Is the vectorial representation of the Lorentzian Group unitary?

1. Aug 10, 2011

IRobot

I am 99% sure it is not, but I would like to hear that from someone else to be more serene.

2. Aug 10, 2011

The_Duck

3. Aug 10, 2011

IRobot

Thanks I already was thinking with that argument, I did the calculation of $\Lambda \Lambda ^\dagger$ for some random Lorentz Matrix. Was just looking for a confirmation.

Last edited: Aug 10, 2011
4. Aug 10, 2011

dextercioby

It's because the Lorentz group (or its component connected to the identity) is non-compact. It's a classical result, a proof of which can be found in Cornwell's compendium.