I would like to verify

Subscript[\[CurlyEpsilon], i\[InvisibleComma]j\[InvisibleComma]k] Subscript[\[CurlyEpsilon], i m n]=Subscript[\[Delta], j m] Subscript[\[Delta], k n]-Subscript[\[Delta], j n] Subscript[\[Delta], k m]

I have defined:

Subscript[\[Delta], i_Integer, j_Integer] := KroneckerDelta[i, j]

Subscript[\[Epsilon], i__Integer] := Signature[{i}]

What is a good way to demonstrate the above identity holds?

Thanks in advance.


1 Answer 1


Avoiding complicated and confusing subscripts and variable names until we have something working ... define

d[i_, j_] := KroneckerDelta[i, j]
e[i__Integer] := Signature[{i}]

The equivalence to prove is

equiv = Sum[e[i, j, k] e[i, m, n], {i, 1, 3}] == d[j, m] d[k, n] - d[j, n] d[k, m];

Check it for all possible values of the free variables

And @@ Flatten[Table[equiv, {n, 3}, {m, 3}, {l, 3}, {k, 3}, {j, 3}]]
(* True *)
  • $\begingroup$ I understand. But, my assignment question tend to come loaded with 'fancy' notation; 're-formatting' it may be tedious unless there are some formatting features by Mathematica that I am unaware of. For that I apologise. I have been called out before for this issue. $\endgroup$
    – Physkid
    Apr 8, 2017 at 11:27

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