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In Peskin Schroeder pag.696 a Higgs mechanism for an $SU(3)$ gauge theory with a scalar field $\phi$ in the adjoint representation is presented. The covariant derivative of $\phi$: $$ D_{\mu}\phi_{a} = \partial_{\mu}\phi_{a}+ g f_{abc}A^{b}_{\mu}\phi_{c}\tag{20.32}$$ defining the quantity $$\Phi=\phi_c t^c\tag{20.34}$$ We consider the expansion of $\Phi$ around two vacuum choice: $$1) \,\,\,\Phi_0 = v \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix}$$ $$2) \,\,\,\Phi_0 = v \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ In the first case the unbroken generators are: $T_1,T_2,T_3,T_8$

In the second case the unbroken generators are: $T_3,T_8$

Where $T_i$ are Gell-Mann matrices according to the normalization which here is $T_i = \lambda_i/2$.

In the book is said that $SU(3)$ breaks spontaneously to $S U (2) \times U(1)$ in the first case and $U(1) \times U(1)$ in the second case.

I really can't understand how to prove that $T_1,T_2,T_3,T_8$ generate $S U (2) \times U(1)$ and $T_3,T_8$ generate $U(1) \times U(1)$

Starting from the simplest how a 3 x 3 matrix $$T_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ can generate $U(1)$, (ot the same for $T_8$)?

A basis for the Lie Algebra of $SU(2)$ (aka its infinitesimal generators):

$$ \left(\ \frac{1}{2} \sigma_{1}, \ \frac{1}{2} \sigma_{2}, \ \frac{1}{2} \sigma_{3} \right) $$
I can recognize that inside the matrices that remain I can found $\sigma_{1}, \sigma_{2}, \sigma_{3}$ $$T_1 = \begin{bmatrix} (\sigma_1)_{00} & (\sigma_1)_{01} & 0 \\ (\sigma_1)_{10} & (\sigma_1)_{11} & 0 \\ 0 & 0 & 0 \end{bmatrix} T_2 = \begin{bmatrix} (\sigma_2)_{00} & (\sigma_2)_{01} & 0 \\ (\sigma_2)_{10} & (\sigma_2)_{11} & 0 \\ 0 & 0 & 0 \end{bmatrix} T_3 = \begin{bmatrix} (\sigma_3)_{00} & (\sigma_3)_{01} & 0 \\ (\sigma_3)_{10} & (\sigma_3)_{11} & 0 \\ 0 & 0 & 0 \end{bmatrix} $$ However the matrices are not $2 \times 2$ but $3 \times 3$. Recalling the definition of $SU(2)$: $$SU(2) \equiv \{ M \in GL(n, \mathbb{C}) | M^{\dagger} \mathbb{1}_2 M = \mathbb{1}_2, \,\, det(M)=1 \} $$ We notice that: $$det(T_1)=det(T_2)=det(T_3)=0$$ Moreover thare is a dimensional problem in performing $T_1^{\dagger} \mathbb{1}_2 T_1,$ $\,\,$ $T_2^{\dagger} \mathbb{1}_2 T_2,$ $\,\,$ $T_3^{\dagger} \mathbb{1}_2 T_3$.

I'm quite lacking in group theory so I would need a step-by-step answer.

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    $\begingroup$ You need consider determinant of group element, not generators!! Group element is exponent of generators with arbitrary coefficients! $\endgroup$ – Nikita Jan 14 at 10:44
  • $\begingroup$ @Nikita thank you very much, my final doubt is that to be SU(2) $e^{- i \alpha_a T^a}$ has to be in $GL(2, \mathbb{C})$, however $e^{- \alpha_a T^a}$ is a $3 \times 3$ matrix. I understood that you can represent the SU (2) group with 3 x 3 matrices but in that case the generators are no longer Pauli's matrices $\endgroup$ – Stefano Barone Jan 14 at 23:18
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Using concrete realization of this generators, you can easily check that $T_1, T_2, T_3$ are exactly $SU(2)$ and $T_8$ commute with them and so gives you $U(1)$.

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  • $\begingroup$ Thank you so much for your answer. Since I still had many doubts I modified the question so that you can see what is not yet clear, as in a comment there was no space. $\endgroup$ – Stefano Barone Jan 14 at 9:38
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    $\begingroup$ @Stefano Barone. Have you reviewed SU(3), to see how the isospin su(2) subalgebra fits into su(3)? You just search for generators commuting with the two alternative vacua, respectively. Generators are traceless, instead of their exponentials which are unimodular. $\endgroup$ – Cosmas Zachos Jan 14 at 15:19

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