# geometric product between nonorthogonal basis and its reciprocal basis

This is related to the question of wedge product between nonorthogonal basis and its reciprocal basis in geometric algebra.

If {$e_i$} is a set of basis that are not necessarily orthogonal, and {$e^i$} is the corresponding reciprocal basis, we have $$e_i \cdot e^j = \delta_j^i$$

why do we also have the geometric product sum of
$$\sum_i e_i e^i = n$$

This would be obvious if $e_i$ are orthogonal, but is it also trivial for any basis? From the definition of reciprocal frame, we only know the inner product, why the outer product part is zero?

The above is used to show (4.106) in the book of Geometric Algebra for Physicists. ($e_i e^i \wedge A_r = e_ie^iA_r - e_ie^i \cdot A_r = (n - r) A_r$. Is this only true for orthogonal basis?)

• You ask later "Is this only true for orthogonal bases?" but earlier you phrase it as "why do we also have..." as if you knew it were true. Am I right in supposing that Doran and Lasenby always use orthonormal bases, and you are just curious about the general case? – rschwieb May 30 '14 at 15:41
• I've also forgotten which inner product is used there. Can you refresh my memory? – rschwieb May 30 '14 at 15:57

## 1 Answer

The geometric product sum equals the inner product sum since the wedge product sum part is zero, as shown in wedge product between nonorthogonal basis and its reciprocal basis in geometric algebra.