$\bf 14.9.$ Lie bracket in local coordinates
Consider the two vector fields $X,Y$ on $\mathbb{R}^n$: $$X=\sum a^i\dfrac\partial{\partial x^i},\qquad Y=\sum b^j\dfrac\partial{\partial x^j},$$ where $a^i(x),b^j(x)$ are $C^\infty$ functions on $\mathbb R^n$. Since $[X,Y]$ is also a $C^\infty$ vector field on $\mathbb R^n.$ $$[X,Y]=\sum c^k\dfrac\partial{\partial x^k}$$ for some $C^\infty$ functions $c^k.$ Find the formula for $c^k$ in terms of $a^i$ and $b^j$.
Can you help for solving this.I have an manıfold exam and ı am working but ı have a problem about lie bracket.
And ı am putting what ı did..
Best Answer
Note that $XY-YX$ means $X\circ Y-Y\circ X$
For each $f\in C^\infty(\mathbb R^n), f:\mathbb R^n\longrightarrow \mathbb R$ wehave : $$[X,Y](f)= (X\circ Y)(f)-(Y\circ X)(f)=$$ $$X(\sum b^i\dfrac{\partial f}{\partial x^i})-Y(\sum a^i\frac{\partial f}{\partial x^i})=$$ $$\sum\left(X(b^i)\dfrac{\partial f}{\partial x^i}+b^iX(\frac{\partial f}{\partial x^i})\right)-\sum\left(Y(a^i)\dfrac{\partial f}{\partial x^i}+a^iY(\frac{\partial f}{\partial x^i})\right)=$$ $$\sum\left(a^j\frac{\partial b^i}{\partial x^j}\frac{\partial f}{\partial x^i}+b^ia^j\frac{\partial^2f}{\partial x^j\partial x^i}\right)-\sum\left(b^j\frac{\partial a^i}{\partial x^j}\frac{\partial f}{\partial x^i}+a^ib^j\frac{\partial^2f}{\partial x^j\partial x^i}\right)=$$ $$\sum\left(a^j\frac{\partial b^i}{\partial x^j}\frac{\partial f}{\partial x^i}-b^j\frac{\partial a^i}{\partial x^j}\frac{\partial f}{\partial x^i}\right)=$$ $$\sum \left(a^j\frac{\partial b^i}{\partial x^j}-b^j\frac{\partial a^i}{\partial x^j}\right)\left(\frac{\partial}{\partial x^i}\right)(f)=\sum c^i\frac{\partial}{\partial x^i}(f)$$ Therefore, $$c^i=\sum \left(a^j\frac{\partial b^i}{\partial x^j}-b^j\frac{\partial a^i}{\partial x^j}\right)$$