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Related to Why symplectic geometry gives Poisson geometry

Is there any common use for the Lie bracket of the gradients of two functions? That is, $[\nabla f, \nabla g]$?

Here, $\nabla f$ denotes the vector field gotten by "raising the index" from $df$. In other words, $(\nabla f)(g) = m(df, dg)$, where $m$ is the Riemannian metric on some manifold.

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  • $\begingroup$ 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. $\endgroup$
    – Ivan Izmestiev
    Commented Dec 1, 2016 at 16:46
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    $\begingroup$ @IvanIzmestiev That doesn't seem to be true; take $f = x, g = x^2$ with the usual 1-dimensional metric. Then $\nabla f = \frac{\partial}{\partial x}, \nabla g = 2x \frac{\partial}{\partial x}, [\nabla f, \nabla g] = -2 \frac{\partial}{\partial x}$. $\endgroup$
    – user44191
    Commented Dec 2, 2016 at 0:10
  • $\begingroup$ Indeed. But if the gradients are linearly independent, they can be seen as coordinate vector fields, and then they do commute. $\endgroup$
    – Ivan Izmestiev
    Commented Dec 2, 2016 at 5:36
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    $\begingroup$ @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. $\endgroup$
    – user44191
    Commented Dec 2, 2016 at 5:41

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