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Apr 17, 2019 at 6:30 vote accept 256ABC
Apr 17, 2019 at 4:12 comment added Hans Moleman Of course the multiplication $c_1 c_2$ is technically fine. The weird part is dividing by $dx^i dx_i$ and turning $dx^\sigma/dx_i$ into a Jacobian. It's not! More generally there's no need to introduce line elements $dx^i$ or $dx^\sigma$ in this proof at all. When you prove that an inner product in $\mathbb{R}^3$ is invariant under $SO(3)$ you don't need it either.
Apr 17, 2019 at 3:56 comment added 256ABC Would you say that if $\vec{a}.{b} = \vec{c}.\vec{d}$ and $\vec{a_1}.{b_1} = \vec{c_1}.\vec{d_1}$, then $(\vec{a}.{b})(\vec{a_1}.{b_1}) = (\vec{c}.\vec{d})(\vec{c_1}.\vec{d_1})$ is true ? And if so, I am just wondering why that doesn't translate when we are talking about $C_1 C_2$ ?
Apr 17, 2019 at 2:04 history answered Hans Moleman CC BY-SA 4.0