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Nov 17, 2015 at 8:34 history edited Jean Duchon CC BY-SA 3.0
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Nov 16, 2015 at 16:51 comment added Jean Duchon You're right, we need to chose $e^\pm$ non unit to satisfy integrability. I guess it's possible. Won't be enough anyway...
Nov 16, 2015 at 16:22 comment added David E Speyer But then you want to integrate their flows to make $X$, right? So you need the fields to be integrable. For the record, I agree that this sounds like a promising approach, I'm just trying to get the details right.
Nov 16, 2015 at 15:08 comment added Jean Duchon I didn't mean "vector fields" as differential operators, just as functions from $\mathbb R^2$ to itself...
Nov 16, 2015 at 12:53 comment added David E Speyer Why can you choose the fields to commute?
Nov 14, 2015 at 15:19 history answered Jean Duchon CC BY-SA 3.0