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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 $[U, V] = 0$ ?

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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. – Yuri Vyatkin 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. – matgaio Apr 22 '12 at 5:32
Note that U is a dilation and V is a rotation. Therefore they commute. – PAD Aug 27 '12 at 23:08
@matgaio that is a wonderful answer! I can't believe I didn't know that before. Is that a common interpretation, or can you recommend a reference which uses it? – Adam Saltz Aug 28 '12 at 1:23
@AdamSaltz, this is a kind of common interpretation and proofs for this can be found in several textbooks on analysis on manifolds, for example Lee's "Smooth Manifolds" (a great book!). What amuses me on this is the fact that it is a really natural question to know when two vector fields commute (in the sense I gave before) and that the answer is a simple algebraic calculation. – matgaio Sep 2 '12 at 3:57

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.

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Can you prove this please? $\Phi_*([U,V])=[\Phi_*(U),\Phi_*(V)]$ – sifsa Mar 29 at 18:42
@sifsa See 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... – Thomas Klimpel Mar 29 at 19:20

matigaio gave the geometric meaning. Guessing if it vanishes or not require some practice by doing computations and drawing the vector fields (locally). You can for instance after drawing, try to see if it could commute or not by "following" the "quadrilateral" (sometimes open) made by integrating (Euler method, for small time) one vector field after the other one, etc.

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