The character tables for D_4 and the quaternions

 We have looked at the character tables for the abelian groups of order 8, ℤ₈, ℤ₄✕ℤ₂ and ℤ₂✕ℤ₂✕ℤ₂. Because they are abelian, each has 8 conjugacy classes and their character tables are distinct. With the two non-abelian groups, D4 and Q, there are five conjugacy classes. The trace of the identity element is the dimension of the irreducible representation, and the sums of the squares of these number must equal 8. There is only one way to do this, 1² + 1² + 1² + 1² + 2².

 

Here are the conjugacy classes for Q.

Class #1: 1

Class #2: -1

Class #3: i and -i

Class #4: j and -j

Class #5: k and -k

 

And here are the conjugacy classes for D4.

Class #1: R0°

Class #2: R180°

Class #3: R90° and R270°

Class #4: M0° and M90°

Class #5: M45° and M135°

 

The entries of the one dimensional representations must be numbers on the complex unit circle, but the two dimensional numbers are the traces of the 2✕2 matrices. In 2x2 matrices, the 180° rotation and the matrix representing -1 are the same.

 

-1  0

 0 -1 

 

The trace here is -2. 

 

Let's start making the table for Q.

 

1  -1   i  -i   j  -j   k  -k 

1   1   1   1   1   1   1   1 

1

1

1

2  -2

All the conjugates are negatives of one another, so for the one dimensional representations f(-1) = 1.

1  -1   i  -i   j  -j   k  -k 

1   1   1   1   1   1   1   1 

1   1

1   1

1   1

2  -2

To make all the one dimensional representations orthogonal to one another, we send two of the conjugacy classes to -1 and the last class to 1.

1  -1   i  -i   j  -j   k  -k 

1   1   1   1   1   1   1   1 

1   1   1   1  -1  -1  -1  -1

1   1  -1  -1   1   1  -1  -1

1   1  -1  -1  -1  -1   1   1

2  -2

Let us call the traces of the two dimensional conjugacy classes a, b and c.

1  -1   i  -i   j  -j   k  -k 

1   1   1   1   1   1   1   1 

1   1   1   1  -1  -1  -1  -1

1   1  -1  -1   1   1  -1  -1

1   1  -1  -1  -1  -1   1   1

2  -2   a   a   b   b   c   c

To make the two dimensional representation orthogonal to the one dimensional representations, we have these four equations.

 2a + 2b + 2c = 0

 2a - 2b - 2c = 0

-2a + 2b - 2c = 0

-2a - 2b + 2c = 0

Adding the first two equations together, we get 4a = 0.

Adding the first and third, we get 4b = 0.

Adding the first and fourth, we get 4c = 0.

1  -1   i  -i   j  -j   k  -k 

1   1   1   1   1   1   1   1 

1   1   1   1  -1  -1  -1  -1

1   1  -1  -1   1   1  -1  -1

1   1  -1  -1  -1  -1   1   1

2  -2   0   0   0   0   0   0

 

The logic for filling in the D4 table is the same, because yet again, each element that has a conjugate is the negative of that conjugate, where -1 is represented by R180°, which commutes with everything.

 

Next, we will look at other small non-abelian groups to determine their character tables. 

The Character tables of the groups of order 8: Z_8

 ℤ8 is generated by a single element, the square root of i, which I will denote by the letter a. Wherever a is sent, all other entries will be to the appropriate power of a. The eight elements will be called 1, a, i, ai, -1, -a, -i and -ai.

 *    1    a    i   ai   -1   -a   -i   -ai

1st | 1    1    1    1    1    1    1    1 

2nd | 1    a    i   ai   -1   -a   -i   -ai

3rd | 1    i   -1   -i    1    i   -1   -i 

4th | 1   ai   -i    a   -1   -ai   i   -a 

5th | 1   -1    1   -1    1    -1    1  -1 

6th | 1   -a    i   -ai  -1     a   -i  ai

7th | 1   -i   -1   i     1    -i   -1   i

8th | 1   -ai  -i  -a    -1    ai    i   a

Checking for orthogonality is a little tougher because you must turn one row into its complex conjugates before doing the dot product, but it does still work. This orthogonality makes group representation theory a useful part of physically describing an object whose group actions are represented by some set of matrices.

 

Tomorrow: The character tables for the quaterions and D4. 

The character tables for the groups of order 8: Z_4 x Z_2

 The group ℤ4 ✕ ℤ2 has is generated by two elements we will call a and b, where a⁴ = 1 and b² = 1. On the complex unit circle, we can send a to powers of i (i, -1, -i and 1) and b to 1 or -1. all the rest of the elements will be defined by the rule of homomorphisms, specifically f(xy) = f(x)f(y). 

Instead of proceeding step by step, I present the entire character table here and color the entries for column a and column b.

    |  1  |  a  |  a² |  a³ |  b  | ba  | ba² | ba³ |

1st |  1  |  1  |  1  |  1  |  1  |  1  |  1  |  1  |

2nd |  1  |  i  | -1  | -i  |  1  |  i  | -1  | -i  |

3rd |  1  | -1  |  1  | -1  |  1  | -1  |  1  | -1  |

4th |  1  | -i  | -1  |  i  |  1  | -i  | -1  |  i  |

5th |  1  |  1  |  1  |  1  | -1  | -1  | -1  | -1  |

6th |  1  |  i  | -1  | -i  | -1  | -i  |  1  |  i  |

7th |  1  | -1  |  1  | -1  | -1  |  1  | -1  |  1  |

8th |  1  | -i  | -1  |  i  | -1  |  i  | -1  | -i  |

The character tables for the groups of order 8: Z_2 x Z_2 X Z_2

 We have been introduced to all the groups of order 8. Three of them are abelian - ℤ8, ℤ4✕ℤ2 and ℤ2✕ℤ2✕ℤ2 - and two are non-abelian, the dihedral 4 group, a.k.a. symmetries of the square, and the quaternions. In an abelian group, every element is in a singleton conjugacy class, so the character table will be 8x8. Our two non-abelian groups both have five conjugacy classes, and since the sum of the squares of the dimensions of the representations must add to 8, there are four one-dimensional irreducible representations and one irreducible representation that is two-dimensional.

 

Let me begin with ℤ2✕ℤ2✕ℤ2, where the elements will be all the bit strings of length 3 - 000, 001, 010, 011, 100, 101, 110 and 111 - and the group operation will be bitwise exclusive or. 000 is the identity element and every element is its own inverse. As usual, we will fill in the first row with all 1 and likewise the first column.

 

xor  000  001  010  011  100  101  110  111 

1st | 1    1    1    1    1    1    1    1 

2nd | 1 

3rd | 1  

4th | 1 

5th | 1 

6th | 1 

7th | 1 

8th | 1 

Because every element is its own inverse, the only entries will be 1 and -1. Once it is decided how to map 001, 010 and 100, all the other entries will be determined. Let's start with 001 sent to -1 and the other strings with a single 1 sent to 1.

 

xor  000  001  010  011  100  101  110  111 

1st | 1    1    1    1    1    1    1    1 

2nd | 1   -1    1    -1   1   -1    1   -1 

3rd | 1  

4th | 1 

5th | 1 

6th | 1 

7th | 1 

8th | 1 

Next, 010 is sent to -1 and the other strings with a single 1 are sent to 1.

 

xor  000  001  010  011  100  101  110  111 

1st | 1    1    1    1    1    1    1    1 

2nd | 1   -1    1   -1    1   -1    1   -1 

3rd | 1    1   -1   -1    1    1   -1   -1 

4th | 1 

5th | 1 

6th | 1 

7th | 1 

8th | 1 

Next, 100 is sent to -1 and the other strings with a single 1 are sent to 1.

 

xor  000  001  010  011  100  101  110  111 

1st | 1    1    1    1    1    1    1    1 

2nd | 1   -1    1   -1    1   -1    1   -1 

3rd | 1    1   -1   -1    1    1   -1   -1 

4th | 1    1    1    1   -1   -1   -1   -1 

5th | 1 

6th | 1 

7th | 1 

8th | 1 

For representations 5 through 7, two of the single 1 bit values are sent to -1.

 

xor  000  001  010  011  100  101  110  111 

1st | 1    1    1    1    1    1    1    1 

2nd | 1   -1    1   -1    1   -1    1   -1 

3rd | 1    1   -1   -1    1    1   -1   -1 

4th | 1    1    1    1   -1   -1   -1   -1 

5th | 1   -1   -1    1    1   -1   -1    1 

6th | 1   -1    1   -1   -1    1   -1    1 

7th | 1    1   -1   -1   -1   -1    1    1

8th | 1 

 

For representation 8, all of the single 1 bit values are sent to -1.

 

xor  000  001  010  011  100  101  110  111 

1st | 1    1    1    1    1    1    1    1 

2nd | 1   -1    1   -1    1   -1    1   -1 

3rd | 1    1   -1   -1    1    1   -1   -1 

4th | 1    1    1    1   -1   -1   -1   -1 

5th | 1   -1   -1    1    1   -1   -1    1 

6th | 1   -1    1   -1   -1    1   -1    1 

7th | 1    1   -1   -1   -1   -1    1    1 

8th | 1   -1   -1    1   -1    1    1   -1 

It is easy to check orthogonality in each case. If we take any two different rows, they will have four entries where the signs disagree, so the dot product will be 4 - 4 = 0. The dot product of any row with itself is 8.

 

Tomorrow, I will present the character table for ℤ4✕ℤ2.