# A Charge in a Lie Group... is it always a projection?

#### arivero

Gold Member
Given a representation of a Lie Group, is there a equivalence between possible electric charges and projections of the roots? For instance, in the standard model Q is a sum of hypercharge Y plus SU(2) charge T, but both Y and T are projectors in root space, and so a linear combination is. But I am not sure if this is a general fact for any representation of any lie algebra.

If it is so, do computer algebra programs have specific functions to search for all the possible different projections? I guess, the thing that Lisi et al "elementary particle explorer" did visually, but in an automated way.

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#### Orodruin

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You are talkig about breaking a larger gauge symmetry to a U(1) gauge symmetry. The charge you get from this is going to depend on the linear combination of gauge fields that corresponds to the U(1) symmetry after symmetry breaking and therefore reflects this fact. In other words, it depends on how you break the symmetry (i.e., typically on the representation you put the symmetry breaking vev in).

#### arivero

Gold Member
You are talkig about breaking a larger gauge symmetry to a U(1) gauge symmetry. The charge you get from this is going to depend on the linear combination of gauge fields that corresponds to the U(1) symmetry after symmetry breaking and therefore reflects this fact. In other words, it depends on how you break the symmetry (i.e., typically on the representation you put the symmetry breaking vev in).
But, is any linear combination of gauge fields a projection vector (a scalar product) in the space of roots of the original representation? Also, it is more about the branching down that the breaking, for instance in the standard model you always get the electric charge as sum of hypercharge and weak isospin, independently of the final combination given from the Weinberg angle. So it seems one does not need all the possible linear combinations over the reals, but just some combination of generators which should be easy to define and to scan.

Of course, a problem of this approach is that I only get a vector in root space, without any indication of the path followed to branch down towards this vector.

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#### arivero

Gold Member
Let me put some code as example of my question.

We get the 27 of E6
we tensor it with its conjugate to get 27 \times \bar 27.
and now we start looking for projections with some conditions:

That we have at least 12 neutrinos
That we have at least four particles with multiplicity greater or equal than 6, so they could be quarks
That if we assume the least charged of this particle is the down quark, there is at least 6 states with the double of this charge, to make the role of the up quarks, and at least two states with three times this charge, to make the role of the electron

Code:
import numpy as np
from math import gcd
from collections import Counter
import itertools
#the 27 of E6
pesos=[[1, 0, 0, 0, 0, 0],[-1, 1, 0, 0, 0, 0],[0, -1, 1, 0, 0, 0],[0, 0, -1, 1, 0, 1],[0, 0, 0, -1, 1, 1], [0, 0, 0, 1, 0, -1],[0, 0, 0, 0, -1, 1], [0, 0, 1, -1, 1, -1],[0, 0, 1, 0, -1, -1], [0, 1, -1, 0, 1, 0],[0, 1, -1, 1, -1, 0], [1, -1, 0, 0, 1, 0],[1, -1, 0, 1, -1, 0], [0, 1, 0, -1, 0, 0], [-1, 0, 0, 0, 1, 0],[-1, 0, 0, 1, -1, 0], [1, -1, 1, -1, 0, 0],[-1, 0, 1, -1, 0, 0], [1, 0, -1, 0, 0, 1],[-1, 1, -1, 0, 0, 1], [1, 0, 0, 0, 0, -1],[0, -1, 0, 0, 0, 1], [-1, 1, 0, 0, 0, -1],[0, -1, 1, 0, 0, -1],[0, 0, -1, 1, 0, 0],[0, 0, 0, -1, 1, 0],[0, 0, 0, 0, -1, 0]]
#we tensor it with its conjugate to get 27 \times \bar 27.
pesos=[x+[0,0,0,0,0,0] for x in pesos]+[[0,0,0,0,0,0]+[x[4],x[2],x[1],x[3],x[0],x[5]] for x in pesos]
pesos=np.array(pesos)
vistos=set()
#and now we start looking for projections
k=0
while True:
k=k+1
print(k)
for y in itertools.product(range(-k,k+1),repeat=6):
#we use the same projector in the 27 and the conjugate 27
x = y + y
#and only evaluate for the new vector in the itertools... this is poor man programming :-(
if max(x)==k or min(x)==-k:
suma=Counter( x @ w for w in pesos)
if suma[0]==12: #Neutrinos: we require the result to have at least 12 charge zero elements
ordered=sorted(list(suma.keys()))
dato=tuple(((k0,suma[k0]) for k0 in ordered))
ok = sum( 1 if j>5 else 0 for j in suma.values()) > 4 #quarks: we require 4 having of each electric charge having at least multiplcity 6.
if ok and  np.gcd.reduce(ordered)==1 and not dato in vistos:
quarkCharges=set( (abs(k0) for k0 in suma if suma[k0]>=6) )
quarkCharges.remove(0)
down=min(quarkCharges)
#condition: the minimum charge quark is the down quark, so we check that
#1) there exists a up quark:
ok = ok and suma[2*down] >=6 and suma[-2*down]>=6
#2) there are electrons:
ok = ok and suma[3*down] >=2 and suma[-3*down]>=6
if ok:
print(min(quarkCharges)/max(quarkCharges),dato[0][0]/dato[-1][0],x,dato)
vistos.add(dato)
Code:
The output of the program is to suggest five different multiplicities:
0.3333333333333333 -1.25 (-2, -2, -1, 1, 0, -2, -2, -2, -1, 1, 0, -2) ((-5, 1), (-3, 6), (-2, 6), (-1, 9), (0, 12), (1, 6), (2, 6), (3, 6), (4, 2))
0.3333333333333333 -1.0 (-2, -2, -1, 1, 1, -2, -2, -2, -1, 1, 1, -2) ((-5, 1), (-4, 1), (-3, 6), (-2, 6), (-1, 7), (0, 12), (1, 7), (2, 6), (3, 6), (4, 1), (5, 1))
0.3333333333333333 -1.2 (-2, -2, -1, 1, 1, 1, -2, -2, -1, 1, 1, 1) ((-6, 1), (-4, 1), (-3, 6), (-2, 7), (-1, 7), (0, 12), (1, 5), (2, 6), (3, 5), (4, 3), (5, 1))
0.3333333333333333 -0.8 (-2, 0, 2, -1, 1, 1, -2, 0, 2, -1, 1, 1) ((-4, 2), (-3, 6), (-2, 6), (-1, 6), (0, 12), (1, 9), (2, 6), (3, 6), (5, 1))
0.3333333333333333 -1.0 (-2, 1, 2, -1, 1, 1, -2, 1, 2, -1, 1, 1) ((-5, 1), (-3, 6), (-2, 7), (-1, 7), (0, 12), (1, 7), (2, 7), (3, 6), (5, 1))
(removing the "vistos" variable we see all the projections producing a given multiplicity, this is of course a wider set).

So I would conclude that, with this kind of down quark, there are five different ways to distribute the expected fourteen exotics in the 27 \times \bar 27:
1) 1 particle -5/3, 4 extra of charge -1, three extra of -1/3, 4 extra of charge +1, two extra of charge +4/3
2) 1 particle -5/3, 1 particle -4/3, 4 extra of charge -1, 1 extra of charge -1/3, and the symetrical of positive charge
3)
4) etc
5)

but:
I do not know if the procedure is correct.
If it is, I do not know if it is exhaustive, nor when to finish the loop.
We only have the proyection vector for this U(1), I have no idea of how to find the branching rule.

And of course we can not grant the existence of SU(3) colour, we only have a "six or more" multiplicity[/code]

#### arivero

Gold Member
Note the bug in the previous code
ok = ok and suma[3*down] >=2 and suma[-3*down]>=6
but well, it is equally ilustrative.

The corrected program finds some extra possible multiplicities:

Code:
1
0.5 -1.0 (-1, -1, -1, 1, 1, 1, -1, -1, -1, 1, 1, 1) ((-4, 1), (-3, 3), (-2, 8), (-1, 9), (0, 12), (1, 9), (2, 8), (3, 3), (4, 1))
0.5 -1.0 (-1, -1, 0, 1, -1, -1, -1, -1, 0, 1, -1, -1) ((-3, 3), (-2, 7), (-1, 11), (0, 12), (1, 11), (2, 7), (3, 3))
0.5 -1.0 (-1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, 1) ((-4, 2), (-3, 3), (-2, 7), (-1, 9), (0, 12), (1, 9), (2, 7), (3, 3), (4, 2))
2
0.3333333333333333 -1.25 (-2, -2, -1, 1, 0, -2, -2, -2, -1, 1, 0, -2) ((-5, 1), (-3, 6), (-2, 6), (-1, 9), (0, 12), (1, 6), (2, 6), (3, 6), (4, 2))
0.3333333333333333 -1.0 (-2, -2, -1, 1, 1, -2, -2, -2, -1, 1, 1, -2) ((-5, 1), (-4, 1), (-3, 6), (-2, 6), (-1, 7), (0, 12), (1, 7), (2, 6), (3, 6), (4, 1), (5, 1))
0.5 -1.0 (-2, -2, 0, -2, 1, -2, -2, -2, 0, -2, 1, -2) ((-5, 1), (-4, 2), (-3, 3), (-2, 7), (-1, 8), (0, 12), (1, 8), (2, 7), (3, 3), (4, 2), (5, 1))
0.3333333333333333 -0.8 (-2, 0, 2, -1, 1, 1, -2, 0, 2, -1, 1, 1) ((-4, 2), (-3, 6), (-2, 6), (-1, 6), (0, 12), (1, 9), (2, 6), (3, 6), (5, 1))
0.3333333333333333 -1.0 (-2, 1, 2, -1, 1, 1, -2, 1, 2, -1, 1, 1) ((-5, 1), (-3, 6), (-2, 7), (-1, 7), (0, 12), (1, 7), (2, 7), (3, 6), (5, 1))
3
0.5 -1.0 (-2, -3, -1, -3, 0, -3, -2, -3, -1, -3, 0, -3) ((-5, 2), (-4, 1), (-3, 3), (-2, 8), (-1, 7), (0, 12), (1, 7), (2, 8), (3, 3), (4, 1), (5, 2))

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"Charge in a Lie Group... is it always a projection?"

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