Symmetry breaking/ degrees of freedom

[Neitrino]

Dear PF...

Please help me with basic question more I think more I get confused...

In O(n) space there are n(n-1)/2 generators...
suppose I have symmetric tensor in O(n) space, it will have n(n+1)/2 independent components... and i am building invariant potential from it (quartic polynomial expression), then I will find which components minimize my potential and will have symmetry breaking.

Broken generators will transform into Goldstone boson.
So initially i have n(n-1)/2 goldstonic degrees of freedom.

But before i minimize my potential i want to diagonalize my tensor...
Does diagonalization consume goldstonic degrees of freedom/consumes generators?

After diagonalisation shall i have again n(n-1)/2 goldstonic degrees of freedom in my bydget or less ?

thank you

 

[LightHero]

in advance.The diagonalization of a symmetric tensor with n(n+1)/2 independent components will not consume any of the n(n-1)/2 generators. After the diagonalization process, you will still have n(n-1)/2 Goldstone bosons, as this is the number of broken generators that are associated with the symmetry breaking of the quartic polynomial expression. However, it is important to note that some of the Goldstone bosons may become absorbed by other particles and cease to exist, depending on the type of symmetry breaking and the structure of the quartic expression.