Broken Symmetries (Weinberg p215)

[Final]

Hi...
A group G is proken to a subgroup H. Let $$t_{\alpha}$$ the generator of G and
$$t_i$$ the generator of H. The t_i form a subalgebra. Take the x_a to be the other indipendent generator of G.
Why any finite element of G may be expressed in the form $$g=exp[i\xi_ax_a]exp[i\theta_i t_i]$$ even if $$[t_i,x_a]\neq0$$?

 

[Avodyne]

Because the http://en.wikipedia.org/wiki/Baker-Campbell-Hausdorff_formula" says that
$$\exp[i\xi_a x_a]\exp[i\theta_i t_i] = \exp[i\tilde\xi_a x_a + i\tilde\theta_i t_i]$$
where the new parameters are complicated functions of the old ones.

 

[nrqed]

Hi...
A group G is proken to a subgroup H. Let $$t_{\alpha}$$ the generator of G and
$$t_i$$ the generator of H. The t_i form a subalgebra. Take the x_a to be the other indipendent generator of G.
Why any finite element of G may be expressed in the form $$g=exp[i\xi_ax_a]exp[i\theta_i t_i]$$ even if $$[t_i,x_a]\neq0$$?

By definition of a group, you can always write the product of two group elements as a third group element. That's all there is to it.

 

[Final]

By definition of a group, you can always write the product of two group elements as a third group element. That's all there is to it.

I don't understand... My problem is to express a generic element of the group $$g=exp[i\xi_ax_a+i\theta_i t_i]$$ as the product of 2 element of the form
$$g_1=exp[i\xi_ax_a] \ g_2=exp[i\theta_i t_i]$$.

Thank you