I am trying to define a tensor $T$ which is antisymmetric under pair-wise exchange of its indices:

$T = \delta_{a],[b}\,\delta_{c],[d}\,\delta_{e],[f}\,\delta_{g],[h}$

where $T$ is antisymmetric under the exchanges $b \leftrightarrow c, d \leftrightarrow e, f \leftrightarrow g ,h \leftrightarrow a$.

I guess that defining $T$ as a Table will take too much memory. Maybe defining it as a function is better. In both cases I have no clue on how to do this in a clever, short way without writing down all possible exchanges by hand. Can you help?

  • 2
    $\begingroup$ $Assumptions = T \[Element] Arrays[{d, d, d, d, d, d}, Reals, Antisymmetric /@ Partition[RotateLeft@Range[6], 2]];? $\endgroup$
    – kglr
    Sep 8, 2020 at 10:34
  • $\begingroup$ It looks good but I'm not skilled with Assumptions. Can you expand a bit more, including (also in a sketchy way) the definition of T with the Kronecker's Deltas? $\endgroup$
    – IgnoranteX
    Sep 8, 2020 at 10:43
  • 1
    $\begingroup$ Is it a mathematica question ? $\endgroup$
    – yarchik
    Sep 8, 2020 at 12:00
  • $\begingroup$ Yes, I would like to see a Mathematica code in which one defines $T$ as I said (with the Kronecker's Deltas) and implementing the trick given by @kgir $\endgroup$
    – IgnoranteX
    Sep 8, 2020 at 15:51

1 Answer 1


Expanding on a comment by @kglr above, if you want a general form of the tensor with only the independent components, you can use SymmetrizedArray as follows:

antiSymmetricTensor[dim_, symbol_ : A] := 
 SymmetrizedArray[pos_ :> symbol @@ pos, ConstantArray[dim, 8],
  {Cycles[{#}], -1} & /@ Partition[RotateLeft@Range[8], 2]]

as a demonstration, here's the $i=1,j=2,k=3,l=2$ for $d=3$: enter image description here

This is indeed in the subdomain of arrays with dimension dim and the specified symmetries:

  dim = 3,
  symbol = A,
  domain = Arrays[ConstantArray[dim, 8], Reals, 
    Antisymmetric /@ Partition[RotateLeft@Range[8], 2]]
 Element[antiSymmetricTensor[dim, symbol], domain];
 Simplify[%, symbol[__] ∈ Reals]
(* True *)
  • $\begingroup$ Thanks a lot! @George Varnavides $\endgroup$
    – IgnoranteX
    Dec 15, 2021 at 17:14

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