The Laplacian of the squared length of a (0,2)-tensor

Solution 1:

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 $2g^{mn} g^{ij} g^{kl} \nabla_m T_{ik} \nabla _n T_{jl}$.