The "geometric product" of a frame and its reciprocal frame

Question

On p.102 of their book "Geometric Algebra for Physicists", Doran and Lasenby start an argument with "Since $e_{i}e^{i} = n$, ...". Here $e_{i}$ is an arbitrary (not necessarily orthonormal) frame in an n-dimensional inner product space of arbitrary nondegenerate signature, $e^{i}$ is the corresponding reciprocal frame, and the summation convention is in force. They have not proved this assertion (they have proved that $e_{i}·e^{i} = n$), and neither can I. Can anyone help, i.e supply a proof or a refutation?

Answer 1

Examine the bivector components of $e_i\wedge e^i$: $$ (e^j\wedge e^k)\cdot(e_i\wedge e^i) = (e^k\cdot e_i)(e^j\cdot e^i) - (e^j\cdot e_i)(e^k\cdot e^i) = e^j\cdot e^k - e^k\cdot e^j = 0. $$ It follows that $e_i\wedge e^i = 0$. Hence $$ e_ie^i = e_i\cdot e^i + e_i\wedge e^i = n + 0 = n. $$