Wilson Loop and spontaneous symmetry breaking.

[askalot]

I would like to ask about the case of:

$SU(2)\otimes U(1) \rightarrow U(1)\otimes U(1),$ spontaneous symmetry breaking.

It is given that the Wilson Loop:
$W \equiv exp[ig \oint dy H T^1]= diag(−1,−1,1).$

Where $y$ is the $S^1/Z^2$ fifth/extra dimension, $H = \frac{1}{g R}$ and $T^1$ is the first $SU(3)$ generator.
As we can check, the Wilson Loop does not commute with $SU(3)$ generators $T^6, T^7$ but it still commutes with $T^3, T^8$.
The non-trivial parity in this case is $P = diag(1,−1,−1)$ and it anti-commutes only with $T^1$.
How does this SSB occur?

 

[fzero]

Since I am not sure what choice you're using for generators, I will use the Gell-Mann matrices, which seem to be different from your choice. $\lambda_8$ is the most convenient choice to use for the Wilson line, since it commutes with itself and $\lambda_{1,2,3}$. From the gauge Lagrangian, we have a potential term $\text{Tr}[A^a_\mu,A^8_5]^2$. The Wilson line corresponds to a constant background field in the 8-component, so the gauge field components that do not commute with $\lambda_8$ get a tree-level mass proportional to the square of this background field ($m_{\text{KK}}$ with the appropriate choice), whereas the massless $A^a_\mu$, $a=1,2,3$ and $A^8_\mu$ generate an unbroken $SU(2)\times U(1)$ gauge group.

If you want to break this further to $U(1)\times U(1)$, a convenient choice would be to put a Wilson line for $A^3_5$ as well.

 

[askalot]

Thank you very much for your answer fzero!
First of all the matrix $T^1$ equals $\lambda_1/2$.
My question is mostly of algebraic nature:
Could you please explain, in algebraic terms, how the massless $A^a_μ, a=1,2,3$ and $A^8_μ$ generate an unbroken $SU(2)\otimes U(1)$?
Or even better, how $U(1) \otimes U(1)$ is realized?

 

[fzero]

.
My question is mostly of algebraic nature:
Could you please explain, in algebraic terms, how the massless $A^a_μ, a=1,2,3$ and $A^8_μ$ generate an unbroken $SU(2)\otimes U(1)$?
Or even better, how $U(1) \otimes U(1)$ is realized?

These fields are massless, so must correspond to some residual gauge symmetry. You can either note that $\lambda_{1,2,3}$ project to the standard Pauli matrices or just verify that they satisfy the $SU(2)$ algebra by hand, so we get a factor of $SU(2)$. Since $\lambda_8$ commutes with them, it is the generator of a separate $U(1)$ factor.

If we put an additional Wilson line for $\lambda_3$, then $A^{1,2}_\mu$ will be massive, but $A^3_\mu$ will remain massless and preserve a $U(1)$ subgroup of $SU(2)$.