Suppose you're given a 4 tuple and told that its scalar product with any 4-vector is a lorentz scalar. How do I show that this implies the 4-tuple is a 4-vector? Thanks
It can't be done unless you're told that its scalar product with any 4-vector is a scalar. If the given 4-tuple is x and the (arbitrary) 4-vector is y, [tex]x_\mu y^\mu=x'_\mu y'^\nu=x'_\mu\Lambda^\mu{}_\nu y^\nu[/tex] [tex]x_\mu=\Lambda^\mu{}_\nu x'_\mu[/tex] Now do some raising and lowering of indices and apply a Lorentz transformation to solve for x', and you're done. This post should help with the notation.
Just apply the definitions on the <scalar> product. Denoting by F the matrix the 4-tuple (index down) uses to transform under a Lorentz group element, you'll end with a matrix equation [tex] \mathbb{F} \Lambda = \mbox{1}_{4\times 4} [/tex]. Since [itex] \Lambda [/itex] is invertible, the conclusion follows easily.
Here's how I would do it in matrix notation: [tex]x^T\eta y=x'^T\eta y'=x'^T\eta\Lambda y[/tex] [tex]x^T\eta=x'^T\eta\Lambda[/tex] [tex]\eta x=\Lambda^T\eta x'[/tex] [tex]x'=\eta^{-1}(\Lambda^T)^{-1}\eta x=\eta^{-1}(\eta\Lambda\eta^{-1})\eta x=\Lambda x[/tex] The fact that [tex](\Lambda^T)^{-1}=\eta\Lambda\eta^{-1}[/tex] follows from the definition of a Lorentz transformation, [tex]\Lambda^T\eta\Lambda=\eta[/tex]. Just multiply both sides with [tex]\eta^{-1}[/tex] from the right. I suggest that you be a bit more careful with the terminology. A 4-tuple can't ever be 4-vector. In order to define a 4-vector you must specify a 4-tuple for each coordinate system. It's the assignment of 4-tuples to coordinate systems that defines a 4-vector, not a single 4-tuple. The assignment is of course usually done by specifying the 4-tuple that you want to associate with a specific coordinate system, and then explicitly stating that the 4-tuples associated with all the other coordinate systems are given by the tensor transformation rule.
Yes, that was sloppy of me. The 4-tuple I had in mind only has to be a continuous function of the coordinate transformation. But I wasn't thinking of it being confined by the tensor transformation rule.