0
$\begingroup$

In an expression how can we replace a group of variables $a, b,c,...$ that satisfy the condition $cond(a,b,c,...)$ with other variables $\alpha,\beta,\gamma,...$? I know how to do this in the case of individual constraints, but not in one where the pattern relies on the whole group of variables. E.g. I want to rename the indices in the formula $$F^{abcdef}\cdot(X_{a}Y_{cd}+X_{d}Y_{ac})\cdot(X_{b}Y_{ef}+X_{e}Y_{fb}),$$ so that it can be put into a standard form: $$\tilde{F}^{\alpha\beta\mu\nu\rho\sigma} X_{\alpha}X_{\beta}Y_{\mu\nu}Y_{\rho\sigma}$$ where $\tilde{F}^{\alpha\beta\mu\nu\rho\sigma}$ is a sum of $Fs$. It can be achieved by replacing $i\_,j\_,k\_,m\_,n\_,l\_$ that appear in the combination $X_{i}X_{j}Y_{km}Y_{nl}$ with $\alpha,\beta,\mu,\nu,\rho,\sigma$ (including all possible matches), so what kind of rule should I apply?

Edits:

Desirable output for the example above:

Taking the second terms in the first parenthesis and in the second parenthesis we get $$K_{22}=F^{abcdef}X_{d}Y_{ac}X_{e}Y_{fb}$$

To change it to the standard form, we can rewrite it as $F^{abcdef}X_{d}X_{e}Y_{ac}Y_{fb}$ and apply the rule: $d\rightarrow\alpha,e\rightarrow\beta,a\rightarrow\mu,c\rightarrow\nu,f\rightarrow\rho,b\rightarrow\sigma$. The result is $F^{\mu\sigma\nu\alpha\beta\rho}X_{\alpha}X_{\beta}Y_{\mu\nu}Y_{\rho\sigma}$. Of course, $K_{22}$ can also be arranged as $F^{abcdef}X_{d}X_{e}Y_{fb}Y_{ac}$, $F^{abcdef}X_{e}X_{d}Y_{ac}Y_{fb}$, and $F^{abcdef}X_{e}X_{d}Y_{fb}Y_{ac}$. They will produce different results, which should be added up to get the final result for $K_{22}$. So $$\tilde{F}^{\alpha\beta\mu\nu\rho\sigma} =F^{\mu\sigma\nu\alpha\beta\rho}+...$$ where the ellipsis stands for the other 15 terms.

$\endgroup$
3
  • $\begingroup$ is f a function or an index? $\endgroup$
    – user42582
    Commented Sep 17, 2016 at 14:27
  • $\begingroup$ @user42582 I've changed $f$ to uppercase to distinguish it from the index $f$. $F$ is a coefficient. $\endgroup$
    – Xavier
    Commented Sep 17, 2016 at 14:41
  • $\begingroup$ please provide the desirable output for $F^{abcdef} X_d Y_{ac} X_e Y_{fb}$ $\endgroup$
    – user42582
    Commented Sep 17, 2016 at 15:01

2 Answers 2

0
$\begingroup$

To begin with, it is easier to work with indexes than with subscripts/exponents, so the expression in the beginning of the initial question should be rewritten as

expr = (X[a] Y[c, d] + X[d] Y[a, c]) (X[b] Y[e, f] + X[e] Y[f, b])

Notice, how I have not included the coefficient $F^{abcdef}$ in the definition of expr above for reasons that will become apparent shortly.

Now, if we expand expr above we get

In[2} := exp=Expand[expr]
Out[2] := X[b] X[d] Y[a, c] Y[e, f] + X[a] X[b] Y[c, d] Y[e, f] + 
+ X[d] X[e] Y[a, c] Y[f, b] + X[a] X[e] Y[c, d] Y[f, b]

Note that all four terms appear in the right order (X[a_]X[b_]Y[c__]Y[d__]) but we are still missing the coefficients.

Define,

rule[a_, b_, c_, d_, e_, f_] := Module[{Aa, Bb, Cc},

 Aa = Permutations[{a, b}];
 Bb = Permutations[{c, d}];
 Cc = Permutations[{e, f}];

 Flatten[
  Outer[
   F[Sequence @@ #1, Sequence @@ #2, Sequence @@ #3] &, Aa, Bb, Cc, 1], 2]

]

and then apply the following rule on exp (see above)

exp /. {
 X[a_] X[b_] Y[c_, d_] Y[e_, f_] :> 
  X[a] X[b] Y[c, d] Y[e, f] Plus @@ rule[a, b, c, d, e, f]
}

to obtain what is (hopefully) the canonical order.

$\endgroup$
1
  • $\begingroup$ This code gives F only a copy of the indices of X and Y, but not the permutation of (a,b,c,d,e,f) under which the indices of X and Y transform. Anyway, its really helpful, thanks a lot! $\endgroup$
    – Xavier
    Commented Sep 18, 2016 at 11:20
0
$\begingroup$

This code solves the problem:

(*assignments*) 
indiceLength = 6; fields = {X, Y};
expr = F[a, b, c, d, e, f] (X[a] Y[c, d] + X[d] Y[a, c]) (X[b] Y[e, f] + X[e] Y[f, b]) //Expand;
indices = Take[CharacterRange["a", "z"], indiceLength] // ToExpression;

(*temporary indices of the fields*)
temporaryIndices[fieldsTemp_] := 
  fieldsTemp /. ((# -> List) & /@ fields) // Flatten;

(*including all possible combinations for the same type of fields*)
interchanges[a_, b_, c_, d_, e_, f_] := Module[{permuteX, permuteY},
   permuteX = Permutations[{a, b}];
   permuteY = Permutations[{{c, d}, {e, f}}];
   Partition[
    Flatten[Outer[Join, permuteX, permuteY, 1]], indiceLength]
   ];
(*set the rule*)
rule[fieldsTemp_] :=
 Thread[indices -> #] & /@
  (Permute[indices, #] & /@
    ((FindPermutation[#, indices]) & /@
      interchanges[Sequence @@ temporaryIndices[fieldsTemp]]))

(*the result*)
(Plus @@ (# /. rule[Cases[#, X[__] | Y[__]]])) & /@ expr;
DeleteCases[%, X[_] | Y[__], Infinity]

The result is

F[a, b, c, d, e, f] + F[a, b, e, f, c, d] + F[a, d, e, f, b, c] + 
 F[a, f, c, d, b, e] + F[b, a, c, d, e, f] + F[b, a, e, f, c, d] + 
 F[b, d, e, f, a, c] + F[b, f, c, d, a, e] + F[c, a, d, b, e, f] + 
 F[c, b, d, a, e, f] + F[c, f, d, a, b, e] + F[c, f, d, b, a, e] + 
 F[e, a, f, b, c, d] + F[e, b, f, a, c, d] + F[e, d, f, a, b, c] + 
 F[e, d, f, b, a, c]
$\endgroup$

Your Answer

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

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