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Nov 16, 2021 at 8:53 comment added Martin Johnsrud As my edit above says, it seems that the original paper only require the weaker claim of "in a neighbourhood of the identity of $G$". I guess the strong claim might not be true, nor needed, but i have to look closer at it.
Nov 15, 2021 at 22:19 comment added Valter Moretti However it should be a known fact if it is true. For instance it is known that if the group is compact then the exponential map is surjective (not necessarily injective outside a neighborhood of the identity).
Nov 15, 2021 at 22:13 comment added Martin Johnsrud Thank you for considering the question, it has been bugging me. It is always good to hear that your issues are not trivial.
Nov 15, 2021 at 22:07 comment added Valter Moretti Have a look at Barut Raczak's book on representation theory.
Nov 15, 2021 at 22:05 comment added Valter Moretti I am not saying that it is not true. Maybe it is. But the proof is not trivial.
Nov 15, 2021 at 22:01 comment added Martin Johnsrud Well, that is not strong enough. Weinberg claims "Because $t_a$ and $x_i$ span the Lie algebra of $G$, any finite element of $G$ may be expressed in the form $g = \exp(i \xi_i x_i)\exp(i \theta_a t_a)$". Are you saying this is not true? As far as i can tell, this is necessary for the construction of the realization of the Goldstone bosons.
Nov 15, 2021 at 21:45 comment added Valter Moretti No, not every element. However every element is a finite product of thes products of two factors only. This is consequence of the fact that the group is connected.
Nov 15, 2021 at 21:39 comment added Martin Johnsrud Is this enough to conclude that any element of $G$ can be written in this way?
Nov 15, 2021 at 21:35 history answered Valter Moretti CC BY-SA 4.0