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I have a function, explicit that takes dot product of two symbols, and replaces it with repeated dummy indices generated by Unique[]. 

explicit[expr_] := 
  expr //. {Dot[a_, b_] :> (Subscript[a, #] Subscript[b, #]) &[
  Unique[]]};

So that if my input is v.w, then the output is $v_{$3}w_{$3}$ which is good.

The problem is that if there is a single term multiplying two pairs of dot products, I don't know how to get Unique[] to generate a new symbol for each pair. That is, if I input c.d e.f the output is $c_{$3}d_{$3}e_{$3}f_{$3}$, which is bad. I want $c_{$3}d_{$3}e_{$4}f_{$4}$, with new repeated subscripts. How do I modify my code?

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  • $\begingroup$ By the way I don't think you need ReplaceRepated here but I left it in my answer since that wasn't the focus. $\endgroup$
    – Mr.Wizard
    Commented Mar 8, 2013 at 3:52

1 Answer 1

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You merely need to watch your operator precedence. Here with brackets to group the RHS properly:

explicit[expr_] := 
  expr //. {Dot[a_, b_] :> ((Subscript[a, #] Subscript[b, #]) &[Unique[]])};

explicit[c.d e.f]

Mathematica graphics

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  • $\begingroup$ How did you know that this was the way to solve the problem? Even with answers to my other questions on SE, I still don't seem to get it... it just seems like magic half the time, and I just move on... $\endgroup$
    – QuantumDot
    Commented Mar 8, 2013 at 5:34
  • $\begingroup$ @QuantumDot Well, I know that Unique[] needs to be evaluated for each occurrence of Dot. For that I need some head that holds it unevaluated. You had two in your original code: RuleDelayed and Function. Function is already being used to distribute a single instance of Unique[] to multiple expressions. That leaves RuleDelayed. (Which is exactly the right tool.) Now I must make sure that Unique[] is part of the RHS of RuleDelayed. I can use these methods to check how Mathematica "sees" this, or just my experience. :-) $\endgroup$
    – Mr.Wizard
    Commented Mar 8, 2013 at 5:40

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