Let us have a second rank tensor $T$. So how the components of tensor Laplacian of $T$ be computed?
I mean if I expand $\nabla^\lambda \nabla_\lambda T_{\mu\nu}$, that would be a monstrous expression! But I'm seeking for if there any compact one $\left[ \text{as for scalar, it is,} ~ \nabla^\lambda \nabla_\lambda \Phi = \dfrac{1}{\sqrt{g}}\partial_\alpha \left(\sqrt{g}~g^{\alpha\beta}~\partial_\beta \Phi\right) \right]$.
Wolfram Mathworld is giving an awesome compact expression for this, but the second and third line of that's derivation is unclear! How they have got the second line from the first one and also the third line from the second one, they haven't explained!
So I'm seeking for if there any alternative or detailed derivation or any other compact expression for it!
Best Answer
As Deane pointed out the formula in MathWorld looks wrong. They use the following derivation \begin{align} {A_{\mu\nu\,;\,\lambda}}^{;\,\lambda} &=\Big(g^{\lambda\kappa}A_{\mu\nu\,;\,\lambda}\Big)_{;\,\kappa}\tag{1}\\ &=g^{\lambda\kappa} \frac{\partial^2A_{\mu\nu}}{\partial x^\lambda\partial x^\kappa} -g^{\color{red}{\mu\nu}}\,{\Gamma^\lambda}_{\color{red}{\mu\nu}} \frac{\partial A_{\color{red}{\mu\nu}}}{\partial x^\lambda}\tag{2}\\ &\dots\\ &=\frac{1}{\sqrt{g}}\Big(\sqrt{g}\,g^{\mu\kappa}\,A_{\mu\nu,\,\kappa}\Big)_{,\,\mu}\,\tag{5} \end{align} where $A_{\mu\nu\,;\,\lambda}$ is the covariant derivative, $g_{\mu\nu}$ is the metric tensor, $g$ its determinant and $A_{\mu\nu\,,\,\kappa}$ is the ordinary partial derivative.
The term (2) has obvious errors that I highlighted in red. Summation convention doubles indices and does not triple them, and the indices $\mu\nu$ cannot be summed away completely.
Even if we try a fix with some dummy indices $$ g^{\color{blue}{\alpha\beta}}\,{\Gamma^\lambda}_{\color{blue}{\alpha\beta}} \frac{\partial A_{\color{green}{\mu\nu}}}{\partial x^\lambda}\tag{2'} $$ it looks not convincing as I will show below.
The whole derivation and the formula (5) are highly questionable.
A correct derivation should start as follows and it will soon become clear that things are not simple: First, \begin{align} A_{\mu\nu\,;\,\lambda}=A_{\mu\nu\,,\,\lambda} \underbrace{-{\Gamma^\rho}_{\mu\lambda}A_{\rho\nu}}_{(*)}\; \underbrace{-{\Gamma^\rho}_{\nu\lambda}A_{\mu\rho}}_{(**)}\,.\tag{A} \end{align} Since the Christoffel connection is metric compatible we have ${g ^{\lambda\kappa}}_{;\,\alpha}=0$ so that $g^{\lambda\kappa}$ can be pulled out of a covariant derivative as if it were a constant. This leads to \begin{align*} {A_{\mu\nu\,;\,\lambda}}^{;\,\lambda}&=g^{\lambda\kappa}\,A_{\mu\nu\,;\,\lambda\kappa}\\ &=g^{\lambda\kappa}\Big\{A_{\mu\nu\,;\,\lambda,\,\kappa}-{\Gamma^\rho}_{\mu\kappa}A_{\rho\nu\,;\,\lambda} -{\Gamma^\rho}_{\nu\kappa}A_{\mu\rho\,;\,\lambda}-{\Gamma^\rho}_{\lambda\kappa}A_{\mu\nu\,;\,\rho}\Big\}\,. \end{align*} Plugging in the three terms for each covariant derivative of $A_{\alpha\beta}$ from (A) leads to a total of, I believe, fourteen terms which contain partial derivatives of Christoffel symbols because the product rule needs to be applied on (*) and (**).