Download Group Theory – Crash Course 1 What is a group?

Group Theory – Crash Course
M. Bauer
1
WS 10/11
What is a group?
In physics, symmetries play a central role. If it were not for those you would not find an
advantage in polar coordinates for the Kepler problem and there would be no seperation
ansatz to help you solving the Schrödinger equation of the Hydrogen atom. The mathematical structures behind symmetries are groups. For more complex theories, such as QCD,
where there is no analytic solution known, identifying the underlying symmetry group helps
a lot for understanding the theory. There have even been Nobel prizes for guessing the right
symmetry group of electroweak interactions1 , at least that should motivate you to understand group theory. Let us start with the definition:
A group (M, ×) is a set M with a group multiplication ×, with the properties
E
A
N
I
Existence and uniqueness (or closure):
Associativity:
Neutral element:
Inverse element:
(C Commutativity
∀g1 , g2 , g3 ∈ M , g1 × g2 = g3 lies in M .
∀g1 , g2 , g3 ∈ M, (g1 × g2 ) × g3 = g1 × (g2 × g3 ).
∃ e ∈ M , so that ∀g ∈ M, g × e = e × g = g.
∀g ∈ M there is a g −1 ∈ M , so that
g −1 × g = g × g −1 = e.
∀g1 , g2 ∈ M, g1 × g2 = g2 × g1 ).
If the last requirement is fulfilled the group is called abelian.
A subgroup H = (N, ×) is a subset N ⊂ M , which is a group with the same group
multiplication as G = (M, ×). One writes
H < G.
Also, Abelian groups can only have abelian subgroups.
Above, Group multiplication does not have to be literally multiplication. One can take any
binary operation (addition, matrix multiplication, composition of functions, etc.). Further,
we distinguish finite and infinite groups:
A finite group is a group with a finite number of elements. The number of elements is
called the order of the group |G|. If the number of elements is infinite, the group is called
infinite group. In this case group elements are labelled by one or more parameters instead
of an index like in the definition above2 ,
gi
−→
1
g(a)
Weinberg, Glashow, Salam 1979
Just like the transition from many particles- to field theory, where you can describe a stretched string
as the limes of infinitely many harmonic oscillators and finally exchange their label by the parameter of
the field.
2
1
The number of (real) parameters is called the dimension of the group.
Z2 = ({1, −1}, ·). The group structure (that is the result of all possible combinations of
elements) can be expressed through a multiplication or Cayley table.
Now for a few examples. Lets consider a unilateral triangle and the symmetry transformations which project it onto itself. Besides the identity, there are rotations about the
central point of 120◦ and 240◦ which leave the triangle invariant. If we name these operations d and d2 , we already found ourselves the alternating group3 A3 = ({e, d, d2 }, ◦),
where ◦ denotes the composition of operations. We can go on and add reflections on the
bisecting lines of the angles, lets call them s1 , s2 , s3 respectively. We have found an even
larger group, namely the symmetric group4 S3 = ({e, d, d2 , s1 , s2 , s3 }, ◦). Both groups are
obviously finite and therefore their respective group structure (that is the instruction how
to combine all possible group elements) can be represented in a so called multiplicative or
Cayley table, see Figure 1. You will have noticed that A3 is a subgroup of S3 .
We won’t find any larger symmetry groups by looking at a triangle. We can of course go
on and draw squares and hexagons and more and more complicated polygons and step by
step classify the finite groups. However, since everyone of us knows how to “compute” π,
it should not be surpising that if we instead draw the outer circle of the triangle above and
look at its symmetry group we will find a larger group which will include many of these
finite polygon groups as subgroups. This group is called unitary group of dimension one or
circle group
U (1) = ({d(φ)}, ◦)
.
Here, d(φ) denotes the generalization of d, d2 above, denoting rotations about any angle φ.
The index has been traded for a parameter. Since it includes rotations about any angle,
U (1) is an infinite group. Most probably you are used to identify
U (1) = ({ei φ }, ·)
and the above notation looks clumsy to you. But this is already a representation of U (1),
which should be well distinguished from the abstract group, as we will learn in Section 3.
Exercise
[1pt] Which of the groups A3 , S3 is a subgroup of U (1) and why?
Instead of drawing the multiplication table, one can define a function
gk = gi × gj
⇒
k = f (i, j) ,
with i, j, k ∈ {1 . . . , |G|}.
Here, f is called composition function and gives the index of the group element which is
given by the product of the ith with the jth group element. This is a non-continuous function mapping {1, . . . , |G|} ∈ R to itself. While multiplication tables cannot be generalized
to infinite groups, the composition function can,
g(c) = g(a) × g(b)
3
4
⇒
c = f (a, b) ,
with a, b, c ∈ R.
The alternating groups An contain the n!/2 even permutations of n elements.
The symmetric groups Sn contain the n! permutations of n elements.
2
e d d2 s 1 s 2 s 3
e e d d2 s 1 s 2 s 3
d d d2 e s 3 s 1 s 2
d2 d2 e d s2 s3 s1
s1 s1 s2 s3 e d d2
s 2 s 2 s 3 s 1 d2 e d
s 3 s 3 s 1 s 2 d d2 e
Figure 1: Unilateral triangle and its symmetry groups A3 and S3 . The red part denotes
the Cayley table of A3 embedded into the Cayley table of S3 .
The properties of the composition function now allow for further classification of infinite
groups. It can be not continuous 5 , continuous or even analytical6 . In the last two cases
one speaks of continuous or Lie groups, respectively.
Let us look at an example:
(Z, +)
<
(Q, +)
<
(R, +)
These groups are all infinite, and Q is in addition continuous, but no Lie group, while R is
a Lie group. Just think of the smoothness of the function which describes the combination
of elements of these groups.
In physics, Lie groups are the most important groups. Because only Lie groups give rise to
conserved Noether currents. In addition, finite groups are typically subgroups of some Lie
group. In the next section we will therefore put our focus on Lie groups. But before that,
I want to fit in one more definition:
Groups with a bounded parameter space are called compact.
Compact groups are in a way generalizations of finite groups (all finite groups are compact)
and their representation theory is very well understood. On the other hand, if a group
is not compact a lot of nice theorems do not hold and since non-compact groups are of
physical relevance it is important to be able to handle them.
5
To be mathematically strict, with a discrete topology even finite groups are continuous(those are then
called discrete groups). What I mean by continuous is continuous with respect to the standard topology
of R.
6
That is, f can be represented by a convergent Taylor series.
3
Let me further introduce the notion of a matrix group. Matrix groups are sets of N × N
dimensional matrices over some field (R, C) with matrix multiplication as group multiplication. All matrix groups are subgroups of GL(N, F), which is the general linear group, the
group of all invertible matrices over the field F. Matrix groups are probably what you are
familiar with and know as a group. The reason is that a lot of groups can be represented
by matrix groups. Namely,
• Every finite group is isomorphic to a matrix group.
• Every compact Lie group is isomorphic to a matrix group (Peter-Weyl Theorem).
We will therefore generalize the definition and speak of a matrix group whenever the group
at hand is isomorphic to a matrix group. This holds also for most non-compact Lie-groups.
There exist counterexamples, but they are rare. I will give one later on, but it needs more
math to understand it. Also, the group multiplication is always matrix multiplication for
matrix groups and therefore I will just define the set of matrices from now on.
2
Lie Groups & Algebras
First of all, Lie groups are differentiable manifolds. Instead of defining them as infinite
groups with an analytic composition function one can define them as manifolds with a
group structure. This endows them with a vivid geometric meaning. Therefore it is worth
to work through a bit more mathematical formalism, in order to understand these statements. Let us start with manifolds. I will not give the rigorous mathematical definition.
For our purpose a manifold is a space M that looks like Rn if one zooms in, but on a large
scale it can be curved. One has than a number of charts, maps φ : M → Rn that map those
locally flat regions to Rn . In this notation n is called the dimension of the manifold.
The dimension of a Lie group, which we have defined earlier as the number of parameters
the group elements depend on, is also the dimension of the corresponding manifold. As an
example let us look at U (1) again. Each element of the U (1) can be identified with a point
on a circle. Since the circle is a one-dimensional manifold, the dimension of U (1) is one.
We might as well count the parameters and find that it depends only on one parameter, the
angle. Dont get confused here, the N in GL(N, F) is not the same as the dimension of the
group/ the corresponding manifold. A general invertible N × N - matrix rather depends
on N 2 parameters, so that n = N 2 .
We can get a better understanding of the connection of these parameters and the dimension
of the group manifold if we introduce the notion of a tangent space.
Imagine a point x on a manifold M . If one collects all possible curves γ : R → M on the
manifold going through that point x, and builds their respective tangential vectors v at
γ(0) = x,
d
v = γ(t) ∈ Tx M
dt
t=0
they span a vector space attached at x with the same dimension as the manifold. That is
the tangential space Tx M . An illustration is given in Figure 2.
4
TxM
υ
x
γ(t)
M
Figure 2: Illustration of the definition of a tangent space.
For all manifolds we care about (that would be semi- Riemann or Riemann manifolds) one
can locally define an exponential map. An exponential map is a map which goes from
the tangent space at some point x to the underlying manifold,
exp : Tx M −→ M
v 7→ expx (v) = γv (0) .
That means, if you have a tangential vector in Tx M the exponential map gives you the
corresponding curve on M in a neighborhood of the point x. For manifolds formed by
matrix groups M = G, the exponential map can be defined as the matrix exponential
exp V =
∞
X
Vk
k=0
1
= 1 + V + V 2 + ...
k!
2
Now you should recall how we represented the elements of U (1) in section 1, which is
isomorphic to the group of 1 × 1 complex matrices with norm 1. For the unit circle in the
complex plane, the tangent space at each point is the imaginary line, which is R (if we take
out an i). Writing a vector in R as its component times the only basis vector 1, we see that
all elements of U (1) can be written as
g(φ) = exp(i φ 1) ,
with φ ∈ R.
More general, elements of an arbitrary matrix group g ∈ G, which are close to the identity
(since exp is only locally defined) can always be written as
g = exp(i
n
X
αa T a ) ,
(T a are matrices here) ,
a=1
where n is the number of basis vectors T a of Te G and equals the dimension of the Lie
group. Because of the closure requirement, every group element can then be generated (by
repeated multiplication) out of those in the vicinity of the identity. Therefore these basis
vectors are called infinitesimal generators of the Lie group.and calls the basis vectors T a
generators.
5
z
z
y
y
x
x
Figure 3: Illustration of the non-commutativity of SO(3). Starting point is the red vector.
The green vector is the result of first performing a rotation of 45◦ about the the z axis and
than a rotation of 45◦ about the y axis, while the blue vector is the result when we first
rotate about the y axis and than about the z axis (all rotations counterclockwise).
Again, for the U (1) that means, that if we choose an element close to the identity, that is
φ = 0 + with 1, we can write
g = exp(i φ) = exp(i ) exp(i ) . . . exp(i) = exp(i [ + + . . . ]) ,
which justifies the statement above, but is pretty boring, since there is just one generator.
Let me quote another example. The group of all rotations in R3 is isomorphic to the matrix
group
SO(3) = R ∈ GL(3, R)|RT R = 1 det(R) = +1 ,
the special orthogonal group in 3 dimensions. As a matrix group it is embedded in GL(3, R).
One calls it special, because it is a subgroup of the orthogonal group in 3 dimensions, O(3),
where the requirement det(R) = 1 is dropped.
We know that any arbitrary rotation in R3 can be described by successively rotating about
all 3 axes (as soon as a coordinate system is chosen):
R(φ, θ, ψ) = Rx (φ)Ry (θ)Rz (ψ) = exp(i φTx ) exp(i θTy ) exp(i ψTz ) = exp(i
3
X
αa T a ) .
a=1
We also know, that rotating first about the x-axes and than about the y-axes is not the
same as rotating first about the y- and than about the x-axes as shown in Figure 2, which
simply translates into SO(3) being non-abelian. In general:
R(α)R(β)R−1 (α)R−1 (β) = R(θ) 6= 1 .
Upon expanding the group elements this yields (from now on I will use Einstein notation),
1 − αa βb (T a T b − T b T a ) + O(α2 , β 2 ) = 1 + iθc T c + O(θ2 ) ,
⇒
[T a , T b ] = i f abc T c .
6
Where [·, ·] is the commutator and the symbols f abc are called structure constants and
are antisymmetric in the first two indices. It should be clear that the commutator only
vanishes if the group is abelian.
Exercise
[3pt] What can you say about the generators of SO(3) just from the group
definition? (Hint: Write RT R = 1 and det(R)=1 in exponential form.) Find real 3 × 3
matrices which have the properties you just derived (choose a basis for Te SO(3)), and compute the structure constants f abc of SO(3).
With the bilinear operation [·, ·] : Te G × Te G → Te G we have found additional structure on
the tangent vectors of the Lie group. A vector space with such an inner product is called an
algebra. The commutator of matrices is in addition antisymmetric and since associativity
of the matrix product on group level implies associativity of multiplication on the level of
the generators, the Jacobi identity holds. We can therefore define:
A Lie algebra is a vector space V which is closed under a bilinear operation [·, ·] : V ×V →
V , such that
1. [v, w] = −[w, v] ∀ v, w ∈ V
2. The Jacobi identity [v, [w, z]] + [z, [v, w]] + [w, [z, v]] = 0 holds.
Viewed in the geometrical context this tells us that not all differentiable manifolds are Lie
groups. Only if the tangent space can be made a Lie algebra, the manifold is a Lie group.
An example for a manifold which is no Lie group is the 2-dimensional sphere pictured in
Figure 2.
Even though the elements of the algebra and the elements of the group are both represented
by n × n matrices you must keep in mind that you are dealing with different algebraic and
geometric objects here. For example the closure requirement makes sure that the matrix
product of two elements of the group does not lead you out of the group, the same does
not hold for the elements of the algebra. On the other hand, since the Lie algebra is a flat
vector space and thus is isomorphic to Rn it is closed under addition. That does not hold
for the Lie group. Just think of two matrices A, B ∈ SO(3) having determinant 1. The
sum of them has determinant 2 and is no element of SO(3).
I want to introduce a few more geometrical terms. One says a manifold M is pathconnected if for every two points x, y ∈ M on a manifold there is a continuous curve
γ : [0, 1] → M with γ(0) = x and γ(1) = y. The number of connected components is
the number of segments of the manifold which cannot be joined by such a curve.
Let us look at the Lie groups we have discussed so far. The circle and therefore U (1) is
connected. What about SO(3)? In this case and for many others where we cannot simply
draw the manifold and look at the picture we will have to find another way to tell about
their properties. A conveniant way is to look at the determinant as a continuous function
from the manifold/group into the real numbers, det : SO(3) → R. One can show that continuous functions preserve the number of connected components (as well as they preserve
compactness) and since det(R) = 1∀R ∈ SO(3) and 1 has one connected component and
is compact we can deduce that the same holds for SO(3).
7
What about orthogonal matrices which are not special? For those we have
O(3) = {M ∈ GL(3, R)M T M = 1}
⇒ det(M T M ) = 1 ⇒ det(M ) = ±1 .
We thus find that O(3) has two connected components (but is still a compact group).
I introduced these concepts because what I said earlier in this chapter is not (entirely)
true. You can only generate those elements of the Lie group from its algebra, which are
in the connected component of the identity. If you don’t believe me try to generate the
parity P = −diag(−1, −1, −1) with infinitesimal rotations using the traceless generators
you found in the last exercise. On the other hand, as long as the manifold is not totally
disconnected (and then its no longer a continuous group), one can generate every element of
the connected component of the one and add the additional discrete transformations to go
from there to any other connected component of the group and thus generate any element
of the group. Therefore one can write
O(3) = SO(3) ∪ P SO(3)
or
O(3) = SO(3) × {1, P } .
Bonus Introduce simply connected,fundamental group, universal cover and how
one lie algebra can generate different Lie groups. Introduce 1 → Z2 → SU (2) → SO(3) →
1.
3
Representations
In physics, when we work with groups we almost always work with representations of groups.
In general, a representation of an abstract group G on some vector space V of dimension
d is defined as a linear map
D : G → GL(d, V ) ,
D(g) ∈ GL(d, V ) : V → V
which satisfies the Homomorphism property (it should preserve the structure of the group)
D(g1 )D(g2 ) = D(g1 × g2 ) .
Note, that GL(d, V ) is not the general linear group in which the matrix group G is imbedded. This would be GL(N, V ), where n = N 2 denotes the number of generators (or the
dimension of the group manifold), while d is the number of basis vectors of the vector space
the group is represented on.
In a representation, the exponential map reads
D(g) = exp i
d
X
αa D(T a )
a=1
where D(T a ) is the representation of the Lie algebra elements.
In addition, a representation is called faithful, if it is injective. That is, D(gi ) 6= D(gj ) ∀gi 6=
gj (the indices are just for conveniance, all this holds also for continuous groups).
8
Let us have a look at a few examples. Consider the 2-dimensional representation of U (1)
on C2 , given by
!
eiφ 0
2
D(g) ∈ GL(2, C ) ,
D(g) =
.
0 eiφ
In this example, the representation is faithful, but n = 1 and d = 2. Another example is
the trivial representation, which sends all group elements to the identity. This can never
be a faithful representation, if the group is anything else besides the trivial group G = {e}.
Let me give another example for a representation which is not faithful, the representation
of O(3) on R,
D(R) ∈ GL(1, R) ,
D(R) = det(R) ,
where n = 3 and d = 1. Every element in one connected component is send to the same
D(g). That would however be a faithful representation of Z2 = {1, −1} .
A very important representation is the fundamental representation, which is the smallest dimensional representation which is faithful. For the fundamental representation, d =
N . The example above is the fundamental representation of Z2 . The fundamental representation for the SO(3) is given by the real 3 × 3 matrices for which RT R = 1 and det R = 1.
Active rotations about the 3 axes are then given by





1 0
0
cos θ 0 − sin θ
cos ψ − sin ψ





Rx (φ) = 0 cos φ − sin φ , Ry (θ) =  0 1 0  , Rz (ψ) =  sin ψ cos ψ
0 sin φ cos φ
sin θ 0 cos θ
0
0

0

0 .
1
Exercise
[3pt] Derive the fundamental representation of the generators with these
matrices via the exponential map,
R = exp(i
n
X
αa D(T a ))
⇒
a=1
D(T a ) = −i
∂
R
∂αa
and check what you found in the last exercise (your result should agree up to a similarity
transformation).
You will probably protest now and say that we defined the abstract matrix groups SO(3) in
section 2 as the same thing which I now try to introduce as the fundamental representation.
And you are right. The general abstract definition would have been: R ∈ SO(N ) are those
matrices which preserve the euclidean scalar product on the vector space RN and lie in the
connected component of the identity,
for
y i → Rij xj
⇒
!
y i δij y j = Ril xl δij Rjk xk = xl (RT )li δij Rjk xk = xi δij xj
⇒ (RT )li δij Rjk = δlk .
If one now chooses a representation7 of RN , these matrices can explicitly be given as a
representation of the group on this vector space. The fundamental representation is one
7
That means, choose a basis for RN .
9
possibility. But since all properties can be derived from the fundamental representation,
it is also called defining representation in the literature. If we define a group via its
fundamental representation, we can write for the according vector space with dimension
!
dim V = d = N
for a vector x ∈ V ,
D(g) x = g x ,
and D(T a ) = T a .
Further, one defines the conjugate representation (or dual representation) D(g), for
which also d = N , by
D(g) x = g ∗ x ,
where the ∗ denotes complex conjugation. It follows immediately for the conjugate representation of the generators,
exp(iαa T a )∗ = exp(−iαa T a ∗ )
⇒
D̄(T a ) = −(T a ∗ ) .
For real groups like SO(3) the fundamental and the conjugate representations are isomorph.
For U (1) the only generator is also real, so nothing new here. How about the unitary group
in N dimensions? In a representation independent way it is defined as the transformations
which leave the scalar product on CN invariant. Defining it via its fundamental representation we can write
U (N ) = {U ∈ GL(N, C)|U † U = 1} .
An arbitrary element of the Lie algebra is given by a complex N × N matrix and therefore
the conjugate and fundamental representations should (in general) not be the same.
Exercise
[1pt] Ask yourself if the structure constants depend on the representation.
What would you expect? (Please look up what Homomorphism property means, if you are
unsure). Compute the commutator of the fundamental [D(T a ), D(T b )], and of the conjugate representation [D(T a ), D(T b )] of the generators. Can you derive if the structure
constants are in general real of complex?
There is yet one more important representation you must know, the adjoint representation. Let me introduce it via an example. Consider the two-dimensional complex vector
space C2 . We denote its elements as ψ ∈ C2 and the corresponding adjoint vector as
ψ ≡ ψ † . The scalar product
L = ψψ
is invariant under the fundamental representation U ∈ U (2). If we put a diagonal, universal
(that means m11 = m22 ) matrix M in there L stays invariant
ψ U †M U ψ
⇒
m U † 1U = M .
How does this change if we replace M by a general matrix A. We can directly read off that
L stays invariant if A transforms as
A −→ U AU † .
10
Since the space of complex 2 × 2 matrices is isomorph to C4 , we can then define the adjoint
~ through the action of
representation as a 4 × 4 matrix Dad (g) acting on the C4 vector A
the fundamental representation on the corresponding 2 × 2 matrix
Dad (g)A = gAg −1 ,
The following steps can be a bit confusing if you have never seen it before, so I recommend
careful reading from now on. Again, A denotes a complex 2 × 2 matrix. Since we know that
the fundamental representation of the generators D(T b ) of the Lie algebra of U (2) form a
basis for M2×2 (C) (the tangent space), we can write A as
A=
4
X
Ab D(T b ) .
i=1
It has nothing to do with the group itself. We just decompose A into components times
basis vectors. On the other hand, A as a C4 column-vector can be decomposed into the
basis vectors of C4 , A = Ai ei , and we can treat it as the same object. Now we put the
latter into the left hand side of the definition of the adjoint representation, the first into
the right hand side and expand both sides (suppressing higher orders and applying Einstein
notation),
Dad (g) Ab eb = U Ab D(T b ) U †
exp i αc Dad (T c ) Ab eb = exp i αa D(T a ) Ab D(T b ) exp − i αa D(T a )
1 + αc Dad (T c ) Ab eb = 1 + i αa D(T a ) Ab D(T b ) 1 − i αa D(T a )
= Ab D(T b ) + i αa Ab [D(T a ), D(T b )]
= A − αa Ab f abc T c
So that
ab
Dad (T c ) = i f abc = −i f bac ,
since the structure constants are real (check your last exercise). We can generalize the
result from U (2) to U (N ) and see that the adjoint representation has dimension d = n,
where n is the dimension of the Lie group.
Let me repeat, the dimension of the fundamental/conjugate representation is given by
d = N , whereas the dimension of the adjoint representation is n, the dimension of the Lie
group (the number of parameters the group depends on). There is a convention to call the
representations by the dimension of the vector spaces they are defined on. The fundamental
representation of the SO(3) is called 3 as is the fundamental representation of SU (3). The
conjugate representation of SU (3) would be 3̄, the adjoint representation 8, and the trivial
1. You get the idea.
A few more definitions: A representation is called unitary, if the representation matrices
are unitary. Note, that that does not mean that the group has to be unitary. Further, two
11
representations D1 and D2 of the same group are called equivalent representations, if
there exists an invertible linear map S, so that
D2 (g) = S D1 (g) S −1 .
This implies for the generators,
D2 (T a ) = S D1 (T a ) S −1 ,
a = 1, . . . , n.
We can then quote the important theorem, that
the representations of every finite group and every compact Lie group are equivalent to a
unitary representation.
Again, compact Lie groups play a special role. Keep in mind that that also implies that we
cannot always find a unitary representation if the group is not compact.
Let me give another example. Consider the special unitary group in two dimensions,
SU (2) = {U ∈ U (2)| det U = 1} .
Analogue to the exercise you did in the last chapter for SO(3), we can use the fundamental
representation to determine number and form of the generators,
U = exp (i
n
X
αa T a ).
a=1
We know that n is less then N 2 = 8, since SU (2) < GL(2, C), which just means that the
generators will be complex 2 × 2 matrices. From
U †U = 1
⇔
exp(i(T a − T a † ))
⇒
Ta = Ta†
det(U ) = 1
⇒
Tr(T a ) = 0 ,
we can deduce that they are hermitian, which takes away 4 parameters (the trace is real
and the non-diagonal entry is self-conjugated), and traceless, which is one more condition,
leaving us with 3 parameters. And everyone of you knows a basis for the space of hermitian,
traceless complex 2 × 2 matrices, the Pauli matrices:
!
!
!
0
1
0
−i
1
0
σ1 =
, σ2 =
, σ3 =
.
1 0
i 0
0 −1
Therefore, every element of the SU (2) in the fundamental representation can be written as
!
3
X
σa
,
D(g) = exp i
θa
2
a=1
where the factor 1/2 is a convention. For the conjugate representation we know that
D(T a ) = (D(T a ))∗ and thus
D(T a ) = (σ a )∗ = σ 1 , −σ 2 , σ 3 .
12
If you know remind that
σ 2 σ 1 , σ 2 , σ 3 σ 2 = σ 1 , −σ 2 , σ 3 ,
you can check that the fundamental and the conjugate representations of SU (2) are equivalent, using S = σ 2 and the equivalence relation
!
3
X
σa
−1
2
S D(g) S = σ exp i
θa
σ 2 = D(g) or 2 ' 2̄ .
2
a=1
This is a speciality of SU (2), which does not hold for higher SU (N )’s.
As promised earlier, every other representation can be constructed from the fundamental
representation. In order to see this we need to generalize the notion of direct sums and
Kronecker (tensor) products from matrices to representations.
A direct sum of a vector space is a way to get a new big vector space from two (or more)
smaller vector spaces. One denotes the direct sum by V = V1 ⊕ V2 . Let me demonstrate
the direct sum of two vector spaces of column vectors. Consider a vector ~x ∈ V1 , which has
dimension n and ~y ∈ V2 which has dimension m. Than, the direct sum ~x ⊕ ~y ∈ V = V1 ⊗ V2
can be written as
   
 
x1 
x1
0


 ..   .. 
 .. 
 . 

.  . 

   
 
 xn   0 
 xn  






~x ⊕ ~y =   +   =   n + m

 0   y1 
 y 1 

 ..   .. 
 .. 

.  . 
 . 



0
ym
ym
and the resulting vector space V has dimension n + m. Consider now a quadratic matrix
A acting on ~x and a quadratic matrix B acting on ~y , via
~x 7→ A~x
and
~y 7→ B~y .
The obvious way to define the direct sum is
A⊕B =
A
0n×m
0m×n
B
!
,
what becomes clear by acting on the direct sum of the vectors,
!
!
A
0n×m
x
(A ⊕ B) (~x ⊕ ~y ) =
= (A~x) ⊕ (B~y ) .
0m×n
B
y
Keep in mind that there are n2 independent matrices on V1 and m2 independent matrices
on V2 , but (n + m)2 independent matrices on V = V1 ⊕ V2 . The remaining 2nm matrices
cannot be written as a direct sum. Now, replace matrices by representations and you know
what the direct sum of two representations is.
13
Excersice Express det(A ⊕ B) and Tr(A ⊕ B) through the determinants and traces of A
and B.
The tensor product is another way to construct a bigger vector space out of two or more
smaller ones. I will also explain this with an example. Let us take the same vector spaces
as above and build W = V1 ⊗ V2 . In order to simplify notation, let me set n = 2 and m = 3.
Then, the direct product of our two vectors reads


x1 y1 


x y 




1
2

!
! 


x1 y3 
x1
x1 y 1 x1 y 2 x 1 y 3

~x ⊗ ~y =
'
y1 y2 y3 =
x y  n × m, ,
x2
x2 y 1 x2 y 2 x 2 y 3
 2 1 




x2 y2 



x2 y3
where the last two notations are equivalent. Proceeding with the latter, the tensor product
(also called Kronecker product) of the two matrices acting on these vectors is defined by


!
b11 b12 b13
a11 a12


⊗ b21 b22 b23 
A2×2 ⊗ B3×3 =
a21 a22
b31 b32 b33


a11 b11 a11 b12 a11 b13 a12 b11 a12 b12 a12 b13

a b
 11 21 a11 b22 a11 b23 a12 b21 a12 b22 a12 b23 


 a11 b31 a11 b32 a11 b33 a12 b31 a12 b32 a12 b33 
,

=

a
b
a
b
a
b
a
b
a
b
a
b
21 12
21 13 22 11
22 12
22 13 
 21 11


 a21 b21 a21 b22 a21 b23 a22 b21 a22 b22 a22 b23 
a21 b31 a21 b32 a21 b33 a22 b31 a22 b32 a22 b33
so that
(A ⊗ B) (~x ⊗ ~y ) = (A~x) ⊗ (B~y ) .
You can convince yourself that the last line holds by explicitly writing it out.
The point is that the direct sum as well as the tensor product of representations are again
representations. Consider now a representation D(g) on some vector space W (this might
well be the tensor product of two vector spaces). If it is possible to find a sub vector space
V ⊂ W , which is left invariant by the operation of D(g),
D(g)~x ∈ V ∀ ~x ∈ V
and ∀g ∈ G,
the representation is said to be reducible. That means it can be reduced. On the level on
the matrix you can find an equivalence transformation, so that

!  DV (g) : V → V

DV (g) α(g)
D(g) =
DW \V (g) : W \ V → W \ V

0 DW \V g

α(g) : W \ V → V
14
If one cannot find such a transformation, that means the vector space on which the representation acts does not have a (non-trivial) invariant subspace, the representation is called
irreducible. And if you can decompose a vector space into a direct sum of sub vector spaces
W = V1 ⊕V2 which all transform under irreducible representations, W is called completely
reducible. For the representation matrices above this implies that α(g) = 0, W \ V = V2
and all representations DV1 (g) and DV2 (g) are irreducible. In general a completely reducible
representation can be brought in a blockdiagonal form,


Da






Db





D1 (g) ⊗ D2 (g) = 




Dc






...
where all representations on the blockdiagonal are irreducible and depend on the same
group element g.
Not all representations are completely reducible. However, there is a powerful theorem
which guarantees that every (reducible) unitary representation is completely reducible.
One more reason why compact groups are so well understood.
Quantum numbers are labels of representations, be it charge or spin or mass. They tell us
under which representation of the corresponding group (through Noethers Theorem) the
field which describes the respective particle transforms. There is an abuse of language since
people say that a particle is in the fundamental representation (for example), which means
that the field is a vector of the vector space on which the fundamental representation acts.
When we want to know the quantum numbers of bound states (protons, atoms, etc.) we
built the tensor product of the representations of the involved fundamental particles (which
transform under irreducible representations) and decompose the product into a direct sum
of irreducible representations,
D1 ⊗ D2 = Da ⊕ Db ⊕ Dc ⊕ . . . ,
where the right hand side decomposes in a unique way. Thus, in some sense, you can see irreducible representations as the prime numbers of representation theory. It makes therefore
sense that fundamental particles transform according to them, since we expect everything
we see to be build up from elementary particles, just as every representation is built up
from irreducible representations (and every number from prime numbers).
When I say, build the tensor product of irreducible representations, I mean irreducible
representations of the group. However, as it is often the case, the whole procedure can be
simplified by going to the level of algebras. It is way easier to identify the generators then
the group element in some representation. We can expand the exponential map again and
15
use that
D1 (g) ⊗ D2 (g) = (1 + i αa D1 (T a )) ⊗ (1 + i αa D2 (T a )) + O(α2 )
= (1 ⊗ D1 (T a ) + D2 (T a ) ⊗ 1) iαa + O(α2 )
which we decompose into a direct sum and write it as an exponential again.
In that sense you can understand the term, addition of angular momenta in the case of
SU (2).
Let me illustrate this with a few examples. If we describe a coupled system of two charged
particles, for example an electron and a proton, we know already (since we consider charge a
conserved quantum number) that the outcome carries charge zero. This symmetry is related
to the invariance of the lagrangian under U (1) through Noethers theorem. In representation
theory what we have done by reading off the possible charges of the final state corresponds to
building the direct product of the representations under which our input particles transform
and compute the possible irreducible representations of this direct product, what gives us
the possible final state charges. All irreducible representations of U (1) are of dimension 1,
because it is an abelian group8 and can therefore be written as
Dl (g) = exp(i l φ)
l ∈ Z.
That l must be in Z follows from the fact that φ is 2 π periodic. Therefore two representations are the the same if
Dl (φ1 ) = exp((i l φ))
and
Dl (φ2 ) = exp((i l φ2 )) = exp(i l(φ1 + k 2 π) with k ∈ Z
⇒ exp(2 π l k) = 0 ⇒ l ∈ Z.
Because every representation is one-dimensional, one does not use the dimension to label
the representations but l itself. Our electron is thus in the representation l = −1 and our
proton transforms under the representation l = 1. the direct product of the generators is
easily computed and confirms our result,
−l ⊗ 1 + 1 ⊗ (l) = −l + l = 0
which is the trivial representation and by itself already irreducible, so we have not to convert it into a direct sum. The right hand side is a one-dimensional block-diagonal matrix,
a number.
Let us now consider the spin of said particles. Electrons and protons carry spin 1/2 and you
know that the resulting system can either have spin 1 or spin 0 from the addition of angular
momenta. This however is another example for representation theory at work. Both fields
transform under the fundamental representation of SU (2) (which is what spin 1/2 means),
and we know its generators are given by the pauli matrices. We can now build the direct
8
This is a consequence of Schurs Lemma. Suppose that we have a representation of an Abelian group
of dimensionality d is greater than one. Suppose furthermore that these matrices are not all unit matrices
(for, if they were, the representation would already be reducible to the d-fold direct sum of the identical
representation.) Then, since the group is Abelian, and the representation must reflect this fact, any nonunit matrix in the representation commutes with all the other matrices in the representation. According to
Schur’s First Lemma, this renders the representation reducible. Hence, all the irreducible representations
of an Abelian group are one-dimensional.
16
product of these generators which gives us the representation under which the final state
transforms,


0 1 1 0
1 0 0 1


Jx =σx ⊗ 1 + 1 ⊗ σx = 

1 0 0 1
0 1 1 0


0 −i −i 0
 i 0 0 −i 


Jy =σy ⊗ 1 + 1 ⊗ σy = 

 i 0 0 −i 
0
i
i
2
0

Jz =σz ⊗ 1 + 1 ⊗ σz = 
0
0
0
0
0
0

0 0
0 0


0 0
0 −2

0
With the help of the a unitary matrix9 U , these can be transformed into a direct sum, that
is we choose the appropriate basis vectors in the product space and get the generators in
blockdiagonal form,


0 0 0 0
0 0 1 0


U Jx U † = 

0 1 0 1


0 0 1 0


0
0

U Jy U † = 
0

0
0 0 0
0 −i 0 


i 0 −i 

0 i 0


0
0

†
U Jz U = 
0

0
0
1
0
0
0 0
0 0


0 0

0 −1
,
which is the generator of the trivial representation in the upper left corner and the generators
of SO(3) in the second block. They agree with what we have found earlier, if you choose a
9
Derived in the appendix 4
17
equivalence transformation and make the Jz generator diagonal. We have found, that the
hydrogen atom can transform as a Singlet or as a Triplet state. In short one writes,
2 ⊗ 2̄ = 1 ⊕ 3 .
For larger systems and groups, this calculation becomes more and more complicated. There
is a systematic way to decompose a tensor product into a direct sum, making use of socalled Young tableaus. I will not introduce them here, but you should definitely read up
on it. We will rather go a more constructive way and see what we can conclude from the
symmetry properties of the product spaces the representations act on. We will restrict
ourself to the orthogonal and unitary groups, since every group we care about here is at
least locally isomorph to those.
An element of the product space of two (or more) copies of the same vector space is the
most common way to define a (second rank) tensor. From the discussion above, you can
see that this is a special case of a tensor product. Take for example a vector space V with
dimV = 3,


 
v1 v
w
v
w
v1w
1 1 1 2
3


 
T = ~v ⊗ w
~ = v2  w1 w2 w3 = v2 w1 v2 w2 v2 w3 
for v, w ∈ V
v3
v3 w1 v3 w2 v3 w3
which is also called a dyadic product in older textbooks. From this expression, we can
understand why people often define tensors as ”objects which transform as tensors”. Or,
as an object which transforms as if it is equal to the product T ij ∼ v i wj . Therefore, let us
assume v 0i = D(g)ij v j transforms under the fundamental representation of SO(N ). Then
T ij → T 0ij = D(g)il D(g)jm T lm .
If we decompose this tensor into a symmetric and an antisymmetric part, S ij = 1/2(T ij +
T ji ) and Aij = 1/2(T ij − T ji ), we see that a transformation of those stays symmetric
or asymmetric 10 . Thus, the product representation acting on T ij is reducible, both the
antisymmetric and the symmetric rank 2 tensors transform among themselves and are thus
invariant subspaces. In addition, it follows from the orthogonality of the representation
matrices that the trace of a symmetric tensor, Tr(T ) = δ ij T ij , also transforms into itself,
δ ij T 0ij = δ ij D(g)il D(g)jm T lm = (D(g)T )li δ ij D(g)jm T lm = δ lm T lm .
The symmetric part can thus be further reduced and we end up with three irreducible
subspaces, the antisymmetric tensors, the symmetric traceless tensors and the trace. As an
example, for SO(3) this tells us
3 ⊗ 3 = 1 ⊕ 3 ⊕ 5,
which reflects the well-known result of two spin-one particles coupling to a spin 0, 1 or 2
object.
10
You can easily read off that D(g)il D(g)kj S kl and D(g)il D(g)kj Akl still have the same symmetry properties.
18
Exercise
[3pt] Find a formula to decompose general products N ⊗ N for SO(N ).
In the case of SU (N ) we have to distinguish between vectors transforming under the fundamental and under the conjugate transformation and tensors in the product space of
combinations of those. We denote a vector in V by v i and
v i → v 0i = D(g)ij v j
and a complex conjugate vector in V ∗ as v ∗ i , transforming as
v ∗ i → v 0∗ i = D(g)ij v ∗ j = (D(g))†i j v ∗ j .
Remember that in terms of the representation matrices D(g) = D(g)∗ . We can thus introduce the notion vi for a vector transforming as v ∗ i . We distinguish them by the position of
the indices. We can now have rank 2 tensors transforming as Tji ∼ v i vj ≡ v i v ∗ j as well as
those which transfrom udner the direct product of two fundamental T ij or two conjugate
representations Tij . The trace is only an invariant subspace of the Tji type tensors. One
can see this by explicite computation, if TrT ≡ δjk Tkj , one has
δjk Tk0j = δjk D(g)jl (D(g)† )hk Thl = D(g)kl (D(g)† )hk Thl = δlh Thl ,
while δ ij T ij is not invariant. Therefore, we have just the symmetric and antisymmetric
tensors as irreducible components of the product representation acting on T ij and consequentially for SU (3),
3 ⊗ 3 = 3 ⊕ 6.
On the other hand. for the representation acting on Tij the trace is the only invariant
subspace. There is really no point in looking for a symmetry by exchanging indices which
belong to different vector spaces. But since we are talking about a compact group, we can
assume a unitary representation and therefore the “rest” is also an invariant subspace. In
fact it turns out to be the adjoint representation. Let us again take SU (3) as an example,
3 ⊗ 3̄ = 1 ⊕ 8 .
We can go on and derive the irreducible components from the symmetry properties of the
tensors on which the product representations act. For example, for SU (3), if we want to
know how 3 ⊗ 3 ⊗ 3 decomopses, we have to consider the object T ijk with 33 independent
components. We can reduce it first in the subspace of tensors which are symmetric or
antisymmetric under exchange of 2 indices,
T ijk = T i[j k] + T i{j k}
which have 3 times the components you found in the last exercise (for every possible value
of i). And we can further take out the space of total antisymmetric tensors as an irreducible
subspace of T i{j k} , which is only one tensor, and the subspace of total symmetric tensors
of T i[j k] with 10 components11 . These have to be substracted from the dimensions of T i[j k]
and T i{j k} and if you made no mistake in the last exercise, we end up with
3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10 .
This can become arbitrarily complicated if you want, but you should really look into Young
tabloux for more complicated products.
11
A general symmetric tensor of rank r over a vector space of dimension N has
19
N +r−1
r
elements.
4
Physics
Explain gauge groups and SU (2) → SO(3) and SL(2, C) → L↑+ .
First and only appendix
Compute the matrix in section 3 and bring counterexamples for section 2.
Questions about the problems: [email protected]
20