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I'm looking for an equation that describes the components of the Laplacian of a general $(r,s)$ tensor over the real numbers. Basically all I know so far is that it should be a linear map $$\Delta:~^r_s\mathbb{R}\to~^r_s\mathbb{R}$$ I.e, it preserves the order of the tensor. However, I can't find any published formulae. Wolfram Mathworld writes the components of the tensor Laplacian using some very weird notation as

$$\left(\Delta \mathbf{T}\right)^{i_1\dots i_r}{}_{j_1\dots j_s}=T^{i_1\dots i_r}{}_{j_1\dots j_s~;k}{}^{;k}$$

My question is simple: what the heck is that?! I know that the lowered semicolon index indicates a component of the covariant derivative. But what about the raised one? Is that the elusive contravariant derivative? Or is it instead a covariant derivative that has been "index raised" by contraction with the inverse metric? I'm afraid the short article on Mathworld is of little clarity and use (not to mention the readability is poor due to the low resolution of the typesetting). I would appreciate if someone could shed some light on this for me. Even only doing some special cases, i.e

  • $(1,0)$ tensors [vectors]
  • $(0,1)$ tensors [covectors]
  • $(1,1)$ tensors [matrices]

Would be just fine as well. Thanks for the help.

K.defaoite
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  • Does this answer (parts) of your question? – peek-a-boo Apr 06 '21 at 14:16
  • @peek-a-boo Your $\operatorname{grad}$ operator added a contravariant index instead of a covariant one. Why is this? Why don't you use the typical covariant derivative? Since your divergence operator contracts along an up index, it would make sense to me that we would want to contract grad over a lower index, for symmetry... can you elaborate on this? – K.defaoite Apr 06 '21 at 17:29
  • I'm not sure I understand; I'm using the usual covariant derivative $\nabla$. Do you agree that for a smooth function $f$, $\nabla f= df$ is a covector field/1-form? Well, what we usually call the gradient vector field is the vector field obtained by "raising the index" i.e by applying the musical isomorphism to convert this covector field into a vector field. I'm just generalizing that idea: given any tensor field $T$, $\nabla T$ introduces an extra lower index, so I'm using the musical isomorphism to raise it. – peek-a-boo Apr 06 '21 at 18:00
  • Next, the divergence of a vector field contracts the upper index of a vector field with the lower "differentiation index". So, I'm just generalizing that idea. Putting the two together gives exactly what we want: given a tensor field $T$, taking grad introuces an upper index, but taking the divergence contracts it, so we're again left with a tensor field of the same type (i.e we go from $(r,s)$ to $(r+1,s)$ using grad then back to $(r,s)$ using div). Said again: we need that upper index from grad so that we can apply div afterward (otherwise there's nothing to take the trace over). – peek-a-boo Apr 06 '21 at 18:04
  • It is in fact just a raised index. – Khalid Wenchao Yjibo Apr 12 '21 at 12:49
  • @peek-a-boo Would it be correct to write $$(\boldsymbol{\triangle}\mathbf{T})^{i_1\dots i_r}{}{j_1\dots j_s}=g^{kl}T^{i_1\dots i_r}{}{j_1\dots j_s~;k;l}=g^{kl}(\nabla(\nabla \mathbf{T}))^{i_1\dots i_r}{}_{j_1\dots j_s~kl}$$ – K.defaoite Apr 13 '21 at 14:14
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    yea sure that looks right (you can think of this as $\nabla \nabla T$ being a "Hessian" of $T$, and then you're taking the trace of the last two entries (which is where the differentiation occurs), so this conforms with the idea that the Laplacian is also the trace of the Hessian). And I think the equivalence with the definition I gave in the other page comes from the metric compatibility: $\nabla g = 0$. – peek-a-boo Apr 15 '21 at 02:35

1 Answers1

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Posting an answer so as to mark this query as resolved. The formula on mathworld is just the trace of a double covariant derivative, the second of which has been raised using the metric. I.e,

$$(\Delta\mathbf{T})^{i_1\dots i_r}{}_{j_1\dots j_s}=T^{i_1\dots i_r}{}_{j_1\dots j_s~;k}{}^{;k}=g^{kl}T^{i_1\dots i_r}{}_{j_1\dots j_s~;k;l}$$ $$=g^{kl}(\nabla(\nabla\mathbf{T}))^{i_1\dots i_r}{}_{j_1\dots j_s~kl}$$

But better still is to just write $$\Delta=\nabla^i\nabla_i$$

K.defaoite
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