Lie bracket exercise

Question

How can I solve the following exercise?

Let $X,Y\in \mathfrak{X}$ $\left ( M \right )$ and $f,g\in C^{\infty }\left ( M \right )$ . Show that

$\left [ fX,gY \right ]=fg\left [ X,Y \right ]+f\left ( Xg \right )Y-g\left ( Yf \right )X $, where $\left [ , \right ]$ is Lie bracket.

Thanks for your help

Answer 1

If we accept the notion that a Lie bracket $[V, W]$ is determined by its action on functions $h \in C^\infty(M)$, we may compute

$[fX, gY] h = fX(gY(h)) - gY(fX(h)) = fX(g)Y(h) + fgXY(h) - gY(f)X(h) - gfYX(h)$ $= fg(XY(h) - YX(h)) + fX(g)Y(h) - gY(f)X(h)$ $= fg[X, Y]h + f(X(g)Y(h) - gY(f)X(h), \tag 1$

where we have used the Leibniz rule

$V(\phi \theta) = \phi V(\theta) + \theta V(\phi) \tag 2$

for vector fields $V$, since they are derivations on $C^\infty(M)$. Since (1) holds for every $h \in C^\infty(M)$, we conclude that

$[fX, gY] h = fg[X, Y] + f(X(g)Y - gY(f)X, \tag 3$

for $X,Y\in \mathfrak X(M)$