I am learning how to compute the character table for few groups. I encountered the example of $S_4$ in Fulton & Harris, Section 2.4.
I was able to understand how to proceed. It says that it will use the relations given by the equation $(2.10)$.
If $V$ is irreducible then by Schur’s lemma $\dim \mathrm{Hom}(V, W)^G$ is the multiplicity of $V$ in $W$; similarly, if $W$ is irreducible, $\dim \mathrm{Hom}(V, W)^G$ is the multiplicity of $W$ in $V$, and in the case where both $V$ and $W$ are irreducible, we have
$$
\dim \mathrm{Hom}_G(V, W)
=
\begin{cases}
1 & \text{if $V \cong W$,} \\
0 & \text{if $V \ncong W$.}
\end{cases}
$$
But now the character $\chi_{\mathrm{Hom}(V, W)}$ of the representation $\mathrm{Hom}(V, W) = V^* \otimes W$ is given by
$$
\chi_{\mathrm{Hom}(V, W)}(g)
= \overline{\chi_V(g)} \cdot \chi_W(g) \,.
$$
We can now apply formula $(2.9)$ in this case to obtain the striking
$$
\frac{1}{|G|} \sum_{g \in G} \overline{\chi_V(g)} \chi_W(g)
=
\begin{cases}
1 & \text{if $V \cong W$,} \\
0 & \text{if $V \ncong W$.}
\end{cases}
\tag{$2.10$}
$$
(Original image)
I was trying to interpret it using the matrix formed by the character table which is a square matrix. I thought that we must have $AA^t=|G| I$ but it is not the case. The character table given is
$$
\begin{array}{r|ccccc}
{} & 1 & 6 & 8 & 6 & 3 \\[0.5em]
S_4 & 1 & (12) & (123) & (1234) & (12)(34) \\
\hline
\text{trivial $U$}
& 1 & \phantom{-}1 & \phantom{-}1 & \phantom{-}1 & \phantom{-}1
\\
\text{alternating $U'$}
& 1 & -1 & \phantom{-}1 & -1 & \phantom{-}1
\\
\text{standard $V$}
& 3 & \phantom{-}1 & \phantom{-}0 & -1 & -1
\\
V' = V \otimes U'
& 3 & -1 & \phantom{-}0 & \phantom{-}1 & -1
\\
\text{Another $W$}
& 2 & \phantom{-}0 & -1 & \phantom{-}0 & \phantom{-}2
\end{array}
$$
but I am not able to see how the orthogonality relation holds. Can anyone explain me the orthogonality relations in terms of this matrix formed by the character table?
I would be grateful as it will help me understand.
Best Answer
The number of irreducible representations of $G = S_4$, which is equal to the number of conjugacy classes of $S_4$, is $n = 5$. The conjugacy classes of $G$ have sizes $$ w_1 = 1 \,, \quad w_2 = 6 \,, \quad w_3 = 8 \,, \quad w_4 = 6 \,, \quad w_5 = 3 \,. $$ On the vector space $ℂ^n$ we can now consider three different inner products:
The standard inner product $$ ⟨x, y⟩ = \sum_{i = 1}^n \overline{x_i} y_i = \overline{x_1} y_1 + \overline{x_2} y_2 + \overline{x_3} y_3 + \overline{x_4} y_4 + \overline{x_5} y_5 \,. $$
The scaled inner product $$ ⟨x, y⟩_{\mathrm{scal}} = \frac{1}{|G|} ⟨x, y⟩ = \frac{1}{|G|} \sum_{i = 1}^n \overline{x_i} y_i = \frac{1}{24} ( \overline{x_1} y_1 + \overline{x_2} y_2 + \overline{x_3} y_3 + \overline{x_4} y_4 + \overline{x_5} y_5) \,. $$
The “characteristic” inner product, which is not only scaled but also incorporates the numbers $w_i$ as weights: $$ ⟨x, y⟩_{\mathrm{char}} = \frac{1}{|G|} \sum_{i = 1}^n w_i \overline{x_i} y_i = \frac{1}{24} ( \overline{x_1} y_1 + 6 \overline{x_2} y_2 + 8 \overline{x_3} y_3 + 6 \overline{x_4} y_4 + 3 \overline{x_5} y_5 ) \,. $$
The condition $A A^{\mathsf{t}} = 𝟙$ mean that the rows of $A$ are orthonormal with respect to the standard inner product $⟨-,-⟩$. Similarly, your proposed condition $A A^{\mathsf{t}} = |G| 𝟙$ means that the rows of $A$ are orthonormal with respect to the scaled inner product $⟨-,-⟩_{\mathrm{scal}}$. However, the orthonormality of irreducible characters tells us that the rows of $A$ are orthonormal with respect to $⟨-,-⟩_{\mathrm{char}}$ instead.
For example, the last two rows of $A$ are given by $x = (3, -1, 0, 1, -1)$ and $y = (2, 0, -1, 0, 2)$. These rows are orthogonal with respect to $⟨-,-⟩_{\mathrm{char}}$ since \begin{align*} ⟨x, y⟩_{\mathrm{char}} &= \frac{1}{24} \Bigl( 1 ⋅ 3 ⋅ 2 + 6 ⋅ (-1) ⋅ 0 + 8 ⋅ 0 ⋅ (-1) + 6 ⋅ 1 ⋅ 0 + 3 ⋅ (-1) ⋅ 2 \Bigr) \\[0.5em] &= \frac{1}{24} ( 6 + 0 + 0 + 0 - 6 ) = 0 \,. \end{align*} Both $x$ and $y$ are normalized with respect to $⟨-,-⟩_{\mathrm{char}}$ because $$ ⟨x, x⟩_{\mathrm{char}} = \frac{1}{24} \Bigl( 3^2 + 6 ⋅ (-1)^2 + 8 ⋅ 0^2 + 6 ⋅ 1^2 + 3 ⋅ (-1)^2 \Bigr) = \frac{1}{24}( 9 + 6 + 0 + 6 + 3 ) = 1 $$ and also $$ ⟨y, y⟩_{\mathrm{char}} = \frac{1}{24} \Bigl( 2^2 + 6 ⋅ 0^2 + 8 ⋅ (-1)^2 + 6 ⋅ 0^2 + 3 ⋅ 2^2 \Bigr) = \frac{1}{24}( 4 + 0 + 8 + 0 + 12 ) = 1 \,. $$