§
Chapter 1: Representations
1.1
Introduction
One of the many very useful mathematical concepts used in physics is a group. Groups have been so thoroughly studied that there is an extraordinary amount that can be said about a particular group’s structure. Groups are wonderful tools for describing symmetries and thus a physicist might be inclined to exploit group structure to extract information about his or her problem. However, the mathematical definition of a group is just a set of elements that obey a few properties and these elements say nothing working in a real world coordinate system. A physicist may thus wish to find a representation for these group elements that is related to the coordinate system.
1.2
Representations
Consider 
, the group of non-singular linear transformations of a
vector space 
. An element 
is a map 
over the complex
numbers. Now if we define an ordered basis 
for 
, then every element 
has an associated
invertible square matrix with coefficients 
. There can be some confusion at this point because some
authors define the General Linear group 
in terms linear
transformations, whereas others authors define the group in terms of their
associated matrices. In this reading we use the former definition.
Definition 1.1 A representation of 
is defined as a
homomorphism 
. If 
is injective (every
element of the group has a unique matrix representation), then 
is called a faithful representation of G. If we let 
be an ordered basis
for 
, we then let 
be the matrix of 
with respect to this
basis, 
. We thus denote the coefficients of 
as 
.
Recall that the homomorphic property just says that 
, a very necessary property if we hope to accomplish
anything. It’s worth going over a few of the highlights of the above definition
for understanding. When we defined a representation 
we did this without
specifying a basis for the vector space 
. We could consider two different basis of 
, say 
and 
. The representation 
can thus have two
very different look associated matrices depending on the chosen basis, to whit 
and 
. The two representations 
and 
are called equivalent if they are related by a
similarly transform 
such that 
, such as in this case.
Example Let’s
consider now an example of two representations of the group 
. This group can be thought of as the group of 90-degree
rotations about the z-axis. We denote its elements by 
. Note also that this group is generated by the element 
and we therefore
often denote a group in terms of its generator, 
.
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The first representation, the unfaithful representation, is aptly called the trivial representation – notice that it does obey the homomorphic property and is a mapping to a one-dimensional vector space. The second representation is faithful because all of the elements are unique – notice that it too obeys the homomorphic property.
1.3
Invariant
Subspaces
From linear algebra we need to recall the idea of a direct
sum. Given a vector space 
and two subspaces 
and 
, then 
if for every 
can be written
uniquely as 
where 
and 
. This means would mean that 
and 
. In terms of basis vectors, this is equivalent to saying
that if 
is spanned by 
and 
by 
then 
is spanned by 
. When 
, we call 
the complement of 
in 
. If 
for all 
and 
, we call 
the orthogonal complement of 
in 
.
Definition Given
a representation 
, then if 
and 
for all 
, then we say 
is invariant under 
. Similarly, if 
is a proper subspace,
and 
for all 
, then 
is an invariant subspace under 
.
Theorem Let 
be a representation
and 
a proper subspace. If
is an invariant
subspace under 
, then the orthogonal complement of 
is also an invariant
subspace under 
.
Proof (sketch of
ideas) For proof of this theorem we consider the representation 
in matrix form 
with respect to some
basis 
. First we let 
be a 
-invariant subspace of 
and 
be an ordered basis
of 
. We extend 
to the ordered basis 
of 
and thus denote the
subspace 
as being spanned by 
. In this case we are not, in particular, looking at the orthogonal complement of 
, but just the complement 
. A representation
takes on the form
Note that I am suppressing some information by simply
writing 
instead of 
for each component.
Operating on a basis vector 
, with respect to the ordered basis, yields the vector
.
However, for the basis vectors 
where 
we know that 
and thus the
components of 
zero outside of the
subspace. We now have a representation of the form
.
Apply the homomorphic property 
for some 
, we find that
.
From this we see that both 
and 
also give us
representations of the group 
because they obey the
homomorphic property. But while 
seems to be invariant
because it doesn’t ‘leak’ into the other subspace (the lower left hand
partition of the matrix is all zeros), 
does leak out of it’s
subspace and is thus not invariant.
We can extended this argument by decomposing 
into its 
-invariant subspace 
and 
’s orthogonal
complement 
which is also a 
-invariant subspace. Thus 
. If 
is an ordered basis
for 
and 
is an ordered basis
for 
, then 
is ordered basis for 
. Extending the same argument as above, we find that we now
have a representation of the form
.
At this stage it should now be clear that both 
and 
are invariant.
So what would happen if we continued to decompose the
representations 
using the same method
until it can be reduced no more? The representation will be in block diagonal
form and this leads us to the concept of irreducible representations.
Definition A
representation 
is considered irreducible if there exists no non-zero
proper subspace of 
invariant under 
.
Theorem 1.2 Every representation is a direct sum of irreducible representations.
Proof Through the
extended proof of the last theorem this should be a pretty logical result.
Proof of this can shown by induction on the dimension of 
.
1.4
Direct
Product
1979 Nobel Prize winner Steven Weinberg wrote, “The universe is an enormous direct product of representation of symmetry groups.” Here we examine the direct product and its representations.
Definition Given
a finite collection of groups 
, then the direct product is 
. We define the binary operation by 
.
Example Consider
the group 
where 
and 
. The group elements are thus 
.
Definition Let 
and 
be two
representations of 
with basis 
and 
. We thus have that
We now define the direct product to be
.
This definition does still satisfy the homomorphic property, you can check this easily. The above definition has so many indices that it’s a little daunting. So note that this is equivalent to saying
if both 
and 
are two dimensional
representations. This is rather easy to calculate.
The direct product of two irreducible representations is not necessary irreducible. The direct product representation may be the direct sum of irreducible representations. To determine exactly how the direct product of a representation decomposes into direct sums, we need a few more tools that we will develop studying characters.
§
Chapter 2: Characters
2.1
Introduction
Having established the basic of representations in the last chapter, we now need a few more tools in order to work with them. Ultimately we would like to be able to take the direct product of two representations and then decompose the direct product into irreducible sums. Solving this problem is equivalent to finding a ‘good’ set of basis vectors that block diagonalize the matrices of the representation. This is exactly the problem that must be solved in order to find Clebsch-Gordon coefficients as we will see later.
This chapter will be presented in a rather unconventional manner. We will begin with a quick review and by presenting the major results of the chapter. This will help to provide structure and motivation for the subsequent proofs and derivation of the results.
2.2
The Tools
Recall the definition of a conjugacy class.
Definition 2.1 Let
be a group. Define 
and 
as the conjugacy
class of 
. The conjugacy class partitions the group 
.
Example 2.2
has three conjugacy
classes:
Definition 2.2 The
character of a representation 
of 
is defined as 
. The ordered collection of the characters of the elements
for a representation is denoted by 
. It may also be convenient to denote this vector as a ket 
.
Theorem 2.3
i. Two equivalent representations have the same character.
ii. Elements in the same conjugacy class have the same character.
iii.
If a representation is unitary, then 
(complex conjugate).
Proof
i.
Recall from linear algebra that 
. Let 
and 
be two equivalent
representations and 
the
basis-transformation matrix such that 
. Notice that 
. In words, two equivalent representations have the same
character. This is a great result because it means that when we use characters,
we do so without having to choose a particular basis for our representation.
ii.
Elements in the same conjugacy class of a group have the same
character. Let’s assume that 
and 
are conjugates given
by 
. Then
and thus all elements in a conjugacy class have the same character.
iii.
If 
is unitary then 
(the conjugate
transpose).
With this, we now present the major results of this chapter.
The Tools
- The
number of irreducible representations of a group
is equal to the
number of its conjugacy classes, 
. - Let
be the dimension
of the irreducible representation 
. Then 
. In words, the sum of the squares of the dimensions of
the irreducible representations is equal to the size of the group. 
. We denote this inner product with 
. Orthogonality of character for representations 
and 
.- The
character of a representation can be expressed as a linear combination of
the characters of the irreducible representations of a group. This is
because the characters
span all of group space.
These are the tools necessary to decompose a representation
into its irreducible parts. If a representation 
has character 
and the irreducible
representations 
have characters 
, then we simply use the inner product. Thus, 
.
If you do not wish to see the derivation of the above results, then skip to the last section of the chapter and start looking at some examples.
1.3
Schur’s Lemmas
Lemma Let 
be an irreducible
representation 
. If 
is a linear
transformation that commutes with 
, 
, then 
is a scalar.
Proof Because we
are working over the algebraically close field of the complex numbers, we know
that every linear transformation has at least one eigenvector and corresponding
eigenvalue. We let 
be an eigenvector of 
with corresponding
eigenvalue 
.
We can now say that,
In other words, we see that 
is also an
eigenvector of 
with eigenvalue of 
. Actually, 
could be more than
one eigenvector because we’re looking at all 
. Additionally, this set of eigenvectors also forms a
subspace 
that must be group
invariant (because we used the group to build it). Because we assumed 
is irreducible, we
are left with two choices, either 
or 
. The latter case is quickly ruled out because we know that
the subspace contains at least one eigenvector, thus we conclude that 
. This implies that 
for all 
. Thus 
.
Lemma Let 
and 
be two inequivalent
representations and 
a mapping between the
two vector spaces. If 
, then 
.
Proof
Case 1 
: Let 
be arbitrary, then 
. This is just a fancy way of saying that 
for all 
. But of course, we know from the definition 
that 
and that the 
is group invariant (created
by 
). We again invoke the irreducibility of the representation 
, and thus either 
or 
. However, the dimension of the image cannot be larger than
the range, thus 
which says that 
. We have reached a contradiction, and thus it must be true
that 
because 
.
Case 2 
: We consider now
the kernel of 
by examining the
relation 
just as we did
before, but instead we use the vectors 
such that 
. Thus we now have that 
. Thus 
for all 
and the kernel of 
is a group invariant
subspace. Invoking irreducibility, we know that either 
or 
. However, we know that there must be something in the kernel
because the
transformation 
reduces
dimensionality (range larger than the domain, 
) and therefore 
. So if the kernel 
is the whole vector
space, then 
.
Case 3 
: Consider the
same argument where we know that 
or 
. This time if 
, then 
is one-to-one and
thus invertible. This would imply that we could write 
and thus 
, a contradiction to our assumption that these are
inequivalent representations. We have thus shown for the third and final time
that 
.
1.4
Orthogonality
Let 
and 
be two
representations and 
a mapping between the
two vector spaces. We now define the operator 
,
This is the same ordered sum that was used to define the
character previously. Notice what we’re trying to do here. The object is to
find an orthogonality relation, 
, between the different irreducible relations and we will do this
by exploiting Schur’s Lemmas. We’re summing over all of the group elements
because, of course, we need to know about the entire groups behavior. We would
expect that if the irreducible representations are equivalent, that 
will just give us the
identity element 
. The representation of 
is, of course the
identity, and we’re summing 
times. So we might
expect the sum, and therefore 
, to collapse to the size of the group 
if the irreducible
representations are equivalent (Schur’s First Lemma) and the sum to be zero if
they are inequivalent (Schur’s Second Lemma). Let’s show this now.
Consider now the elements 
and 
of 
with the
representations 
and 
. Multiply the sum with these elements and we have that
We now use the homomorphic property and find that
Recall that 
and thus
But the right side of the equation is just equal to 
and we therefore have
the relation
Now it should be clear that we can use Schur’s Lemmas. 
is zero if the
irreducible representations are inequivalent, but is some scalar times the
identity if they are equal. We write this as
We can actually figure out what 
is by taking the
trace of both sides. We find that
From which it is clear that 
. We now have the following relation
(2.1)
To proceed any further we are going to have to write the sums with the representations in terms of their components. However, before we do that it is useful to review summation of matrices.
Review
Let 
and 
be two matrices whose
product, 
, can be written as 
. The product 
, written with components, reads 
.
Let 
, 
and 
. The left side of equation 2.1 becomes
A fancy way of writing the trace would be to say 
. Additionally, the identity 
can be written as 
. Thus, equation 2.1 now becomes
Equating the coefficients of 
two of the sums are
gone and we are left with
(2.2)
This is what many authors refer to as “The Fundamental
Orthogonality Theorem” or sometimes even “The Great Orthogonality Theorem.”
Let’s take a look and review what this theorem tells us about irreducible
representations. The first delta function 
says that we’d better
be dealing with the same representation or else we’re going to get zero. The
next two delta functions 
tell us that the
representation of 
and its inverse had
better be the transpose of each other, otherwise we’re going to get zero.
To find the orthogonality of the characters we just need to
take the appropriate traces. We want the trace of the two representations, so
we can do this by simply multiplying by the delta functions 
. We thus have the following mess
The delta functions 
collapse to 
and we can also write
the traces in terms of characters, thus
However, 
is simply 
which is just 
. Recall also that 
. Thus,
(2.3)
This shows the orthogonality of characters. We therefore define the inner product
This is a very important result that makes dealing with representations a very simple task.
1.5
Number of Irreducible Representations
In addition to orthogonality of characters for each element,
we can also show that the conjugacy classes characters are orthogonal. Let’s
denote the conjugacy classes by 
with 
elements – with a
total of 
classes. It should be
pretty clear that we can rewrite the orthogonality theorem as
.
This implies that there can be at most 
mutually orthogonal
vectors (characters). But of course, this also just means that that there can
be at most 
irreducible
representations. Let’s denote the number of irreducible representation by 
, thus 
.
Alternatively, it is also possible to rewrite the orthogonality theorem as
Using the same argument as before, this result implies that 
. From which we conclude that the number of irreducible
representations is equal to the number of conjugacy classes.
Theorem Let 
be a representation
with character 
and let 
be the irreducible
representations with character 
. Then 
decomposes in the
direct sum of irreducible representations
where 
is the number of
occurrences of 
.
Proof Think characters!
The last item for us to show is that the sum of the squares of the dimensions of the irreducible representations is equal to the size of the group.
Theorem Let 
be the dimension of
the irreducible representation 
. Then 
.
Proof Ich habe kein Bock – will das nicht beweisen…
1.6
Examples
Example
Consider now the group 
. We know the group 
has three irreducible
representations from fact (1) above. Exploiting fact (2) we know that 
where 
is the dimension of
the representation. The only integer solution to this equation is 
and we thus have
three one-dimensional representations of the group. Inserting the trivial representation,
the character table looks the table below.
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It to find the remaining pieces we use the fact that the characters of 1-dimensional representations are the representations themselves. Thus we know that
where 
.
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At this point it is also important to
notice that 
and thus we can again
write the character table as
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Now let’s consider the
three-dimensional Euclidean vector space 
and the standard
basis. If we wish to consider 
as rotations about
the z-axis, we already know how to write one such three dimensional
representation, 
.
From this we can we see that 
with respect to the
basis of 
in group space.
Dotting this with the three basis vectors 
it should be clear
that 
. From this we now know that 
.
We now need to assign a basis vector
to each dimension of each irreducible representation that makes up 
. For example, knowing that we want the z-axis fixed under
this group we can assign the basis 
for 
, respectively. We now observe what a group element does to
each of these vectors.
In the standard basis of 
, we now have that
Something a little bit more fancy
might be to use the basis that will give us the rotation matrices we are used
to seeing. Out of thin air we find and then decide to use the basis 
. Now we have
and the last term is in the form 
denoting that
The basis we chose for 
yields the same
result as 
. We have determined the components of 
in this new basis,
namely,
.
It can be shown that the character
table for 
where 
denotes the nth root
of 1, is the following:
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So now if you’ll notice, the
character in the nth representation is the complex conjugate of the
character in the 2nd representation. It turns out that the direct
sum of 
in the basis 
will give you the
following representation,
§
Chapter 3: Infinite Rotational Group
3.1
Introduction
The object now is to examine the three-dimensional rotation
group, commonly denoted by 
-- the Special Orthogonal group of three dimensions and determinant 1. The idea
here is to look at this group the same way we looked at the other simpler
groups, such as 
, by finding the representations and their characters of the
group. Ultimately, because this group is infinite, it will prove to be much
more difficult to achieve our goals. However, there are many rewards that we
will pick up along the way.
3.2
Generators
Often groups are defined in terms of their generators such
as the dihedral-3 group 
where 
and 
are the generators of
the group. We will now try to find the generators of 
. We use the notation 
to denote the
rotation by 
about the 
axis. 
represents an
infinitesimally small rotation about the 
axis. From basic
calculus, we know that
. (2.1)
This is great, but there are two things to note at this
point. First, as a matter of convention and convenience we will denote the
above limit by 
instead of 
. Second, note that 
is actually just the
identity matrix, i.e., we don’t rotate at all. We now have
where 
represents the
identity matrix. Rearranging the above equation, for small 
we can write
.
Here’s a little bit of trickery: to rotate by angle 
we now rotate by 
a total of 
times. Hence,
.
If we take this limit we no longer have an approximation.
The last step follows from the definition of the exponential
function from calculus. Great. So now we have a way rotation about an axis in
terms of the operator 
. But just what is this operator and what does it look like
in the standard basis for 
? As you intuitively suspect, we can represent 
in terms a linear
combination of 
, 
and 
. The rotations about the x, y and z axis can be represented
by
From equation 2.1 we can find the infinitesimal rotations for this representation
.
Now we can write
in terms a linear combination of 
, 
and 
, 
. From inspection of
the coordinate system we can determine the missing coefficients (Heine, 53). If
is the angle of the 
vector from the
z-axis down to the x-y plane and 
is it angle from the
x-axis, then 
. We have now completed the first of our goals. We can write
any rotation in 
in terms of the three
generators 
, 
and 
.
3.3
Commutation Relations
The next thing that is useful to consider is the commutations between the three generators. It should be pretty intuitively clear the three operators don’t commute. The simplest method for determining the commutation relations is to just calculate them using the representations from above. We could look at the general case, but this representation works just fine too.
If we use this same procedure we can find the two other relations,
These relations turn out to be somewhat useful later. Otherwise they are said to define an algebra, whatever that means…
3.4
The Irreducible Representations
If you recall from the last
chapter, when we were trying to find the irreducible representations of some
group we looked for invariant subspaces of the group. We looked for subspaces, 
, that contained no proper group invariant subspaces. A set
of basis vectors of 
could be used to make
that particular irreducible representation of that group. Our approach is going
to be the same here. We are going to search for the basis vectors of subspace
invariant under the full rotation group. Because any element of the full rotation
group can be represented by a linear combination of the three generators, we
will use the generators to find the invariant subspaces.
A very nice
set of vectors to work with, are the eigenvectors of the rotation operators.
However, from the commutation relations, we know that the generators cannot
have simultaneous eigenvectors. By convention we will work with the
eigenvectors of the 
operator. Let’s start
by working in some finite n-dimensional subspace 
that we assume is
group invariant. We can now assume that 
has a least one
eigenvector in 
, let’s denote it by 
.
Let’s see
if we can find the other eigenvectors of 
. The recommend way of doing this is to create an operator,
call it 
, that moves from the one eigenvector that we know exists, 
, to the other eigenvectors 
. Symbolically we want 
. Amazingly, just by examining the commutation relations we
can determine what 
actually has to be.
Using 
as our test function,
From which we see that 
for some 
. Of course, everything in the group can be written as a
linear combination of the generators, so we can write 
. Let’s try the commutation relation one more time.
But since we know from above that 
, we can equate the coefficients. Doing this we find that 
, 
and 
. The only values of 
that satisfy the
relation 
are 
. From this we now see that 
. Letting 
we now denote the two
operators as 
and 
with commutation
relations 
.
Let’s
examine what 
do to the eigenvector
of 
.
So it seems that 
moves the eigenvector
to an eigenvector
with an eigenvalue 1 higher. Let’s write this as 
where the constant 
is yet to be
determined. Similarly, 
lowers the
eigenvector. This is a little bit of a problem because we want to deal with
finite vector spaces, so we can just arbitrarily say that the 
vector is the last
one and thus 
. At this point it’s thus probably good to change our
notation a little and write our vectors as 
so we know that we
can only raise 
up to 
before we’re done and
hit the wall.
We need to
take care of one thing that will soon be useful. We need to determine 
.
Given that 
, we will similarly define 
. Let’s see how 
and 
relate.
Now we’re set to determine 
.
but we also know that 
. Setting the component equal we see that 
.
We’re almost there! We now have a recursive equation that
establishes the missing coefficients. Solutions to this type of equation are
often solved in numerical analysis books, see Stegner’s Diskrete Strukturen for example of this. We start by substituting 
and reducing it to a
linear equation. Now,
The difference of these last two equations gets rid of that
pesky 
.
Now we just need to get rid of the 2, so we use the same process one more time.
Finally we can write and solve the characteristic equation for the recursion.
Thus closed form solutions to the recursion take the form
or simply,
.
with still to be determined coefficients 
, 
and 
. To determine these coefficients we can plug in the first
three iterations of the recursions. We know that we 
because that’s to be
the end of the ladder. Thus,
Similarly,
We can write that,
Three equations and three unknowns can be easily solved. In matrix form,
From which it easily determined that 
, 
and 
. Thus,
We can use the exact same process to determine the
coefficient for the 
operator acting on an
eigenvector of 
. Our final results are thus,
From this we can see that not only does 
, but 
. So it would seem that for some value of 
, 
can take on values
ranging from 
to 
. Thus for some value of 
there are 
eigenvectors of 
. Because we’re working in some n-dimensional vector space
and n is a whole number, this places a limit on what values 
can assume.
So it seem that 
can assume any
half-integer value, 
.
So what have we done? We have found what the irreducible
representations of the rotation group look like using the eigenvectors of 
as a basis. Right now
we have everything in terms of 
, 
and 
, but because we know how 
relate to 
and 
, we can easily recover their form. Also remember that
because we can write
in terms a linear combination of 
, 
and 
, we have the irreducible representations of and arbitrary
rotation 
. Let’s explicitly write out the first two irreducible
representations for 
, 
, 
, 
and 
,
For the 
representation we
have two eigenvectors of 
, 
and 
. With respect to those as an ordered basis we can now write,
.
We also know that 
and 
, thus
For the 
representation we
have the three eigenvectors 
forming basis. Thus,
3.5
Characters of the irreducible representation
Just as we did in the last chapter, we can compute the characters of the irreducible representations. Finding the characters, as you recall, is very useful in determining which irreducible representations compose the reducible representation.