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Symmetry Breaking in Terms of Roots and Weights?

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308 views

I'm currently searching, for quite a while now, for a paper/book that discusses Higgs symmetry breaking in terms of roots and weights.

Concretely I have in mind a discussion of what happens when we give a vev to some Higgs field, represented by a weight of some representation.

I know how to compute which generators (=roots) remain unbroken, but my problem is determining which group this implies after symmetry breaking.

Any suggestions would be much appreciated!

asked Jul 7, 2015 in Mathematics by Jakob [ revision history ]
recategorized Jul 7, 2015 by dimension10

This recent paper might help: http://arxiv.org/abs/1504.08113.

Roots are not the same thing as generators.Usually a $U(N)$ the theory is broken down to $U(1)^N$ if there is a full-symmetry breaking as in the cases or to a subgroup of $U(M) \subset U(N)$ times priducts of the remaining $U(1)^{N-M}$'s. For theories with gauge groups other than $U(N)$ it might be a bit different but qualitatively there is no big difference. 

1 Answer

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So basically you know how to find the unbroken Lie algebra but do not know how to find the associated Lie group.

To a given Lie algebra $\mathfrak g$ there exists a unique group $\tilde G$, called the universal covering group, with the property of being [simply connected][1]. For example, the covering group of the algebra $\mathfrak{su}(2)$ is $SU(2)$. 

The other groups, $\{G\}$,  associated to the same algebra can be obtained from the covering group in the following way
$$G=\frac{\tilde G}{Ker(\rho)},$$
where $Ker(\rho)$ is the kernel of the group homomorphism $\rho:\tilde G\rightarrow G$. Once you have defined a particular representation by choosing a particular highest weight, you are able to compute this kernel. For example, you start with an $\mathfrak{su}(2)$ algebra. Then if you choose the adjoint representation (the highest weight being the highest root) you can show that $Ker(\rho)=\mathbb Z_2$ and the group will be $G=SU(2)/\mathbb Z_2=SO(3)$. On the other hand, if you choose the defining representation you get $Ker(\rho)=\mathbb 1$ and $G=SU(2)/\mathbb 1=SU(2)$.

There are some technical details needed to compute those kernel but in general,
$$Ker(\rho)\subset\mathcal Z(\tilde G),$$
where $\mathcal Z(\tilde G)$ is the center of $\tilde G$, and this center is a finite group which can be obtained from the extended Dynkin diagram.

Same references:
Cornwell, group theory in physics, 1984;
Olive, Turok, Nucl Phys B215, 1983, p470;

answered Jul 11, 2016 by Diracology (90 points) [ no revision ]

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