Q:

Bi-linearity of Lie Bracket

The Lie Bracket of 2 Vector Fields is given by: $$[X,Y](p) = X(p) \cdot Y - Y(p) \cdot X$$

Show that for smooth functions f and g that:

$[fX,gY] = fg[X,Y] + f(X \cdot g)Y - g(Y \cdot f)X$

So I was going to go about this by using the fact that:

$[X,gY] = X(g) \cdot Y + g([X,Y])$ and try by extrapolating out to have
$$[fX,Y] = fX \cdot Y + f([X,Y])$$ and then bring these two expressions together to show the bilinearity above but can't seem to do this. Anyone have any suggestions how to proceed?

A:

By anti-symmetry, it suffices to check that $[fX,Y] =f[X,Y] - Y(f)X$. Take a "test function" $\phi$. We have:$$\begin{align} [fX,Y](\phi) &= (fX)(Y(\phi)) - Y((fX)(\phi)) \\ &= f X(Y(\phi)) - Y(fX(\phi)) \\ &= fX(Y(\phi)) - Y(f)X(\phi) - fY(X(\phi)) \\ &= f(X(Y(\phi))-Y(X(\phi))) -
 Y(f)X(\phi) \\ &= f[X,Y](\phi) - Y(f)X(\phi) \\ &= (f[X,Y]-Y(f)X)(\phi).\end{align}$$Since $\phi$ was arbitrary, we're done.

Just to make sure we're in the same page here: when I dismiss $g$ by calling anti-symmetry, I mean the following argument: $$[X,gY] = -[gY,X] = -(g[Y,X] - X(g)Y) = g[X,Y] + X(g)Y.$$

