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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 enter image description here

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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.

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