# Broken Symmetries (Weinberg p215)

1. Aug 16, 2008

### 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 $$[t_i,x_a]\neq0$$?

2. Aug 16, 2008

### Avodyne

Because the http://en.wikipedia.org/wiki/Baker-Campbell-Hausdorff_formula" [Broken] 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.

Last edited by a moderator: May 3, 2017
3. Aug 17, 2008

### nrqed

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.

4. Aug 17, 2008

### Final

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

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