I have struggled for weeks thinking how to introduce the topic of group representation theory, and my best idea is to look at it as a game with rules. We want to define the character table of a group, which is a grid of the traces of matrices of irreducible representations.
For any finite group G, we can define a representation as a homomorphism h, h:G→K, where K is a group of invertible nxn matrices.
Stated without proof: the number of irreducible representations is equal to the number of conjugacy classes.
Stated without proof: The order of the group is equal to the sum of the squares of the dimensions of the representations.
Corollary: A finite abelian group of order n has n one-dimensional representations, and all of the entries are on the complex unit circle.
One more definition is needed: The complex conjugate of a+bi = a-bi.
Corollary #1: The complex conjugate of a real number a is a.
Corollary #2: On the complex unit circle, the multiplicative inverse of any number is its complex conjugate, (a+bi)(a-bi) = 1.
Let's start small, and ℤ2 is the smallest interesting case. Let's call the elements 1 and -1.
| 1 | -1
1 | |
-1 | |
I will fill in the character table in red.
Rule #1: The homomorphism of sending everything to 1 is always an irreducible representation.
| 1 | -1
1 | 1 | 1
-1 | |
Rule #2: The identity is always sent to the identity, and the trace of the identity is equal to the dimension of the representation.
| 1 | -1
1 | 1 | 1
-1 | 1 |
Rule
#3: Every row of the character table can be looked upon as a vector and the dot product of one row with the complex conjugate of a different row will always be zero. The dot product of a row with itself will be the order of the group.
A dot product of zero indicates that the two vectors are orthogonal, which is the same as perpendicular, the two vectors meet at a 90° angle. If we are dealing in two or three dimensions, it's easy to show 90° angles, but as we get to vectors in larger dimensions, visualization becomes difficult and the dot product is our best way to determine orthogonality.
| 1 | -1
1 | 1 | 1
-1 | 1 | -1
Next, we look at ℤ2✕ℤ2. I will call the elements 1, a, b and ab, where the square of every element is equal to 1.
| 1 | a | b | ab
1 | | | |
a | | | |
b | | | |
ab | | | |
First row and first columns are all 1s.
| 1 | a | b | ab
1 | 1 | 1 | 1 | 1
a | 1 | | |
b | 1 | | |
ab | 1 | | |
To make orthogonal vectors, we map two of the letter variables to -1, and their product to 1.
| 1 | a | b | ab
1 | 1 | 1 | 1 | 1
a | 1 |-1 |-1 | 1
b | 1 |-1 | 1 |-1
ab | 1 |-1 |-1 | 1
One more small group to look at, ℤ4. I will call the elements 1, i, -1 and -i.
| 1 | i | -1 | -i
1 | | | |
i | | | |
-1 | | | |
-i | | | |
Yet again, all 1s in the first row and column.
| 1 | i | -1 | -i
1 | 1 | 1 | 1 | 1
i | 1 | | |
-1 | 1 | | |
-i | 1 | | |
Second row, send every element to itself.
| 1 | i | -1 | -i
1 | 1 | 1 | 1 | 1
i | 1 | i | -1 | -i
-1 | 1 | | |
-i | 1 | | |
Third row, send every element to its square.
| 1 | i | -1 | -i
1 | 1 | 1 | 1 | 1
i | 1 | i | -1 | -i
-1 | 1 |-1 | 1 | -1
-i | 1 | | |
Fourth row, send every element to its cube.
| 1 | i | -1 | -i
1 | 1 | 1 | 1 | 1
i | 1 | i | -1 | -i
-1 | 1 |-1 | 1 | -1
-i | 1 |-i | -1 | i
This is the first case where we have to consider complex conjugates. The dot product of rows 2 and 4 becomes
1x1 + (i)(i) + (-1)(-1) + (-i)(-i) = 1 + -1 + 1 + -1 = 0.
In the next posts, we will look at all the groups of order 8, three of them abelian - ℤ8, ℤ4✕ℤ2 and ℤ2✕ℤ2✕ℤ2 - and two non-abelian, D4 and the quaternions.
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