# About the Lorentz transformation in Spacetime and Geometry

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?

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.