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?)
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.