Determine all characters of the representations of $T_d$ in ${\cal H}_l$ for $l=0,1,2$ discussed in subsection.

For $l=1$ we get: $\chi_4(E)=3$, $\chi_4(C_3)=0$, $\chi_4(C_2)=-1$, $\chi_4(S_4)=-1$, $\chi_4(\s)=1$.
For $l=2$ we get two characters, but the first one is again $\chi_4$. The seconde is given by $\chi_3(E)=2$, $\chi_3(C_3)=-1$, $\chi_3(C_2)=2$, $\chi_3(S_4)=0$, $\chi_3(\s)=0$.

Find all characters of $T_d\simeq S(4)$ by employing orthogonality of the columns of the character table and the previous example.

Inserting the trivial character $\chi_1$ and the sign $\chi_2$ we therefore have the following table $$ \begin{array}{c|rrrrr} T_d&1E&8C_3&3C_2&6S_4&6\s\\ \hline \chi_1&1&1&1&1&1\\ \chi_2&1&1&1&-1&-1\\ \chi_3&2&-1&2&0&0\\ \chi_4&3&0&-1&1&-1\\ \chi_5&?&?&?&?&? \end{array} $$ Since $\sum\chi_j(E)^2=24$ we get: $\chi_5(E)=3$. Now orthogonality of the columns yields: $\chi_5(C_3)=0$, $\chi_5(C_2)=-1$, $\chi_5(S_4)=-1$, $\chi_5(\s)=1$, i.e.: $$ \begin{array}{c|rrrrr} T_d&1E&8C_3&3C_2&6S_4&6\s\\ \hline \chi_1&1&1&1&1&1\\ \chi_2&1&1&1&-1&-1\\ \chi_3&2&-1&2&0&0\\ \chi_4&3&0&-1&1&-1\\ \chi_5&3& 0&-1&-1&1 \end{array} $$