in the context of geometric algebra, what's the wedge product between basis and reciprocal basis? say, if {$e_i$} is a set of basis that are not necessarily orthogonal, and {$e^i$} is the corresponding reciprocal basis, we have $$e_i \cdot e^j = \delta_j^i$$
Do we have $$e_i \wedge e^i = 0$$ for any $i$? (EDIT: No. It has to be sum over $i$).
Or $$\sum_i e_i \wedge e^i = 0 $$
I tried to use $$e^i = (-1)^{i-1}e_1 \wedge e_2 \wedge ... \wedge \hat{e_i} \wedge ...\wedge e_nE_n^{-1}$$ or $$e^i = g^{ik}e_k$$ But cannot see it through via either way.
(this question is from the geometric algebra answer of why not the Ricci tensor is the contraction of first and second indices of Riemann tensor. )
We will use the summation convention. Since $\{e^i\}$ is a basis, we have $e^i=e^i\cdot e^j e_j$. Then \begin{align*} e^i\wedge e_i &=(e^i\cdot e^j e_j)\wedge e_i\\ &=e^i\cdot e^j e_j \wedge e_i\\ &=-e^i\cdot e^j e_i\wedge e_j\\ &=-e^j\cdot e^i e_j\wedge e_i\\ &=-e^i\wedge e_i. \end{align*} We must conclude that $e^i\wedge e_i=0$.