I have the following 'homework style' question:
Complete the construction of the $C_{4v}$ character table started in the class.
Just to be clear, this question is about all the symmetry operations regarding a square-based pyramid (or if you prefer molecules then please see this instead):
I have essentially 2 questions regarding the solution given by the author, which in part, states that:
The group $C_{4v}$ has 8 elements arranged in 5 conjugate classes: $E\,, \color{red}{2C_4}\,, C_2\,, \color{red}{2\sigma_v}\,, \color{red}{2\sigma_d}$. The number of classes equals the number of irreps, so the group has 5 inequivalent irreps.
First we determine the dimensions of these irreps. We know that $$\sum_i n_i^2=m\tag{1}$$ where $n_i$ is the dimension of the ith irrep. For $C_{4v}$, we have $m=8$, so we are looking for five positive integers whose squares sum to 8, and after some thought we realise that $$8=1 + 1 + 1 + 1 + 4 = 1^2+1^2+1^2+1^2+2^2\tag{2}$$ so the group has 4 one-dimensional irreps and 1 two-dimensional irrep. $\color{red}{\text{This allows us to complete the first column of the character table}}$, corresponding to the characters for the identity element since these are equal to the dimensions of the irrep. For any group, there is also the trivial irrep, where $\color{red}{\chi(C) = 1}$ for all classes. This is usually the first row of the character table. So far we have: $$ \begin{array}{c|c|c|c} C_{4v} & E & 2C_4 & C_2 & 2\sigma_v & 2\sigma_d \\\hline \chi^{(1)} & 1 & 1 & 1 & 1 & 1 \\\hline \chi^{(2)} & 1 \\\hline \chi^{(3)} & 1 \\\hline \chi^{(4)} & 1 \\\hline \chi^{(5)} & \color{red}{2} \end{array} $$
I have marked in red the parts of the authors' solution for which I do not understand. I first need to understand why the classes are $E\,, \color{red}{2C_4}\,, C_2\,, \color{red}{2\sigma_v}\,, \color{red}{2\sigma_d}$, so starting from the Cayley table for $C_{4v}$:
The first problem is that the matrices as I have seen them on page 6 of this are, $$C_4=\begin{bmatrix} \cos 2\pi/4 & -\sin 2\pi/4 & 0 \\ \sin 2\pi/4 & \cos 2\pi/4 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$C_4^3=\begin{bmatrix} \cos 3\pi/2 & -\sin 3\pi/2 & 0 \\ \sin 3\pi/2 & \cos 3\pi/2 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$C_2=\begin{bmatrix} \cos \pi & -\sin \pi & 0 \\ \sin \pi & \cos \pi & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\sigma_x=\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\sigma_y=\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\sigma_d=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
$$\sigma_{d^\prime}=\begin{bmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
and $$E=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\tag{!}$$
But, what I understood is that only the last IRREP, is two-dimensional, but in order for the group symmetry elements to be compatible, they must have the same dimension. Is there a problem here?
Using what I learnt in my previous question, I'm going to compute the red conjugacy classes for $C_{4v}$ and see if they match up with $E\,, \color{red}{2C_4}\,, C_2\,, \color{red}{2\sigma_v}\,, \color{red}{2\sigma_d}$.
But to make things easier I first note that: $$ \begin{array}{c|c|c|c} C_{4v} & E & C_4 & C_2 & C_4^3 & \sigma_x & \sigma_y & \sigma_d & \sigma_{d^\prime} \\\hline \mathrm{Order} & 1 & 4 & 2 & 4 & 2 & 2 & 2 & 2 \\\hline \mathrm{Inverse} & E & C_4^3 & C_2 & C_4 & \sigma_x & \sigma_y & \sigma_d & \sigma_{d^\prime} \\\hline \end{array} $$
Now the class of $C_4$ is: $$C_2C_4{C_2}^{-1}=C_2C_4C_2=C_2C_4^3=C_4$$ $$\sigma_d C_4 \sigma_d^{-1}=\sigma_d C_4 \sigma_d=\sigma_d\sigma_x=C_4^3$$ I won't include the other 6 elements as it is tedious to write them all (which I have done on paper), and they are all either $C_4$ or $C_4^3$. So the class of $C_4$ is $\{C_4, C_4^3\}$.
The class of $C_4^3$ is: $$C_2C_4^3{C_2}^{-1}=C_2C_4^3C_2=C_2C_4=C_4^3$$ $$\sigma_yC_4^3{\sigma_y}^{-1}=\sigma_yC_4^3\sigma_y=\sigma_y\sigma_{d^\prime}=C_4$$ again excluding the other 6 elements the class of $C_4^3$ is $\{C_4^3, C_4\}$.
So the conclusion is that the elements $C_4$ and $C_4^3$ belong to the same class, namely, $\{C_4, C_4^3\}$. I could carry out the same calculation for the other two classes I marked red ($\color{red}{2\sigma_v}\,, \color{red}{2\sigma_d}$), I'd like to avoid typing this out but what I find is that the class of $\sigma_x$ (or ${\sigma_y}$) is $\{\sigma_x, \sigma_y\}$ and the class of $\sigma_d$ (or $\sigma_{d^\prime}$) is $\{\sigma_d, \sigma_{d^\prime}\}$.
Now here is the problem:
Why is the character table not being written like this:
$$ \begin{array}{c|c|c|c} C_{4v} & E & \{C_4, C_4^3\} & C_2 & \{\sigma_x, \sigma_y\} & \{\sigma_d, \sigma_{d^\prime}\} \\\hline \chi^{(1)} & 1 & 1 & 1 & 1 & 1 \\\hline \chi^{(2)} & 1 \\\hline \chi^{(3)} & 1 \\\hline \chi^{(4)} & 1 \\\hline \chi^{(5)} & \color{red}{2} \end{array} \tag{?}$$
Put simply, I don't understand how this $\{C_4, C_4^3\}$ class is the same as writing the class as "$2C_4$"? The same question also applies to "$2\sigma_v$" and "$2\sigma_d$"; is this some kind of special notation? I can only presume that the factor of 2 in front of these classes comes into it each time as there are 2 elements in these classes?
Moving onto the second question I have, which, namely, is how the first row and column of the character table were obtained. I know a character is the sum of its diagonal elements of one of the matrices of a representation $D$, $$\chi(g)=\sum_i D_{ii}(g)=\mathrm{Tr}\Big(\underline{\underline{D}}(g)\Big)\tag{3}$$ for some group element $g$.
I also know that the following orthogonality condition holds for characters: $$\sum_g{\chi^{(j)}}^*(g)\chi^{(i)}(g)=m\delta_{ij}\tag{4}$$ where $m$ is the order of the group. But, what is the link between equations $(1)$ and $(2)$ and characters, $\chi(g)$? Put another way, the author so readily writes
"$\color{red}{\text{This allows us to complete the first column of the character table}}$", but how does this relate to characters $\chi(g)$, as given in the two formulae I stated, $\big((3)$ and $(4)\big)$?
Update:
This is an update in response to this comment:
Look at the last row and column of those matrices. They can all be written as a sum of a $2×2$ matrix and a $1×1$ matrix. The $3$-dimensional representation is the sum of the trivial and the $2$-dimensional
This has lead me to a new point of confusion:
In one of the earlier quotes from my notes, I wrote that
For any group, there is also the trivial irrep, where $\color{red}{\chi(C) = 1}\,$ $\color{red}{\text{for ALL classes}}$. This is usually the first row of the character table.
The trace of all of the matrices (which are listed below the $C_{4v}$ Cayley table) for the group elements of $C_{4v}$ all give $1$ (Except $E$ where the trace is $3$). So by my logic the (partial) character table for $C_{4v}$ should look like
$$ \begin{array}{c|c|c|c} C_{4v} & E & \{C_4, C_4^3\} & C_2 & \{\sigma_x, \sigma_y\} & \{\sigma_d, \sigma_{d^\prime}\} \\\hline \chi^{(1)} & 1 & 1 & 1 & 1 & 1 \\\hline \chi^{(2)} & 1 \\\hline \chi^{(3)} & 1 \\\hline \chi^{(4)} & 1 \\\hline \chi^{(5)} & \color{blue}{3} \end{array} \tag{$\color{blue}{?}$}$$
But according to the comment quoted the above (partial) character table cannot be correct. So looking at say, the $\{C_4, C_4^3\}$ class from the character table above, this class in its $3\times 3$ form does indeed have a trace of 1.
However, since these matrices are in block-diagonal form they are also reducible. For example, $$C_4=\begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ can be reduced to IRREPs $\begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix}$ and the trivial IRREP $1$. I did this by considering the '$z$-component' of the original $3\times 3$ matrix then the remaining $x,y$ component block.
My question is: Are the $3\times 3$ matrices I gave below the Cayley table for point group $C_{4v}$ of the correct dimensions for the group elements $\big(E,\,C_4,\, C_4^3,\, C_2, \,\sigma_x,\, \sigma_y,\, \sigma_d,\, \sigma_d^{\prime}\big)$?
If these are the correct matrices then why must we reduce these matrices into irreducible forms and then calculate the trace, $\chi^{(i)}$?
But more importantly, when we reduce the identity group element, $E$, to it's sum of IRREPS: $E=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ to $\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ and the trivial IRREP, $1$. Why do we then discard the trivial IRREP $1$ and put a trace of $2$ in the character table?
The reason I'm asking these questions is that it just confused me a lot when I saw that the trace of the $3\times 3$ group element matrices gave the same trace as the IRREPS (with the exception of $E$).
Edit:
Before the chat was cleared I was given some useful information from @David A. Craven who mentioned that:
The blue $3$ shouldn't be in your partial character table. Your 3-dimensional rep is not irreducible, so has no place there. The $3$-dimensional is the sum of the trivial and the 2-dimensional, and by looking at those matrices you can fill in the fifth row. It should be $2,\, 0,\, -2,\, 0, \,0$. Notice that the sum of this and the first row gives you the traces of your $3\times 3$ matrices.
&
A representation is irreducible if there is no subspace of the vector space that is stabilized by your group, or equivalently, there is no basis with respect to which the matrices of the group action can be written as block-diagonal matrices (this equivalence is only valid over fields of characteristic $0$). Your matrices are already in block-diagonal form, therefore the representation is not irreducible.
From the first quote, I notice that the sum of the first and fifth rows are $3,\,1,\,-1,\,1,\,1$ and these indeed are the traces of the $3 \times 3$ matrices belonging to classes $\{E\}$, $\{C_4,\, C_4^3\}$, $\{C_2\}$, $\{\sigma_x,\,\sigma_y\}$, and $\{\sigma_d,\,\sigma_{d^\prime}\}$ respectively.
The last row of the table as mentioned in the quote above is $2,\, 0,\, -2,\, 0, \,0$, and I notice that these traces were obtained by summing together the first 2 diagonal elements of the classes $\{E\}$, $\{C_4,\, C_4^3\}$, $\{C_2\}$, $\{\sigma_x,\,\sigma_y\}$, and $\{\sigma_d,\,\sigma_{d^\prime}\}$ respectively. This is now starting to make sense to me as I see that the last row of the character table corresponds to the only 2-dimensional IRREP.
So from this, I now understand why the 3 (which I marked blue) must not appear in the character table. I also understand the meaning of the beginning of the second sentence in the first quote above by David: "Your 3-dimensional rep is not irreducible, so has no place there." - I understand this because the matrix for $E$ is in block-diagonal form, and as I understand it, a block-diagonal matrix is reducible.
Now here is the only part I cannot understand, in the last sentence of the second quote by David where he mentions that "Your matrices are already in block-diagonal form, therefore the representation is not irreducible."
But, it was my understanding that block-diagonal matrices only have entries on the leading diagonal (such as the $3\times 3$ matrix for identity, $E$) and zeroes elsewhere. But some of those matrices given below the $C_{4v}$ Cayley table have off-diagonal (non-zero) components (such as in the matrices for $C_4$ and $\sigma_d$, etc.).
So, in short, are block-diagonal matrices reducible or irreducible?
OK, so this author has some really wonky notation, but it's all fine in the end.
The first comment is that the pyramid is a red herring. All symmetries of the pyramid leave the peak fixed, so actually this is just symmetries of the square. This is called the dihedral group $D_8$ (some people write $D_4$), and that is what you are trying to do.
Now, on to the notation. The notation the author chooses for conjugacy classes is weird, but actually has some sense. Normally we simply write a group element from each class (a class representative) at the top of each column, but including the size of the class, like in $2C_4$, is useful. This is because the size of the class is important for what we will discuss later.
What one puts as the column headings is unimportant, as long as the reader knows which class is being referred to. Common examples are $C_1$, $C_2$, ... ($C$ for 'class'), $x_1$, $x_2$, ..., where the $x_i$ are group elements belonging to each class, 1A, 2A, 2B, 3A, ..., where 1A is the first class of elements of order 1, 2A and 2B are classes of elements of order 2, and so on. The author has chosen $2C_4$ to denote the class $\{C_4,C_4^3\}$. I would just write $C_4$ at the top instead.
Obviously the author failed, because the point is that the notation doesn't matter as long as the reader knows what it means. Since you are the reader and you didn't get it, the notation wasn't right.
Computing the conjugacy classes takes a long time, and you are right, the classes are $\{E\}$, $\{C_4,C_4^3\}$, $\{C_2\}$, $\{\sigma_x,\sigma_y\}$, $\{\sigma_d,\sigma_d'\}$. There are faster ways to do this than to simply compute all conjugates, but that's group theory rather than character theory.
On to the first column. You have mentioned the orthogonality relation, that the rows of a character table are orthogonal with respect to a certain product. That's not what is used here at all. This is the column orthogonality relation. This states that summing the products of the entries from column A and the complex conjugates of the entries from column B gives $0$ if A and B are different, and gives $|G|$ divided by the size of the class if A=B.
Specialized to the case where A=B is the column $\{E\}$, which simply records the size of the matrices (and is therefore real), we obtain that $\sum_\chi \chi(E)^2=|G|$. There is only one way to write $8$ as the sum of five squares, and we therefore have $1,1,1,1,2$ as the values for $\chi(E)$ for the various rows. The convention is to order them in increasing value, but it's not a requirement in the table.
For the first row, that is the trivial character, which sends every element $g\in G$ to $1$. That satisfies $\chi(g)=1$, so every column has entry $1$. By convention we put this as the first row, but again it doesn't have to be. But you will never find a character table that doesn't put it at the start.
I would recommend reading and doing exercises from another source. James and Liebeck's book on representations and characters of finite groups is probably the best.