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In Spacetime and Geometry by Sean Carroll, page 18, he said

"We will therefore introduce a somewhat subtle notation, by using the same symbol for both matrices, just with primed and unprimed indices switched. That is, the Lorentz transformation specified by ${\Lambda^{\mu '}}_{\nu}$ has an inverse transformation written as ${\Lambda^{\rho }}_{\sigma'}$. Operationally this implies ${\Lambda^{\mu }}_{\nu'}{\Lambda^{\nu '}}_{\rho}=\delta^{\mu}_{\rho}.$"

I have systematically learned about special relativity before and this really confuses me. Shouldn't the inverse of Lorentz transformation be like $\Lambda^{-1}=\eta\Lambda^{T}\eta$ ? Is me misunderstanding things or this book defines the inverse transformation in a weird way?

Qmechanic
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Lagrange629
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1 Answers1

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The effect of general coordinate transformation (passive point of view), or diffeomorphism (active point of view) can be written as $$ g^{\mu' \nu'} = \Lambda^{\mu'}_{\;\alpha}\Lambda^{\nu'}_{\;\beta} g^{\alpha\beta} $$ Multiplying left and right hand by $g_{\lambda'\mu'}$ $$ \delta_{\lambda'}^{\;\nu'} = \Lambda_{\lambda'}^{\;\beta} \Lambda^{\nu'}_{\;\beta} $$ So, $$ \Lambda_{\lambda'}^{\;\beta} = \left(\Lambda^{-1}\right)^{\beta}_{\;\lambda'} $$

For the specific case that $\Lambda$ is a Lorentz transformation (which actis on flat spacetime) then $$ g^{\mu' \nu'} = g^{\mu \nu} $$ and we can discard the primes altogether.