$\bf 14.9.$ 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..

enter image description here


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, $$c^i=\sum \left(a^j\frac{\partial b^i}{\partial x^j}-b^j\frac{\partial a^i}{\partial x^j}\right)$$

  • $\begingroup$ Thank you ı am so happy now :) @Dimitris Dallas $\endgroup$
    – Aera
    Apr 23 '13 at 22:01
  • 2
    $\begingroup$ meta.stackexchange.com/questions/104227/… $\endgroup$ Apr 23 '13 at 22:02
  • $\begingroup$ @ZevChonoles You are right, I am going to be careful with this. $\endgroup$
    – Dimitris
    Apr 23 '13 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.