Weinberg quick question on chapter 2 zero mass

[Heffernana]

In Weinberg QFT vol 1, page 69, when deriving the little group structure for the case of zero mass can anyone explain the following:

The transformation W by definition leaves k^{\mu}=(0,0,1,1) invariant, i.e. W^{\mu}_{\nu}k^{\nu} = k^{\mu}. Why can you immediately deduce from this that for a time-like vector t^{\mu}=(0,0,0,1) whose length (noting Weinberg's bizarre placing of the time component at the end of the 4-vector) is t^2=-1 is invariant under the previous W transformation, i.e. that:

(Wt)^{\mu}(Wt)_{\mu}=t^{\mu}t_{\mu}=-1

Cheers

 

[Fredrik]

W is defined to be a member of a certain subgroup of the Lorentz group, so it's a Lorentz transformation. By definition, a Lorentz transformation W satisfies W^T\eta W=\eta, and this implies (Wt)^2=(Wt)^T\eta(Wt) =t^TW^T\eta Wt=t^T\eta t=t^2=-1. You don't have to use the assumption about the components of t for anything other than the final "=-1".