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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. )

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  • $\begingroup$ I'm not sure how you could try and not see how it works... are you rusty with wedge multiplication, maybe? $\endgroup$ – rschwieb May 27 '14 at 20:32
  • $\begingroup$ it is trivial for orthogonal basis. But it has to work for nonorthogonal basis to answer the symmetry of riemann tensor indices as in the linked question. $\endgroup$ – ahala May 27 '14 at 20:48
  • $\begingroup$ Personally, I would've asked the question, "Why is $\nabla \wedge x = 0$?" as it suggests an approach through calculus. $\endgroup$ – Muphrid May 27 '14 at 20:50
  • $\begingroup$ @ahala Ah, since you proposed the orthogonal definition of the dual basis, I thought you were more interested in why it worked. If you want one for arbitrary nonorthogonal bases it seems hopelessly complicated. $\endgroup$ – rschwieb May 27 '14 at 20:58
  • $\begingroup$ @Muphrid, thanks. I thought your comments in another question deserves a separate question so I asked here. Are you saying $\sum_i e^i \wedge e_i = \nabla \wedge x$, therefore it is zero? $\nabla \wedge x = 0$ is the starting point rather than the results? $\endgroup$ – ahala May 27 '14 at 21:15
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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 $e^i\wedge e_i=0$.

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  • $\begingroup$ I don't see the meaning of these equations without parentheses. The dot product and geometric product don't associate. $\endgroup$ – mr_e_man Apr 20 '18 at 4:50

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