# Covariant derivative for symbolic tensors

I want to define a "prefix" (D_i) covariant derivative operator CD[] for symbolic tensors in form of a function, i.e. for

$Assumptions={Element[g,Arrays[{3,3},Reals,Symmetric[{1,2}]]]}  also Element[CD[g],Arrays[{3,3,3},Reals,Symmetric[2,3]]]  should be made an assumption, an additional slot is created on the left. So far the only way I found to make this possible is defining a function $Assumptions = {Element[v,Vectors[3, Reals]]}

CDdef[a_] := (AppendTo[$Assumptions, Element[CD[a],Arrays[{3}~Join~TensorDimensions[a]],Reals, TensorSymmetry[TensorProduct[v, a]]]];$Assumptions =  DeleteDuplicates[$Assumptions]; CD[a])  and effectively writing all the assumption by hand into$Assumptions. But this could generate a huge list in bigger calculations. All my attempts of the kind

$Assumptions={Element[CD[a_?SomeSymmetryQ],Arrays["one more slot"]]}  to use pattern that are at least valid for all tensors with the same symmetries have failed. Is there any effective way to proceed? Many thanks, Hendrik • Syntax! for starters: not $Assumptions={Element[g,Arrays[{3,3},Reals],Symmetric[{1,2}]]} but $Assumptions = Element[g, Arrays[{3, 3}, Reals, Symmetric[{1, 2}]]] Mar 31, 2014 at 12:59 • I think this depends on if you are planning to add further assumptions later on. Or do you think starters can only use one assumption at a time? ;-) Mar 31, 2014 at 13:36 • that's not it, rather the closed bracket after Reals : the argument Symmetric belongs inside Arrays. Mar 31, 2014 at 14:20 • sorry, this was just a typo that does not appear in my actual code, i will correct Mar 31, 2014 at 14:54 ## 1 Answer Good question; the notion of a tensorial (covariant) derivative is something that is missing in Mathematica AFAIK. I can think of two ways to proceed: ## Option 1 One way is to overload the TensorRank, TensorDimensions, and TensorSymmetry functions for patterns that have head CD: CD /: TensorRank[CD[tensor_]] := TensorRank[tensor] + 1 CD /: TensorDimensions[CD[tensor_]] := Join[{First[#]}, #]& @ TensorDimensions[tensor] CD /: TensorSymmetry[CD[tensor_]] := TensorSymmetry[tensor] /. (h : (Cycles | Symmetric | Antisymmetric))[list_List] :> h[list + 1]  Note that this is a bit rough on the edges and gives incorrect results if tensor hasn't been defined as a tensor or is a scalar, but it shouldn't be too hard to improve the code. The following then works as expected: $Assumptions = m ∈ Matrices[{4, 4}, Reals, Symmetric[{1, 2}]];

TensorRank[CD@m]

3

TensorDimensions[CD@m]

{4, 4, 4}

TensorReduce@TensorTranspose[CD@m, {1, 3, 2}]

CD[m]


## Option 2

However, if your covariant derivatives is commutative (like for instance the partial derivative), a better way would be to simply define it as a vector:

\$Assumptions = CD ∈ Vectors[4, Reals] && m ∈ Matrices[{4, 4}, Reals, Symmetric[{1, 2}]]


If you like the CD[tensor] notation, you may even define

CD[tensor_] := TensorProduct[CD, tensor]


All the above commands then give the same result. In addition, multiple derivatives are symmetric:

TensorSymmetry[CD@CD@m]

{{Cycles[{{1, 2}}], 1}, {Cycles[{{3, 4}}], 1}}


So only use this if your derivatives actually commute!

• Since I want to consider non-commuting derivatives I favour option one. Many thanks for this very useful solution!!! Any details about exceptions I will hopefully be able to figure out on my own. Jul 22, 2014 at 13:40
• I think there is an issue with option 2, even if the derivative can be assumed commutative: It will always act on everything on the right. This is because the tensor product is associative, and brackets have no effect. So there is no way to distinguish between D_i A_j B_k = (D_i A_j) B_k + A_j (D_i B_k) and (D_i A_j) B_k Aug 7, 2014 at 13:24
• @Hendrik Yes, that's indeed an issue with option 2. So you can't write CDTCD*T (where * is the TensorProduct) to mean (CD@T)*(CD@T) -- yet another reason to opt for option 1! Aug 7, 2014 at 13:36