# Geometric intuition behind the Lie bracket of vector fields

I understand the definition of the Lie bracket and I know how to compute it in local coordinates.

But is there a way to "guess" what is the Lie bracket of two vector fields ? What is the geometric intuition ?

For instance, if we take $U = x \frac{\partial}{\partial x} + y \frac{\partial}{\partial y}$ and $V = -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}$, should it be obvious that $[U, V] = 0$ ?

• In your case, you can notice that v.f. $V$ does not change the direction relative to the flow of v.f. $U$, and moreover the magnitude of $V$ changes at the same rate as the one of $U$, so the derivative of $V$ w.r.t. $U$ will not see any change at all, i.e. it will give 0. Apr 17 '12 at 23:38
• More generally, one of the geometrical meanings of commutativity (zero bracket) of vector fields relies on the flows generated by these two vectors. Two vector fields are commutative if and only if its flows are too, in the sense that there is no difference starting at one point $p$, traveling a time $t_1$ over the flow of, lets say, $X_1$ and then a time $t_2$ over the flow of, say, $X_2$, or, instead, traveling first $t_2$ over the flow of $X_2$ and then $t_1$ over the flow of $X_1$. You end up at the same point if and only if these two vector fields have zero bracket. Apr 22 '12 at 5:32
• Note that U is a dilation and V is a rotation. Therefore they commute.
One way to get a geometric intuition for the Lie bracket is to note $\Phi_*([U,V])=[\Phi_*(U),\Phi_*(V)]$, i.e. the Lie bracket transforms canonically under the diffeomorphism $\Phi$. Now if we have a straightening $\Phi^U$ of the vector field $U$ (such that $\Phi^U_*(U)$ is constant in our coordinate system), then $[U,V]$ is just the derivative of $V$ along (the constant direction) $U$ in that coordinate system.
• Can you prove this please? $\Phi_*([U,V])=[\Phi_*(U),\Phi_*(V)]$ Mar 29 '16 at 18:42
• @sifsa See drive.google.com/folderview?id=0B6x6GQ82vuH_YnE2STNKZXU1cUk page 19 at the bottom. The proof exploits a definition of the Lie bracket by its effect on a test function $\varphi$ via $L_v(L_w \varphi) - L_w(L_v \varphi) = L_{[v,w]}\varphi$. Then it is sufficient to know that $L_v \varphi$ transforms canonically... Mar 29 '16 at 19:20