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This is probably very trivial, but I can't find an argument, why the orthochronal transformations (i.e. those for which [itex]\Lambda^0{}_0 \geq 1[/itex]) form a subgroup of the Lorentz group, i.e. why the product of two orthochronal transformations is again orthochronal?

Since when you multiply two orthochronal matrices [itex]\Lambda[/itex] and [itex]\overline{\Lambda}[/itex] the term [tex]\Lambda\overline{\Lambda})^0{}_0 = \Lambda^0{}_0\overline{\Lambda^0{}_0} + \Lambda^0{}_1\bar{\Lambda^1{}_0} + \Lambda^0{}_1\overline{\Lambda^2{}_0} + \Lambda^0{}_3\overline{\Lambda^3{}_0},[/tex] where I can only say that [itex]\Lambda^0{}_0\overline{\Lambda^0{}_0} \geq 1[/itex] but I don't really know much about the other terms.

Since when you multiply two orthochronal matrices [itex]\Lambda[/itex] and [itex]\overline{\Lambda}[/itex] the term [tex]\Lambda\overline{\Lambda})^0{}_0 = \Lambda^0{}_0\overline{\Lambda^0{}_0} + \Lambda^0{}_1\bar{\Lambda^1{}_0} + \Lambda^0{}_1\overline{\Lambda^2{}_0} + \Lambda^0{}_3\overline{\Lambda^3{}_0},[/tex] where I can only say that [itex]\Lambda^0{}_0\overline{\Lambda^0{}_0} \geq 1[/itex] but I don't really know much about the other terms.

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