Understanding how to compute character table 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

$$
  \begin{array}{r|ccccc}
    {}  & 1 &    6 &     8 &      6 &        3 \\[0.5em]
    S_4 & 1 & (12) & (123) & (1234) & (12)(34) \\
    \hline
    \text{trivial $U$}
    & 1 & \phantom{-}1 & \phantom{-}1 & \phantom{-}1 & \phantom{-}1
    \\
    \text{alternating $U'$}
    & 1 &           -1 & \phantom{-}1 &           -1 & \phantom{-}1
    \\
    \text{standard $V$}
    & 3 & \phantom{-}1 & \phantom{-}0 &           -1 &           -1
    \\
    V' = V \otimes U'
    & 3 &           -1 & \phantom{-}0 & \phantom{-}1 &           -1
    \\
    \text{Another $W$}
    & 2 & \phantom{-}0 &           -1 & \phantom{-}0 & \phantom{-}2
  \end{array}
$$
(Original image)

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.
 A: 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}} = $ mean that the rows of $A$ are orthonormal with respect to the standard inner product $⟨-,-⟩$.
Similarly, your proposed condition $A A^{\mathsf{t}} = |G| $ 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
$$
  ⟨y, y⟩_{\mathrm{char}}
  =
  \frac{1}{24}
  \Bigl( 2^2 + 6 ⋅ 0^2 + 8 ⋅ (-1)^2 + 6 ⋅ 0^2 + 3 ⋅ 2^2 \Bigr)
  = \frac{1}{24}( 4 + 0 + 8 + 0 + 12  )
  = 1 \,.
$$
