Relating the Lie derivative to inner product of 2-tensors Let $(M,g)$ be a Riemannian manifold. Let $f \in C^\infty(M)$, $X$ be a vector field and $h$ be a (symmetric) covariant $2$-tensor. Denote by $\langle \cdot, \cdot \rangle$ the inner product induced by the metric $g$ on the space of covariant $2$-tensors. I would like to know why the following formula holds:
$$\langle df \otimes df, \mathcal{L}_X(h) 
\rangle = (\mathcal{L}_X(h))(\nabla f, \nabla f).$$
Here, $\mathcal{L}_X$ is the Lie derivative in the direction of $X$ and $\nabla f$ is the gradient of $f$.
I am sorry if this is a silly question, but I have never seen such identity.
 A: By an equivalent definition, inner product (contraction) of two $(0,2)$ tensors $A_{ij}$ and $B_{ij}$ is as follows:
$$\langle A,B\rangle= A_{i}{\ }^{j}B_{j}{\ }^{i}.$$ and it is also equal to $A^{i}{\ }_{j}B^{j}{\ }_{i}$. One can show that
$$\langle A,B\rangle= A(e_i,e_j)B(e_i,e_j)$$
for arbitrary orthonormal frame $\{e_1,\dots,e_n\}$.
By this we have
$$\langle df\otimes df,B\rangle=(df\otimes d f)(e_i,e_j)A(e_i,e_j)=df(e_i)df(e_j)A(e_i,e_j)=(e_if)(e_jf)A(e_i,e_j)=A((e_if)e_i,(e_jf)e_j)=A(\nabla f,\nabla f).$$
Note that in general by the definition of $df$ $$df(X)=X.f=\nabla_Xf$$
and $(e_if)e_i=\nabla f=\mathsf{grad} f=df^\sharp$. Now let $A=\mathcal{L}_Xh$.
Now can you prove @Jesse Madnick claim in comment? (Proof is same as above.) $$\langle \alpha \otimes \alpha, \beta \rangle = \beta(\alpha^\sharp, \alpha^\sharp)$$
A: If you were using the abstract index notation, you would not even need to ask this question:
$$
\langle df \otimes df, \mathcal{L}_X(h) \rangle  = g^{a c} g^{b d} (\mathcal{L}_X h)_{a b} (\nabla_c f) \nabla_d f = \\ (\mathcal{L}_X h)_{a b} (\nabla^a f) \nabla^b f = (\mathcal{L}_X(h))(\nabla f, \nabla f)
$$
where all the equalities are just expanded definitions.
