The Laplacian of the squared length of a (0,2)-tensor Let $(M,g)$ be a smooth Riemannian manifold and $\nabla$ be the Levi-Civita connection. Suppose $T$ is a smooth $(0,2)$-tensor field on $M$, and it is given by $$T=T_{ij}dx_i\otimes dx_j$$ in local coordinates, where we adopt the summation convention on repeated indices.
The squared length of $T$ is given by
$$|T|^2=\langle T,T\rangle=T_{ij}T_{kl}\langle dx_i,dx_k\rangle\langle dx_j,dx_l\rangle=T_{ij}T_{kl}g^{ik}g^{jl},$$
where $(g^{ij})$ is the inverse of $(g_{ij}).$
Let $\Delta_M$ be the Laplace-Beltrami operator on $M.$ I am now puzzled at the calculation of $\Delta_M|T|^2.$
In the spirit of Page 134, Line 17 of Han's Book Nonlinear Elliptic Equations of the Second Order, we should have $$\frac{1}{2}\Delta_M|T|^2=\frac{1}{2}\Delta_M(g^{ik}g^{jl}T_{ij}T_{kl})=g^{ik}g^{jl}T_{kl}\Delta_MT_{ij}+g^{ik}g^{jl}g^{pq}T_{kl;p}T_{ij;q},$$ where we use the notation that $$T_{ij;k}:=\nabla_{\partial_k}T(\partial_i,\partial_j).$$
At this stage, I can understand the appearance of the term $g^{ik}g^{jl}T_{kl}\Delta_MT_{ij}.$ Probably by using $g^{ij}_{;k}=0,$ those terms with $T_{kl}T_{ij}$ may disappear. However, I do not know why do the covariant coefficients $T_{kl;p}$ and $T_{ij;q}$ appear, and I think it should be directional derivatives $T_{kl,p}$ and $T_{ij,q}.$
 A: One can write $|T|^2 = \mathrm{L}( g^{-1} \otimes g^{-1} \otimes T\otimes T)$ where $\mathrm{L}$ is taking traces of some indices. By defintion, for any tensor $G$,
$$ \Delta G = \mathrm{tr} \nabla ^2 G.$$
So to calculate $\Delta |T|^2$, we first calculate $\nabla ^2 |T|^2$. Since $\nabla g^{-1} = 0$ and that $\nabla $ commutes with $\mathrm{L}$,
$$ \nabla |T|^2 = L(g^{-1} \otimes g^{-1}\otimes \nabla T \otimes T) + L(g^{-1} \otimes g^{-1}\otimes T \otimes \nabla T)$$
and
$$\nabla ^2|T|^2 = L(g^{-1} \otimes g^{-1}\otimes \nabla^2 T \otimes T) + 2L(g^{-1} \otimes g^{-1}\otimes \nabla T \otimes \nabla T) + L(g^{-1} \otimes g^{-1}\otimes T \otimes \nabla^2 T)$$
This implies
\begin{align} 
\Delta |T|^2 &= \mathrm{tr} \nabla ^2 |T|^2 \\
&=L(g^{-1} \otimes g^{-1}\otimes \Delta T \otimes T) + 2 \mathrm{tr} L(g^{-1} \otimes g^{-1}\otimes \nabla T \otimes \nabla T) + L(g^{-1} \otimes g^{-1}\otimes T \otimes \Delta T).
\end{align}
The first and third terms give $2g^{ij} g^{kl} T_{ik} (\Delta T)_{jl}$ and the second term gives $2g^{mn} g^{ij} g^{kl} \nabla_m T_{ik} \nabla _n T_{jl}$.
