# Character Table Soduku

1. Mar 5, 2014

### ferret123

1. The problem statement, all variables and given/known data

Complete the character table giving a brief explanation of each step.
Hint: You should use the fact that some representations are one dimensional.

2. The attempt at a solution
Reading through Mathematical Methods for Physics and Engineering by Riley, Hobson & Bence section 25.8 is on the construction of a character table. It gives a set of properties of a group that can be used to determine the character table.

One of these properties is that the identity irreducible representation is always present in a character table and all it's entries are 1. This implies that the 4th irreducible representation must be this identity by elimination.

Another is that the ‘vector’ formed by the characters from a given irreducible representation is orthogonal to the ‘vector’ formed by the characters from a different irreducible representation. Using this and performing a dot product between the newly determined 4th row and the 5th row, which is 0 by orthogonality, I determine that the missing entry must be 1.

Is this a correct method that I can use to determine the rest of the unknowns?

Thanks

2. Mar 7, 2014

### micromass

Staff Emeritus
Is there nothing you know about the group? Like the orders of the group and conjugacy classes?

3. Mar 8, 2014

### johnqwertyful

doesn't matter

The way you got row 5 is how you get the rest. Just dot them.

4. Mar 8, 2014

### micromass

Staff Emeritus
It's been years since I did character tables, but I doubt this approach is correct. The correct formula of orthogonality is

$$\sum_{x\in G}\chi_i(x)\overline{\chi_j(x)} = |G| \delta_{ij}$$

But many $x$ will lie in the same conjugacy class. So if you let $\{C_1,...,C_n\}$ be the conjugacy class such that $|C_j| = c_j$, then

$$\sum_{k = 1}^n c_k \chi_i(C_k)\overline{\chi_j(C_k)} = |G| \delta_{ij}$$

So it's not exactly like the different rows are orthogonal. Rather, the rows are orthogonal if we take into account the cardinalities of the conjugacy classes.

So I don't see how the cardinalities of the conjugacy classes is not important here. Also, if you could just dot in everything, the hint about one-dimensional stuff wouldn't make much sense.

Please correct me if I'm wrong.

5. Mar 8, 2014

### CAF123

Hi micromass,
There exists an inner product on the space of characters yes, but I don't think you necessarily need to use that here (and cannot, since the cardinalities are not given.)
I believe the one-dim stuff is meant to hint that the trivial representation is always contained in the set of irreducible representations of $G$. So, from the table, this has to be $\rho_4$. Then there also exists an orthonormal inner product on the class space, so by dotting the columns should give the remaining entries.

6. Mar 8, 2014

### micromass

Staff Emeritus
I'm not talking about the space of characters.

It doesn't work this way. For example, the character table of $S_4$ is

$$\left(\begin{array}{ccccc} 1 & 1 & 1 & 1 & 1\\ 1 & -1 & 1 & -1 & 1\\ 2 & 0 & -1 & 0 & 2\\ 3 & 1 & 0 & -1 & -1\\ 3 & -1 & 0 & 1 & -1 \end{array}\right)$$

For example, if you take the dot product of the last two rows then you get:

$$3\cdot 3 + 1\cdot (-1) + 0\cdot 0 + (-1)\cdot 1 + (-1)\cdot (-1) = 9 -1 - 1 + 1 = 8$$

which is nonzero.

However, if you take into account the cardinalities of the conjugacy classes, which are $1$, $6$, $8$, $6$, $3$, then you get

$$1\cdot 3\cdot 3 + 6\cdot 1\cdot (-1) + 8\cdot 0\cdot 0 + 6\cdot (-1)\cdot 1 + 3\cdot (-1)\cdot (-1) = 9 -6 - 6 + 3 = 0$$

So in order to get orthogonality, you absolutely need to take into account the cardinalities of the conjugacy classes.

7. Mar 8, 2014

### CAF123

When I say there is an inner product on the space of characters, I mean exactly what you write in that to ensure the rows are orthogonal, one needs to write out all the characters of each element in each conjugacy class.

However, there is also an inner product on the class space. Which means by the way the character table is constructed, the columns are necessarily orthogonal. Notice that by taking the dot product of any two different columns of the matrix, you obtain zero. This is all that is needed to construct the table.