How Do You Make a Character Table?

Date: 10/15/2008 at 01:27:16
From: Amy
Subject: How do you make a character table (representation theory)

How do you make a character table?  The course is Representation 
Theory and we're studying representations of finite groups, 
specifically character theory.  The example in the text book is of 
the Symmetric group with 3 elements S_3.  It is 

             | 1    3    2
S_3          | 1  (12)  (123)
________________________________
trivial      | 1    1     1
alternating  | 1   -1     1
standard     | 2    0    -1

I don't get where the inside numbers come from.

I get that the inside numbers are the character of the 
representative.  My understanding of this is that I take the trace 
(sum of the diagonal elements) of the matrix that the element gets 
sent to in the homomorphism that defines the representation.  I'm 
not seeing how this idea gives me the numbers in this table.  The 
book seems to act as if I should know these without any reference to 
matrices.  

My understanding of this is that I take the trace (sum of the 
diagonal elements) of the matrix that the element gets sent to in 
the homomorphism that defines the representation.  I'm not seeing 
how this idea gives me the numbers in this table.  The book seems to 
act as if I should know these without any reference to matrices.

Date: 10/16/2008 at 05:11:40
From: Doctor Jacques
Subject: Re: How do you make a character table (representation theory)

Hi Amy,

Usually, you don't need to use the actual matrices of the 
representations to compute the characters.  You can fill most of the 
character table simply by using general properties of characters 
(this is the subject matter of character theory). For example:

* The number of irreducible characters is equal to the number of
  conjugacy classes (it is the dimension of the center of the group
  algebra).

* There are orthogonality relations between characters;
  specifically, between the rows and the columns of the table.  (As
  the character values are in general complex, these are usually not
  sufficient to determine the table completely).

* The characters of an Abelian group have degree 1; their values are
  roots of unity, and the character group is isomorphic to the
  original group.  This means that you can write them down easily.

* If the group G contains a normal subgroup H, any character of the
  quotient group G/H gives a character of G by composition with the
  canonical homomorphism G -> G/H.  If the quotient happens to be
  Abelian, you can use the previous remark.  In particular, the
  number of characters of degree 1 is the index of the derived group.

* The degrees of the irreducible characters divide the order of the
  group, and the sum of the squares of these degrees is equal to the
  order of the group.

* Characters come in conjugate pairs, because it is not possible to
  distinguish i from -i.  In particular, if there is only one
  character of a given degree, its values must be real.

* If c_1 is a character of degree 1, and c is any character, there
  is also a character c_1*c (corresponding to the tensor product of
  the representations).

* In any representation, the character of an element is the complex
  conjugate of the character of the inverse; in particular, if the
  element and its inverse belong to the same conjugacy class, the
  character is real.

In most "simple" cases, you can fill the character table completely 
by using all those properties.

Let us look, for example, at S_3.  There are 3 conjugacy classes, and 
therefore there are three irreducible characters.  The degrees of 
these characters are divisors of 6, and the sum of their squares 
equals 6.  It is easy to see in this case that the only possibility 
is (1,1,2).  As the trivial character is always present, we can 
already fill part of the character table:

         1      3       2
         ()   (1,2)  (1,2,3)
  ----------------------------
  X1     1      1       1
  X2     1      a       b
  X3     2      c       d

S_3 contains a normal subgroup of order 3, and the quotient subgroup 
is isomorphic to C_2 (an Abelian group).  This means that we must 
have 2 characters of degree 1, containing 1 (the degree) in the 
column of the kernel (the third column).  The first one is the 
trivial character, and the second one must be X2.  This means that we 
have b = 1, and, as (1,2) has order 2 in the quotient group, the 
value of X2 at (1,2) is a primitive square root of 1, i.e., a = -1.  
The table looks now like this:

         1      3       2
         ()   (1,2)  (1,2,3)
  ----------------------------
  X1     1      1       1
  X2     1     -1       1
  X3     2      c       d

Now, orthogonality of the columns gives:

  1*1 + 1*(-1) + 2*c = 0
  1*1 + 1*1    + 2*d = 0

from which we obtain c = 0 and d = -1, and this completes the table.

For a slightly more elaborated example, let us now consider A_4. 
There are 4 conjugacy classes, represented by (), (1,2)(3,4), 
(1,2,3), and (1,3,2).  This means that we should have 4 irreducible 
characters.

A_4 contains a normal subgroup N isomorphic to C2 x C2; this subgroup 
consists of the classes represented by () and (1,2)(3,4).  The 
quotient group is C_3; we should therefore find three characters of 
degree 1, which take the value 1 in the classes corresponding to N:

          I       A       B       C
          1       3       4       4
          () (1,2)(3,4) (1,2,3) (1,3,2)
  -------------------------------------
  X1      1       1       1       1
  X2      1       1       ?       ?
  X3      1       1       ?       ?
  X4      ?       ?       ?       ?

(The first row is used to label the classes, this makes it easier to 
refer to entries in the table).

Now, class B has order 3 in the quotient group, therefore X2(B) must 
be a primitive cube root of unity. Let us denote such a cube root by 
w = (1+sqrt(-3))/2. The other cube root is w^2, and w and w^2 are 
conjugate.

As the classes B and C are inverses of each other, X2(B) and X2(C) 
are complex conjugates. We can take X2(B) = w and X2(C) = w^2.  As 
characters come in conjugate pairs, we have X3(B) = w^2 and
X3(C) = w.  The table now looks like this:

          I       A       B       C
          1       3       4       4
          () (1,2)(3,4) (1,2,3) (1,3,2)
  -------------------------------------
  X1      1       1       1       1
  X2      1       1       w      w^2
  X3      1       1      w^2      w
  X4      ?       ?       ?       ?

As the sum of the squares of the characters must equal 12, we find 
that X4 has degree 3 = X4(I).  Now, multiplying X4 by X2 (of degree 
1) must give a character of degree 3, which must be X4, since it is 
the only character of degree 3.  In particular, we must have:

  X4(B) = X4(B)*X2(B) = X4(B)*w

and this implies that X4(B) = 0; by the same argument, X4(C) = 0.

Orthogonality of the first two columns gives:

  1*1 + 1*1 + 1*1 + 3*X4(A) = 0

which gives X4(A) = -1. The whole table is:

          I       A       B       C
          1       3       4       4
          () (1,2)(3,4) (1,2,3) (1,3,2)
  -------------------------------------
  X1      1       1       1       1
  X2      1       1       w      w^2
  X3      1       1      w^2      w
  X4      3      -1       0       0

(we could also have found X4(B) and X4(C) by orthogonality).

Note that the representation X4 has a natural geometric 
interpretation, because A4 is the group of rotations of a regular 
tetrahedron.  If we take the center of the tetrahedron as origin, and 
draw unit vectors to three vertices, the fourth vertex has 
coordinates (-1,-1,-1) (because the sum of the vectors from the 
center must be 0).  Each element of A4 induces a permutation of the 
vertices, which is a linear transformation of R^3.  You can represent 
those transformations by 3*3 matrices, and check that the traces of 
those matrices correspond to the values in the table above.

In summary, the use of the above techniques should allow you to 
construct character tables by hand, at least for simple cases (the 
ones you should expect to find as exercises in a textbook...).

Note that there are general algorithms for computing character 
tables, although these algorithms are more suited to a computer 
implementation.  These algorithms do not need the explicit matrices 
of the representations (of course, they need some information about 
the group, like the conjugacy classes).

I hope this helps.  Please feel free to write back if you require 
further assistance.


- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/