# Lie bracket of vector fields is a vector field

Let $$X$$ and $$Y$$ two vector field on the manifold $$M$$ (dim($$M$$)= $$m$$). Show that the Lie bracket $$[X, Y] := XY - YX$$ is a vector field on $$M$$.

I tried to compute using local coordinates, so for $$X = \sum_{i=1}^{m} X_i \frac{\partial}{\partial x_i}$$ and $$Y = \sum_{i=1}^{m} Y_i \frac{\partial}{\partial x_i}$$ I obtain after computation that \begin{align} [X, Y] = \sum_{i=1}^{m} \left( \sum_{j=1}^{m} X_j \frac{\partial Y_i}{\partial x_j} - Y_j \frac{\partial X_i}{\partial x_j} \right) \frac{\partial}{\partial x_i} \end{align}

But how can I conclude now that the Lie bracket $$[X,Y]$$ is a vector field on $$M$$? Thanks in advance!

• This is a typical homework. See for example here. So follow the hints given there. Oct 7 '18 at 19:37

$$\textbf{Lemma}.$$ Let $$M$$ be a smooth manifold. The map $$D: C^{\infty}(M) \to C^{\infty}(M)$$ is a derivation if and only if, $$Df = Xf$$ for some smooth vector field $$X$$.
In particular, this says that derivations of $$C^{\infty}(M)$$ can be identified with smooth vector fields. So, it suffices to show that the Lie bracket is a derivation of $$C^{\infty}(M)$$.
$$\textbf{Sketch.}$$ \begin{align*} [X,Y](fg) &= X(Y(fg) - Y(X(fg)) \\ &= X(fYg + gYf) - Y(fXg + gXf) \\ &= \ldots \end{align*}