What Does SU(2) and SU(3) Symmetry Mean in Particle Physics?

[jeebs]

I keep seeing these things being mentioned called SU(2) and SU(3) symmetry in particle physics literature. However, I don't know what they mean, and googling around just brings up pages of weird looking mathematics that I don't have the time to sift through and learn right now.
Can anyone give me, or direct me to, a brief explanation of these things, or a summary of what their significance is in particle physics?

 

[Kevin_Axion]

It's something called gauge-symmetry and it's the foundations of particle physics. Here is a a good article on it: http://en.wikipedia.org/wiki/Introduction_to_gauge_theory.

 

[tom.stoer]

Do you knwo what the rotational symmetry SO(n) in n dimensions means?

 

[jeebs]

nope I can't say I'm familiar with that.

 

[tom.stoer]

OK; it's easy. Suppose you have vectors in n-dim. real vector space, i.e.

$$\vec{x} = (x_1, x_2, \ldots x_n)$$

with real numbers x_n

The scalar product between two vectors is

$$\vec{x}^t\vec{y} = \sum_{k_1}^n x_k y_k$$

If one rotates both vectors with an n-dim. rotation matrix (which is an element of the Lie group SO(n)) one finds

$$(\vec{x}^\prime)^t\vec{y}^\prime = (D\vec{x})^t(D\vec{y}) = \vec{x}^t D^t D \vec{y} = \vec{x}^t \vec{y}$$

where one uses

$$D^t D = DD^t = 1$$

which means that the scalar product is invariant w.r.t. a rotation D.

Now replace real numbers x_n with complex numbers z_n plus complex conjugation ^* in the scalar product, and replace the matrix transposition ^t with the hermitean conjugation ^t*. Then the new scalar product allows for a larger symmetry group where D is now an element of SU(n) instead of SO(n).

That means that SO(n) consists of the rotations acting on a real vector space, whereas SU(n) consists of the rotations acting on a complex vector space.

So much for an brief motivation of SU(n) ...

 

[Kevin_Axion]

I actually saw a video where Freddy Cachazo, a String Theorist, explains it rather easily: http://pirsa.org/C08028/4. It's the last video, Symmetry to String Theory.

 

[apeiron]

Now replace real numbers x_n with complex numbers z_n and the matrix transposition ^t with the hermitean conjugation ^t*.

Is there a next step up from this that involves even more "complex" vector rotations? So perhaps from complex numbers to quaternions and octonions?

 

[Kevin_Axion]

$$U(1)\cong SO(2)$$, $$SU(2)\cong SO(3)$$, I think $$SU(2)\times SU(2)\cong SO(3,1) = SL(2,C)$$, where $$C$$ is the set of complex numbers.

$$U(1)$$ is a rotation of the form $$e^{i\theta} = cos\theta + isin\theta$$
$$SU(2)$$ is a rotation of the form $$e^{\alpha\sigma_1 \beta\sigma_2 \zeta\sigma_3$$ where $$\alpha,\beta, \zeta$$ are angles and $$\sigma_1, \sigma_2\sigma_3$$ are Pauli matrices.

$$\sigma_1 =$$ $$\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$$

$$\sigma_2 =$$ $$\left[ \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right]$$

$$\sigma_3 =$$ $$\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]$$

$$SL(2,C)$$ is a rotation of the form $$e^{\alpha\sigma_1 \beta\sigma_2 \zeta\sigma_3$$ but now $$\alpha,\beta, \zeta$$ are complex numbers of the form $$z=a+bi$$ and $$\sigma_1, \sigma_2\sigma_3$$ are Pauli matrices.
This now relates to Lorentz Transformations as the complex part is equal to Lorentz Boosts.

If I have this wrong then someone can correct me.

 

[tom.stoer]

Is there a next step up from this that involves even more "complex" vector rotations? So perhaps from complex numbers to quaternions and octonions?

Yes; you get the symplectic groups Sp(n) which are certain 2n*2n matrices for the quaternions.

But for the octonions it's rather mysterious. They are related to the 5 exceptional groups (only 5!). There is especially the E-series E(6), E(7) and E(8) which are related to octonionic symmetries. Going to E(5), E(4), ... correspond to well-known Lie groups described above, but the series terminates with E(8), which means there are no higher dimensional octonionic spaces for which Lie groups do exist as symmetry groups. I guess this is related to the fact that octonions are not associative. In additoon I do not know how to construct the inner product which is symmetric w.r.t. E(6), E(7) or E(8).

 

[apeiron]

Yes; you get the symplectic groups Sp(n) which are certain 2n*2n matrices for the quaternions.

Thanks. I'd only seen garbled explanations of what symplectic groups were about and this shows the story is quite simple.

To clarify, are you saying that the octonions have no defined higher-D spin symmetry (in principle, because we know they lack sufficient "structure")? Or that the construction is just going to be more complex and has not yet been achieved?

 

[tom.stoer]

First of all: for the E(n) please do not confuse this n with the dimension; it's the rank! For E(8) the rank is 8, the dimension is 248.

And I do not know what you mean by D-spin symmetry.

The strange structure of the octonionic algebra does not allow "simply adding one octonionic dimension". Only certain dimensions are allowed.

Do you now what a Lie algebra is? It is simply the tangent vector space of the Lie group at the identity with the generators serving as basis. Therefore for every Lie group X one can find a Lie algebra x. There is one preferred representation of a Lie group where it acts on its own Lie algebra as vector space (look at my example for SO(n); it does not act on so(n) but on the n-dim. real Euclidean space).

For SU(2) the Lie agebra su(2) is defined via the Pauli matrices (times 1/2).

If you look at E(8) the only way to define it is via its fundamenal rep. = the adjoint rep. = acting on e(8) as the vector space. That means E(8) is defined via e(8), so essentially via itself.

 

[apeiron]

First of all: for the E(n) please do not confuse this n with the dimension; it's the rank! For E(8) the rank is 8, the dimension is 248.

And I do not know what you mean by D-spin symmetry.

The strange structure of the octonionic algebra does not allow "simply adding one octonionic dimension". Only certain dimensions are allowed.

Do you now what a Lie algebra is? It is simply the tangent vector space of the Lie group at the identity with the generators serving as basis. Therefore for every Lie group X one can find a Lie algebra x. There is one preferred representation of a Lie group where it acts on its own Lie algebra as vector space (look at my example for SO(n); it does not act on so(n) but on the n-dim. real Euclidean space).

For SU(2) the Lie agebra su(2) is defined via the Pauli matrices (times 1/2).

If you look at E(8) the only way to define it is via its fundamenal rep. = the adjoint rep. = acting on e(8) as the vector space. That means E(8) is defined via e(8), so essentially via itself.

I've seen statements that the base symmetry for octonions would in fact be 14 dimensional G2 matrices. E8 would be an expansion of this base symmetry surely?

But I am trying to get a more intuitive understanding of how the various levels of symmetry relate. And I'm not sure if octonions have a definite matrice classification to add a fourth to the sequence of SO, SU and SP. So is there indeed a fourth here and what is it called?

I've also read that quarternions make the jump from the one spin plane of complex i to three axes of spin. So octonions would make a further jump to ... planes of spin?

 

[tom.stoer]

My idea was the same, namely to get an intuitive feeling (not Dynkin diagrams) what these E(n) means physically, especially as it is remarkable that E(5), E(4) etc. are standard Lie groups.

My first idea was to see how an invariant inner product would like. I failed completely. Then I asked John Baez; his response was that it is not clear to him how such an inner product would like :-(

b/c the octonions cannot be represented as matrices, all our "intuition" breaks down somehow.

 

[apeiron]

My idea was the same, namely to get an intuitive feeling (not Dynkin diagrams) what these E(n) means physically, especially as it is remarkable that E(5), E(4) etc. are standard Lie groups.

My first idea was to see how an invariant inner product would like. I failed completely. Then I asked John Baez; his response was that it is not clear to him how such an inner product would like :-(

b/c the octonions cannot be represented as matrices, all our "intuition" breaks down somehow.

If you asked Baez, he is certainly the guy who should know! The speculative understanding I am trying to express here is that all spin symmetries can be interpreted as localised resonances. If you constrain dimensionality to a point-like location, you cannot stop the point itself rotating. It still retains internal degrees of freedom. And lie algebras appear to define all the resonant modes of such spin.

The simplest modes are SO, then SU and SP. But by the time we get up to octonions, the mathematical structure - the resonant states - gets too sparse to settle perhaps into simple crisp modes. You get only a fuzzy or patchy structure at best.

So this is why I was hoping for a clear answer. Octonions might in fact have an exact symmetry structure like the lower ranked algebras and I've just not hear about it. Or it might have an exact, but complex, symmetry story. Or it could be a somewhat sparse and patchy symmetry story. I'm interested to know which is the case because the implications for physical theory seems clear enough.

 

[tom.stoer]

Why not aks John again? Perhaps there was some progress in the meantime.

 

[lpetrich]

John Baez has helpfully put his treatise on octonions online: Octonions

As to octonions and G2, G2 is the automorphism group of the octonions. An automorphism group is a group of automorphisms, of course, mappings of a set of entities onto themselves that preserve the entities' interrelationships.

The automorphism group of the real numbers is the trivial one: the identity group.

The automorphism group of the complex numbers has two members: identity and complex conjugation. It's easy to show that it's a version of Z(2); its members are the identity and reversing the sign of i.

The automorphism group of the quaternions is SO(3), acting on i,j,k in the traditional representation of them. It makes 3D rotations on them.

The automorphism group of the octonions is G2, acting on all the units other than 1: i,j,k, ... It makes a subset of 7D rotations on them.

-

SO(n) - n*n matrices of real numbers
SU(n) - n*n matrices of complex numbers
Sp(2n) - n*n matrices of quaternions

BTW, you can represent quaternions in matrix form, using Pauli matrices:
Q = qreal + i*qimag.(pauli-matrices)

However, you can't represent octonions in matrix form, because their multiplication is not associative.

 

[tom.stoer]

exactly; that's what I now

the funny thing is that E(6), E(7) and E(8) is related to the octonions as well

 

[lpetrich]

John Baez even ties in F4 to octonions, thus tying in all 5 exceptional Lie algebras.

As to how to picture these exceptional Lie algebras, that has to be difficult. The smallest nontrivial irreps of the exceptional ones are, along with the adjoints:
G2 - 7, 14
F4 - 26, 52
E6 - 27, 78
E7 - 56, 133
E8 - 248, 248 (the same!)

The only halfway-easy one is G2, which is a subalgebra of SO(7). The smallest SO(7) irreps break down into G2 ones as follows:
Vector: 7 -> 7 (fundamental)
Spinor: 8 -> 7 + 1 (fund, scalar)
Adjoint: 21 -> 7 + 14 (fund, adjoint)
The G2 ones break down into SU(3) ones:
7 -> 3 + 3* + 1 (fund, fund conj, scalar)
14 -> 3 + 3* + 8 (fund, fund conj, adjoint)

The F4 ones break down into SO(9) ones:
26 -> 16 + 9 + 1 (spinor, vector, scalar)
52 -> 16 + 36 (spinor, adjoint)

The E6 ones break down into SU(3)*SU(3)*SU(3) ones:
27 -> (3,3*,1) + (1,3,3*) + (3*,1,3)
27* -> (3*,3,1) + (1,3*,3) + (3,1,3*)
78 -> (8,1,1) + (1,8,1) + (1,1,8) + (3,3,3) + (3*,3*,3*)

The E7 ones break down into SU(8) ones:
56 -> 28 + 28* (antisymmetric pair, conj of it)
133 -> 63 + 70 (adjoint, antisymmetric quadruplet)

The E8 ones break down into SU(9) ones:
248 -> 80 + 84 + 84* (adjoint, antisymmetric triplet, conj of it)

Or alternately into SO(16) ones:
248 -> 120 + 128 (adjoint, spinor)

 

[lpetrich]

Here's an interesting curiosity about the E series of Lie algebras, and how it relates to particle-physics gauge-symmetry groups.

One gets the Dynkin diagrams for E7 and E8 from E6 by adding a simple root to the previous one. Subtracting simple roots will get E5, E4, ... Let's see what we get.

E3: o - o . o
A2 * A1 / SU(3) * SU(2) -- Standard Model

E4: o - o - o - o
A4 / SU(5) -- Georgi-Glashow GUT, the simplest single group that includes the SM

E5: o - o - o [- o] - o
D5 / SO(10) -- elementary fermions are one multiplet per generation

E6: o - o - o [- o] - o - o
Elementary fermions can be unified with Higgses

E7: o - o - o [- o] - o - o - o
?

E8: o - o - o [- o] - o - o - o - o
A symmetry of the HE heterotic superstring. It's possible to get all the Standard-Model multiplets, complete with multiple generations, from a single HE-superstring gauge multiplet.

Not possible to go to E9 and beyond, however; the simple roots lose linear independence.

 

[thoms2543]

May I ask something else here?
I saw somewhere mention about the physics model i.e $$SU(3)_c \times SU(2)_L \times SU(2)_R \times U(1)_{B-L}$$ left right symmetry model...then they say the left handed doublet $$\psi_L(1,2,1,-1)$$ what does the number (1,2,1,-1) mean? And what does the doublet or singlet mean in that group? And what does the number 3,1 mean in $$SO(3,1)$$ mean?

 

[lpetrich]

Here's what the numbers mean:

1 -- state multiplicity in the algebra SU(3)_c
2 -- state multiplicity in the algebra SU(2)_L
1 -- state multiplicity in the algebra SU(2)_R
-1 -- electric-charge-like multiplier for U(1)_B-L

The field X has component fields X_ijk, where
i -- index in SU(3)_c multiplet (only 1 here)
j -- index in SU(2)_L multiplet (1, 2 here)
k -- index in SU(2)_R multiplet (only 1 here)

X is called a doublet because its overall multiplicity is 1*2*1 = 2.

-

As to what SO(3,1) is, it's a variation on SO(4).

The group SO(4) has members R that satisfy R.R^T = I, the identity matrix.

The group SO(3,1) has members R that satisfy R.g.R^T = g

where g is the metric tensor, which can be set to be
diag(+1, +1, +1, -1)
Notice the 3 +1's and 1 -1 -- that's what makes the 3 and the 1 in SO(3,1)

A group SO(2,2) is like SO(3,1) but with g = diag(+1, +1, -1, -1)

It's usually easier to work with Lie algebras than with Lie groups, so let's see how that works out:

R = I + e*L

where L is a linear combination of Lie-algebra basis members and e is small.

For SO(4), and SO(n) in general:
(I + e*L).(I + e*L^T) = I + e*(L + L^T)
yielding
L + L^T = 0

For SO(3,1) and SO(m,n) in general:
(I + e*L).g.(I + e*L^T) = g + e*(L.g + g.L^T)
yielding
L.g + g.L^T = 0

 

[thoms2543]

I have another question.
what is meant by the custodial SU(2) approximate symmetry as a 5, a 3?