I am working on non linear realization of Goldstone bosons, as is done by Weinberg in section 19.6 of Quantum theory of fields, volume II.

We have a real, compact and connected Lie group $G$ with as subgroup $H$. Let $t_a$ be the generators of $H$, and $x_i$ be the generators left over (broken generators) such that they together span the Lie group of $G$, $\mathfrak g$. A general element of $G$ can then, to my understanding, be written $$ g = \exp\{i(\xi_i x_i + \theta_a t_a)\}. $$ However, Weinberg claims we can write (eq 19.6.12) $$ g = \exp\{i \xi_i' x_i \} \exp\{i\theta_a' t_a\}. $$ This is also used in other sources that tackles the same material. Why is this true? It seems plausible to me, by looking at $SO(3)$ as an example, however I have not seen a proof nor any real justification for this form.

Edit: So I investigated the origin of the claim, which as Cosmas states in the comments is this paper by CWZ. The statement here is qualified as "in some neighbourhood of $G$, any element $g \in G$ can be uniquely decomposed as $g = \exp(i \xi_i xi)\exp(i \theta_a t_a)$. This is a weaker statement, and seems straight forward from the inverse function theorem.

  • $\begingroup$ The statement entered the mainstream of physics in CWZ. $\endgroup$ Nov 15, 2021 at 23:19
  • $\begingroup$ You are asking for a proof the the CBH map $(\xi',\theta ')\mapsto (\xi,\theta)$ is invertible, right? $\endgroup$ Nov 15, 2021 at 23:39
  • $\begingroup$ Or that $(\xi', \theta') \rightarrow G$ is surjective, I think. $\endgroup$ Nov 16, 2021 at 6:38
  • $\begingroup$ Right. But this is a strictly math question, best covered in the MSE. If you've got the stomach for it, you ought to go through Fulton & Harris. $\endgroup$ Nov 16, 2021 at 14:21

1 Answer 1


Both are true actually. In a neighborhood of the identity both maps define local diffeomorphisms from the tangent space at the identity element to the Lie group. Notice that the maps are different: the same element of the group is determined by two different sets of values. One speaks of coordinates of first and second type. The proof relies on the inverse function theorem: in both cases the differential of the map at the origin of the tangent space is non singular and then the map is a local diffeomorphism.

  • $\begingroup$ Is this enough to conclude that any element of $G$ can be written in this way? $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ Nov 15, 2021 at 22:05
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    $\begingroup$ 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). $\endgroup$ Nov 15, 2021 at 22:19

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