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Let be $G$ a compact group, symmetry of the theory I am working with. $G$ is broken into one subgroup $H$. I define the generators of G as $T_A = \{T_a,T_\hat{a}\}$, where the first are the unbroken generators (they are generators of $H$), while the latter are the broken ones.

The commutation relations are $$ [T^a,T^b] = if^{ab}_cT^c\qquad (1)\\ [T^a,T^\hat{b}] = if^{a\hat{b}}_\hat{c}T^\hat{c}\qquad (2)\\ [T^\hat{a},T^\hat{b}] = if^{\hat{a}\hat{b}}_\hat{c}T^\hat{c} + if^{\hat{a}\hat{b}}_cT^c\qquad (3) $$

I can rewrite $(1)$ using the adjoint representation $$ [T^a,T^b] = T^c(t^a_\text{Ad})_c^{\ b}\qquad (1.bis) $$ and (2) using a not specified representation of $H$ $$ [T^a,T^\hat{b}] = T^\hat{c}(t_\pi^a)_\hat{c}^{\ \hat{b}}\qquad (2.bis) $$

$t^a_\pi$ is a representation of the generators of $H$ since it respects commutation relations of $H$ (use Jacobi identity to see it). We call this representation r$_\pi$.

I am reading this reference where there is written (pag. 35) that r$_\pi$ can be identified by looking at the decomposition under $H$ of the adjoint of $G$, namely at

Ad$_G$ = Ad$_H \oplus$ r$_\pi$

Can someone explain me this decomposition? Why can we do it?

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Let $\mathfrak{g,h}$ denote the respective Lie algebras of $G,H$. Note that $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{g}/\mathfrak{h}$ as vector spaces.

Obviously, there is the adjoint representation of $\mathfrak{h}$ on the whole of $\mathfrak{g}$. Also, the adjoint action of $\mathfrak{h}$ on itself is a subrepresentation, since the algebras of Lie subgroups must be Lie subalgebras. Furthermore, your eq. (2) means that $\mathfrak{g}/\mathfrak{h}$ is also closed under the action of $\mathfrak{h}$ by the Lie bracket, as it implies $[k,h]\in\mathfrak{g}/\mathfrak{h}$ for every $k\in\mathfrak{g}/\mathfrak{h}$ and every $h\in\mathfrak{h}$. Hence, the map $$ r_\pi : \mathfrak{h}\to\mathrm{End}(\mathfrak{g}/\mathfrak{h}), h\mapsto [\dot{},h]$$ is a subrepresentation of the total representation of $\mathfrak{h}$ on $\mathfrak{g}$. Since $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{g}/\mathfrak{h}$, we have that that representation indeed decomposes as $\mathrm{ad}_\mathfrak{h}\oplus r_\pi$.

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  • $\begingroup$ Why can you write $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{g}/\mathfrak{h}$ ? Due to the commutation relations, $\mathfrak{g}/\mathfrak{h}$ is not a sub-algebra... $\endgroup$ – apt45 Apr 1 '16 at 12:36
  • $\begingroup$ @FrancescoS: I said as vector spaces, not as algebras. $\endgroup$ – ACuriousMind Apr 1 '16 at 13:09

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