TITLE: Ricci identity in Lectures on Ricci Flow
QUESTION [5 upvotes]: So in the book, Lectures on Ricci flow, the identity is given as 
$$-\nabla^2_{X,Y}A(W,Z,\ldots)+\nabla_{Y,X}^2A(W,Z,\ldots)=-A(R(X,Y)W,Z,\ldots)-A(W,R(X,Y)Z,\ldots)$$
where 
$$R(X,Y)=\nabla^2_{Y,X}-\nabla_{X,Y}^2$$
But the way how covariant derivatives distributes over a tensor field I think it should be 
$$-\nabla^2_{X,Y}A(W,Z,\ldots)+\nabla_{Y,X}^2A(W,Z,\ldots)=R(X,Y)(A(W,Z,\ldots))-A(R(X,Y)W,Z,\ldots)-A(W,R(X,Y)Z,\ldots)-\cdots$$
Is there something I am misunderstanding here?

REPLY [6 votes]: You're correct, but so is Topping - the curvature operator on functions is just zero! Remember that $$-R(X,Y)A = \nabla_X (\nabla_Y A) - \nabla_Y (\nabla_X A) - \nabla_{[X,Y]}A,$$ so when $A = f$ is a function you get $XYf - YXf - [X,Y]f = 0$. Since $A(W,Z,\ldots)$ has all its slots filled, it is a scalar function, and thus this curvature term vanishes.