Timeline for Lie bracket of gradients

7 events

when toggle format	what		by	license	comment
Apr 13, 2017 at 12:58	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Dec 5, 2016 at 20:04	history	edited	user44191	CC BY-SA 3.0	Expanding some definitions in response to comment
Dec 2, 2016 at 5:41	comment	added	user44191		@IvanIzmestiev That still isn't true; take $f = x, g = x^2 + y$, and the usual metric. Then $\nabla f = \frac{\partial}{\partial x}, \nabla g = 2x \frac{\partial}{\partial x} + \frac{\partial}{\partial y}$, which are generally linearly independent, and the Lie bracket is the same. Linear independence is not enough; the vector fields will generally not commute.
Dec 2, 2016 at 5:36	comment	added	Ivan Izmestiev		Indeed. But if the gradients are linearly independent, they can be seen as coordinate vector fields, and then they do commute.
Dec 2, 2016 at 0:10	comment	added	user44191		@IvanIzmestiev That doesn't seem to be true; take $f = x, g = x^2$ with the usual 1-dimensional metric. Then $\nabla f = \frac{\partial}{\partial x}, \nabla g = 2x \frac{\partial}{\partial x}, [\nabla f, \nabla g] = -2 \frac{\partial}{\partial x}$.
Dec 1, 2016 at 16:46	comment	added	Ivan Izmestiev		If the gradient is a $1$-form, then how do you define the Lie bracket of $1$-forms? And if you use a Riemannian metric to turn $1$-forms into vector fields, then these two will commute.
Nov 30, 2016 at 23:31	history	asked	user44191	CC BY-SA 3.0