2
$\begingroup$

I need have to create a real antisymmetric tensor whose elements are given as $p_{abcd}$ $${\displaystyle p_{abcd}={\begin{cases} +p_{\sigma\left(abcd\right)} & {\text{if }}\sigma(a,b,c,d){\text{ is an even permutation }}\\ -p_{\sigma\left(abcd\right)} & {\text{if }}\sigma(a,b,c,d){\text{ is an odd permutation }}\\ \;\;\,0 & {\text{otherwise}.} \end{cases}}}$$

For example if I need to create a random real antisymmetric tensor of order 4, then:first step, $$p_{cd}=\begin{bmatrix}p_{11,cd} & p_{12,cd} & p_{13,cd} & p_{14,cd}\\ p_{21,cd} & p_{22,cd} & p_{23,cd} & p_{24,cd}\\ p_{31,cd} & p_{32,cd} & p_{33,cd} & p_{34,cd}\\ p_{41,cd} & p_{42,cd} & p_{43,cd} & p_{44,cd} \end{bmatrix}$$

Then $$p=\begin{bmatrix}\begin{pmatrix}p_{11,11} & p_{12,11} & p_{13,11} & p_{14,11}\\ p_{21,11} & p_{22,11} & p_{23,11} & p_{24,11}\\ p_{31,11} & p_{32,11} & p_{33,11} & p_{34,11}\\ p_{41,11} & p_{42,11} & p_{43,11} & p_{44,11} \end{pmatrix} & \begin{pmatrix}p_{11,12} & p_{12,12} & p_{13,12} & p_{14,12}\\ p_{21,12} & p_{22,12} & p_{23,12} & p_{24,12}\\ p_{31,12} & p_{32,12} & p_{33,12} & p_{34,12}\\ p_{41,12} & p_{42,12} & p_{43,12} & p_{44,12} \end{pmatrix} & \begin{pmatrix}p_{11,13} & p_{12,13} & p_{13,13} & p_{14,13}\\ p_{21,13} & p_{22,13} & p_{23,13} & p_{24,13}\\ p_{31,13} & p_{32,13} & p_{33,13} & p_{34,13}\\ p_{41,13} & p_{42,13} & p_{43,13} & p_{44,13} \end{pmatrix} & \begin{pmatrix}p_{11,14} & p_{12,14} & p_{13,14} & p_{14,14}\\ p_{21,14} & p_{22,14} & p_{23,14} & p_{24,14}\\ p_{31,14} & p_{32,14} & p_{33,14} & p_{34,14}\\ p_{41,14} & p_{42,14} & p_{43,14} & p_{44,14} \end{pmatrix}\\ . & . & . & .\\ . & . & . & .\\ \begin{pmatrix}p_{11,41} & p_{12,41} & p_{13,41} & p_{14,41}\\ p_{21,41} & p_{22,41} & p_{23,41} & p_{24,41}\\ p_{31,41} & p_{32,41} & p_{33,41} & p_{34,41}\\ p_{41,41} & p_{42,41} & p_{43,41} & p_{44,41} \end{pmatrix} & . & . & \begin{pmatrix}p_{11,44} & p_{12,44} & p_{13,44} & p_{14,44}\\ p_{21,44} & p_{22,44} & p_{23,44} & p_{24,44}\\ p_{31,44} & p_{32,44} & p_{33,44} & p_{34,44}\\ p_{41,44} & p_{42,44} & p_{43,44} & p_{44,44} \end{pmatrix} \end{bmatrix}$$

But we can consider the above tensor as a 16 X 16 matrix also. But in that case we will index as: $$p=\begin{bmatrix}p_{11} & p_{12} & p_{13} & . & . & . & & & & & & & & & & p_{1,16}\\ p_{21}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ p_{16,1} & & & & & & & & & & & & & & & p_{16,16} \end{bmatrix}$$ If I define p in Mathematica as:

Clear[p]
p[arg__] /; ! OrderedQ@{arg} := Signature@{arg} p @@ Sort@{arg}
p[___, j_, j_, ___] = 0;
Format[p[arg__]] := Subscript[p, arg]

and created a 16 X 16 matrix in Mathematica as:

pmat = Table[
   If[i > j, RandomReal[], -RandomReal[]], {i, 16}, {j, 16}];

How can I map elements from tensor to pmat? For example what is p[1,1,3,4] in pmat?

$\endgroup$
1
  • $\begingroup$ Is this Transpose[Array[Subscript[p, ##] &, {4, 4, 4, 4}], 2 <-> 4] // MatrixForm what you want? BTW, your tensor p might be LeviCivitaTensor[4]. $\endgroup$ Aug 3, 2021 at 7:11

1 Answer 1

1
$\begingroup$
ArrayFlatten[LeviCivitaTensor[4]] // Normal

$$ \left( \begin{array}{cccccccccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.