Let X be a manifold and $v, w \in C^\infty(TX)$. Suppose (x_1, ... x_n) are local coordinates on an open set $U \subseteq X$, so that we may write $v = \sum_{i} v_i \frac{\delta}{\delta x_1}$ and $w = \sum_{i} \frac{\delta}{\delta x_i}$ on U for $v_i, w_j: U \rightarrow \mathbb{R}$ smooth. Define the Lie bracket $[v,w] \in C^\infty(TX)$ by
$$[v,w] = \sum\limits^{n}_{i,j=1} \bigg(v_i \frac{\delta w_j}{\delta x_i} \cdot\frac{\delta}{\delta x_j} - w_j \frac{\delta v_i}{\delta x_j} \cdot\frac{\delta}{\delta x_i} \bigg)$$ on U.
Suppose that $(y_1, ... y_n)$ is another local coordinate system on $V \subseteq X$, then consider the analogue of the above definition in terms of $(y_1, ..., y_n)$. How would you show that the two experssions define the same vector field on $U\cap V$?
What I tried:
This is my first course in geometry, so I don't yet have much intuition here. I tried to write the following:
$$\frac{\delta}{\delta x_i} = \sum\limits_{j} \frac{\delta y_j}{\delta x_i} \frac{\delta}{\delta y_j}$$
I am not exactly sure if or why this is sensible to write. (Is it? Why/not?) So I plugged this into the above definition of the Lie bracket, which gave me the following expression: $$[v,w] = \sum\limits^{n}_{k,l=1} \bigg(\big(\sum\limits_{i} v_i \frac{\delta y_k}{\delta x_i}\big) \frac{\delta \big(\sum\limits_{i} w_i \frac{\delta y_l}{\delta x_i}\big)}{\delta y_l} \cdot\frac{\delta}{\delta y_k} - \big(\sum\limits_{i} w_i \frac{\delta y_k}{\delta x_i}\big) \frac{\delta \big(\sum\limits_{i} v_i \frac{\delta y_l}{\delta x_i}\big)}{\delta y_l} \cdot\frac{\delta}{\delta y_k} \bigg)$$
I guess I should somehow show that this equals the defining equations above... This is where I lost it. How could this be finished? Also, does it make any sense to show the equality "component-wise"?
NB the indices in the second term in parentheses in your definition aren't quite right---the fact that one index occurs once and the other thrice is a sign something has gone awry. It's convenient to use the equation $$\boxed{[v, w] = \sum_{i, j} \left(v^j \frac{\partial w^i}{\partial x^j} - w^j \frac{\partial v^i}{\partial x^j} \right) \frac{\partial}{\partial x^i}}$$ for the Lie bracket.
Hint We want to show that this equation is independent of the choice of coordinates, that is, if we change coordinates from $(x^i)$ to $(y^i)$ then $$\sum_{i, j} \left(\tilde v^j \frac{\partial \tilde w^i}{\partial y^j} - \tilde w^j \frac{\partial \tilde v^i}{\partial y^j} \right) \frac{\partial}{\partial y^i} = \sum_{i, j} \left(v^j \frac{\partial w^i}{\partial x^j} - w^j \frac{\partial v^i}{\partial x^j} \right) \frac{\partial}{\partial x^i},$$ where $\tilde v, \tilde w$ are the components of $v, w$ w.r.t. the frame $(\frac{\partial}{\partial y^i})$. Now substitute your transformation formulas and compare. In particular, after substituting $\frac{\partial}{\partial x^i} = \sum_j \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^i}$, both sides will be a linear combination of the vector fields $\frac{\partial}{\partial y^i}$, but because these fields are independent, this is equivalent to checking whether the coefficients of both sides with respect to each $\frac{\partial}{\partial y^i}$ agree.