- #51

strangerep

Science Advisor

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- 1,130

Corrected, thanks.There's a sign error in the "positive alpha" case.

If the transformations form a group which is continuous and differentiable in a real parameter (i.e., ##\eta## here), we must allow ##\eta## to takes arbitrary real values.I'm not convinced that this is accurate (without an assumption like my 1b). Is {0} really the only subset of ℝ that's closed under the operation ##(u,v)\mapsto (u+v)/(1-uv)##?

Consider:

$$

e^{\eta K} e^{\eta K} ~=~ e^{2\eta K}

$$ so by composing transformations I can get arbitrarily large rapidities. Any restriction on the values of ##\eta## must also be compatible with arbitrarily many such compositions, else we don't have a group. The only such valid restriction (afaict) is the restriction to ##\eta=0##, i.e., a group consisting trivially of the identity and nothing else.