# An Compact Expression for the Tensor Laplacian

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!

• $T_{\mu\nu}$ is a scalar and $\nabla^\lambda\nabla_\lambda$ is an operator being applied to that scalar. The expression you link to is exactly the same as the expression for $\nabla^\lambda\nabla_\lambda\Phi$ you wrote with $\Phi = T_{\mu\nu}$. Commented Apr 19, 2023 at 19:07
• I think not. By $\nabla^\lambda \nabla_\lambda T_{\mu\nu} \equiv \displaystyle\sum_{\lambda} \nabla^\lambda \nabla_\lambda T_{\mu\nu}$, I or the link, I've given, meant to understand the $(\mu,\nu)$ component of the rank-$2$ tensor, $\Delta T$. Whereas, $\nabla^\lambda \nabla_\lambda \Phi \equiv \displaystyle\sum_{\lambda} \nabla^\lambda \nabla_\lambda \Phi$ is entirely a scalar, not a component of any tensor. This fact, I think, matters when deriving the expression for $\nabla^\lambda \nabla_\lambda \Phi$, that I wrote. Hence both the cases are different!
– SCh
Commented Apr 19, 2023 at 19:15
• @NicholasTodoroff What are you talking about? The individual components of a tensor, though real numbers, are not scalars, i.e, they don't transform as such. Commented Apr 19, 2023 at 20:03
• Unfortunately there is no compact expression for a general tensor. You just have to do all the tedious computations with the covariant derivative. I have a Mathematica program somewhere that does it automatically, if you would be interested in that. Commented Apr 19, 2023 at 20:04
• MathWorld's formula for the tensor Laplacian appears incorrect to me. I have never seen a compact expression for it, either. The formula contains not only Christoffel symbols but also derivatives of the Christoffel symbols. The scalar Laplacian is simpler, because $\nabla_af = \partial_a$ contains no Christoffel symbols and therefore $\Delta f$ contains Christoffel symbols but no derivatives of Christoffel symbols. Commented Apr 20, 2023 at 6:08

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.
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 (**).