Is the vectorial representation of the Lorentzian Group unitary?

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