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  • $\begingroup$ Is this enough to conclude that any element of $G$ can be written in this way? $\endgroup$
    – Martin Johnsrud
    Commented Nov 15, 2021 at 21:39
  • $\begingroup$ 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. $\endgroup$
    – Valter Moretti
    Commented Nov 15, 2021 at 21:45
  • $\begingroup$ 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. $\endgroup$
    – Martin Johnsrud
    Commented Nov 15, 2021 at 22:01
  • $\begingroup$ I am not saying that it is not true. Maybe it is. But the proof is not trivial. $\endgroup$
    – Valter Moretti
    Commented Nov 15, 2021 at 22:05
  • 2
    $\begingroup$ 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). $\endgroup$
    – Valter Moretti
    Commented Nov 15, 2021 at 22:19