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Consider the spacetime bivectors $\gamma_{12}$ and $\gamma_{30}$ with the metric $\eta=(1, -1, -1, -1)$. I would calculate their dot product to be

$\gamma_{12} \cdot \gamma_{30} = \frac{1}{2}\left(\gamma_{12} \gamma_{30} + \gamma_{30} \gamma_{12}\right) = \frac{1}{2}\left(\gamma_1 \wedge \gamma_2 \wedge \gamma_3 \wedge \gamma_0 + \gamma_3 \wedge \gamma_0 \wedge \gamma_1 \wedge \gamma_2 \right) = \frac{-I_{0123} - I_{0123}}{2} = -I_{0123}$, where $I_{0123}$ is the pseudoscalar.

However, as the two bivectors should be orthogonal I would have expected the result to be zero, which is the case for bivector pairs sharing an index like $\gamma_{12}$ and $\gamma_{23}$. Is my calculation wrong or what is wrong with my reasoning?

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This is very incorrect. First, this dot product should yield a scalar, not a pseudoscalar. Second, you're probably getting the formula $$ a\cdot b = \frac12(ab + ba) $$ from the specific case that $a$ and $b$ are vectors.

If $a_1,\dotsc,a_k$ and $b_1,\dotsc,b_k$ are vectors, then for two $k$-blades we can find that $$ (a_k\wedge a_{k-1}\wedge\dotsb\wedge a_1)\cdot(b_1\wedge b_2\wedge\dotsb\wedge b_k) = \det\bigl(a_i\cdot b_j\bigr)_{i,j=1}^k. $$ So when $k=2$ $$ (a_1\wedge a_2)\cdot(b_1\wedge b_2) = (a_2\cdot b_1)(a_1\cdot b_2) - (a_1\cdot b_1)(a_2\cdot b_2). $$

In your particular case though its even simpler. For $\gamma_{12}\cdot\gamma_{30}$ the result is supposed to be a scalar, and in fact $$ \gamma_{12}\cdot\gamma_{30} = \langle\gamma_1\gamma_2\gamma_3\gamma_0\rangle. $$ But clearly $\gamma_1\gamma_2\gamma_3\gamma_0$ is a pseudoscalar and has no scalar part, so $\gamma_{12}\cdot\gamma_{30} = 0$.


I can't say much more without knowing your particular definition of $\cdot$ or what you're learning from.

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  • $\begingroup$ I am studying 'Spacetime algebra as a powerful tool for electromagnetism' (doi.org/10.1016/j.physrep.2015.06.001). On p. 15 they define the bivector dot product of two bivectors F and G: $$F \cdot G = \frac{FG + GF}{2}$$. So, with $\gamma_{12}$ and $\gamma_{30}$ I am referring to grade 2 spacetime bivectors (not vectors) constructed from orthonormal Minkowski 4-vectors $\gamma_{\mu\nu}=\gamma_\mu \wedge \gamma_\nu$, where $\gamma_{\mu} \cdot \gamma_{\nu} = \eta_{\mu\nu}$ with the metric $\eta_{\mu\nu}$ with a signature (+, -, -, -), such that $\gamma_0^2 = 1$ and $\gamma_j^2 = -1$. $\endgroup$ Commented Apr 14, 2023 at 10:47
  • $\begingroup$ @JürgenAnklam (I don't see it until p. 26) Something is wrong in this paper. That "dot product" is the symmetric part of the product of two bivectors, which is equal to the sum of their scalar product and wedge product $\langle FG\rangle + F\wedge G$. This is nonstandard notation, but from skimming the paper it seems to be what they intend. However, in Eq. 3.22 they are clearly using the dot product of two bivectors to represent the scalar product (or perhaps a contraction, which is the same in this case), which is inconsistent. $\endgroup$ Commented Apr 14, 2023 at 15:25
  • $\begingroup$ It seems that your expectation is also that it should be the scalar product. Using the given definition though your calculation is correct, it just doesn't mean what you thought it meant. $\endgroup$ Commented Apr 14, 2023 at 15:26
  • $\begingroup$ @JürgenAnklam The footnote on p. 17 is incorrect. They say they prefer the signature $({+}{-}{-}{-})$ of $\mathrm{Cl}_{1,3}$ because it reproduces $\mathrm{Cl}_{3,0}$ as the even subalgebra $\mathrm{Cl}^+_{1,3}$; however $\mathrm{Cl}_{3,0}$ is easily seen to also be the even subalgebra of $\mathrm{Cl}_{3,1}$, and furthermore it is a general fact of Clifford algebras that $\mathrm{Cl}^+_{p,q} \cong \mathrm{Cl}^+_{q,p}$. $\endgroup$ Commented Apr 14, 2023 at 16:06
  • $\begingroup$ Sorry, the page reference was incorrect. I meant the table on page 26 that you found. $\endgroup$ Commented Apr 16, 2023 at 6:54

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