A quick introduction to continuous groups

M. Dugger

Arizona State University, Tempe, AZ 85287-1504

An introduction to continuous groups is presented.

I. INTRODUCTION

The most common groups you encounter as a nuclear/particle researcher will be what are known as “continuous groups”. Of these, the groups SU(N) and SO(N) play an especially important role in particle physics.

This document is not intended to offer a complete, or mathematically rigorous, treatment of the subject. Instead, the intention of this document is to introduce continuous groups to students of particle physics so that they will gain a level of comfort with the subject that they can then utilize in furthering their understanding with future study.

II. DEFINITION OF A GROUP

To define what a group is I will use the two-dimensional rotation group, SO(2), with group elements R_{θ} as an example, where

R_{θ} ≡ [\begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix}] .

We say that elements g_{i} form a group if they obey the following

1. Closure: For any group members g_{i}, and g_{j}, then g_{i} · g_{j} is also a member of the group.

In our SO(2) example we have R_{α} · R_{β} = R_{α+β}, which is clearly within the group.

2. Associativity: g_{i} · (g_{j} · g_{k}) = (g_{i} · g_{j}) · g_{k}

In our SO(2) example we have R_{α}·(R_{β}·R_{γ}) = (R_{α} · R_{β}) · R_{γ}. And from our experience with R_{θ} elements, we know that associativity holds.

3. Identity: There exists an identity element I such that g_{i} · I = I · g_{i} = g_{i}

In our SO(2) example we have I = R_{θ=0}.

4. Inverse: For every group element g_{i} there exists an element g_{i}^{−1} within the group such that g_{i} · g_{i}^{−1} = I.

In our SO(2) example we have R_{θ}^{−1} = R_{−θ}.

III. DEFINITIONS OF THE GROUPS SU(N) AND SO(N)

Special unitary matrices of dimension N are the group of all N by N unitary matrices (i.e. U^{†}U = 1), represented here as U, that have det(U) = 1 (special) This group is denoted as SU(N).

Similarly, the group of all N by N orthogonal matrices (i.e. O^{T} O = 1), represented here as O, that have det(O) = 1 are called the group of special orthogonal matrices of dimension N. This group is denoted as SO(N). In short:

• S → unit determinate

• O → orthogonal matrix

• U → unitary matrix

• N → dimension of the group.

What do these groups do? The group members of SO(N) can be used to transform N-component real vectors, while group members of SU(N) are used to transform N-component complex vectors. It is important to note that both of these transformations will maintain an invariant length (i.e. v⃗ · v⃗ = constant, where v⃗ · v⃗ ≡ v^{†}v⃗ for the complex case).

IV. THE GROUPS U(1), SO(2), SU(2), SO(3), AND SU(3)

A. U(1)

One of the simplest groups that can be studied is the U(1) group. This group is comprised by all unitary matrices of dimension 1. Since this group is dimension 1, we do not need to worry about the determinate. For this reason, we do not talk of SU(1), instead we will only consider U(1) and find that the unitarity constraint “fixes” the overall size of these complex numbers.

We start out by letting U be a group member of U(1). Furthermore, since U is just some complex number, we can set U = re^{iθ}, with r and θ being free real parameters that are consistent with the unitarity requirement U^{†}U = 1. When employing the unitary requirement it is easy to see that r^{2} = 1, and since r ∈ the group of purely real numbers ℜ, then r = ±1. Thus, we take

U = e^{iθ}.

What does the U(1) group do? A U(1) group member simply takes a complex number and changes the phase of that number.

The group U(1) is the gauge group that is responsible for electrodynamics. It is beyond the scope of this document to fully describe what is meant by a “gauge symmetry”, however, I will briefly sketch the highlights: When the Lagrangian for a spin-1/2 Dirac-field is demanded to be invariant under a local phase transformation of that Dirac-field (ψ → e^{iθ(x)}ψ, where θ(x) means θ is position dependent), extra terms must be introduced. These extra terms are identical to the terms one would include if the original Lagrangian included, in addition to the spin-half Dirac field, the photon field (and the interaction between the two). For this reason, it is sometimes said that electrodynamics is a result of U(1) gauge symmetry.

This would be a mere curiosity if it were not for the fact that the weak and strong forces can be similarly generated by gauge groups. The strong force is generated by a SU(3) gauge symmetry and the weak force is generated by a SU(2) gauge symmetry. This is why you will sometimes see the entire physics of these forces in the Standard Model described simply as

SU(3) × SU(2) × U(1).

B. SO(2)

The most intuitive group for us to study is SO(2). Undoubtedly, you have worked with this group many many times, but did not know the group theory terminology for it. Because of your familiarity with calculating quantities using this group, this group provides a great opportunity to build up your knowledge of group theory terminology.

The group SO(2) is the group of all 2 × 2 orthogonal matrices that have unit determinate. The orthogonality requirement is defined as O^{T} O = 1, where O^{T} is the transpose of O (i.e. for each o_{i,j} matrix element of O, (o_{i,j})^{T} = o_{j,i}).

Since this group is so familiar and is of low dimension, it will not be too bothersome to present the details of how one might derive the group members of SO(2) in two different ways. In the first method we will derive the group members by directly calculating the individual matrix elements of O. In the second method we will derive the group members by finding the “generator” of the group SO(2).

In the first method we want to calculate the matrix elements of O. If we write the matrix O as

O = [\begin{matrix}o_{11} & o_{12} \\ o_{21} & o_{22} \end{matrix}]

then the condition O^{T} O = 1 reads

O^{T} O = [\begin{matrix}o_{11} & o_{21} \\ o_{12} & o_{22} \end{matrix}] [\begin{matrix}o_{11} & o_{12} \\ o_{21} & o_{22} \end{matrix}] = [\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}] .

This implies that

o_{11}^{2} + o_{21}^{2} = 1  (1)
o_{11}o_{12} + o_{21}o_{22} = 0  (2)
o_{12}^{2} + o_{22}^{2} = 1.  (3)

While the condition det(O) = 1 gives

o_{11}o_{22} − o_{12}o_{21} = 1.  (4)

Eqns. 1 and 3 tells us that −1 ≤ o_{11}, o_{12}, o_{21}, o_{22}, ≤ +1 . This means that we can parameterize one of these matrix elements by a function that spans from -1 to +1. In particular, we can choose o_{11} ≡ cos θ, where θ is a free parameter (that may end up being restricted in value when satisfying Eqns. 1 through 4). After performing some rudimentary algebra we have

O = [\begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix}] ,

and we see there was no need to make any restrictions on the θ parameter. We can now identify the group members of SO(2) as the rotation matrices for two-dimensional space.

Now, we will derive the group members of SO(2) a second way. This derivation is easier but will use some more advanced matrix algebra. I’ll try to be careful about introducing the matrix algebra. First let me define a matrix A and the matrix e^{A}. The matrix e^{A} is defined as follows,

e^{A} = ∑_{j=0}^{∞} \frac{A^{j}}{j!} = I + \frac{A}{1!} + \frac{A^{2}}{2!} + \frac{A^{3}}{3!} + ⋯

We will need a couple more matrix identities:

(A + B)^{T} = A^{T} + B^{T}
(AB)^{T} = B^{T} A^{T}

This means that

(e^{A})^{T} = ∑_{j=0}^{∞} \frac{(A^{j})^{T}}{j!} = ∑_{j=0}^{∞} \frac{(A^{T})^{j}}{j!} = e^{A^{T}}

The orthogonality condition can now be written as:

O^{T} O = (e^{A})^{T} e^{A} = e^{A^{T}} e^{A}

At this point is is tempting to write e^{A^{T}} e^{A} = e^{(A^{T} +A)}, but we need to be careful! In general, e^{A}e^{B} = e^{A+B}, ONLY IF [A, B] = 0. For this case we are in luck, [A^{T} , A] = 0.

We can now write the constraint

e^{(A^{T} +A)} = 1 ⇒ A^{T} + A = 0

Therefore A is anti-symmetric. This means that we can write A as

A = [\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}] θ,

where θ is a free real parameter. Also, we can now clearly see that [A^{T} , A] = 0.

At this point we should check that e^{A} is equal to the rotation matrix we obtained earlier. Also, this is a good opportunity to introduce a new group theory term: generator. Let us define J such that

A = iJθ = [\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}] θ.

This J is called the generator of the SO(2) group and is given by

J = [\begin{matrix}0 & −i \\ i & 0 \end{matrix}] .

Now lets see how e^{iJθ} is equal to the 2-dimensional rotational matrix:

O = e^{iJθ} = I + i\frac{Jθ}{1!} − \frac{(Jθ)^{2}}{2!} − i\frac{(Jθ)^{3}}{3!} + \frac{(Jθ)^{4}}{4!} ⋯

We notice that

J^{2} = [\begin{matrix}0 & −i \\ i & 0 \end{matrix}] [\begin{matrix}0 & −i \\ i & 0 \end{matrix}] = [\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}] = I.

Which allows us to write

O = e^{iJθ} = I + iJ \frac{θ}{1!} − \frac{θ^{2}}{2!} − iJ \frac{θ^{3}}{3!} + \frac{θ^{4}}{4!} ⋯
= I \{1 − \frac{θ^{2}}{2!} + ⋯ \} + iJ \{\frac{θ}{1!} − \frac{θ^{3}}{3!} + ⋯ \}
= I cos θ + iJ sin θ
= [\begin{matrix}cos θ & sin θ \\ − sin θ & cos θ \end{matrix}] ,

as we saw above.

We have now seen two ways to calculate the group members of SO(2): By direct calculation, and by determination and use of the group generator.

C. A relationship between U(1) and SO(2)

What about the generator for U(1)? Well, it is simply equal to 1. One interesting detail regarding the generators for SO(2) and U(1) is that there is exactly one generator for each of these continuous groups. Knowing this, we might expect there to be some connection between these two groups. There is, in fact, a connection.

Let me define a complex vector V with a real xcomponent V_{x} and a real y-component V_{y} such that V = V_{x} +iV_{y}. What does a U(1) transformation V → V′ look like?

V′ = e^{−iθ}(V_{x} + iV_{y})
= (V_{x} cos θ + V_{y} sin θ) +
i(V_{y} cos θ − V_{x} sin θ)

This implies that

V′_{x} = V_{x} cos θ + V_{y} sin θ
V′_{y} = V_{y} cos θ − V_{x} sin θ.

We obtain a 2-dimensional rotation using the U(1) group when we define a vector as V = V_{x} + iV_{y}. This example nicely illustrates the connection between the U(1) and SO(2) groups ☺.

D. SU(2)

SU(2) is another group with which you have some experience. We will investigate this group by looking at the generators of the group. One nice feature of this group is that we will have three separate generators that will allow us to talk about the concept of a Lie algebra.

Let U be a group member of SU(2) and define the matrix A such that U = e^{A}. In what follows we will need the useful matrix identity:

det(e^{A}) = e^{tr(A)},

where the trace of A is the sum of the diagonal elements (i.e. tr(A) = ∑_{i} A_{i,i}). Now, if we set an arbitrary group object U of SU(2) as U = e^{A}, then the requirement that det(U) = 1 becomes

det(U) = det(e^{A}) = e^{tr(A)} = 1.

This implies that tr(A) = 0.

Note: Since, in the previous subsection, the A matrix of SO(2) was antisymmetric, we did not have to worry about tr(A) = 0. An antisymmetric matrix automatically has diagonal elements all equal to zero.

The unitarity condition U^{†}U = 1 implies that A^{†}+A = 0 (i.e. A is anti-hermitian). The condition tr(A) = 0 gives us

a_{11} + a_{22} = 0.  (5)

That A is anti-hermitian tells us that

a_{11}^{∗} + a_{11} = 0  (6)
a_{21}^{∗} + a_{12} = 0  (7)
a_{22}^{∗} + a_{22} = 0.  (8)

From equation 6 and 8 we can see that a_{11} and a_{22} are completely imaginary. The other two elements a_{12} and a_{21} can have both real and imaginary parts. From this we can see that there are six independent quantities. In addition to a_{11} and a_{22} being imaginary, Eqn 5 tells us that a_{11} = −a_{22}. This reduces the number of independent elements to five. The last equation at our disposal, Eqn. 7, tells us that the real part of a_{21} equals the real part of −a_{12}, and that the imaginary part of a_{21} is equal to the imaginary part of a_{12}. We have exhausted all of our equations (Eqns 5 through 8), and have three independent matrix elements. Summarizing we have:

Im(a_{12}) = Im(a_{21})  (9)
−Re(a_{12}) = Re(a_{21})  (10)
−Im(a_{11}) = Im(a_{22})  (11)
Re(a_{11}) = Re(a_{22}) = 0.  (12)

We have a matrix A with three independent matrix elements. This means that we can construct the matrix A as a linear combination of three independent matrices. Let me define three such independent matrices as A_{1}, A_{2}, and A_{3}. A convenient choice is:

A_{1} = \frac{1}{2} [\begin{matrix}0 & −i \\ −i & 0 \end{matrix}] , A_{2} = \frac{1}{2} [\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}] , A_{3} = \frac{1}{2} [\begin{matrix}i & 0 \\ 0 & −i \end{matrix}] ,

where we have utilized Eqn. 9 to construct A_{1}, Eqn. 10 to construct A_{2}, and Eqns. 11 and 12 to construct A_{3}. Note: The factor \frac{1}{2} is not absolutely necessary, but introduced for later convenience. In fact, there are an infinite number of ways one could define A_{1}, A_{2}, and A_{3}.

The linear combination can be written

A = A_{1}θ_{1} + A_{2}θ_{2} + A_{3}θ_{3},

where the θ_{i} are independent parameters.

If we write e^{A} = e^{iS⃗·θ⃗}, then we find that the generators of the SU(2) group are the spin matrices in the Pauli basis:

S_{1} = \frac{1}{2} [\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}] , S_{2} = \frac{1}{2} [\begin{matrix}0 & −i \\ i & 0 \end{matrix}] , S_{3} = \frac{1}{2} [\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix}] .

This means that the SU(2) group is the group that deals with spin. It should be noted that at no time did we use quantum mechanics. The mathematics of spin in quantum mechanics is a natural consequence of group theory!

We can now introduce the concept of Lie algebra. The Lie algebra is simply the commutation relations between the generators of the group. For SU(2) we have as the Lie algebra

[S_{i}, S_{j}] = iϵ_{ijk}S_{k}.  (13)

The constants ϵ_{ijk} are called the “structure constants.”

Previously, I mentioned that the electromagnetic field could be seen as a consequence of U(1) gauge symmetry and that the weak interaction was a consequence of SU(2) gauge symmetry. Now that we have seen that there is one generator for the U(1) group and three generators for SU(2), it is natural to ask how the number of generators is related to the produced fields. It turns out that for each generator there is a field. In the case of SU(N), there are N^{2} − 1 generators. For electromagnetism (U(1)) we have one generator and one field (photon). For the weak interaction (SU(2)) we have three generators and three fields (W^{+}, W^{−}, and Z^{0}). For the SU(3) gauge-field (QCD, or the strong interaction) there are eight generators and eight colored gluon fields.

E. SO(3)

As you probably have suspected, SO(3) is the group that is responsible for rotations in three dimensional space. Because you already have worked with this group many times and are comfortable with it, I will not attempt to construct the group members but merely state a few results.

If we write the group member O of SO(3) as e^{iJ⃗θ⃗}, the generators of the SO(3) group can be written as

J_{1} = [\begin{matrix}0 & 0 & 0 \\ 0 & 0 & i \\ 0 & −i & 0 \end{matrix}] , J_{2} = [\begin{matrix}0 & 0 & −i \\ 0 & 0 & 0 \\ i & 0 & 0 \end{matrix}] , J_{3} = [\begin{matrix}0 & i & 0 \\ −i & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] .

The Lie algebra for SO(3) is

[J_{i}, J_{j}] = iϵ_{ijk}J_{k}.  (14)

We also know that B⃗ ×C⃗ transforms as a vector under an SO(3) transformation. Because of this, we have grown accustomed to writing quantities like L⃗ = r⃗ × p⃗. A different way to write this is L_{k} = ∑_{i,j} ϵ_{ijk} r_{i} p_{j} (or simply L_{k} = ϵ_{ijk} r_{i} p_{j}, if we suppress the summation notation). NOTE: Being able to write antisymmetric combinations of vector quantities ϵ_{ijk} A_{i} B_{j}, that also transform as vectors, will be very useful when we talk about multiplets under SU(3).

F. A relationship between SU(2) and SO(3)

The case for there to be a relationship between SU(2) and SO(3) is very strong. We can see from Eqns. 13 and 14 that SU(2) and SO(3) share the same Lie algebra! To clearly see the connection between the two groups, let me define the vectors V⃗ and V⃗′ as

V⃗ ≡ V_{x}σ_{x} + V_{y}σ_{y} + V_{z}σ_{z},  (15)
V⃗′ ≡ e^{iS⃗θ̇}V e^{−iS⃗θ̇}  (16)

To make life easier, I will only consider rotations about the z-axis:

V⃗′ = [cos(\frac{θ}{2}) + iσ_{z}sin(\frac{θ}{2})] V⃗ [cos(\frac{θ}{2}) − iσ_{z}sin(\frac{θ}{2})] (17)
= [V_{x} cos θ + V_{y} sin θ]σ_{x} + [V_{y} cos θ − V_{x} sin θ]σ_{y}
+V_{z}σ_{z},  (18)

where a whole bunch of algebra was performed between Eqn. 17 and Eqn. 18. We can now make the identifications:

V′_{x} = V_{x} cos θ + V_{y} sin θ
V′_{y} = V_{y} cos θ − V_{x} sin θ.

Therefore, if we make the definitions shown in Eqns. 15 and 16, we can perform a three-vector spatial rotation by using SU(2) group members. Sometimes you will hear this type of relation referred to as an isomorphism. For this particular situation the terminology is: SU(2) is isomorphic to SO(3).

You will probably hear that SU(2) “double covers” SO(3). This terminology is due to the fact that the group members of SU(2) are given by U = e^{iS⃗θ̇} = e^{iσ⃗ \frac{θ⃗}{2}} . This means that a full rotation in SU(2) requires θ = 4π. If we had, instead, defined U = e^{iσ⃗θ⃗} (i.e. without the factor \frac{1}{2} on σ⃗), the transformation V⃗ → V⃗′ would result in

V′_{x} = V_{x}cos(2θ) + V_{y}sin(2θ)
V′_{y} = V_{y}cos(2θ) − V_{x}sin(2θ).

At this point we would not be able to associate θ as a spatial angle. Instead, we would have to define the spatial angle associated with the SU(2) transformation as θ/2. This is why the factor of \frac{1}{2} was included in the definition of the SU(2) generator. This factor allows the free parameters of SU(2) to be equal to the angles of SO(3).

G. SU(3)

I will conclude this section with a very brief look at SU(3). As stated earlier, this group has eight generators. Because of the large number of generators it would be good to have some sort of simple algorithm to find them. One such algorithm relates the generators of SU(N) to those of SU(N − 1). I will describe this algorithm for obtaining the SU(3) generators using the SU(2) generators.

Let the generators of SU(3) be denoted as λ_{i}, where i = 1, 2, 3, ..., 8. For each generator of SU(2) (in this case, we are using the Pauli matrices to be the generators), create a matrix that includes that SU(2) generator in the upper left-hand part of the matrix. This gives us λ_{1}, λ_{2}, and λ_{3}.

λ_{1} = [\begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] , λ_{2} = [\begin{matrix}0 & −i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] , λ_{3} = [\begin{matrix}1 & 0 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & 0 \end{matrix}] .

The next two matrices we construct will be obtained by taking a 3 × 3 matrix that has all elements equal to zero, and then placing in the last column and last row, the numbers 1 and 1 symmetrically

λ_{4} = [\begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}] , λ_{6} = [\begin{matrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}] .

We then construct the next two matrices by placing −i and i symmetrically in the last row and column:

λ_{5} = [\begin{matrix}0 & 0 & −i \\ 0 & 0 & 0 \\ i & 0 & 0 \end{matrix}] , λ_{7} = [\begin{matrix}0 & 0 & 0 \\ 0 & 0 & −i \\ 0 & i & 0 \end{matrix}] .

Finally, we put the 2×2 unit matrix in the left top corner of the 3 × 3 matrix and demand that the matrix is traceless and that the trace of the matrix-squared is equal to 2:

λ_{8} = \frac{1}{\sqrt{3}} [\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & −2 \end{matrix}] .

These matrices are called the Gell-Mann “λ-matrices” and can be found in most particle physics books. One of the nice features about this simple algorithm is that it allows an easy way to remember the form of the λmatrices. Note: We could use this same algorithm to construct the generators of SU(4) by starting with our newly created SU(3) generators (if interested, see pages 372-373 of “Quantum Field Theory”, by Michio Kaku). For completeness, the Lie algebra for SU(3) is

[\frac{λ_{i}}{2} , \frac{λ_{j}}{2} ] = if_{ijk} \frac{λ_{k}}{2} ,

where the structure constants f_{ijk} are

f_{123} = 1,
f_{147} = −f_{156} = f_{246} = f_{257} = f_{345} = −f_{367} = \frac{1}{2},
f_{458} = f_{678} = \frac{\sqrt{3}}{2}.

It is worth noting that these Gell-Mann λ matrices were constructed in a manner that left, explicitly, SU(2) as a subgroup. The way to see this is by noticing that λ_{1}, λ_{2}, and λ_{3} are just the Pauli matrices with an extra row and column of zeros thrown in. This means that we can use the λ_{1}, λ_{2}, and λ_{3} generators to form the SU(2) group. Thus, if we restricted ourselves to transformations that only utilized the λ_{1}, λ_{2}, and λ_{3} generators, we would only be mixing two out of the three complex vector components.

V. GROUP MULTIPLICATION

A. SO(2) × SO(2)

Let the vectors A⃗ and B⃗ be arbitrary two-component real vectors. We know that each these vectors can be transformed such that the scalar length of each vector is invariant under the transformation. In particular, the transformation for real two-component vectors that preserves length is provided by the two-dimensional rotation group. This group, as we have already seen, is called SO(2).

We are interested in constructing something called the “fundamental representation” of SO(2) × SO(2). This fundamental representation will simply be the direct product of A⃗ and B⃗, and will be defined as

T^{i,j} = A^{i}B^{j}.

Since the representation is rather small, it takes little effort to write out the matrix form of T and count the number of independent elements:

T^{i,j} = [\begin{matrix}A^{x}B^{x} & A^{x}B^{y} \\ A^{y}B^{x} & A^{y}B^{y} \end{matrix}]: 4 independent elements.

We are now interested in finding ways of “breaking up” the fundamental representation into parts that will transform only among themselves. To accomplish this break-up, we will notice that we can create an antisymmetric tensor A out of our fundamental representation T by the simple construction A^{i,j} = (T^{i,j} −T^{j,i})/2. We can also make a symmetric tensor S^{i,j} in a similar manner, S^{i,j} = (T^{i,j} + T^{j,i})/2. Another quantity that is easy to construct is the trace (dot product): tr(T^{i,j}). Since A is antisymmetric it has no trace, but S has a trace, and if we want to break-out the trace from T so that it is independent of the other quantities, we will have to remove this trace from S^{i,j}. If we let Q be the symmetric and traceless matrix corresponding to T, then Q^{i,j} = S^{i,j}− \frac{1}{2}tr(T). In summation, we have a fundamental representation T that is broken into three independent parts: An antisymmetric tensor A; a traceless symmetric tensor S, and a trace tr(T). We can write out these tensors and count the number of independent elements:

A = \frac{1}{2} [\begin{matrix}0 & (A^{x}B^{y} − A^{y}B^{x}) \\ (A^{y}B^{x} − A^{x}B^{y}) & 0 \end{matrix}] : 1.
Q = \frac{1}{2} [\begin{matrix}(A^{x}B^{x} − A^{y}B^{y}) & (A^{x}B^{y} + A^{y}B^{x}) \\ (A^{y}B^{x} + A^{x}B^{y}) & (A^{y}B^{y} − A^{x}B^{x}) \end{matrix}] : 2.
tr(T) =  A^{x}B^{x} + A^{y}B^{y}  : 1.

We cannot decompose this down any further. We say that the fundamental representation has been “reduced” into its “irreducible representations”. We count the number of independent elements of each irreducible representation and write

SO(2) ⊗ SO(2) = 2 ⊕ 1 ⊕ 1.

We say that we have the decomposition of SO(2)⊗SO(2).

It will be instructive to find the explicit transformation rules for these tensors. Immediately, we can see that the antisymmetric tensor represents the cross product, and that the trace represents the dot product. Since we know that the cross product (in 2-dimension) and dot product are invariant under rotation, then tr(T) and the antisymmetric tensor A are invariants (as is required by them having only one independent element). We only need to worry about how the components of the traceless antisymmetric tensor Q^{i,j} transform:

(A^{x}B^{x} − A^{y}B^{y})′ =
C^{1}(A^{y}B^{y} − A^{x}B^{x}) + C^{2}(A^{y}B^{x} + A^{x}B^{y})
(A^{y}B^{x} + A^{x}B^{y})′ =
C^{2}(A^{y}B^{x} − A^{x}B^{y}) + C^{1}(A^{y}B^{y} + A^{x}B^{x})

where C^{1} = (cos^{2}(θ) − sin^{2}(θ)), and C^{2} = 2 cos θ sin θ. If we now define the vector V⃗_{irreducible} as

V⃗_{irreducible} = ( \begin{matrix}V^{1} \\ V^{2} \\ V^{3} \\ V^{4} \end{matrix} ) ≡ ( \begin{matrix}A^{x}B^{x} − A^{y}B^{y} \\ A^{y}B^{x} + A^{x}B^{y} \\ |A⃗ × B⃗| \\ A⃗ · B⃗ \end{matrix} ) , (19)

then the transformation V⃗_{irreducible} → V⃗′_{irreducible} can be written

V⃗′_{irreducible} = [\begin{matrix}[\begin{matrix}C^{1} & C^{2} \\ C^{2} & C^{1} \end{matrix}] & 0 0 0 0 \\ 0 0 0 0 & 1 0 0 1 \end{matrix}] V⃗_{irreducible}.

Thus, we can write the transformation of an irreducible representation as a block diagonal matrix.

Hopefully, you can now see that writing SO(2) ⊗ SO(2) = 2⊕1⊕1 makes a lot of sense. The term 2⊕1⊕1 tells us how big the block diagonals are in the irreducible representation.

In short, if we had two real two-component vectors, then we could create the vector V⃗_{irreducible} and any coordinate rotation would result in mixing only the V_{1} and V_{2} components in Eqn. 19. For two-dimensional real vectors all this group theory is overkill. You have gotten along just fine without all of this mathematical jargon and machinery. The only real reason I wrote all of this stuff for SO(2) ⊗ SO(2), is that it is easy and intuitive. In particle physics we classify the hadrons by there irreducible representations. Later, you will see how all of this machinery relates to the quark composition of the π mesons. First, however, we will see how the decomposition we performed here relates to SO(N) ⊗ SO(N).

B. SO(N) ⊗ SO(N)

To see that the reduction we used for SO(N)⊗SO(N) (traceless-symmetric, trace, and antisymmetric decomposition) applies generally for SO(N) ⊗ SO(N), we just need to show that, in general, Q^{i,j}, tr(T), and A^{i,j} transform among themselves.

Since the transformation of symmetric and antisymmetric tensors is written as,

(S′)^{i,j} = ∑_{l}^{N} ∑_{m}^{N} O^{i,l}O^{j,m}S^{l,m}

(A′)^{i,j} = ∑_{l}^{N} ∑_{m}^{N} O^{i,l}O^{j,m}A^{l,m},

and S^{l,m} = S^{m,l}, and A^{l,m} = −A^{m,l}. We can rewrite these as

(S′)^{i,j} = ∑_{l}^{N} ∑_{m}^{N} O^{i,l}O^{j,m}S^{m,l} = (S′)^{j,i}
(A′)^{i,j} = − ∑_{l}^{N} ∑_{m}^{N} O^{i,l}O^{j,m}A^{m,l} = −(A′)^{j,i}.

Thus, the symmetric and antisymmetric tensors transform only among themselves. Also, since the trace transforms as

tr(T′) = ∑_{i}^{N}(T′)^{i,i} = ∑_{i,j,k}^{N} O^{j,i}O^{k,i}T^{j,k}

= ∑_{j,k}^{N} T^{j,k}(O^{T} O)^{j,k} = ∑_{j}^{N} T^{j,j}
= tr(T),

then we have shown that the reduction into symmetric traceless, trace, and antisymmetric tensors form valid representations of SO(N) ⊗ SO(N).

C. SU(2) ⊗ \overline{SU(2)} flavor symmetry

It is probably more common to see SU(2) ⊗ SU(¯2) (or even 2 ⊗ ¯2) than SU(2) ⊗ \overline{SU(2)}. The reason I titled this subsection as SU(2) ⊗ \overline{SU(2)}, rather than the more common notation, is to make it very clear that there is an important distinction being made. In particle physics the over-bar on a particle represents anti-particle. (So, for example, the u quark has as its anti-particle ū.)

The SU(N) transformation properties are different for particles and anti-particles. To see that this should be true, you need to know that the anti-particle scalar-field is obtained by complex conjugation of the particle scalarfield. (For a Dirac field you need, in addition to the conjugation, a factor of the gamma matrix γ^{2}.) For a particle that respects SU(N) symmetry we have the transformation for the field ϕ^{i} representing that particle as

ϕ^{i} → ϕ^{′ i} = U_{j}^{i}ϕ^{j},

where we have used the convention that A^{i}B_{i} ≡ ∑_{i}^{N} A^{i}B_{i} (i.e. upper and lower indices that match are automatically summed). For a scalar-antiparticle we get

ϕ^{∗i} → (U_{j}^{i})^{∗}ϕ^{∗j} = ∑_{i}^{N}(U^{†})_{i}^{j}ϕ^{∗j},

This is a bit of a mess! One way to make this a bit cleaner is to define the anti-particle field for ϕ^{i} as having a lower index instead and defining the transformation as

ϕ_{i} → ϕ′_{i} = (U^{†})_{i}^{j}ϕ_{j}.

For right now, we just want to know that the SU(2) transformation properties of a particle are different than that of an anti-particle, and that we can keep track of this difference by using an upper index for particles and a lower index for anti-particles.

Lets define two arbitrary two-component complex vectors A⃗ and B̄ that have components

A⃗ = ( \begin{matrix}A^{u} \\ A^{d} \end{matrix} ) , B̄ = ( \begin{matrix}B_{u} \\ B_{d} \end{matrix} ) ,

where we are using the upper index to define a particle and the lower index to define an anti-particle.

The fundamental representation of SU(2)⊗SU(2) can be written

T_{j}^{i} = [\begin{matrix}A^{u}B_{u} & A^{u}B_{d} \\ A^{d}B_{u} & A^{d}B_{d} \end{matrix}]: 4 independent elements.

In contrast to SO(N) ⊗ SO(N), we can no longer form symmetric and antisymmetric tensors that transform only among themselves. We do, however, have the trace that transforms correctly:

tr(T′) = (T′)_{i}^{i} = U_{j}^{i}(U^{†})_{i}^{k}T_{k}^{j} = (UU^{†})_{j}^{k}T_{k}^{j} = T_{j}^{j} = tr(T).

This means that we can only form the trace (denoted as ϕ) and traceless (ϕ_{j}^{i} = T_{j}^{i} − \frac{1}{2}tr(T)) tensors:

ϕ =  A^{u}B_{u} + A^{d}B_{d}  : 1
ϕ_{j}^{i} = [\begin{matrix}(A^{u}B_{u} − A^{d}B_{d})/2 & A^{u}B_{d} \\ A^{d}B_{u} & (A^{d}B_{d} − A^{u}B_{u})/2 \end{matrix}] : 3.

In group theory terminology we say SU(2) ⊗ \overline{SU(2)} = 3 ⊕ 1.

If we make the substitutions A^{u} → u, A^{d} → d, B_{u} → ū, B_{d} → d̄, make the meson identifications as given in the PDG, and normalize, then the triplet gives us

π^{0} = (uū − dd̄)/\sqrt{2}
π^{+} = ud̄
π^{−} = dū

The pions are constructed as elements of the irreducible triplet representation of SU(2)⊗SU(\overline{2}) flavor symmetry. We say that in isospin space, the pions transform among each other. We can make the irreducible representation vector as

V⃗_{irreducible} = ( \begin{matrix}π^{+} \\ π^{−} \\ π^{0} \\ singlet \end{matrix} )

What about the singlet? At first glance it looks to be the ω! Well, we don’t have the complete story yet. To have a better picture of what is going on, we will need to look at SU(3) ⊗ SU(\overline{3}).

D. SU(3) ⊗ SU(\overline{3})

We can construct SU(3)⊗SU(\overline{3}) in a similar manner to that of SU(2)⊗SU(\overline{2}). In fact we can start out by writing SU(3) ⊗ SU(\overline{3}) = ¯8 ⊕ 1, since we know that there will be nine states and the only thing we can do is form a singlet representation that is the trace, and a representation that is traceless. Immediately we can write

ϕ =  uū + dd̄+ ss̄
α_{j}^{i} = [\begin{matrix}\frac{1}{3}(−2uū+dd̄+ss̄) & ud̄ & us̄ \\ dū & \frac{1}{3}(uū−2dd̄+ss̄) & ds̄ \\ sū & sd̄ & \frac{1}{3}(uū+dd̄−2ss̄) \end{matrix}] .

We can identify the singlet as the η′ = (uū+dd̄+ss̄)/\sqrt{3}, but where did our π^{0} = (uū + dd̄)/\sqrt{2} go?

We know that SU(2) flavor symmetry is a “good” symmetry. It certainly is better than SU(3) (since m_{u} = 0.002, m_{d} = 0.005, and m_{s} = 0.1 GeV). It seems reasonable to see if we can form the SU(2) ⊗ SU(\overline{2}) subgroup explicitly in our octet representation. In fact, we explicitly constructed our Gell-Man λ matrices to maintain 2 ⊗ ¯2 as a subgroup. It would be somewhat unusual if we could not construct 3 ⊗ ¯3 with a 2 ⊗ ¯2 subgroup!

To make 3⊗¯3 with an explicit 2⊗¯2 subgroup we simply define

ϕ_{j}^{i} = [\begin{matrix}\frac{1}{3}(uū−dd̄) & ud̄ & us̄ \\ dū & \frac{−2}{3}(uū−ss̄) & ds̄ \\ sū & sd̄ & \frac{1}{3}(uū+dd̄−2ss̄) \end{matrix}]

Once we make the meson identifications as given in the PDG, and normalize, then the octet gives us

η = (uū+dd̄−2ss̄)/\sqrt{6}
π^{0} = (uū − dd̄)/\sqrt{2}
π^{+} = ud̄
π^{−} = dū
K^{+} = us̄
K^{−} = sū
K^{0} = ds̄
K̄^{0} = sd̄.

Note: Since \frac{2}{3}(uū−ss̄) = π^{0}+η, then the \frac{−2}{3}(uū−ss̄) term is not independent, and does not belong in our particle listing.

E. The SU(3) ⊗ SU(3) ⊗ SU(3) baryons

So far we have only concerned ourselves with the qq̄ mesons. If we take three quarks we can construct baryons.

Just as in the case of SO(3) ⊗ SO(3) we can form a vector from a rank 2 tensor. In the case of SU(3) ⊗ SU(3) the resulting vector will have the transformation characteristics of SU(3). In particular, we can form the vectors

ϕ_{k} = ϕ^{i,j}ϵ_{ijk}

and

ϕ^{k} = ϕ_{i,j}ϵ^{ijk}.

These vectors within SU(3) are the complex analog of A⃗ × B⃗ within SO(3). Note, this only gives a non-zero result when ϕ^{i,j} or ϕ_{i,j} is antisymmetric in i ↔ j.

One thing that is important for you to keep in mind: The vector ϕ_{k} = ϕ^{i,j}ϵ_{ijk}, transforms like an antiparticle transforms. We are NOT saying that the vector ϕ_{k} represents an antiparticle. We are just saying that ϕ_{k} has the same transformation properties under SU(3) as an antiparticle would have.

We can always form objects that have definite symmetry properties under interchange of the upper indices, and definite symmetry properties under interchange of the lower indices. Pay careful attention, this is the important part: for any two upper (lower) indices that are antisymmetric under interchange, we can reduce the number of indices by one and lower (raise) that index by the operation ϕ_{k} = ϕ^{i,j}ϵ_{ijk} (ϕ^{k} = ϕ_{i,j}ϵ^{ijk}). This means that we only have to consider irreducible representations that are totally symmetric in lower and upper indices!

Let me form the direct product representation of SU(3) ⊗ SU(3) ⊗ SU(3) as

T^{i,k,j} = ϕ^{i}ϕ^{j}ϕ^{k}

We can obtain an irreducible representations that are

• totally symmetric: ϕ^{i,j,k}

• ϕ = ϕ_{k}^{k} = ϕ^{i,j,k}ϵ_{ijk} (antisymmetric trace)

• antisymmetric in i ↔ j: ϕ_{m}^{k} = ϕ^{i,j,k}ϵ_{ijm} traceless

• antisymmetric in i ↔ k: ϕ_{n}^{j} = ϕ^{i,j,k}ϵ_{ink} traceless

But what about antisymmetric in j ↔ k? It turns out that the states that are antisymmetric in j ↔ k can be made out of the other states. (See handout.)

In short we can write

ϕ^{i} ⊗ ϕ^{j} ⊗ ϕ^{k} = ϕ^{i,j,k} ⊕ ϕ_{m}^{k} ⊕ ϕ_{n}^{j} ⊕ ϕ,

or

3 ⊗ 3 ⊗ 3 = 10_{S} ⊕ 8_{M} ⊕ 8_{M} ⊕ 1_{A},

where we have a symmetric decuplet, two mixed symmetry octets, and an antisymmetric singlet. We can even write:

(1, 0) ⊗ (1, 0) ⊗ (1, 0) = (3, 0) ⊕ (1, 1) ⊕ (1, 1) ⊕ (0, 0),

where the notation (m, n) means m upper indices and n lower indices.

Note: We already found that ϕ_{m}^{k} → 8. To determine ϕ^{i,j,k} → 10 we simply count the possible quark states:

• no indices matching → 1 state: |uds >

• two indices matching → 6 states: |uud >, |uus >, |ddu >, |dds >, |ssu >, |ssd >

• all indices matching → 3 states: |uuu >, |ddd >, |sss >.

F. A couple more SU(3) decompositions

Let’s do a couple more decompositions. If I want to do a group multiplication of 3 ⊗ 3 I get

ϕ^{i} ⊗ ϕ^{j} = ϕ^{i,j} ⊕ ϕ_{k}.

Since ϕ^{i,j} is symmetric under i ↔ j, there are six independent states. Thus,

3 ⊗ 3 = 6 ⊕ ¯3.

Note: we obtain a ¯3 decomposition instead of 3 since this triplet transforms the same way as an antiparticle triplet.

If I want to do a group multiplication of the baryon octet with the meson octet, I get

ϕ_{j}^{i} ⊗ ϕ_{l}^{k} = ϕ_{j,l}^{i,k} ⊕ ϕ^{i,k,m} ⊕ ϕ_{j,l,n} ⊕ ϕ_{o}^{m} ⊕ ϕ_{n}^{p} ⊕ ϕ.

To count the number of states within ϕ_{j,l}^{i,k} we notice that there are six ways to have (i, k) symmetric and six ways to have (j, l) symmetric. This implies 6 · 6 = 36 states symmetric in i ↔ k and j ↔ l. We just have to subtract out all of the traces to find the number of states within ϕ_{j,l}^{i,k}.

To count the number of traces we notice that if one upper and one lower index are summed to form a trace, there are three ways to choose the remaining upper index and three ways to choose the remaining lower index. This means that there are 3 · 3 = 9 states that include at least one trace. Now, to find the number of states within ϕ_{j,l}^{i,k} we just need to subtract the nine traces from the 36 states symmetric in i ↔ k and j ↔ l. Thus ϕ_{j,l}^{i,k} contains 27 states.

We can now make the identification:

8 ⊗ 8 = 27 ⊕ 10 ⊕ ¯10 ⊕ 8 ⊕ 8 ⊕ 1.

So, finding multiplets within SU(3) is a fairly straightforward process. The most challenging part is counting the number of states within any given irreducible representation.

VI. CONCLUSION

In this document, an introduction to continuous groups has been presented. The symmetry groups have many applications in particle physics, particularly SU(2) flavor symmetry (isospin), SU(3) flavor symmetry (the quark model with u, d and s quarks), and SU(3) color symmetry (with R, G, and B color charges).

VII. ACKNOWLEDGMENTS

This work was supported at Arizona State University by the National Science Foundation award PHY-0653630.