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This is a follow up to this question.

A generator $T_a$ of a given gauge group $G$ remains unbroken after some Higgs field $\Phi$ gets a vev if

$$ T_a \langle\Phi\rangle =0 $$

I'm trying to understand how I can check this explicitly using matrices for the vev and the generators in different representations (i.e. non-adjoint Higgs).

Let's assume we have Higgs fields in some representation $R$, of a gauge group $G$ as for example $SO(10)$, that can be written as a matrix. Then given a vev written as a matrix how can I compute which generators remain unbroken after the Higgs fields get a vev?

For concreteness, let's consider the $SO(10)$ example. The gauge bosons live in the adjoint 45-dim rep. Because $10 \otimes 10= (1 \oplus 54)_S \oplus 45$, we can write each gauge boson as a (antisymmetric) $10 \times 10$ matrix $M_{45}$. Further let's assume we have Higgs fields in the $54$ dimensional representation. For the same reason we can write these 54 Higgs fields as (symmetric) $10 \times 10 $ matrices $M_{54}$.

Then how can I compute which generators remain unbroken when some of the Higgs fields (or just one) gets a vev? (The vev is then a $10 \times 10 $ matrices $M_{54VEV}$)

Three possibilities that come to my mind:

  • $M_{45}M_{54VEV}=0$
  • $M_{45}M_{54VEV}M_{45}^T=0$

  • $[M_{45},M_{54VEV}]=0$

Which one is correct or is it something completely different?

EDIT: Related MathStackexchange questions:

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  • $\begingroup$ The vev should be a column vector $|\Phi\rangle_{VEV}$, no? $\endgroup$ – SuperCiocia Aug 1 '15 at 13:50
  • $\begingroup$ No, see for example this question physics.stackexchange.com/q/196988 or the paper quoted there. In addition a vev in matrix form is what one usually uses to break $SU(5)$ to the standard model gauge group. $\endgroup$ – jak Aug 1 '15 at 14:28
  • $\begingroup$ But you are not in the adoint, you are in the defining of SO(10), so $\Phi$ is a column vector. $\endgroup$ – Diracology Jan 9 '16 at 17:51

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