Lie bracket between two vector fields in $\mathbb{R}^2$ Consider the fields $\xi = \left(\frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right), \eta = (-y, x)$ in $\mathbb{R}^2$. The problem I am set is computing the Lie Bracket of these two fields: $[\xi, \eta]$. When doing so, I get a non-trivial answer:
$$[\xi,\eta] = \left(-\frac{y}{\sqrt{x^2+y^2}},  \frac{x}{\sqrt{x^2+y^2}}\right),$$
but in the solutions I am given, the apparent correct answer is $[\xi, \eta] = 0$. I don't see how this is so because I have these non-vanishing terms appearing in my calculations giving my answer above - they don't vanish because the derivative I am multiplying them by (by the definition of a Lie bracket) does not vanish. Who is correct?
 A: (Note: in what follows I am going to  use the usual identification between vectors  and tangent vectors of $\mathbb{R}^n$ and matrices and linear maps.)
You want to compute $[\eta,\xi]= \nabla_\eta \xi - \nabla_\xi \eta$ ($\nabla$ here is the Jacobian matrix $\nabla F (x) = \frac  {\partial F}{\partial x}(x)$ so $\nabla_V$ is the vector of derivatives in direction $V$, $\frac{\partial F}{\partial x}\cdot V$).
So the Jacobian of $\xi(x)= \frac x {||x||}$ at $x$ is given by 
$$ \nabla \xi(x)= \frac {Id} {||x||}- \frac{x}{||x||^3}\cdot\langle x,\cdot \rangle$$ (where $Id$ is the identity map of $\mathbb{R}^n$). The Jacobian of $\eta(x)= J x$ (where $J$ is the symplectic matrix $J=\begin{bmatrix}
0 & I \\
-I & 0 \\
\end{bmatrix}$) at $x$ is given by
$$\nabla\eta(x)= J.$$
So the Lie parenthesis is 
$$\nabla_\eta \xi(x) - \nabla_\xi \eta(x) = \frac {Jx} {||x||}- \frac{x}{||x||^3}\cdot\langle x,J x \rangle - J\frac x {||x||}
=\frac {Jx} {||x||}-J\frac x {||x||}=0$$
Where $\langle x,J x \rangle=0$ because $J$ is symplectic.
