# How does following formula of the Lie bracket of two vector fields arise?

Let X be a smooth manifold and $$v, w \in \Gamma(TX)$$ be two vector fields. Under the natural correspondence of vector fields and derivations on $$C^\infty(X)$$ let $$\delta, \epsilon: C^\infty(X) \rightarrow C^\infty(X)$$ be the corresponding derivations.

Then the commutator $$[\delta, \epsilon] = \delta \circ \epsilon - \epsilon \circ \delta$$ is another derivation. Define the Lie bracket $$[v,w]$$ as the vector field corresponding to this commutator. Using local coordinates on X, we can express the action of the two derivations on some $$a \in C^\infty(X)$$ as $$\delta(a) = \sum v_i \frac{\partial a}{\partial x_i} \ \text{ and } \ \epsilon(a) = \sum w_i \frac{\partial a}{\partial x_i}$$

Using this how can $$[v,w]$$ be expressed using the same local coordinates? I suspect the answer is the following, but I don't know how to derive it.

$$[v,w] = \sum_{i,j} \bigg(v_i \frac{\partial w_j}{\partial x_i}- w_i \frac{\partial v_j}{\partial x_i}\bigg) \frac{\partial}{\partial x_j}$$

Your suspicion is correct. In the same local coordinates you have \begin{equation*} \begin{aligned} \ [v,w](a)&=\delta\circ \epsilon(a)-\epsilon\circ \delta(a)\\ &=\Big(\sum_i v_i\frac{\partial}{\partial x_i}\Big(\sum_j w_j\frac{\partial a}{\partial x_j}\Big)\Big)-\Big(\sum_j w_j\frac{\partial}{\partial x_j}\Big(\sum_i v_i\frac{\partial a}{\partial x_i}\Big)\Big)\\ &=\sum_{i,j} \Big(v_i\frac{\partial w_j}{\partial x_i}\frac{\partial a}{\partial x_j}+v_iw_j\frac{\partial^2a}{\partial x_ix_j}\Big)-\sum_{j,i} \Big(w_j\frac{\partial v_i}{\partial x_j}\frac{\partial a}{\partial x_i}+w_jv_i\frac{\partial^2a}{\partial x_jx_i}\Big)\\ &=\sum_{i,j}\Big(v_i\frac{\partial w_j}{\partial x_i}\frac{\partial a}{\partial x_j}-w_j\frac{\partial v_i}{\partial x_j}\frac{\partial a}{\partial x_i}\Big)+\sum_{i,j} \Big(v_iw_j\frac{\partial^2a}{\partial x_ix_j}-w_jv_i\frac{\partial^2a}{\partial x_jx_i}\Big)\\ &=\sum_{i,j}\Big(v_i\frac{\partial w_j}{\partial x_i}-w_i\frac{\partial v_j}{\partial x_i}\Big)\frac{\partial a}{\partial x_j}, \end{aligned} \end{equation*} where we smoothness of $$a$$ to cancel $$\frac{\partial^2a}{\partial x_ix_j}-\frac{\partial^2a}{\partial x_jx_i}=0.$$ Therefore $$[v,w]=\sum_{i,j}\Big(v_i\frac{\partial w_j}{\partial x_i}-w_i\frac{\partial v_j}{\partial x_i}\Big)\frac{\partial}{\partial x_j}.$$
Associating the vector field $$v_i$$ with the differential operator $$v:=v_i\partial^i$$ gives $$vwf=v_i\partial^i(w_j\partial^jf)$$ for a sufficiently nice function $$f$$. The product rule, plus the symmetry of derivatives to delete the second-order terms, gives $$[v,\,w]f=Rf$$, where $$R$$ is the expression for $$[v,\,w]$$ you desire. The rest is an exercise.