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?
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Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 26 '23 at 19:23
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$e_{i}.e^{i}=\sum_{k=1}^{n} e_{ik}e^{ik}$ but note that $<(0,...,1,...,0),(0,...,1,...,0)>=[0,...,1,...0].[0,...,1,...,0]^{T}=1=e_{ik}e^{ik}$ for all $k=1, 2,...,n$ – Safal Das Biswas May 26 '23 at 21:03
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More generally $e_{i}.e^{i}=\sum_{j=1}^{n} \sum_{k=1}^{n} \delta_{jk} e_{ijk}.e^{ijk}$ which can be renamed upto variable as above. – Safal Das Biswas May 26 '23 at 21:07
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@HappieHappieHappie I don't see how this is relevant. – Nicholas Todoroff May 26 '23 at 23:48
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Sir, I am assuming it's dot product. – Safal Das Biswas May 26 '23 at 23:50
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I wedge product is anyways 0 – Safal Das Biswas May 26 '23 at 23:50
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@HappieHappieHappie The question is about why $e_ie^i = n$; Dullard says they already understand why $e_i\cdot e^i = n$, so they need help understanding why $e_i\wedge e^i = 0$ and just saying "it does" is unhelpful. You also seem to be assuming that $e_i$ is the standard basis, but it's not, it's an arbitrary basis. – Nicholas Todoroff May 27 '23 at 05:25
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May be I am wrong for sure, Beacuse I won't read the question yet properly and I am telling inspite if I now read the question, I cannot answer the question becauss, I need to read those which I will learn from Answers here. – Safal Das Biswas May 27 '23 at 06:09
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I have no idea about these stuff, I saw "e" so I thought may be some running idex of e, since it saying "n" lol, that's my guess only. I still yet not read the question lol... – Safal Das Biswas May 27 '23 at 06:11
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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 $$ e_ie^i = e_i\cdot e^i + e_i\wedge e^i = n + 0 = n. $$
Nicholas Todoroff
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I note, by way of thanks, that Doran and Lasenby only extend the reciprocal frame to all grades after their "Since $e_{i}e^{i}=n$". Nor, I believe, do they mention any result equivalent to Theorem 6.22(c) of Macdonald's Linear and Geometric Algebra, which seems to be necessary to obtain the middle member of your answer. – Dullard May 29 '23 at 19:46
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@Dullard Oh! I was just checking Doran and Lasenby again to make sure they really don't give any tools to derive this, but actually Eq. 4.105 directly preceeding the $e_ie^i = n$ statement says $$e_i,e^i\cdot A_r = rA_r.$$ Now choose $A_r$ to be a pseudoscalar :) I also missed this the first time. – Nicholas Todoroff May 29 '23 at 21:42
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