# Understanding symmetry and super symmetry

shounakbhatta
Hello,

I am quiet new to this subject. I am just repeating in a few words, what I have learned so far: The 4 fundamental forces of nature, strong, weak, electromagnetism and gravity. Physicists are trying to re-generate a condition, a very high temperature during the Big Bang, to find out which force actually works during that time. The Weinberg-Glashow and Abdus Salam model unified weak and electromagnetism to electro-weak force. Gravity still lays undefined in the Standard Model. At Planck's scale, physicists observe that the 4 forces of nature unit to One, single force. We have not yet been able to define that force as yet but the quest is on.

Now the U(1), SU(2), SU(3)........What are these? Can anyone please help me understand, sequentially how the theories evolved? The Grand Unified theory is one which is SU(5)? Does SU stand for special unitary group?

Kindly help.

Mentor
At Planck's scale, physicists observe that the 4 forces of nature unit to One, single force.
There is no way to observe this (yet?). It is just a hypothesis.

Now the U(1), SU(2), SU(3)........What are these?
The special unitary groups that correspond to the gauge symmetries of those interactions.

The Grand Unified theory is one which is SU(5)?
That is unknown, SU(5) is one of multiple options.

1 person
Gold Member
Dearly Missed
Hello Shounakbhatta,
MFB gave you a complete answer, but in addition if you are a novice you might want to know what "unitary" and "special" mean.

You should learn what a matrix is. Already know? If not it's easy. Know the complex number plane?

Know how to take the complex CONJUGATE of a complex number
x + iy goes to x - iy
it is flipping the complex number plane over the horizontal axis, the x-axis

Know about the TRANSPOSE of a matrix? Flipping the matrix over its main diagonal, the diagonal from upper left to lower right.

Know about the INVERSE of a matrix?

You could keep on questioning. You could ask for people to explain to you what a unitary matrix is.

How,when you apply a unitary matrix to a vector it does not change the "length" of the vector (by length I mean a generalization of the length idea to vectors of complex numbers---sometimes called the "norm" of the vector. It is kind of a general idea of size (not just of one number but of a string of number, like 3 or 4 numbers). A unitary matrix, working as a transformation of vectors, does not change their size.

Special basically means it does not change volumes either! Several vectors taken together can define a box-like volume. You need to ask more questions if you aren't familiar with this. I'm speaking a sloppy way. Special unitary matrices, working as transformations, are very nice, the keep sizes and volumes (properly interpreted) the same. It is as if they simply ROTATED the world by some angle and kept it looking basically the same. Not enlarged or shrunk or skewed or stretched in some direction. Nicely behaved.

SU(2) and SU(3) are groups of 2x2 and 3x3 matrices, especially nicely behaved ones.

shounakbhatta
Hello Mfb,

Thank you very much for the replies.

Thank you Marcus.

My question to Marcus: Yes, I have gone through linear algebra and unitary matrix. My questions:

(a) If we consider SU(5) as the highest level of special unitary, which is a superset of SU(3) and SU(2) and U(1), which is shown as SU(3)xSU(2)xU(1) does it imply that SU(2) is the electroweak unification force and SU(3) and SU(3) is something else and U(1) as the unitary group and ALL of them is unified into SU(5), which is the grand unified theory?

Secondly, the matrices, what do they represent? Forces? Like strong weak?

From what you have replied above, the unitary matrices, once rotated, turned, flipped, does not change their size, just like a tensor? It maintains the vectors?

is that the reason, the fundamental forces of nature are put into unitary matrix to maintain their symmetry? Am I getting close or a dead end?

Gold Member
Dearly Missed
Yes, I have gone through linear algebra and unitary matrix.
Good. Then the person you should be asking your questions of is Mfb.

BTW you know that for a unitary matrix U, the matrix inverse U-1 is equal to the conjugate transpose, that is often written with a dagger. You might want to learn LaTex for typing math notation. But if you want to write U-dagger without bothering with LaTex you can (at least if your keyboard is like mine) hold down the option key and type a t.

option-t gives †
So if you use that as a superscript on a matrix U, you get U
Then the equation defining unitary matrices is just U-1 = U

I was simply worried that you might not have taken a course in linear algebra and might not have gone through the definitions of the special unitary matrices. But you have!

It turns out that I needn't have worried! Apologies for the interruption.

shounakbhatta
Hello Marcus,

I have got your point, but my question still remains unanswered. If the GUT, SU(5) is a super set of SU(3) x SU(2) x U(1), then how the fundamental forces are related to special unitary group?

lpetrich
Gauge Theory 101.

Elementary particles' symmetries can be expressed as matrices operating on their multiplets:

ψ -> R.ψ

where ψ is the field and R the matrix. For groups like U(n) and SU(n), R is unitary. An easy case is electric charge, which is U(1):
R = exp(i*q*a)

where q is the charge and a is an overall variable for the different particles. For electric charge, it's easy to see that
Product of incoming particles' R's = product of outgoing particles' R's
One gets it from
Sum of incoming particles' q's = sum of outgoing particles' q's

Larger groups have several a's in their R's.

If a is independent of space-time, then the symmetry is global. But if it depends on space-time, then the symmetry is local. But one gets some complications with the fields' kinetic terms. They contain differential operators: D(ψ)

D(R.ψ) = D(R).ψ + R.D(ψ)

The D(R) terms are what cause the trouble, since they disrupt the symmetry of the fields. But if one adds a "gauge field" A, then it can absorb the D(R) terms.

Let the "gauge covariant" derivative be DA(ψ) = D(ψ) - i*g*A.ψ

Let A transform as (R.A - (i/g)*D(R)).R-1

Then DA(R.ψ) = R.DA(ψ)

The gauge field has absorbed the awkward local-symmetry term, and the symmetry is restored.

In the electromagnetic case, we get
A -> A + (q/g)*D(a)
the familiar gauge symmetry the electromagnetic potentials.

One can construct a better-behaved quantity from the A's:
DAi(DAj(ψ)) - DAj(DAi(ψ)) = - i*g*Fij

One gets
F -> R.F.R-1
under this symmetry, without weird derivative terms.

For electromagnetism, F is the familiar electromagnetic-field tensor, a tensor which contains the electric and magnetic fields.

I've fuzzed over a lot of details here, but I hope that I've made the overall ideas reasonably clear.

lpetrich
I'll continue.

Continuous groups can be constructed from generators, and these generators from elements of those groups that are close to the identity element:

R = 1 + i*ε*L + O(ε2)

for small ε and generator L. A group may have several generators, and their effects can interact with each other. The generators form a "Lie algebra", and there's a well-developed theory of Lie algebras, like which ones there can possibly be. It's often much easier to use a continuous group's generator algebra to find various results about that group than the group's elements themselves.

For a unitary group U(n), there are n2 generators, and for SU(n), n2 - 1 generators. The group of orthogonal matrices SO(n) has (1/2)n(n-1) generators, and the spinor group Spin(n) shares its generator algebra. There's also a group of "symplectic" matrices, Sp(2n), with n(2n+1) generators. The algebras for SU(n), SO(n), and Sp(2n) are all "simple" Lie algebras, and there are five additional ones, the exceptional ones G2, F4, E6, E7, and E8.

Back to gauge theories.

In a gauge theory, each gauge-field mode corresponds to a generator of the theory's symmetry group. If some generators interact with each other, then the corresponding gauge-field modes also interact with each other.

For electromagnetism, it's easy. The photon corresponds to the electric charge, with symmetry U(1).

For quantum chromodynamics, it's much more complicated. A quark has three states, red, green, and blue, and an antiquark has three states, antired or cyan, antigreen or magenta, and antiblue or yellow. The gluon has color-anticolor states, and at first sight, one expects 9 states. But one can form a colorless state from those, so the gluon only has 8 states. Working it out mathematically, QCD has color symmetry group SU(3), 8 generators and all.

For electroweak unification, one gets two symmetry groups with their associated particles, "weak isospin" with group SU(2), sort of like QCD with only 2 "colors", and "weak hypercharge", with group U(1), just like electric charge.

Electroweak symmetry is broken, of course, and the surviving symmetry is the U(1) of electric charge.

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lpetrich
So for the Standard Model, we have symmetry group SU(3)*SU(2)*U(1)

SU(3) = QCD = colors red, green, blue
SU(2) = weak isospin = "colors" up, down, like the spin states of a spin-1/2 particle
U(1) = weak hypercharge -- like electric charge

In fact, (electric charge) = (projected weak isospin) + (weak hypercharge)

There's been a lot of effort expended on Grand Unified Theories, and I'll briefly mention some of them.

A simple one is the Georgi-Glashow SU(5) theory. It has 5 "colors":
QCD red, green, blue; weak-isospin up, down

One gets weak hypercharge from a generator that responds 1/3 to the QCD "colors" and -1/2 to the WIS "colors".

Thus, 12 of the 24 SU(5) generators become the QCD, WIS, and WHC generators; 8, 3, and 1 each. This leaves 12 additional ones, and they correspond to gauge particles that make isolated protons decay.

Proton decay has been searched for, and a recent lower limit to the proton's mean life is around 1030 years or so. This corresponds to a SU(5) symmetry-breaking energy of about 1016 GeV. That's about how massive the proton-decayer particles have to be.

But the SU(5) theory has the plus of putting each generation of elementary fermions into two multiplets, three if one counts right-handed neutrinos.

Some GUT's go even farther, like SO(10). They put each generation of elementary fermions into 1 multiplet.

How much further can one go? According to string theory, one can fit *all* of the Standard Model's particles into *one* multiplet of a big symmetry group called E8. The smallest one, with size 248. It would include all the generations of elementary fermions, as well as the gauge and Higgs particles.

friend
Back to gauge theories.

In a gauge theory, each gauge-field mode corresponds to a generator of the theory's symmetry group. If some generators interact with each other, then the corresponding gauge-field modes also interact with each other.

...

Electroweak symmetry is broken, of course, and the surviving symmetry is the U(1) of electric charge.

Thanks Ipetrich.

Perhaps would like to field a few questions I've been having.

1) Is it true that each generator corresponds to a separate particle field? Is there a coupling constant even in the interaction of fields of the same symmetry group?

2) Is it only the Higgs mechanism that breaks the SM symmtries? Or are there other ways of breaking the symmtries?

Thanks.

lpetrich
1) Is it true that each generator corresponds to a separate particle field? Is there a coupling constant even in the interaction of fields of the same symmetry group?
Separate particle field, yes, though a member of a gauge-particle multiplet.

When self-interacting, the fields share the coupling constant that they have when interacting with other fields.

The gluon is a color-state multiplet with states sort of like this:
red-yellow
red-magenta
green-yellow
green-cyan
blue-magenta
blue-cyan
(red-cyan - green-magenta)/sqrt(2)
(red-cyan + green-magenta - 2*blue-yellow)/sqrt(6)

The ninth possible state is colorless:
(red-cyan + green-magenta + blue-yellow)/sqrt(3)

2) Is it only the Higgs mechanism that breaks the SM symmtries? Or are there other ways of breaking the symmtries?
The Standard Model's electroweak symmetry is indeed broken by the Higgs particle.

There are other ways of breaking symmetries, but I'd have to research them.

lpetrich
Before I get into supersymmetry, I'll mention space-time symmetries.

One handles them in roughly the same way as one handles gauge symmetries. Their generators have some connections with some conserved quantities:

Space translations - Momentum
Time translation - Energy
Rotations - Angular Momentum
Boosts - Centroid Position

Something like EM gauge symmetry - electric charge

In general, continuous symmetries and conserved quantities are connected, and that connection is Noether's theorem.

Now for supersymmetry.

It's a symmetry that relates particles with different spins. More precisely, spins differing by 1/2. A SUSY generator applied to a particle field turns it into one with spin differing by 1/2. SUSY generators turn bosons into fermions and fermions into bosons. They are related to space-time ones, as one would expect from their changing of spin, though not to gauge ones.

So all members of a SUSY multiplet would have the same gauge interactions, like the same electric charges. This is true of their interactions in general, and also of their masses.

But we don't observe SUSY partners of known particles at those particles' masses, so SUSY must be broken at energies of at least about a TeV.

But it's nevertheless interesting, because it could help explain why the Standard Model has particles with spins 0 and 1/2, particles with no clear connection to some symmetry.

friend
When self-interacting, the fields share the coupling constant that they have when interacting with other fields.

I have a hard time assimilating this statement. I assume there is a set of coupling constants between the fields of one group, say SU(3), that multiply each other when fields of only that symmetry group interact with each other. Then there is another set of coupling constants associated with the fields of another group, say SU(2), that are used when only the SU(2) fields interact with each other. But when fields of SU(2) interact with fields of SU(3), what coupling constants are used then, a multiple of coupling constants used when the fields interacted only with there own kind?

The Standard Model's electroweak symmetry is indeed broken by the Higgs particle.

When you say the Higgs mechanism breaks the symmetry of the electroweak symmetry, does that mean it breaks it into the weak symmetry, SU(2), and the electromagnetic symmetry, U(1)? Or does the Higgs mechanism break the electroweak symmetry into the mass generations, or both? I think you mean that the Higgs breaks things into there mass generations, since the Higgs mechanism determines the mass. Whereas I believe it is the energy level that causes the coupling constants to diverge from their values at the electroweak scale to produce the separate EM and Weak fields.

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Gold Member
The Higgs breaks the electroweak symmetry by giving mass to the electroweak gauge bosons. Instead of the original massless SU(2) x U(1) gauge bosons, all that's left massless is a new U(1)' gauge boson - I have written in U(1)' because it's different from the original U(1). The original U(1) is the hypercharge boson, but the new U(1) is the photon of electromagnetism.

lpetrich
I have a hard time assimilating this statement. I assume there is a set of coupling constants between the fields of one group, say SU(3), that multiply each other when fields of only that symmetry group interact with each other. Then there is another set of coupling constants associated with the fields of another group, say SU(2), that are used when only the SU(2) fields interact with each other. But when fields of SU(2) interact with fields of SU(3), what coupling constants are used then, a multiple of coupling constants used when the fields interacted only with there own kind?
The SU(3) ones don't interact with the SU(2) ones, and neither of them interact with the U(1) one.

SU(3) - g (gluon) - QCD - color
SU(2) - W - weak isospin
U(1) - B - weak hypercharge

When you say the Higgs mechanism breaks the symmetry of the electroweak symmetry, does that mean it breaks it into the weak symmetry, SU(2), and the electromagnetic symmetry, U(1)? Or does the Higgs mechanism break the electroweak symmetry into the mass generations, or both? I think you mean that the Higgs breaks things into there mass generations, since the Higgs mechanism determines the mass. Whereas I believe it is the energy level that causes the coupling constants to diverge from their values at the electroweak scale to produce the separate EM and Weak fields.
The two electroweak ones can be interpreted as W+, W-, W0, and B.

Their interaction with the Higgs particle is |G.H|2 where

G = g2 * {{W0, W+}, {W-, -W0}} + g1 * B * {{1,0},{0,1}}

H = {H+, H0}

where H+ and H0 are both complex. I'm waving my hands about factors of 2 and the like.

Due to its self-interaction, |H| becomes nonzero. Its resolution into components is essentially arbitrary, but I'll follow the usual convention and use H = {0, v}, where |H|2 = v2. The observed Higgs particle is a field that gets added to v.

That turns G.H into {g2*W+, - g2*W0 + g1*B}*v

and its absolute square into

(g2*v)2 * (W+*W-) + (g2*W0 - g1*B)2*v2

The first term gives the mass of the charged W particle: mW = g2*v

The second term is more complicated, since it mixes the W0 and the B. The mixing angle is called the "Weinberg angle", aW:

g2 = sqrt(g22 + g12) * cos(aW)
g1 = sqrt(g22 + g12) * sin(aW)

Let
W0 = Z*cos(aW) + A*sin(aW)
B = - Z*sin(aW) + A*cos(aW)
A = photon field

Then the second Higgs-interaction term gives the mass of the Z: mZ = sqrt(g22 + g12) * v

The photon field drops out, which is why the photon stays massless.

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lpetrich
Let's see what the photon and Z interactions look like. I'll set
g12 = sqrt(g22 + g12)
g2 = g12*cos(aW)
g1 = g12*sin(aW)
mZ = g12*v

In general, an interaction between a field and the neutral electroweak gauge particles is
(I3*g2*W0 + Y*g1*B) * field
I3 = projected weak-isospin component
Y = weak hypercharge

Plugging in the Z-photon resolution:

g12 * (I3*cos(aW)*(Z*cos(aW) + A*sin(aW)) + Y*sin(aW)*(-Z*sin(aW) + A*cos(aW)))

The photon interaction is the simplest: g12*cos(aW)*sin(aW)*(I3 + Y) * A

So the particle's electric charge is Q*e, where
Q = I3 + Y
e = g12*cos(aW)*sin(aW)
is the elementary charge

This is the same for both left-handed and right-handed parts of elementary fermions, a side effect of how the Higgs mechanism generates mass.

The Z particle is another story: g12*(I3*cos(aW)2 - Y*sin(aW)2) * Z
or
g12*(I3 - Q*sin(aW)2)

For the elementary fermions, the left-handed part has I3 nonzero but the right-handed part has I3 zero. Writing the left-handed projection operator as (1+h)/2 and the right-handed one as (1-h)/2, with h.h = 1, we get for the interactions:

Photon:
e*Q*(1+h)/2 + e*Q*(1-h)/2 = e*Q -- no chirality!

Z:
g12*(I3 - Q*sin(aW)2)*(1+h)/2 + g12*( - Q*sin(aW)2)*(1-h)/2 =
g12*(I3*(1+h)/2 - Q*sin(aW)2)

Like the charged weak interaction, which is proportional to (1+h)/2, the neutral one is chiral.

friend
The SU(3) ones don't interact with the SU(2) ones, and neither of them interact with the U(1) one.

I have to say I don't understand this statement. The link below shows which particles interact with which others.

http://en.wikipedia.org/wiki/Standard_Model#Gauge_bosons

If I'm reading this right, quarks, SU(3), interact with weak bosons, SU(2), and weak bosons interact with leptons, U(1). So I assume there is a coupling constant associated with each of these cross-symmetry interactions. I think this means my original question goes unanswered so far: Are the coupling constants associated with the cross-symmetry interactions expressed in terms of the coupling constants associated with intra-symmetry interactions? Or are all the coupling constants simply put in by hand after being determined by experiment without regard to whether there is any kind of relationship between the constants?

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shounakbhatta
Hello lpetrich.

Sorry for the late reply, but I really have to take some time out in understanding the concepts; even I don't know

whether I have understood it not. 'To err is to human'.

I am unable to understand the equation R=exp(i*q*a)

Well, speaking of space-time symmetry, I understand, the 4 areas including the Lorentz boost. For super symmetry,

SUSY it is that every boson changes into fermions and vice-versa.

shounakbhatta
If I don't want to understand the maths part, is it possible to understand what super symmetry is all about?

Jim Kata
I'll give it a shot, but I must warn you I haven't read the other responses so I'm probably being redundant. In Physics there is this theorem, Noether's theorem, which says for each symmetry of your action, the thing that determines the dynamics of your system, you get a conserved charge. For example, if your action has translation symmetry you get conservation of momentum, or if it has rotational symmetry you have conservation of angular momentum and so on. Physicist were interested in all the types of symmetry a physically suitable action could have and have tried to classify them all. These were known as the No Go theorems. The most famous being the one of Coleman and Mandula. Which basically said the only symmetries a physical system could have were the generators of the Poincare group and internal symmetries. There is however one possible extension to this theorem, for point particles, this extension is supersymmetry. In the standard model there are two types of particles fermions and bosons. The fermions are particles of half integer spin and are like electrons, protons, things like that. The bosons are particles with integer spin and are the messenger particles, the particles that send the force messages. These are things like the photon, the graviton, and the gluons. Supersymmetry is a symmetry that exchanges bosonic variables with fermionic ones and vice versa. Physically what this predicts, if it is correct, is that the standard model is only half of the story. For every boson there is a corresponding super partner fermion, sparticle, and vice versa for the fermions.

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shounakbhatta
Hello Jim,

Thanks for the wonderful answer. Now that you have started explaining things in a better manner, let me ask you few more questions to clear out some of my doubts:

(1) The grand unified theory; is it a type of super symmetry, where all the 4, now 3 forces of nature are uniting at the Planck scale? Are they the same?

(2) I sometimes find that SU(5) $\supsetx$SU(3) x SU(2) x U(1). Does this mean that the SU(5) contains both SU(3) and SU(2) and U(1)?

(3) What I understand that SU is special unitary group. Now U(1) being a unitary matrix, resembles a circle group, hence when we try to include the forces, like electromagnetism it falls under this unitary matrix right?

(4) Now, when we come to SU(2), we find a matrix which has 2 rows and 2 columns. What do the physicists fit inside this matrix? I am not talking about dimensions. The particles that we write in SU(2), what are they? What about SU(3)?

(5) How are matrices related to the particles?

May be some question are irrelevant, but while understanding the concept, you understand, things first becomes hazy and then clearer.

Thanks.

Jim Kata
Remember in my previous response where I said a physical theory could also have internal symmetries? Well these are the internal symmetries. The internal symmetry group of the standard model is SU(3)XSU(2)XU(1) and like you'd expect they lead to conserved charges by Noether's theorem. For example, a U(1) gauge symmetry is what's responsible for conservation of electric charge. SU(3) is what's responsible for conservation of color charge, the chromo in chromodynamics, and SU(2) is what's responsible for isospin charge. To be technical the U(1) in SU(3)XSU(2)XU(1) is actually hypercharge, but SU(2)XU(1) is broken down to the U(1) of electromagnetism by symmetry breaking. The matrices SU(2), SU(3) are observables. This might get confusing. For example, SU(2) has three matrices representing the three directions in isospin space, like x, y, z direction in 3D. So the matrix t_z is the operator for isospin in the z direction in isospin space. It acts on your particle states to tell you what their isospin is in the z direction. As far as G.U.Ts are concerned, a natural question is why SU(3)XSU(2)XU(1)? People then went about finding bigger groups which they could embed the standard model in SU(5), SO(10), E_6, and various other groups.

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lpetrich

I am unable to understand the equation R=exp(i*q*a)
R multiplies an elementary-particle field value
q is the particle's electric charge
a is a parameter. Its value is arbitrary

The arbitrariness of a is why this is a symmetry.

1 person
shounakbhatta
Hello Jim,

Thank you very much for your wonderful answer. I would look into it tonight and would let you know.

friend
Remember in my previous response where I said a physical theory could also have internal symmetries? Well these are the internal symmetries. The internal symmetry group of the standard model is SU(3)XSU(2)XU(1) and like you'd expect they lead to conserved charges by Noether's theorem. For example, a U(1) gauge symmetry is what's responsible for conservation of electric charge. SU(3) is what's responsible for conservation of color charge, the chromo in chromodynamics, and SU(2) is what's responsible for isospin charge.

We have fields associated with the U(1) symmetry; we have other fields associated with the SU(2) symmetry, and we have still other fields associated with the SU(3) symmetry. IIRC we have a particular algebra of multiplication between the fields of the U(1) symmetry; we have a different algebra of multiplication between the fields of the SU(2) symmetry, and we have still another algebra of multiplication between the fields of the SU(3) symmetry. Is it true that the algebra of U(1) is complex, the algebra of SU(2) is quaternions, and the algebra of SU(3) is octonions? I think I heard something like this at one time in my studies.

Jim Kata
We have fields associated with the U(1) symmetry; we have other fields associated with the SU(2) symmetry, and we have still other fields associated with the SU(3) symmetry. IIRC we have a particular algebra of multiplication between the fields of the U(1) symmetry; we have a different algebra of multiplication between the fields of the SU(2) symmetry, and we have still another algebra of multiplication between the fields of the SU(3) symmetry. Is it true that the algebra of U(1) is complex, the algebra of SU(2) is quaternions, and the algebra of SU(3) is octonions? I think I heard something like this at one time in my studies.

That's an interesting question, and I doubt I'm in a position to answer it, but Geoffry Dixon does work on this very concept. A little numerology that's always fascinated me is
\begin{align} S^1 \hookrightarrow S^3 \rightarrow S^2 \\ S^3 \hookrightarrow S^7 \rightarrow S^4 \\ S^7 \hookrightarrow S^{15} \rightarrow S^8 \end{align}

notice that the fiber of the complex hopf fibration is $$S^1 \cong U(1)$$ and the fiber of the quatronic hopf fibration is $$S^3 \cong SU(2)$$ however the fiber of the octonic hopf fibration $$S^7$$ isn't even a group. Technically it's a Moufang loop. I think what Dixon does is take the automorphism group of the octonions which is G_2 and get's SU(3) out of that. That's according to something I read on John Baez's this weeks finds.

friend
We have fields associated with the U(1) symmetry; we have other fields associated with the SU(2) symmetry, and we have still other fields associated with the SU(3) symmetry. IIRC we have a particular algebra of multiplication between the fields of the U(1) symmetry; we have a different algebra of multiplication between the fields of the SU(2) symmetry, and we have still another algebra of multiplication between the fields of the SU(3) symmetry. Is it true that the algebra of U(1) is complex, the algebra of SU(2) is quaternions, and the algebra of SU(3) is octonions? I think I heard something like this at one time in my studies.

That's an interesting question, and I doubt I'm in a position to answer it, but Geoffry Dixon does work on this very concept.

I'm sorry, but I thought this would be a straightforward question. Could it still be that we don't know how the fields of the SU(3) symmetry multiply each other? It sounds like a yes or no question. I don't know what could be so complicated about that.

lpetrich
The theory of the SU(3) part is quantum chromodynamics or QCD.

There isn't any way to mark out a red quark as distinct from a green one or a blue one. They all look alike from outside. The same is true of the gluon color states. So one has to use more indirect methods, like comparing quark-quark-gluon and gluon-gluon-gluon overall interaction strengths. But the results are consistent with the QCD symmetry group being SU(3).

One can approach the question in another way. What symmetry groups can there be? We look among the "Lie groups", those whose elements are some differentiable function of some parameters. These groups get very complicated very quickly, so we look at their generators, the "Lie algebras", which are often much simpler to work with. If you've ever done operator algebra with quantum-mechanical angular momentum, you've worked with a Lie algebra.

Also, the particle multiplets are in "representations" of the group and its algebra. Each representation or rep has a matrix realization, and if the matrices cannot all be turned into the same block-diagonal form, then the rep is irreducible, making an irrep.

There's a Lie-algebra rep that's formed from the algebra's structure and that uses the generators as its basis space: the adjoint rep.

The interactions in QCD must all be intertwined and not separable, otherwise it could be decomposed into some subsets that are independent or more-or-less independent of each other. Quarks and gluons must all be in irreps, because otherwise, they'd split into different sets.

As is true of gauge theories in general, the gluons must be in the adjoint rep of the QCD symmetry group, because each gluon mode is associated with a generator of that group.

So we look in the simple Lie algebras. Which ones have 3D reps among all their reps? Not very many. Of the four infinite families and five exceptional ones, only SU(3) and SO(3) have 3D reps. SO(3) is the 3D rotation group, and it and SU(2) have isomorphic algebras. However, the 3D rep of SO(3) is a real rep, and thus not very suited. So we are left with SU(3). Its 3D rep is complex, and it has a conjugate one. Just what one needs for quarks and antiquarks. One can create a colorless state from 3 quarks in both SU(3) and SO(3), which is further support for these solutions.

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shounakbhatta
Hello Jim,

Thanks for the wonderful reply. It helped me to clear up the concepts.

What I got, is that each internal symmetry is responsible for conservation of charges. Conservation is a fundamental property of Nature.