Lie bracket in local coordinates. Find the formula $c^{k}$ in terms of $a^{i}$ and $b^{j}$

Question

This is from T.U Loring's manifold book. I tried. But I didnt do the question. Please show me how to solve instructively and explicitly. I want to learn this topic. Thank you for help.

Answer 1

For each $f\in C^\infty(\mathbb R^n), f:\mathbb R^n\longrightarrow \mathbb R$ wehave : $$[X,Y](f)= (X\circ Y)(f)-(Y\circ X)(f)=$$ $$X(\sum b^i\dfrac{\partial f}{\partial x^i})-Y(\sum a^i\frac{\partial f}{\partial x^i})=$$ $$\sum\left(X(b^i)\dfrac{\partial f}{\partial x^i}+b^iX(\frac{\partial f}{\partial x^i})\right)-\sum\left(Y(a^i)\dfrac{\partial f}{\partial x^i}+a^iY(\frac{\partial f}{\partial x^i})\right)=$$ $$\sum\left(a^j\frac{\partial b^i}{\partial x^j}\frac{\partial f}{\partial x^i}+b^ia^j\frac{\partial^2f}{\partial x^j\partial x^i}\right)-\sum\left(b^j\frac{\partial a^i}{\partial x^j}\frac{\partial f}{\partial x^i}+a^ib^j\frac{\partial^2f}{\partial x^j\partial x^i}\right)=$$ $$\sum\left(a^j\frac{\partial b^i}{\partial x^j}\frac{\partial f}{\partial x^i}-b^j\frac{\partial a^i}{\partial x^j}\frac{\partial f}{\partial x^i}\right)=$$ $$\sum \left(a^j\frac{\partial b^i}{\partial x^j}-b^j\frac{\partial a^i}{\partial b^j}\right)\left(\frac{\partial}{\partial x^i}\right)(f)=\sum c^i\frac{\partial}{\partial x^i}(f)$$ Therefore, $$c^i=\sum \left(a^j\frac{\partial b^i}{\partial x^j}-b^j\frac{\partial a^i}{\partial b^j}\right)$$

Answer 2

Since it's a homework problem, I'm not going to give you an explicit solution. However, I will give you a

Hint/walkthrough:

Write down $X = \sum a_i\partial_i$ and $Y = \sum b_j\partial_j$. There is only one thing you can do: write down the definition of the Lie bracket, plug in the local coordinate expressions of $X$ and $Y$, and chug-chug-chug. \begin{align*}[X,Y] &= XY - YX \\ &= \sum a_i\partial_i\sum b_j\partial_j - \sum b_j\partial_j\sum a_i\partial_i \\ \end{align*} Now distribute, differentiate with the product rule, and use the fact that mixed partial commute. You will end up with a sum of the form $$\sum \big(\mbox{stuff}\big)\partial_k.$$ The "stuff" are your coefficients $c_k$.