# A question about expanding $(L_{fV}g)(X,Y)$.

Let us consider $$(L_{fV}g)(X,Y)$$. Why is this equal to $$f(L_Vg)(X,Y)+L_V(X(f),Y)+L_V(X,Y(f))$$ and not just $$fL_Vg(X,Y)$$

The context is the following calcualation: the second to third step

This is basically the Lie bracket, which is not a tensor (i.e. when you plug in $$fV$$ you cannot simply pull out $$f$$). From the definition, $$L_Vg(X, Y)=V(g(X, Y))-g([V, X], Y)-g(X, [V, Y]),$$ you can use the identity $$[V, X]=\nabla_VX-\nabla_XV$$ as well as $$V(g(X, Y))=g(\nabla_VX, Y)+g(X, \nabla_VY)$$ to get $$L_Vg(X, Y)=g(\nabla_XV, Y)+g(X, \nabla_YV).$$ Here $$\nabla$$ is the Levi-Civita connection. Replace $$V$$ by $$fV$$, we get $$L_{fV}g(X, Y)=g\big((Xf)V+f\nabla_XV, Y\big)+g\big(X, (Yf)V+f\nabla_YV\big),$$ write this out you will get the desired identity.