How is one supposed to carry sign in Lie brackets? Is it bilinearity? How is one supposed to carry sign in Lie brackets?
E.g. if one has:
$X=x \frac{\partial}{\partial y}, Y= y\frac{\partial}{\partial z} - z \frac{\partial}{\partial y}$
Then is $[X,Y]$:
$=[x \frac{\partial}{\partial y}, y \frac{\partial}{\partial z}]+[x \frac{\partial}{\partial y},- z \frac{\partial}{\partial y}]$
and particularly is this:
$=[x \frac{\partial}{\partial y}, y \frac{\partial}{\partial z}]-[x \frac{\partial}{\partial y},z \frac{\partial}{\partial y}]$
since I could not find a rule that would answer this.
Or is this bilinearity?
 A: Yes, the Lie bracket is bilinear. So in particular $[x \frac{\partial}{\partial y}, -z \frac{\partial}{\partial y}] = -[x \frac{\partial}{\partial y}, z \frac{\partial}{\partial y}]$.
A: Remember that $XY-YX$ means $X\circ Y-Y\circ X$
So in particualr we have $f\in C^\infty(\mathbb R^n), f:\mathbb R^n\longrightarrow \mathbb R$ we have :
\begin{align}[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)
\end{align}
Therefore, $$c^i=\sum \left(a^j\frac{\partial b^i}{\partial x^j}-b^j\frac{\partial a^i}{\partial b^j}\right)$$
and sign will just naturally pull through.
