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I've symmetric tensor $h(z)$ with rank $s$.

Instead of writing symmetric tensor with its components I use the following notation

$h^{(s)}(z;a) = \sum_{\mu_i}(\prod_{i=1}^{s}a^{\mu_i})h^{(s)}_{\mu_1\mu_2 ... \mu_s}(z)$

And I use the following $*_{a}$ operator for contracting tensors in this notation

$*_{a} = \frac{1}{(s!)^2}\prod_{i=1}^{s} \overset{\leftarrow}{\partial^{\mu_i}_{a}} \overset{\rightarrow}{\partial_{\mu_i}^{a}} $

Arrows on derivatives show the direction of differentiation.

with this operator, tensor contraction can be written without components.

$g^{(s)}(z;a) *_a h^{(s)}(z;a) = g^{\mu_1\mu_2...\mu_s} h_{\mu_1\mu_2...\mu_s} = gh$

Now I'm trying to implement this operator in Wolfram Mathematica, but all my attempts failed. Is it possible to define such an operator in Mattematica, and if yes how can I achieve that?

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  • $\begingroup$ "I use the following notation" - how is it represented in Mathematica? $\endgroup$ – J. M.'s technical difficulties Sep 26 '18 at 6:18
  • $\begingroup$ @Melik Karapetyan Do you have a clear example of using this definition? $\endgroup$ – Alex Trounev Sep 26 '18 at 6:31
  • $\begingroup$ I don't get it. Your method is a very obfuscated way of writing Flatten[g].Flatten[h]. $\endgroup$ – Henrik Schumacher Sep 26 '18 at 8:03
  • $\begingroup$ I've took the notation from arxiv.org/pdf/1002.1358.pdf I'm trying to automate the computation with this notation using Mathematica and searching a way to represent it in Matematica. The clear example of using this definition can be found on Section 1 of that paper. $\endgroup$ – Melik Karapetyan Sep 26 '18 at 9:43

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