of solution (with enough regularity in the hypothesis of course) where charts overlap so this should be a fine procedure. $\endgroup$ Aug 5 at 22:44
  • $\begingroup$ @MatíasMatteini Wouldn't one need to check for topological constraints too? I would imagine that different manifolds which admit the "same" solutions on a chart may or may not admit global solutions $\endgroup$ Aug 5 at 22:48
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    $\begingroup$ Yes, of course. But the problem should show in any chart you are considering around the "constraints" and carry over with partition of unity. Think of $\mathbb{R}^{2}$ minus the origin and extending the constant field in any direction, you notice the obstruction via a local (hence in chart) observation, not with any global consideration. (please notice that I'm genuinely not an expert on the topic and trying to work out an answer with you, I don't want to lead you on any wrong road!) $\endgroup$ Aug 5 at 22:53
  • $\begingroup$ @MatíasMatteini I think I see what you mean, but the example confuses me. What is the obstruction here? And thanks for the help! $\endgroup$ Aug 5 at 23:01
  • $\begingroup$ This is a first order PDE $PX = Y$, where the differential operator $P = -L_Y - 1$ is a first order differential operator mapping vector fields to vector fields. Perhaps existence theory for first order PDEs is well studied? $\endgroup$
    – Mason
    Aug 7 at 5:36


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