2
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If I use a replacement rule like

Times[2, Plus[Times[AuRr, BuRb], Times[AuRb, BuRr]], h10] /. 
  {Times[r_, Plus[Times[x_, y_], Times[v_, w_]], h10] :> 
     r*TensorProduct[(TensorProduct[x, y] + TensorProduct[v, w]), h10]}

I get correctly

(* 2 (AuRb \[TensorProduct] BuRr + AuRr \[TensorProduct] BuRb) \[TensorProduct]h10 *)

Nevertheless, if I try to do the same thing more general, i.e. allowing for a general last coefficient, not specifically h10:

Times[2, Plus[Times[AuRr, BuRb], Times[AuRb, BuRr]], h10] /. 
  {Times[r_, Plus[Times[x_, y_], Times[v_, w_]], h_] :> 
    r*TensorProduct[(TensorProduct[x, y] + TensorProduct[v, w]), h]}

I get the wrong result

 (* 2 h10 (AuRb \[TensorProduct] BuRr + AuRr \[TensorProduct] BuRb) *)

Any ideas about what is going wrong here, would be much appreciated!

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  • $\begingroup$ Please don't add bugs tag until the community has conformed it. BTW it's almost impossible to find a bug in core language of Mathematica. $\endgroup$ – xzczd Dec 29 '15 at 9:27
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Mathematica automatically factors out constants from TensorProducts:

TensorProduct[a, 2 b] (* returns 2 a\[TensorProduct]b *)

If you consider

Times[2, Plus[Times[AuRr, BuRb], Times[AuRb, BuRr]], h10] /. 
 {Times[r_, Plus[Times[x_, y_], Times[v_, w_]], h_] 
  :> 
  Echo[r]*TensorProduct[(TensorProduct[x, y] + TensorProduct[v, w]), Echo@h]
 }

you will see that the pattern-matcher sets r = h10 and h = 2, which accounts for why the constants end up outside the expression. Then it ends up with a TensorProduct of an expression with 1, and that gets rid of the outermost TensorProduct.

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  • $\begingroup$ Thanks a lot for the explanation! I now simply add the condition /; Element[r, Reals] and everything works perfectly $\endgroup$ – jak Dec 29 '15 at 9:13

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