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Last modified: December 6, 2019
Basic Representation Theory
---------------------------
In essence, a representation makes an abstract object
like a group more concrete. It is the algebraic action
of the group elements on a set (basis) where the
'things' that perform the action are invertible n x n
matrices. These matrices 'act' on the set by matrix
multiplication. A representation gives a matrix for
each element, and so another representation with a
an element acting on a different basis is possible.
We will use the dihedral Group, D4 to illustrate
the concepts involved.
The Dihedral Group
------------------
The dihedral group is the group of symmetries of
a regular polygon, which includes rotations and
reflections. Dihedral groups play an important
role in group theory, geometry, and chemistry.
Consider an arbitrary basis set {1,2,3,4} (i.e.,
elements on which symmetry operations are performed).
2 1
-------
| |
| O |
| |
-------
3 4
There are 8 symmetry operations that we can perform.
Do nothing.
Rotations by 90°, 180° and 270°
Relections about the horizontal axix through O.
Relections about the vertical axix through O.
Relections about the right diagonal through O.
Relections about the left diagonal through O.
These operations can be represented by the following
matrices:
Identity:
- - - - - -
| 1 0 0 0 || 1 | | 1 |
I = | 0 1 0 0 || 2 | = | 2 |
| 0 0 1 0 || 3 | | 3 |
| 0 0 0 1 || 4 | | 4 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (1)(2)(3)(4)
1234
Rotation by 90°:
- - - - - -
| 0 0 0 1 || 1 | | 4 |
r = | 1 0 0 0 || 2 | = | 1 |
| 0 1 0 0 || 3 | | 2 |
| 0 0 1 0 || 4 | | 3 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (1432)
4123
Rotation by 180°:
- - - - - -
| 0 0 1 0 || 1 | | 3 |
r2 = | 0 0 0 1 || 2 | = | 4 |
| 1 0 0 0 || 3 | | 1 |
| 0 1 0 0 || 4 | | 2 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (13)(24)
3412
Rotation by 270°:
- - - - - -
| 0 1 0 0 || 1 | | 2 |
r3 = | 0 0 1 0 || 2 | = | 3 |
| 0 0 0 1 || 3 | | 4 |
| 1 0 0 0 || 4 | | 1 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (1234)
2341
Reflection about horizontal axis:
- - - - - -
| 0 0 0 1 || 1 | | 4 |
dr2 = | 0 0 1 0 || 2 | = | 3 |
| 0 1 0 0 || 3 | | 2 |
| 1 0 0 0 || 4 | | 1 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (14)(23)
4321
Reflection about vertical axis:
- - - - - -
| 0 1 0 0 || 1 | | 2 |
d = | 1 0 0 0 || 2 | = | 1 |
| 0 0 0 1 || 3 | | 4 |
| 0 0 1 0 || 4 | | 3 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (12)(34)
2143
Reflection about left diagonal:
- - - - - -
| 0 0 1 0 || 1 | | 3 |
dr3 = | 0 1 0 0 || 2 | = | 2 |
| 1 0 0 0 || 3 | | 1 |
| 0 0 0 1 || 4 | | 4 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (13)(2)(4) ≡ (13)
3214
Reflection about right diagonal:
- - - - - -
| 1 0 0 0 || 1 | | 1 |
dr = | 0 0 0 1 || 2 | = | 4 |
| 0 0 1 0 || 3 | | 3 |
| 0 1 0 0 || 4 | | 2 |
- - - - - -
1 2 3 4
Cycle notation: 1234 => (1)(24)(3) ≡ (24)
1432
The Cayley Table
----------------
The Cayley table for D4 is:
Conjugacy classes defines a partition of the group.
To find the conjugacy classes we compute π-1σπ = ρ
π-1 is found by looking along the row π and reading
off the column corresponding to the element I
(i.e. ππ-1 = I). Therefore:
D4 has a total of 5 conjugacy classes:
{I} = (1234)
{r2} = (13)(24)
{r,r3} = (1234),(1234)
{d,dr2} = (12)(34),(14)(23)
{dr,dr3} = (24),(13)
If 2 matrices are in the same conjugacy class
they are similar, have the same trace, eigenvalues,
period (order) and determinant. Therefore;
{I} {r2} {r,r3} {d,dr2} {dr,dr3}
-------------------------------------
Evalues | 1 1,-1 1,-1,i,-i 1,-1 1,-1 } In
Determ. | 1 1 -1 1 -1 } 4D
Trace | 4 0 0 0 2 } rep.
Period* | 1 2 4 2 2 }
* often called the order of an element = m = gm = e
Reducibility of a Representation
--------------------------------
One of the biggest topics in representation theory
is the reducibility of a representation. In other
words, given a representation, could we encode the
same information about the group with less dimensions
(smaller basis). We will present the critical
formulas and derive some.
Let us label the different representations by a
superscript (r), (s), .... Therefore D(r)(g)
represents the element g in the representation r.
The Character of a Representation
---------------------------------
A given group can have many different representations.
The matrices of a representation are basis dependent.
We can eliminate this basis dependence by defining
the CHARACTER Χ(r) of D(r)(g) as:
χ(r) = Tr(D(r)(g))
The independency can be shown by using the cyclicity
of the trace:
Tr(D'(g)) = Tr(SD(g)S-1) = Tr(D(g)SS-1) = Tr(D(g)
Therefore, all elements in the same conjugacy class
have the same character that is independent of the
basis chosen.
For D4 we had, from above:
{I} {r2} {r,r3} {d,dr2} {dr,dr3}
-------------------------------------
χ(g) | 4 0 0 0 2
Block Diagonal Form
-------------------
A representation is COMPLETELY REDUCIBLE if it can
be reduced to a representation whose matrix elements
have the form:
- -
| D(r)(g) |
D(g) ≡ | D(s)(g) |
| ... |
| D(z)(g) |
- -
Where D(r)(g), D(s)(g) etc. are n x n matrices
that are irreducible representations that cannot
be broken up into smaller pieces. This is called
the BLOCK DIAGONAL form.
The square matrices representing the diagonal
elements can be of any size including 1 × 1.
A representation in block diagonal form can be
written as the DIRECT SUM:
D(r)(g) ⊕ D(s)(g) ⊕ ... ⊕ D(z)(g)
Block Diagonalization
---------------------
Consider the matrices:
- -
| 1 1 1 1 |
S = (1/2)| 1 -1 i -i |
| 1 1 -1 -1 |
| 1 -1 -i i |
- -
and,
- -
| 1 1 1 1 |
S-1 = (1/2)| 1 -1 1 -1 |
| 1 -i -1 i |
| 1 i -1 -i |
- -
We won't discuss how S is determined but instead
focus on the end results. Block diagonalization
is accomplished by applying the transformation
S-1AS = AB
We get:
r' = S-1rS:
- - - -
| 0 0 0 1 | | 1 0 0 0 |
S-1| 1 0 0 0 |S = | 0 -1 0 0 |
| 0 1 0 0 | | 0 0 -i 0 |
| 0 0 1 0 | | 0 0 0 i |
- - - -
r2' = S-1r2S:
- -
| 1 0 0 0 |
| 0 1 0 0 |
| 0 0 -1 0 |
| 0 0 0 -1 |
- -
r3' = S-1r3S:
- -
| 1 0 0 0 |
| 0 -1 0 0 |
| 0 0 i 0 |
| 0 0 0 -i |
- -
d' = S-1dS:
- - - -
| 0 1 0 0 | | 1 0 0 0 |
S-1| 1 0 0 0 |S = | 0 -1 0 0 |
| 0 0 0 1 | | 0 0 0 -i |
| 0 0 1 0 | | 0 0 i 0 |
- - - -
dr' = S-1(dr)S:
- -
| 1 0 0 0 |
| 0 1 0 0 |
| 0 0 0 1 |
| 0 0 1 0 |
- -
dr2' = S-1(dr2)S:
- -
| 1 0 0 0 |
| 0 -1 0 0 |
| 0 0 0 i |
| 0 0 -i 0 |
- -
dr3' = S-1(dr3)S:
- -
| 1 0 0 0 |
| 0 1 0 0 |
| 0 0 0 -1 |
| 0 0 -1 0 |
- -
It is easy to verify that these matrices follows
the D4 rules of the Cayley table. For example,
dr'.dr' = I
r'.r' = r2'
and so on.
Therefore, the matrices after the transformation
still form a representation of the same group.
The 1's on the diagonal are just the trivial
representation. It is the one dimensional
representation where all elements of G act as
the identity.
Great Orthogonality Theorem
---------------------------
The Great Orthogonality Theorem states:
(1/|G|)Σχ(r)†(g)χ(s)(g) = δrs ≡ <χ(r)(g)|χ(s)(g)>
g
or,
(1/|G|)Σncχ(r)†(c)χ(s)(c) = δrs ≡ <χ(r)(c)|χ(s)(c)>
c
Where nc is the number of elements in the conjugacy
class, c.
Consider the block diagonal matrices for D4 from
above.
D(s)(g)
|
D(r)(g) | D(t)(g)
| | |
- v v v -
| 1 |
| ±1
| - - |
| | a b | |
| | c d | |
| - - |
- -
Orthogonality of χ(r)(g) and χ(s)(g):
I r r2 r3 d dr dr2 dr3
(1/8){1.1 + 1.(-1) + 1.1 + 1.(-1) + 1.(-1) + 1.1 + 1.(-1) + 1.1} = 0
Orthogonality of χ(r)(g) and χ(t)(g):
I r r2 r3 d dr dr2 dr3
(1/8){1.2 + 1.0 + 1.(-2) + 1.0 + 1.0 + 1.0 + 1.0 + 1.0} = 0
Orthogonality of χ(t)(g) and χ(t)(g):
I r r2 r3 d dr dr2 dr3
(1/8){2.2 + 0.0 + (-2).(-2) + 0.0 + 0.0 + 0.0 + 0.0 + 0.0} = 1
Test of Reducibility
--------------------
If nr = the number of times the irreducible
representation Dr(c) appears in the block diagonal
matrix representation we can write:
χ(g) = Σnrχr(g)
r
For example,
χ(r') = 1.(1).+ 1.(-1) + 1.(0) = 0
χ(dr') = 1.(1).+ 1.(1) + 1.(0) = 2
etc.
Therefore, we can write:
<χ(g)|χ(g)> = (1/|G|)ΣΣnrχ(r)†(g)nsχ(s)(g)
rs
Now,
ΣΣnrχ(r)†(g)nsχ(s)(g) = |G|ΣΣnrnsδrs
rs rs
Therefore,
<χ(g)|χ(g)> = |G|ΣΣnrnsδrs
rs
For r = s this becomes:
<χ(g)|χ(g)> = |G|Σ(nr)2
r
If Σ(nr)2 = 1 then we know that only one of the
r
nr's is equal to 1 with all others equal to 0,
the given representation D(g) contains the irreducible
representation r once and only once. In other words,
D(g) is irreducible; in fact, it is just D(r)(g).
In contrast, if Σ(nr)2 > 1 then we know that D(g) is
r
reducible. For example, if Σ(nr)2 = 3 then the only
r
possibility is for 3 different irreducible
representations to each occur once.
We can also write the above equation as:
Σncχ†(c)χ(c) = |G|Σ(nr)2
c r
Where nc is the number of elements in each conjugacy
class.
For D4 we get:
(1/|G|)Σχ†(g)χ(g) = Σ(nr)2
g r
LHS = (1/8)[{(Tr(I))2} + {Tr(r2))2} + {(Tr(r))2 + (Tr(r3))2|
+ {(Tr(d))2| + (Tr(dr2))2} + {(Tr(dr))2 + (Tr(dr3))2}]
= (1/8)[{42} + {02} + {02 + 02} + {02 + 22} + {02 + 22}]
= 24/8
= 3
= 12 + 12 + 12
= 1 dimensional + 1 dimensional + 2 dimensional
Since the dimension = 4 the original representation
must break apart into 2 one dimensional and 1 two
dimensional irreducible representations, i.e.,
= 1 dimensional + 1 dimensional + 2 dimensional
Schur's Lemma
-------------
Schur's Lemma dictates how the blocks should be
arranged for the representation to be irreducible.
The lemma states that for D'(g) to be an irreducible
representation of a finite group G there needs to be
some matrix A such that:
AD(g) = D(g)A for all g
Where A = λI for some constant λ.
Consider a group of matrices. All of these matrices
will naturally commute with the identity matrix.
However, it may also be possible to find another
matrix, A, that commutes with all of the matrices
in the group. If this is the case then the
representation is reducible. If the only matrix
that works is the identity then the representation
is irreducible.
Now we can find a unitary matrix, U, which transforms
A into a diagonal matrix, AD i.e.
AD = U-1AU and DD(g) = U-1D(g)U
If we apply this to both sides of ADD(g) = DD(g)A we
get:
U-1ADUU-1DD(g)U = U-1DD(g)UU-1ADU
U-1ADDD(g)U = U-1DD(g)ADU
Multiplying by U from the left and U-1 from the
right gives:
ADDD(g) = DD(g)AD
Now if DD(g) is irreducible it is of block diagonal
form. So we have a diagonal matrix AD multiplying
a block diagonal matrix. The only way the equality
holds is if AD is some multiple of the identity
matrix.
Finding Other Irreducible Representations
----------------------------------------
Postulate:
# of inequivalent irreps = # of conjugacy classes.
Proof:
Any finite group G has an h dimensional representation,
known as the REGULAR representation.
To obtain the regular representation we start
with the Cayley table and rearrange it so that
the group elements are listed in the first column
and the corresponding inverses are listed in the
first row.
The matrix of the regular representation of the
element is then obtained from this table by replacing
the corresponding every appearance of g with 1, and
filling up the rest of the corresponding matrix
with zeros. That is, for example:
What does this mean?
By left multiplication, each element of G acts as a
permutation of the elements of G.
We can write this as:
φ(g): x -> gx for all x ∈ g
For r we have:
r.{I,r3,r2,r,d,dr,dr2,dr3} which is:
- - - - - -
| 0 0 0 1 0 0 0 0 || I | | r |
| 1 0 0 0 0 0 0 0 || r3 | = | I |
| 0 1 0 0 0 0 0 0 || r2 | | r3 |
| 0 0 1 0 0 0 0 0 || r | | r2 |
| 0 0 0 0 0 0 0 1 || d | | dr3 |
| 0 0 0 0 1 0 0 0 || dr | | d |
| 0 0 0 0 0 1 0 0 || dr2 | | dr |
| 0 0 0 0 0 0 1 0 || dr3 | | dr2 |
- - - - - -
Therefore, φ(g) is a homomorphism (a structure
preserving map between two algebraic structures of
the same type).
The characters for the regular representation are easy
to compute since none of the matrices except I have any
diagonal entries. The character χ(I) is equal to
the dimension of the representation, which in our case
is 8 for the regular representation. Therefore,
χ(reg)(g) = { 0 g ≠ e
{ |G| g = e
We can now apply the reducibility test from before:
Σχ†(g)χ(g) = |G|Σ(nr)2
g r
or,
Σncχ†(c)χ(c) = |G|Σ(nr)2
c r
Thus,
Σncχ†(c)χ(c) = (χ(I))2 = |G|Σ(nr)2
c r
From which we get:
Σ(nr)2 = 8
r
Therefore, the regular representation is reducible.
8 can be factores as follows:
8 = 12 + 12 + 12 + 12 + 22 + 22
Multiplicity
------------
The number of times an irreducible representation
D(s)(g) appears in the decomposition of D(reg) is
called the multiplicity. It is given by
<χ(reg)(g)|χ(s)(g)>.
<χ(reg)(g)|χ(s)(g)> = (1/|G|)Σχ(reg)†(g)χ(s)(g)
g
= (1/|G|)χ(reg)†(I)χ(s)(I)
= (|G|/|G|)(I)χ(s)(I)
= χ(s)(I)
= Tr(I(s))
= dim(D(s)(g))
Therefore, the number of times the same irreducible
representation can appear in the block diagonal
form of the regular representationis is equal to
its dimension.
We know that the regular representation can be
decomposed into a direct sum of all the distinct
irreducible representations of the group, D(si)(g),
and their multiplicity is equal to the number of
times they appear in the direct sum. Therefore, we
can write the character of the regular representation
as a sum of the irreducible characters, weighted by
their multiplicity:
χ(reg)(g) = Σdim(D(si)(g))χ(si)(g)
i
We have found from before that the number of
irreducible representations in the regular
representation can be factored as:
8 = 12 + 12 + 12 + 12 + 22 + 22
Using the above equation we can now write the
the above in terms of their multiplicity as:
8 = 1(12) + 1(12) + 1(12) + 1(12) + 2(22)
Therefore, there are 5 inequivalent irreducible
representations of the regular representation.
By induction, we see that these 5 inequivalent
irreducible representations correspond to the
5 conjugacy classes of D4. This enables us to
deduce:
The number of irreducible representations of
a group is equal to the number of conjugacy
classes of the group.
The sum of the squares of the dimensions of all
distinct irreducible representations is equal
to the order of the group:
Σ(dr)2 = |G|
r
Proof:
<χ(reg)(I)|χ(reg)(I)> = (χ(reg)(I))2
= |G|2
But,
<χ(reg)(I)|χ(reg)(I)> also = |G|Σ(nr)2
Therefore, by comparison,
Σ(nr)2 = |G|
r
Now, neglecting the multiplicity:
n1 n2 n3 n4 n5 n6
Σ(nr)2 = 12 + 12 + 12 + 12 + 22 + 22 = 8
r
Now write the dimensions of each matrix in
the same way:
d1 d2 d3 d4 d5 d6
Σ(dr)2 = 12 + 12 + 12 + 12 + 22 + 22 = 8
r
We conclude that dr = nr
Character Tables
----------------
If we consider the irreducible representations as
vectors whose components are characters of the
representation then 2 different IRREDUCIBLE
representations are orthogonal. This dictates
the construction of character tables. We know
that the number of conjugacy classes is equal to
the number of irreducible representations.
Therefore, character tables are always square.
The Character table of D4 turns out to be:
{I} {r2} {r,r3} {d,dr2} {dr,dr3}
--------------------------------
D1(g) | 1 1 1 1 1 }
D2(g) | 1 1 1 -1 -1 } In arbitrary
D3(g) | 1 1 -1 1 -1 } basis (A,B,C,D)
D4(g) | 1 1 -1 -1 1 }
D5(g) | 2 -2 0 0 0 }
^ ^ ^ ^ ^
/ | \ | |
/ | \ |____ |______
/ | \ | |
- - - - - - - - - -
| 1 0 | | -1 0 | | -i 0 | | 0 -i | | 0 1 |
| 0 1 | | 0 -1 | | 0 i | | i 0 | | 1 0 |
- - - - - - - - - -
- - - - - -
| i 0 | | 0 i | | 0 -1 |
| 0 -i | | -i 0 | | -1 0 |
- - - - - -
To see the row and column orthogonalities we
need to expand the table as follows:
I r2 r r3 d dr2 dr dr3
-------------------------------
D1(g) | 1 1 1 1 1 1 1 1
D2(g) | 1 1 1 1 -1 -1 -1 -1
D3(g) | 1 1 -1 -1 1 1 -1 -1
D4(g) | 1 1 -1 -1 -1 -1 1 1
D5(g) | 2 -2 0 0 0 0 0 0
Using the Great Orthogonality Theorem,
<χ(r)(g)|χ(s)(g)> = (1/|G|)Σχ(r)†(g)χ(s)(g) = δrs
g
we get:
1st and 2nd rows: (1/8)(1.1 + 1.1 + 1.1 + 1.1 + 1.(-1))
+ 1.(-1) + 1.(-1) + 1.(-1))
= 0
2nd and 3rd rows: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1)
+ (-1).1 + (-1).1 + (-1)(-1)
+ (-1).(-1))
= 0
1st and 3rd rows: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1) + 1.1
+ 1.1 + 1.(-1) + 1.(-1))
= 0
and so on.
1st and 2nd cols: (1/8)(1.1 + 1.1 + 1.1 + 1.1 + 2.(-2)) = 0
2nd and 3rd cols: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1) + (-2).0)
= 0
1st and 3rd cols: (1/8)(1.1 + 1.1 + 1.(-1) + 1.(-1) + 2.0)
= 0
and so on.
Note if Dr(g) = Ds(g) we get:
Σncχ(r)†χ(s) = (1/8)(1.(2)2 + 1.(-2)2 + 2.(0)2 + 2.(0)2
c
+ 2.(0)2)
= 1
We can test the reducibility of this representation
as follows:
(1/|G|)Σncχ†(c)χ(c) = Σ(nr)2
c r
LHS = (1/8)[(2)2 + (-2)2]
= 1
RHS = 12
Since the dimension = 2 the 2 x 2 representation
cannot be broken down further and is therefore
irreducible.
As an alternative proof of this we can use Schur's
Lemma to show that AD(g) = D(g)A if and only if
A = λI. For example consider:
- - - - - - - -
| a b || 0 -i | - | 0 -i || a b |
| c d || i 0 | | i 0 || c d |
- - - - - - - -
- -
= | (bi + ci) (-ai + di) |
| (di - ai) (-ci - bi) |
- -
For the RHS to equal 0 we must have: a = d and b = -c.
Therefore, we can write:
I:
- - - - - - - -
| a b || 1 0 | - | 1 0 || a b | = 0 for
| -b a || 0 1 | | 0 1 || -b a | any b
- - - - - - - -
r:
- - - - - - - -
| a b || -i 0 | - | -i 0 || a b | = 0 for
| -b a || 0 i | | 0 i || -b a | b = 0 only
- - - - - - - -
r2:
- - - - - - - -
| a b || -1 0 | - | -1 0 || a b | = 0 for
| -b a || 0 -1 | | 0 -1 || -b a | any b
- - - - - - - -
r3:
- - - - - - - -
| a b || i 0 | - | i 0 || a b | = 0 for
| -b a || 0 -i | | 0 -i || -b a | b = 0 only
- - - - - - - -
d:
- - - - - - - -
| a b || 0 -i | - | 0 -i || a b | = 0 for
| -b a || i 0 | | i 0 || -b a | any b
- - - - - - - -
dr:
- - - - - - - -
| a b || 0 1 | - | 0 1 || a b | = 0 for
| -b a || 1 0 | | 1 0 || -b a | b = 0 only
- - - - - - - -
dr2:
- - - - - - - -
| a b || 0 i | - | 0 i || a b | = 0 for
| -b a || -i 0 | | -i 0 || -b a | any b
- - - - - - - -
dr3:
- - - - - - - -
| a b || 0 -1 | - | 0 -1 || a b | = 0 for
| -b a || -1 0 | | -1 0 || -b a | b = 0 only
- - - - - - - -
Since this has to work for ALL of the matrices
we conclude that A is λI and Schur's Lemma holds.
Ultimately, the elements in the 2 x 2 matrices
of the irreducible representation will depend
on the basis chosen. There are many possible
bases that we could use. However, if we choose
the normal Cartesian basis and pick a (-1,1),
the matrices take the form:
{I} {R2} {R,R3} {dL,dR} {rx,ry}
-------------------------------
D5(g) | 2 -2 0 0 0
^ ^ ^ ^ ^
/ | \ | |
/ | \ |____ |______
/ | \ | |
- - - - - - - - - -
| 1 0 | | -1 0 | | 0 1 | | 0 -1 | | 1 0 |
| 0 1 | | 0 -1 | | -1 0 | | -1 0 | | 0 -1 |
- - - - - - - - - -
- - - - - -
| 0 -1 | | 0 1 | | -1 0 |
| 1 0 | | 1 0 | | 0 1 |
- - - - - -
As in the arbitrary basis case, these elements can
be written as products of the following matrices.
- -
R = | 0 1 | ≡ r
| -1 0 |
- -
and,
- -
D = | 0 -1 | ≡ dL
| -1 0 |
- -
For example,
- - - - - -
DR = | 0 -1 || 0 1 | = | 1 0 | ≡ rx
| -1 0 || -1 0 | | 0 -1 |
- - - - - -
DR2 = dR
DR3 = ry
We can now think of the transformation matrices as
acting on the vector, (-1,1), in the Cartesian basis
(x,y) as follows:
(-1,1) (1,1)
2 1
-------
| |
I = | o |
| (0,0) |
-------
3 4
(-1,-1) (1,-1)
Rotation by 90°:
1 4
-------
| |
R = | o |
| |
-------
2 3
- - - - - -
| 0 1 || -1 | = | 1 |
| -1 0 || 1 | | 1 |
- - - - - -
Cycle notation: 1234 => (1234)
2341
Rotation by 180°:
4 3
-------
| |
R2 = | o |
| |
-------
1 2
- - - - - -
| -1 0 || -1 | = | 1 |
| 0 -1 || 1 | | -1 |
- - - - - -
Cycle notation: 1234 => (13)(24)
3412
Rotation by 270°:
3 2
-------
| |
R3 = | o |
| |
-------
4 1
- - - - - -
| 0 -1 || -1 | = | -1 |
| 1 0 || 1 | | -1 |
- - - - - -
Cycle notation: 1234 => (1234)
2341
Reflection about x axis:
3 4
-------
| |
rx = | o |
| |
-------
2 1
- - - - - -
| 1 0 || -1 | = | -1 |
| 0 -1 || 1 | | -1 |
- - - - - -
Cycle notation: 1234 => (14)(23)
4321
Reflection about y axis:
2 3
-------
| |
ry = | o |
| |
-------
1 4
- - - - - -
| -1 0 || -1 | = | 1 |
| 0 1 || 1 | | 1 |
- - - - - -
Cycle notation: 1234 => (13)(2)(4)
3214
Reflection about right diagonal:
4 1
-------
| |
dR = | o |
| |
-------
3 2
- - - - - -
| 0 1 || -1 | = | 1 |
| 1 0 || 1 | | -1 |
- - - - - -
Cycle notation: 1234 => (1)(24)(3)
1432
Reflection about left diagonal:
3 2
-------
| |
dL = | o |
| |
-------
4 1
- - - - - -
| 0 -1 || 1 | = | -1 |
| -1 0 || 1 | | -1 |
- - - - - -
Cycle notation: 1234 = (13)(2)(4)
3214
The matrices in each class are similar. They have
the same eigenvalues, determinant, trace and period.
{I} {R2} {R,R3} {dL,dR} {rx,ry}
-------------------------------------
Evalues | 1 -1 i,-i -1,1 -1,1
Determ. | 1 -1 1 -1 1
Trace | 2 -2 0 0 0
Period* | 1 2 4 2 2
* often called the order of an element = m = gm = e
As before, by applying Schur's Lemma it is easy
to show that this representation is irreducible.
Change of Basis
---------------
Consider from before the reflection about right
diagonal:
4 1
-------
| |
dR = | o |
| |
-------
3 2
- - - - - -
| 0 1 || -1 | = | 1 |
| 1 0 || 1 | | -1 |
- - - - - -
Consider now if we rotate the coordinate system
clockwise by 45° so that x -> x' and y -> y'.
In the x', y' basis the point (-1,1) becomes (-1,0)
and its reflection across the right diagonal becomes
(1,0). We now look for a 2 x 2 matrix that accomplishes
this. We get:
- - - - - -
| -1 0 || -1 | = | 1 |
| 0 1 || 0 | | 0 |
- - - - - -
Therefore, a reflection across the right diagonal
in the x,y basis becomes a reflection across y'
in the x',y' basis.
- - - -
dR 1 : | 1 | -1 : | -1 |
| 1 | | 1 |
- - - -
P(λ) = det(dR - λI)
= | 0 1 - λ | = -λ2 + 2λ - 1
| 1 - λ 0 |
- - - -
rY' 1 : | 0 | -1 : | 1 |
| 1 | | 0 |
- - - -
P(λ) = det(dR - λI)
= | -(1 - λ) 0 | = -λ2 + 2λ - 1
| 0 1 - λ |
The matrices have the same rank, the same trace,
the same determinant, the same eigenvalues, and
the same characteristic polynomial. This means
that the matrices are similar and, therefore,
the representations are indeed equivalent.