[GiNaC-list] Are Clifford units of different bases equal?

[Vladimir Kisil]

		Dear Javier,

		I am using 1.4 branch of GiNaC on my computer and your example
  compiles but do not run on it throwing the exception:

> terminate called after throwing an instance of 'std::invalid_argument'
>  what():  clifford_unit(): metric for Clifford unit must be of type
>  tensor, matrix or an expression with two free indices
> Aborted

 The reason of this (and misunderstanding which you expressed) is your
  calls:

> ex vector1 = lst_to_clifford(lst(a1,b1,c1), mu, basis1),
>    vector2 = lst_to_clifford(lst(a1,b1,c1), nu, basis2);

 Here how GiNaC tutorial describes *two* versions of lst_to_clifford:

>         ex lst_to_clifford(const ex & v, const ex & mu,  const ex &  metr,
>                            unsigned char rl = 0);
>         ex lst_to_clifford(const ex & v, const ex & e);
>
> which converts a list or vector `v = (v~0, v~1, ..., v~n)' into the
> Clifford number `v~0 e.0 + v~1 e.1 + ... + v~n e.n' with `e.k' directly
> supplied in the second form of the procedure. In the first form the
> Clifford unit `e.k' is generated by the call of `clifford_unit(mu,
> metr, rl)'. 

  In other words if the code which you provide is runnable on your
  (presumably old) version of GiNaC it calls the first version. Then a
  *new* clifford_units is created for both vector1 and vector1. Both new
  clifford_units have the same metric which they obtain from basis1 and
  basis2, and both they the same representation labels 0 (the default
  value). The values of representation labels in basis1 and basis2 are
  not consulted in this form of the call.

  If you replace your calls to:

> ex     vector1 = lst_to_clifford(lst(a1,b1,c1), basis1),
>        vector2 = lst_to_clifford(lst(a1,b1,c1), basis2);

  then both metrics and representation labels are inherited from basis1
  and basis2 and output of your example will be:

> [[a1],[b1],[c1]].nu*e~nu+2*e~mu*[[a1],[b1],[c1]].mu
> [[a1],[b1],[c1]].nu*e~nu+2*[[a1],[b1],[c1]].nu*e~nu
> e~0*a1+2*e~2*c1+2*e~0*a1+2*b1*e~1+b1*e~1+e~2*c1
> e~0*a1+2*e~2*c1+2*e~0*a1+2*b1*e~1+b1*e~1+e~2*c1

 as you seems initially expected.

 Best wishes,
 Vladimir
-- 
Vladimir V. Kisil     email: kisilv at maths.leeds.ac.uk
--                      www: http://maths.leeds.ac.uk/~kisilv/